with the first phasor. The sum E of £ , and £ 2 is the sum of the projections of the two phasors on the vertical axis. This is revealed more clearly if we redraw the phasors, as in Fig. 9c, placing the foot of one arrow at the head of the other, maintaining the proper phase difference, and letting the whole assembly rotate counterclockwise about the origin. In Fig. 9c, E can also be regarded as the projection on the vertical axis of a phasor o f length Eg, which is the vector sum o f the two phasors of magnitude Eq. From that
figure, we see that the projection can be written
E = Eg sin {(ot
P),
in agreement with Eq. 6. Note that the (algebraic) sum of the projections o f the two phasors is equal to the projec tion of the (vector) sum of the two phasors. In most problems in optics we are concerned only with the amplitude Eg of the resultant wave disturbance and not with its time variation. This is because the eye and other common measuring instruments respond to the re sultant intensity of the light (that is, to the square o f the amplitude) and cannot detect the rapid time variations that characterize visible light. For sodium light (A = 589 nm), for example, the frequency v (= oi/ln) is 5.1 X 10'“’ Hz. Often, then, we need not consider the ro tation o f the phasors but can confine our attention to finding the amplitude o f the resultant phasor. In Fig. 9c the three phasors form an isosceles triangle whose sides have lengths E q , E q , and E g . In any triangle, an exterior angle (
It is also clear from Fig. 9c that the length o f the base of this triangle is Eg = 2E
q
cos P
These results are identical with Eqs. 7 and 8. In a more general case we might want to find the result ant o f more than two sinusoidally varying wave distur bances. The general procedure is the following: 1. Construct a series of phasors representing the func tions to be added. Draw them end to end, maintaining the proper phase relationships between adjacent phasors. 2. Construct the sum of this phasor array, analogous to a sum of vectors. The length o f the resulting phasor gives the amplitude of the electric field. The angle between it and the first phasor is the phase o f the resultant with respect to this first phasor. The projection of this phasor on the vertical axis gives the time variation o f the resultant wave disturbance.
Figure 9 (a) A time-varying wave £ , is represented by a rotating vector or phasor. (b) Two waves El and £2 differing in phase by
Section 45-4
Interference from Thin Films
955
Figure 11 A soapy water film on a wire loop, viewed by re flected light. The black segment at the top is not a tom film. It occurs because the film, by drainage, is so thin there that destmctive interference occurs between light reflected from the front and back surfaces of the film. Figure 10 Sample Problem 3. Four waves are added graphi cally, using the method of phasors.
Air
Air
Sample Problem 3 Find graphically the resultant E{t) of the following wave disturbances: = E q sin cu/,
£2 = F q sin (cot + 15®), £3 = £0 sin (cot + 30®), £4
= £0 sin (cot + 45®).
Solution Figure 10 shows the assembly of four phasors that represent these functions. The phase angle
£2
+
£3
+
£4
= 3.8£o sin (cot -h 22.5®). Check this result by direct trigonometric calculation or by geo metric calculation from the phasor diagram of Fig. 10.
45-4 INTERFERENCE FROM TH IN FILM S___________________ The colors that we see when sunlight falls on a soap bub ble, an oil slick, or a ruby-throated hummingbird are caused by the interference o f light waves reflected from the front and back surfaces o f thin transparent films. The film thickness is typically o f the order o f magnitude o f the wavelength o f light. Thin films deposited on optical com-
Figure 12 A thin film is viewed by light reflected from a source S. Waves reflected from the front and back surfaces enter the eye as shown, and the intensity of the resultant light wave is determined by the phase difference between the com bining waves. The medium on either side of the film is as sumed to be air.
ponents, such as camera lenses, can reduce reflection and enhance the intensity of the transmitted light. Thin coat ings on windows can enhance the reflectivity for infrared radiation while having less effect on the visible radiation. In this way it is possible to reduce the heating effect of sunlight on a building. Depending on its thickness, a thin film can be perfectly reflecting or perfectly transmitting for light o f a given wavelength, as shown in Fig. 11. These effects result from constructive or destructive interference. Figure 12 shows a transparent film of uniform thick ness d illuminated by monochromatic light o f wavelength A from a point source S, The eye is positioned so that a particular incident ray i from the source enters the eye as
956
Chapter 45
Interference
ray r, after reflection from the front surface of the film at a. The incident ray also enters the film at a as a refracted ray and is reflected from the back surface of the film at b\ it then emerges from the front surface o f the film at c and also enters the eye, as ray rj. The geometry of Fig. 12 is such that rays r^ and rj are parallel. Having originated in the same point source, they are also coherent and so are capable o f interfering. Because these two rays have trav eled over paths of different lengths, have traversed differ ent media, and— as we shall see— have suffered different kinds o f reflections at a and b, there is a phase difference between them. The intensity perceived by the eye, as the parallel rays from the region ac of the film enter it, is determined by this phase difference. For near-normal incidence (0j « 0 in Fig. 12) the geo metrical path difference for the two rays from S is close to 2d. We might expect the resultant wave reflected from the film near a to be an interference maximum if the distance 2d is an integral number of wavelengths. This statement must be modified for two reasons. First, the wavelength must refer to the wavelength of the light in the film and not to its wavelength Ain air; that is, we are concerned with optical path lengths rather than geometrical path lengths. The wavelengths A and A„ are related by Eq. 13 of Chapter 43, A, = A/n,
(15)
where n is the index of refraction of the film. To bring out the second point, let us assume that the film is so thin that 2d is very much less than one wave length. The phase difference between the two waves would be close to zero on our assumption, and we would expect such a film to appear bright on reflection. How ever, it appears dark. This is clear from Fig. 11, in which the action o f gravity produces a wedge-shaped film, ex tremely thin at its top edge. As drainage continues, the dark area increases in size. To explain this and many similar phenomena, one or the other of the two rays of Fig. 12 must suffer an abrupt phase change of tt(= 180°) when it is reflected at the air-film interface. As it turns out, only the ray reflected from the front surface suffers this phase change. The other ray is not changed abruptly in phase, either on transmission through the front surface or on reflection at the back surface. In Section 19-9 we discussed phase changes on reflec tion for transverse waves in strings. To extend these ideas, consider the composite string o f Fig. 13, which consists of two parts with different masses per unit length, stretched to a given tension. A pulse in the heavier string moves to the right in Fig. 13u, approaching the junction. Later there will be reflected and transmitted pulses, the reflected pulse being in phase with the incident pulse. In Fig. 13^> the situation is reversed, the incident pulse now being in the lighter string. In this case the reflected pulse differs in phase from the incident pulse by n (= 180°). In each case the transmitted pulse is in phase with the incident pulse.
A
Initial Final
A
"a (a)
Initial
A
Final
V (W
Figure 13 Phase changes on reflection at a junction between two strings of different linear mass densities. The wave speed is greater in the lighter string, (a) The incident pulse is in the heavier string, (b) The incident pulse is in the lighter string.
Figure 13a suggests a light wave in glass, say, approach ing a surface beyond which there is a less optically dense medium (one of lower index of refraction) such as air. Figure 13b suggests a light wave in air approaching glass. To sum up the optical situation, when reflection occurs from an interface beyond which the medium has a lower index of refraction, the reflected wave undergoes no phase change', when the medium beyond the interface has a higher index, there is a phase change o f n.* The transmit ted wave does not experience a change o f phase in either case. We are now able to take into account both factors that determine the nature of the interference, namely, differ ences in optical path length and phase changes on reflec tion. For the two rays of Fig. 12 to combine to give a maximum intensity, assuming normal incidence, we must have
2 d = (m + i)A„
m = 0, 1, 2, . . . .
The term iA„ is introduced because upon reflection there is a phase change o f 180 °, equivalent to half a wavelength. Substituting A/n for A„ yields finally
2dn = (m + i)X
w = 0, 1,2, . . . (maxima).
(16)
The conditions for a minimum intensity are
2dn = mX
m = 0 ,1 ,2 , . . .
(minima). (17)
These equations hold when the index o f refraction o f the film is either greater or less than the indices o f the media on each side o f the film. Only in these cases will there be a relative phase change o f 180° for reflections at the two surfaces. A water film in air and an air film in the space between two glass plates provide examples o f cases to
• These statements, which can be proved rigorously from Max well’s equations (see also Section 45-5), must be modified for light falling on a less dense medium at an angle such that total internal reflection occurs. They must also be modified for reflec tion from metallic surfaces.
Section 45-4
which Eqs. 16 and 17 apply. Sample Problem 5 provides a case in which they do not apply. If the film thickness is not uniform, as in Fig. 11 where the film is wedge shaped, constructive interference occurs in certain parts o f the film and destructive interference occurs in others. Bands of maximum and of minimum intensity appear, called fringes of constant thickness. The width and spacing o f the fringes depend on the variation o f the film thickness d. If the film is illuminated with white light rather than monochromatic light, the light reflected from various parts o f the film is modified by the various constructive or destructive interferences that occur. This accounts for the brilliant colors of soap bubbles and oil slicks. Only if the film is “thin” (d being no more than a few wavelengths o f light) is it possible to obtain these types of fringes, that is, fringes that appear localized on the film and are associated with a variable film thickness. For very thick films (say d « 1 cm), the path difference between the two rays o f Fig. 12 is many wavelengths, and the phase difference at a given point on the film changes rapidly as we move even a small distance away from a. For “thin” films, however, the phase difference at a also holds for reasonably nearby points; there is a characteristic “patch brightness” for any point on the film, as Fig. 11 shows. Interference fringes can be produced for thick films; they are not localized on the film but are at infinity (see Section 45-6).
Sample Problem 4 A water film (n = 1.33) in air is 320 nm thick. If it is illuminated with white light at normal incidence, what color will it appear to be in reflected light?
Interference from Thin Films
Air Mgp2 n = 1.00 « = 1.38
957
Glass
n = 1.50
Figure 14 Sample Problem 5. Unwanted reflections from glass can be suppressed (at a chosen wavelength) by coating the Rlass with a film of the proper thickness.
reflection from the glass surface. How thick a coating is needed to produce a minimum reflection at the center of the visible spectrum (A = 550 nm)? Solution We assume that the light strikes the lens at near normal incidence (6 is exaggerated for clarity in Fig. 14), and we seek destructive interference between rays r, and Tj . Equation 17 does not apply because in this case a phase change of 180® is associated with each ray, for at both the upper and lower surfaces of the MgFj film the reflection is from a medium of greater index of refraction. There is no net change in phase produced by the two reflec tions, which means that the optical path difference for destruc tive interference is (m + i)A„ (compare Eq. 16), leading to Idn = (m -h i)A
m = 0 , 1, 2, . . .
(minima).
Solving for d and putting m = 0 we obtain Solution
Solving Eq. 16 for A, we obtain
^
Idn _ (2X320 nmXl.33) _ 851 nm
(maxima).
From Eq. 17 the minima are given by 851 nm A= m
(minima).
Maxima and minima occur for the following wavelengths: m A(nm)
0 (max)
1 (min)
1 (max)
2 (min)
2 (max)
1702
851
567
426
340
Only the maximum corresponding to m = 1 lies in the visible region (between about 400 and 700 nm); light of wavelength 567 nm appears yellow-green. If white light is used to illuminate the film, the yellow-green component is enhanced when viewed by reflection. What is the color of the light transmitted through the film?
(m -h i)A _ A _ 550 nm _ 2n An (4X1.38)
__
Sample Problem 6 Figure 15 shows a plano-convex lens of radius of curvature R resting on an accurately plane glass plate and illuminated from above by light of wavelength A. Figure 16 shows that circular interference fringes (called Newton’s rings) appear, associated with the variable thickness air film between the lens and the plate. Find the radii of the circular interference maxima. Solution Here it is the ray from the bottom of the (air) film rather than from the top that undergoes a phase change of 180 ®, for it is the one reflected from a medium of higher refractive index. The condition for a maximum remains unchanged (Eq. 16), however, and is
2^ /= (m + i)A
m = 0 , 1, 2, . . . ,
assuming w = 1 for the air film. From Fig. 15 we can write Sample Problem 5 Lenses are often coated with thin films of transparent substances such as MgFj (n = 1.38) to reduce the
<
/ =
=
/ ?
- / ?
11 -
(18)
958
Chapter 45
Interference which gives the radii of the bright rings. If white light is used, each spectrum comp)onent produces its own set of circular fringes, and the sets all overlap. Note that r > 0 for /w = 0. That is, the first bright ring is at r > 0 , and consequently the center must be dark, as shown in Fig. 16. This observation can be taken as experimental evidence for the 180'' phase change upon reflection used to obtain Eq. 18.
45-5 OPTICAL REVERSIBILITY AND PHASE CHANGES ON REFLECTION (Optional)
Figure 15 Sample Problem 6. The apparatus for observing Newton’s rings.
G. G. Stokes (1819 -1903) used the principle of optical revers ibility to investigate the reflection of light at an interface between two media. The principle states that if there is no absorption of light, a light ray that is reflected or refracted will retrace its original path if its direction is reversed. This reminds us that any mechanical system can run backward as well as forward, pro vided there is no dissipation of energy such as by friction. Figure 17a shows a wave of amplitude E reflected and re fracted at a surface separating media 1and 2, where /I2 > i •The amplitude of the reflected wave is rjjE, in which r ,2 is an amplitude reflection coefficient. The amplitude of the refracted wave is tx2E, where r ,2 is an amplitude transmission coefficient. The sign of the coefficient indicates the relative phase of the reflected or transmitted component. If we consider only the possibility of phase changes of 0 or 180®, then if r^2 = +0.5, for example, we have a reduction in amplitude on reflection by one-half and no change in phase. For r ,2 = —0.5 we have a phase change of 180® because E sin {(ot + 180®) = —E sin cot. In Fig. \lby the rays indicated by r, 2E and txiE have been reversed in direction. Ray r^iEy identified by the single arrows in the figure, is reflected and refracted, producing the rays of ampli tudes r\2E and r, 2^i2^ . Ray txiE, identified by the triple arrows, is also reflected and refracted, producing the rays of amplitudes /, 2/2i ^ and txirixE as shown. Note that r ,2 describes a ray in medium 1 reflected from medium 2, and T2i describes a ray in medium 2 reflected from medium 1. Similarly, r,2describes a ray that passes from medium 1 to medium 2; ^21 describes a ray that passes from medium 2 to medium 1. Based on the reversibility principle, we conclude that the two rays in the upper left of Fig. 17^ must be equivalent to the incident ray of Fig. 17a, reversed; the two rays in the lower left of Fig. 1lb must cancel. This second requirement leads to ^XlixiE + tx2^2\E = 0 ,
Figure 16 Circular interference fringes (Newton’s rings) ob served with the apparatus of Fig. 15.
If r/R c 1, we can expand the square bracket by the binomial theorem, keeping only two terms, or
Combining with Eq. 18 yields r = V(m 4-
w
= 0, 1,2, . . . (maxima).
or /•i2 = - r 2, This result tells us that if we compare a wave reflected from medium 1 with one reflected from medium 2, they behave dif ferently in that one or the other undergoes a phase change of 180®. Experiment shows that the ray reflected from the more opti cally dense medium suffers the change in phase of 180®. This can be demonstrated using the setup shown in Fig. 18, which is called the Lloyd’s mirror experiment. Interference occurs on the screen at an arbitrary point F as a result of the overlap of the direct and
Section 45-6
Michelson's Interferometer
959
Figure 17 (fl) A ray is reflected and refracted at an interface, {b) The optically reversed situation; the two rays in the lower left must cancel.
( 6)
(a)
Screen
Mirror
shows that one of the interfering beams has been shifted in phase by 180®. Since there is nothing to change the phase of the direct beam SP, it must be the reflected beam that experiences the change in phase. This shows that reflection from a more opti cally dense medium involves a change in phase of 180®. ■
(fl)
45-6 M ICHELSON’S INTERFEROMETER*___________
Figure 18 {a) The experimental setup for Lloyd’s mirror. Fringes appear on the screen as a result of interference be tween the direct and reflected beams, {b) Fringes observed in the Lloyd’s mirror experiment.
reflected beams. We can analyze this experiment as two-source interference, in which one of the sources (5") is the virtual image of S in the plane mirror. However, there is one important differ ence between the apparatus of Fig. 18 and the double-slit experi ment: the light from the virtual source 5 ' has been reflected from the mirror and has undergone a phase change of 180®. As a result of this phase change, the lower edge of the screen (at O) shows a dark fringe, instead of the bright fringe that appears at the corre sponding point (the center of the screen) in the double-slit exper iment. Put another way, the appearance of the dark fringe at O
An interferometer is a device that can be used to measure lengths or changes in length with great accuracy by means of interference fringes. We describe the form originally built by A. A. Michelson (1852-1931) in 1881. Consider light that leaves point P on extended source S (Fig. 19) and falls on half-silvered mirror Af (sometimes called a beam splitter). This mirror has a silver coating just thick enough to transmit half the incident light and to reflect half; in the flgure we have assumed for conve nience that this mirror has negligible thickness. At A/the light divides into two waves. One proceeds by transmis sion toward mirror M ,; the other proceeds by reflection toward M2 . The waves are reflected at each o f these mirrors and are sent back along their directions o f inci dence, each wave eventually entering the eye E. Because the waves are coherent, being derived from the same point on the source, they interfere. If the mirrors A/, and M 2 are exactly perpendicular to each other, the effect is that of light from an extended source S falling on a uniformly thick slab of air, between glass, whose thickness is equal to ~ ^ 1 • Interference
• See “ Michelson: America’s First Nobel Prize Winner in Science,” by R. S. Shankland, The Physics Teacher, January 1977, p. 19. See also “Michelson and his Interferometer,” by R. S. Shankland, Physics Today, April 1974, p. 36.
960
Chapter 45
Interference Movable
growing requirements o f science and technology and was replaced by a new standard based on a defined value for the speed of light.
Sample Problem 7 Yellow light (A = 589.00 nm) illuminates a Michelson interferometer. How many bright fringes will be counted as the mirror is moved through 1.0000 cm? Solution Each fringe corresponds to a movement of the mirror through one-half wavelength. The number of fringes is thus the same as the number of half wavelengths in 1.0000 cm, or 1.0000 X 10-2’m = 33,956 fringes. K589.00X 10-’ m) Figure 19 Michelson’s interferometer, showing the path of a ray originating at point P of an extended source S. The ray from P splits at M\ the two rays are reflected from mirrors A/, and M l and then recombine at M. Mirror A/j can be moved to change the path difference between the combining rays.
45-7 MICHELSON’S INTERFEROMETER AND LIGHT PROPAGATION (O ptional)
fringes appear, caused by small changes in the angle of incidence of the light from different points on the ex tended source as it strikes the equivalent air film. For thick films a path difference of one wavelength can be brought about by a very small change in the angle of incidence. If Ml is moved backward or forward, the effect is to change the thickness of the equivalent air film. Suppose that the center of the (circular) fringe pattern appears bright and that Mi is moved just enough to cause the first bright circular fringe to move to the center of the pattern. The path o f the light beam traveling back and forth to Mi has been changed by one wavelength. This means (be cause the light passes twice through the equivalent air film) that the mirror must have moved through a distance o fiA .
The interferometer is used to measure changes in length by counting the number of interference fringes that pass the field of view as mirror Mi is moved. Length measurements made in this way can be accurate if large numbers o f fringes are counted. Michelson measured the length of the standard meter, kept in Paris, in terms of the wavelength of a certain monochromatic red light emitted from a light source con taining cadmium. He showed that the standard meter was equivalent to 1,553,163.5 wavelengths of the red cad mium light. For this work he received the Nobel prize in 1907. Michelson’s work laid the foundation for the even tual abandonment (in 1961) of the meter bar as a standard o f length and for the redefinition of the meter in terms of the wavelength o f light. In 1983, as we have seen, even this wavelength standard was not precise enough to meet the
In Chapter 2 1 we presented Einstein’s hypothesis, now well veri fied, that in free space light travels with the same speed c no matter what the relative velocity of the source and the observer may be. We pointed out that this hypothesis contradicted the views of 19th-century physicists regarding wave propagation. It was difficult for these physicists, trained as they were in the classical physics of the time, to believe that a wave could be propagated without a medium. If such a medium could be estab lished, the speed c of light would naturally be regarded as the speed with respect to that medium, just as*the speed of sound always refers to a medium such as air. Although no medium for light propagation was obvious, physicists postulated one, called the ether, and hypothesized that its properties were such that it was undetectable by ordinary means such as weighing. In 1881 (24 years before Einstein’s hypothesis) A. A. Michel son set himself the task of direct physical verification of the existence of the ether. In particular, Michelson, later joined by E. W. Morley, tried to measure the speed u with which the Earth moves through the ether. Michelson’s interferometer was their instrument of choice for this now-famous Michelson-Morley experiment. The Earth together with the interferometer moving with veloc ity u through the ether is equivalent to the interferometer at rest with the ether streaming through it with velocity —u, as shown in Fig. 20. Consider a wave moving along the path A/A/, A/and one moving along M M iM . The first corresponds classically to a per son rowing a boat a distance d downstream and the same dis tance upstream; the second corresponds to rowing a boat a dis tance d across a stream and back. Based on the ether hypothesis the speed of light on the path A/A/, is c + w; on the return path A/, A/ it is c — u. The time required for the complete trip is =
w
2c -h= d- .2 _ . c —u
2d
1
C
1 —(u/cY '
The speed of light, again based on the ether hypothesis, for path M M i is 4c^ — as Fig. 20 suggests. This same speed holds
Questions
961
the “cross-stream” path and A/A/jA/the “downstream and up stream” path. The time difference between the two waves enter ing the eye is also reversed; this changes the phase difference between the combining waves and alters the positions of the interference maxima. The experiment consists of looking for a shift of the interference fringes as the apparatus is rotated. The change in time difference is 2Ar, which corresponds to a phase difference of A0 = cu(2A0, where o) (= InclX) is the angu lar frequency of the light wave. The expected maximum shift in the number of fringes on a 90® rotation is ^
Figure 20 The “ether” streaming with velocity —u through Michelson's interferometer. The speeds shown are based on the (incorrect) ether hypothesis.
Id ^
Id
1
c Vl —{u /cf
The difference of time for the two paths is
Assuming w/c c 1, we can expand the quantities in the square brackets by using the binomial theorem, retaining only the first two terms. This leads to
-
t
I K - : ) '! - "
Now let the entire interferometer be rotated through 90 ®. This interchanges the roles of the two light paths, A/A/, A/ now being
co{2At)
2cAt
( 20)
where we have used Eq. 19 for At, In the Michelson-Morley interferometer let = 11 m (ob tained by multiple reflection in the interferometer) and X = 5.9 X 10“ ^ m. If u is assumed to be roughly the orbital speed of the Earth, then w/c * 10“^. The expected maximum fnnge shift when the interferometer is rotated through 90® is then AN
for the return path A/j A/, so that the time required for this com plete path is
A0
_ 2 d ( wV _ (2X11 m) (10-^)2 = 0.4. " X U / " 5 .- 9 X 10-^ m
Even though a shift of only about 0.4 of a fringe was expected, Michelson and Morley were confident that they could observe a shift of 0.01 fringe. They foundfrom their experiment, however, that there was no observable fringe shift! The analogy between a light wave in the supposed ether and a boat moving in water, which seemed so evident in 1881, is sim ply incorrect. The derivation based on this analogy is incorrect for light waves. When the analysis is carried through based on Einstein’s hypothesis, the observed null result is clearly pre dicted, the speed of light being c for all paths. The motion of the Earth around the Sun and the rotation of the interferometer have, in Einstein’s view, no effect whatever on the speed of the light waves in the interferometer. It should be made clear that although Einstein’s hypothesis is completely consistent with the null result of the MichelsonMorley experiment, this experiment standing alone does not serve as a proof for Einstein’s hypothesis. Einstein said that no number of experiments, however large, could prove him right but that a single experiment could prove him wrong. Our present-day belief in Einstein’s hypothesis rests on consistent agreement in a large number of experiments designed to test it. The “single experiment” that might prove Einstein wrong has never been found. ■
QUESTIONS 1. Is Young’s experiment an interference experiment or a dif fraction experiment, or both? 2. In Young’s double-slit interference experiment, using a monochromatic laboratory light source, why is screen A in Fig. 6 necessary? If the source of light is a laser beam, screen A is not needed. Why? 3. What changes, if any, occur in the pattern of interference fringes if the apparatus of Fig. 4 is placed under water?
4. Do interference effects occur for sound waves? Recall that sound is a longitudinal wave and that light is a transverse wave. 5. It is not possible to show interference effects between light from two separate sodium vapor lamps but you can show interference effects between sound from two loudspeakers that are driven by separate oscillators. Explain why this is so. 6 . If interference between light waves of different frequencies is
962
7.
8. 9.
10.
Chapter 45
Interference
possible, one should observe light beats, just as one obtains sound beats from two sources of sound with slightly differ ent frequencies. Discuss how one might experimentally look for this possibility. Why are parallel slits preferable to the pinholes that Young used in demonstrating interference? Is coherence important in reflection and refraction? Describe the pattern of light intensity on screen C in Fig. 4 if one slit is covered with a red filter and the other with a blue filter, the incident light being white. If one slit in Fig. 4 is covered, what change would occur in the intensity of light at the center of the screen?
11. We are all bathed continuously in electromagnetic radia tion, from the Sun, from radio and TV signals, from the stars and other celestial objects. Why do these waves not interfere with each other? 12. In calculating the disturbance produced by a pair of super imposed wavetrains, when should you add intensities and when amplitudes? 13. In Young’s double-slit experiment suppose that screen A in Fig. 6 contained /wo very narrow parallel slits instead of one. (o) Show that if the spacing between these slits is properly chosen, the interference fringes can be made to disappear. (b) Under these conditions, would you call the beams emerging from slits Si and 5*2in screen B coherent? They do not produce interference fringes, (c) Discuss what would happen to the interference fringes in the case of a single slit in screen A if the slit width were gradually increased. 14. Defend this statement: Figure 7o is a sine (or cosine) wave but Fig. lb is not. Indeed, you cannot assign a unique fre quency to the curve of Fig. lb. Why not? {Hint: Think of Fourier analysis.) 15. Most of us are familiar with rotating or oscillating radar antennas that produce rotating or oscillating beams of mi crowave radiation. It is also possible to produce an oscilla ting beam of microwave radiation without any mechanical motion of the transmitting antenna. This is done by periodi cally changing the phase of the radiation as it emerges from various sections of the (long) transmitting antenna. Con vince yourself that, by constructive interference from various parts of the fixed antenna, an oscillating microwave beam can indeed be so produced. 16. What causes the fluttering of radio reception when an air plane flies overhead? 17. Is it possible to have coherence between light sources emit ting light of different wavelengths? 18. An automobile directs its headlights onto the side of a bam. Why are interference fringes not produced in the region in which light from the two beams overlaps? 19. Suppose that the film coating in Fig. 14 had an index of refraction greater than that of the glass. Could it still be nonreflecting? If so, what difference would the coating make? 20. What are the requirements for maximum intensity when viewing a thin film by transmitted light? 21. Why does a film (for example, a soap bubble or an oil slick) have to be “thin” to display interference effects? Or does it? How thin is “thin” ?
22. Why do coated lenses (see Sample Problem 5) look purple by reflected light? 23. Ordinary store windows or home windows reflect light from both their interior and exterior plane surfaces. Why then do we not see interference effects? 24. If you wet your eyeglasses to clean them you will notice that as the water evaporates the glasses become markedly less reflecting for a short time. Explain why. 25. A lens is coated to reduce reflection, as in Sample Problem 5. What happens to the energy that had previously been re flected? Is it absorbed by the coating? 26. Consider the following objects that produce colors when exposed to sunlight: ( 1) soap bubbles, (2) rose petals, (3) the inner surface of an oyster shell, (4) thin oil slicks, (5) nonre flecting coatings on camera lenses, and (6) peacock tail feathers. The colors displayed by all but one of these are purely interference phenomena, no pigments being in volved. Which one is the exception? Why do the others seem to be “colored ”? 27. A soap film on a wire loop held in air appears black at its thinnest portion when viewed by reflected light. On the other hand, a thin oil film floating on water appears bright at its thinnest portion when similarly viewed from the air above. Explain these phenomena. 28. Very small changes in the angle of incidence do not change the interference conditions much for “thin” films but they do change them for “thick” films. Why? 29. An optical flat is a slab of glass that has been ground flat to within a small fraction of a wavelength. How may it be used to test the flatness of a second slab of glass? 30. In a Newton’s rings experiment, is the central spot, as seen by reflection, dark or light? Explain. 31. In connection with the phase change on reflection at an interface between two transparent media, do you think that phase shifts other than 0 or ttare possible? Do you think that phase shifts can be calculated rigorously from Maxwell’s equations? 32. The directional characteristics of a certain radar antenna as a receiver of radiation are known. What can be said about its directional characteristics as a transmitter? 33. A person in a dark room, looking through a small window, can see a second person standing outside in bright sunlight. The second person cannot see the first person. Is this a failure of the principle of optical reversibility? Assume no absorption of light. 34. Why is it necessary to rotate the interferometer in the Michelson - Morley experiment? 35. How is the negative result of the Michelson - Morley experi ment interpreted according to Einstein’s theory of relativ ity? 36. If the pathlength to the movable mirror in Michelson’s inter ferometer (see Fig. 19) greatly exceeds that to the fixed mirror (say, by more than a meter) the fringes begin to disappear. Explain why. Lasers greatly extend this range. Why? 37. How would you construct an acoustical Michelson interfer ometer to measure wavelengths of sound? Discuss differ ences from the optical interferometer.
Problems
963
PROBLEMS Section 45-1 Double-Slit Interference
1. Monochromatic green light, wavelength = 554 nm, illumi
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3.
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5.
6.
7.
8.
9.
10.
11. 12.
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nates two parallel narrow slits 7.7 /im apart. Calculate the angular deviation of the third-order, w = 3, bright fringe {a) in radians and (b) in degrees. In a double-slit experiment to demonstrate the interference of light, the separation d of the two narrow slits is doubled. In order to maintain the same spacing of the fringes on the screen, how must the distance D of the screen from the slits be altered? (The wavelength of the light remains un changed.) A double-slit experiment is performed with blue-green light of wavelength 512 nm. The slits are 1.2 mm apart and the screen is 5.4 m from the slits. How far apart are the bright fringes as seen on the screen? Find the slit separation of a double-slit arrangement that will produce bright interference fringes 1.00 ° apart in angular separation. Assume a wavelength of 592 nm. A double-slit arrangement produces interference fringes for sodium light (A = 589 nm) that are 0.23° apart. For what wavelength would the angular separation be 10% greater? Assume that the angle 0 is small. A double-slit arrangement produces interference fringes for sodium light (A = 589 nm) that are 0.20° apart. What is the angular fringe separation if the entire arrangement is im mersed in water {n = 1.33)? In a double-slit experiment the distance between slits is 5.22 mm and the slits are 1.36 m from the screen. Two interfer ence patterns can be seen on the screen, one due to light with wavelength 480 nm and the other due to light with wave length 612 nm. Find the separation on the screen between the third-order interference fringes of the two different pat terns. In an interference experiment in a large ripple tank (see Fig. 3), the coherent vibrating sources are placed 120 mm apart. The distance between maxima 2.0 m away is 180 mm. If the speed of ripples is 25 cm/s, calculate the frequency of the vibrators. If the distance between the first and tenth minima of a dou ble-slit pattern is 18 mm and the slits are separated by 0.15 mm with the screen 50 cm from the slits, what is the wave length of the light used? A thin flake of mica {n = 1.58) is used to cover one slit of a double-slit arrangement. The central point on the screen is occupied by what used to be the seventh bright fringe. If A= 550 nm, what is the thickness of the mica? Sketch the interference pattern expected from using two pinholes, rather than narrow slits. Two coherent radio point sources separated by 2.0 m are radiating in phase with A= 0.50 m. A detector moved in a circular path around the two sources in a plane containing them will show how many maxima? In the front of a lecture hall, a coherent beam of monochro matic light from a helium -neon laser (A = 632.8 nm) illu minates a double slit. From there it travels a distance of 20.0 m to a mirror at the back of the hall, and returns the
same distance to a screen, (a) In order that the distance between interference maxima be 10.0 cm what should be the distance between the two slits? (b) State what you will see if the lecturer slips a thin sheet of cellophane over one of the slits. The path through the cellophane contains 2.5 more waves than a path through air of the same geometric thick ness. 14. One slit of a double-slit arrangement is covered by a thin glass plate of index of refraction 1.4, and the other by a thin glass plate of index of refraction 1.7. The point on the screen where the central maximum fell before the glass plates were inserted is now occupied by what had been the m = 5 bright fringe before. Assume A = 480 nm and that the plates have the same thickness t and find the value of t. 15. Two point sources, 5, and S i in Fig. 21, emit coherent waves. Show that curves, such as that given, over which the phase difference for rays r, and Tj is a constant, are hyperbo las. {Hint: A constant phase difference implies a constant difference in length between r, and Tj .) The OMEGA sys tem of sea navigation relies on this principle. S, and 5*2 are phase-locked transmitters. The ship’s navigator notes the received phase difference on an oscilloscope and locates the ship on a hyperbola. Reception of signals from a third trans mitter is needed to determine the position on that hyper bola.
Figure 21
Problem 15.
16 Sodium light (A = 589 nm) falls on a double slit of separa tion = 0.180 mm. A thin lens ( / = 1.13 m) is placed near the slit as in Fig. 5. What is the linear fringe separation on a screen placed in the focal plane of the lens? 17. Sodium light (A = 589 nm) falls on a double slit of separa tion d = 2.0 mm. The slit-screen distance D is 40 mm. What fractional error is made by using Eq. 1 to locate the tenth bright fringe on the screen? Section 45-2 Coherence 18. The coherence length of a wavetrain is the distance over which the phase constant is the same, {a) If an individual
964
Chapter 45
Interference
atom emits coherent light for 1 X 10“ ®s, what is the coher ence length of the wavetrain? {b) Suppose this wavetrain is separated into two parts with a partially reflecting mirror and later reunited after one beam travels 5 m and the other 10 m. Do the waves produce interference fringes observable by a human eye?
26. One of the slits of a double-slit system is wider than the other, so that the amplitude of the light reaching the central part of the screen from one slit, acting alone, is twice that from the other slit, acting alone. Derive an expression for the intensity I in terms of 6. Section 45-4 Interference from Thin Films
Section 45~3 Intensity in Double-Slit Interference 19. Source A of long-range radio waves leads source 5 by 90®. The distance r^ to a detector is greater than the distance r^ by 100 m. What is the phase difference at the detector? Both sources have a wavelength of 400 m. 20 Find the phase difference between the waves from the two slits arriving at the m th dark fringe in a double-slit experi ment. 21. Light of wavelength 600 nm is incident normally on two parallel narrow slits separated by 0.60 mm. Sketch the in tensity pattern observed on a distant screen as a function of angle 0 for the range of values 0 ^ ^ 0.0040 radians. 22 Find the sum of the following quantities (a) graphically, using phasors, and (b) using trigonometry: y, = 10 sin cot, y 2 = 8.0 sin {cot + 30®). 23. Si and S 2 in Fig. 22 are effective point sources of radiation, excited by the same oscillator. They are coherent and in phase with each other. Placed a distance d = 4 Al m apart, they emit equal amounts of power in the form of 1.06-m wavelength electromagnetic waves, (a) Find the positions of the first (that is, the nearest), the second, and the third max ima of the received signal, as the detector D is moved out along O x. (b) Is the intensity at the nearest minimum equal to zero? Justify your answer.
27. We wish to coat a flat slab of glass {n = 1.50) with a transpar ent material (n = 1.25) so that light of wavelength 620 nm (in vacuum) incident normally is not reflected. What mini mum thickness could the coating have? 28. A thin film in air is 410 nm thick and is illuminated by white light normal to its surface. Its index of refraction is 1.50. What wavelengths within the visible spectrum will be inten sified in the reflected beam? 29. A disabled tanker leaks kerosene {n = 1.20) into the Persian Gulf, creating a large slick on top of the water (n = 1.33). (a) If you are looking straight down from an airplane onto a region of the slick where its thickness is 460 nm, for which wavelength(s) of visible light is the reflection the greatest? (b) If you are scuba-diving directly under this same region of the slick, for which wavelength(s) of visible light is the trans mitted intensity the strongest? 30. In costume jewelry, rhinestones (made of glass with n = 1.5) are often coated with silicon monoxide (w = 2.0 ) to make them more reflective. How thick should the coating be to achieve strong reflection for 560-nm light, incident nor mally? 31. If the wavelength of the incident light is A = 572 nm, the rays A and B in Fig. 23 are out of phase by 1.50A. Find the thickness d of the film.
Figure 23 Figure 22
Problem 31.
Problem 23.
24. Add the following quantities graphically, using the phasor method (see Sample Problem 3), and algebraically: y, = 10 sin cot, y 2 = 14 sin {cot + 26®), yj = 4.7 sin {cot — 41®). 25. Show that the half-width fringes is given by
of the double-slit interference
if 6 is small enough so that sin 0 ^ 6 . The half-width is the angle between the two points in the fringe where the inten sity is one-half that at the center of the fringe.
32. Light of wavelength 585 nm is incident normally on a thin soapy film (n = 1.33) suspended in air. If the film is 0.0012 1 mm thick, determine whether it appears bright or dark when observed from a point near the light source. 33. A plane wave of monochromatic light falls normally on a uniformly thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Com plete destructive interference of the reflected light is ob served for wavelengths o f485 and 679 nm and for no wave lengths between them. If the index of refraction of the oil is 1.32 and that of the glass is 1.50, find the thickness of the oil film. 34. White light reflected at perpendicular incidence from a soap film has, in the visible spectrum, an interference maximum at 600 nm and a minimum at 450 nm with no minimum
Problems
35.
36.
37.
38.
in-between. If « = 1.33 for the film, what is the film thick ness, assumed uniform? Two pieces of plate glass are held together in such a way that the air space between them forms a very thin wedge. Light of wavelength 480 nm strikes the upper surface perpendicu larly and is reflected from the lower surface of the top glass and the upper surface of the bottom glass, thereby produc ing a series of interference fringes. How much thicker is the air wedge at the sixteenth fringe than it is at the sixth? A sheet of glass having an index of refraction of 1.40 is to be coated with a film of material having an index of refraction of 1.55 such that green light (wavelength = 525 nm) is pref erentially transmitted, (a) What is the minimum thickness of the film that will achieve the result? (b) Why are other parts of the visible spectrum not also preferentially transmit ted? (c) Will the transmission of any colors be sharply re duced? A thin film of acetone (index of refraction = 1.25) is coating a thick glass plate (index of refraction = 1.50). Plane light waves of variable wavelengths are incident normal to the film. When one views the reflected wave, it is noted that complete destructive interference occurs at 600 nm and contructive interference at 700 nm. Calculate the thickness of the acetone film. An oil drop (n = 1.20) floats on a water (n = 1.33) surface and is observed from above by reflected light (see Fig. 24). (a) Will the outer (thinnest) regions of the drop correspond to a bright or a dark region? (b) How thick is the oil film where one observes the third blue region from the outside of the drop? (c) Why do the colors gradually disappear as the oil thickness becomes larger? Incident light
. W ater;
Figure 24
Problem 38.
39. A broad source of light (A = 680 nm) illuminates normally two glass plates 120 mm long that touch at one end and are separated by a wire 0.048 mm in diameter at the other end (Fig. 25). How many bright fringes appear over the 120-mm distance? I Incident light
965
( 6)
Figure 26
Problem 40.
26a. They touch at A. Light of wavelength 600 nm is inci dent normally from above. The location of the dark fringes in the reflected light is shown on the sketch of Fig. 26b. (a) How thick is the space between the glass and the plastic at B2 (b) Water (n = 1.33) seeps into the region between the glass and plastic. How many dark fringes are seen when all the air has been displaced by water? (The straightness and equal spacing of the fringes is an accurate test of the flatness of the glass.) 41. Light of wavelength 630 nm is incident normally on a thin wedge-shaped film with index of refraction 1.50. There are ten bright and nine dark fringes over the length of film. By how much does the film thickness change over this length? 42. In an air wedge formed by two plane glass plates, touching each other along one edge, there are 4(X)1 dark lines ob served when viewed by reflected monochromatic light. When the air between the plates is evacuated, only 4000 such lines are observed. Calculate the index of refraction of the air from these data. 43. In a Newton’s rings experiment the radius of curvature R of the lens is 5.0 m and its diameter is 20 mm. (a) How many rings are produced? (b) How many rings would be seen if the arrangement were immersed in water (n = 1.33)? Assume that A = 589 nm. 44. The diameter of the tenth bright ring in a Newton’s rings apparatus changes from 1.42 to 1.27 cm as a liquid is intro duced between the lens and the plate. Find the index of refraction of the liquid. 45. A Newton’s rings apparatus is used to determine the radius of curvature of a lens. The radii of the nth and (n + 20)th bright rings are measured and found to be 0.162 cm and 0.368 cm, respectively, in light of wavelength 546 nm. Cal culate the radius of curvature of the lower surface of the lens. 46. In the Newton’s rings experiment, show (a) that the differ ence in radius between adjacent rings (maxima) is given by A r= r„ + ,
0.048 mm
____ L
assuming m and (b) that the area between adjacent rings (maxima) is given by A = nXR,
-120 mm Figure 25
Problem 39.
40. A perfectly flat piece of glass {n = 1.5) is placed over a per fectly flat piece of black plastic (n = 1.2) as shown in Fig.
assuming m 1. Note that this area is independent of m. 47. In Sample Problem 5 assume that there is zero reflection for light of wavelength 550 nm at normal incidence. Calculate the factor by which the reflection is diminished by the coat ing at {a) 450 nm and {b) 650 nm. (Hint: Calculate (f> in Eq. 13.)
966
Chapter 45
Interference
48. A ship approaching harbor is transmitting at a wavelength of A= 3.43 m from its antenna located /z = 23 m above sea level. The receiving station antenna is located / / = 160 m above sea level. What is the horizontal distance D between ship and receiving tower when radio contact is momentarily lost for the first time? Assume that the calm ocean reflects radio waves perfectly according to the law of reflection. See Fig. 27.
Figure 27
51. An airtight chamber 5.0 cm long with glass windows is placed in one arm of a Michelson interferometer as indi cated in Fig. 28. Light of wavelength A = 500 nm is used. The air is slowly evacuated from the chamber using a vac uum pump. While the air is being removed, 60 fringes are observed to pass through the view. From these data, find the index of refraction of air at atmospheric pressure.
Problem 48. Figure 28
Problem 51.
Section 45~6 Michelson*s Interferometer 49. If mirror Afj Michelson’s interferometer is moved through 0.233 mm, 792 fringes are counted with a light meter. What is the wavelength of the light? 50. A thin film with n = \ A2 for light of wavelength 589 nm is placed in one arm of a Michelson interferometer. If a shift of 7.0 fringes occurs, what is the film thickness?
52. Write an expression for the intensity observed in Michel son’s interferometer (Fig. 19) as a function of the position of the movable mirror. Measure the position of the mirror from the point at which d i = d 2.
CHAPTER 46 DIFFRACTION
Diffraction is the bending or spreading o f waves that encounter an object (a barrier or an opening) in their path. This chapter considers only diffraction o f light waves, but diffraction occurs for all types o f waves. Sound waves, for example, are diffracted by ordinary objects, and as a result we can hear sounds even though we m ay not be in a direct line to their source. For diffraction to occur, the size o f the object must be o f the order o f the wavelength o f the incident waves; when the wavelength is much smaller than the size o f the object, diffraction is ordinarily not observed and the object casts a sharp shadow. Diffraction patterns consist o f light and dark bands similar to the interference patterns discussed in Chapter 45. By studying these patterns, we can learn about the diffracting object. For example, diffraction o f x rays is an important method for the study o f the structure o f solids, and diffraction o f gamma rays is used to study nuclei. Diffraction also has unwanted effects, such as the spreading o f light as it enters the aperture o f a telescope, which limits its ability to resolve or separate stars that appear to be close to one another. These various effects o f diffraction are considered in this chapter and the following one.
46-1 DIFFRACTION AND THE WAVE THEORY OF LIGHT W hen light passes through a narrow slit (o f w idth com pa rable to the wavelength o f the light; see Fig. 1 o f C hapter 43), the light beam s not only flare o u t far beyond the geom etrical shadow o f the slit; they also give rise to a series o f alternating light an d dark bands th at resem ble interfer ence fringes (Fig. 1). In C hapter 45, we argued th at the appearance o f interference fringes provides strong evi dence for the wave n ature o f light. W e can also argue th at the appearance o f diffraction p atterns sim ilarly requires th at light m ust travel as waves.
A lthough diffraction was already know n at the tim e o f H uygens and N ew ton, neither o f them believed th at it provided evidence th at light m ust be a wave. N ew ton in particular believed th at light traveled as a stream o f p arti cles. A strong p ro ponent o f the wave theory o f light was the French engineer A ugustin Fresnel (1 7 8 8 -1 8 2 7 ). Fresnel explained diffraction based on the wave theory, which was not widely accepted even after T hom as Y oung’s ex perim ents on double-slit interference. In 1819, Fresnel subm itted a paper on his theory o f diffraction in a com pe tition sponsored by the French A cadem y o f Sciences. O ne o f the m em bers o f the A cadem y, Sim eon-D enis Poisson (a strong opponent o f the wave theory o f light), ridiculed
Figure 1 The diffraction pattern produced when light passes through a narrow slit.
967
968
Chapter 46
Diffraction
Figure 2 The diffraction pattern of a disk. Note the bright Poisson spot at the center of the pattern.
Fresnel’s theory because, as Poisson himself showed, Fresnel’s diffraction theory led to the “absurd” prediction that the shadow o f an opaque object should have a bright spot at its center. Figure 2 shows the diffraction pattern of a disk; the clearly visible bright spot at its center (known as the Poisson spot) supports Fresnel’s interpretation. Figure 3 shows the diffraction pattern produced when an ordinary object is illuminated by monochromatic light. Actually, you don’t need special apparatus to ob serve diffraction. Hold two fingers so that there is a narrow slit between them, and look at a light bulb through the slit. The dark lines you see in the slit are caused by diffraction. Another common example of diffraction is the “floaters” that many people can observe in their field of view. Floaters are translucent dots or tiny chains that appear to float and drift. They can be seen by focusing the eyes at a distance while staring at a brightly illuminated piece of white paper. Floaters are caused by blood cells and other microscopic debris in the fluid of the eyeball; what we observe is the diffraction pattern on the retina. Figure 4 shows the generalized diffraction situation. The curved surfaces on the left represent wavefronts of the incident light. The light falls on the diffracting object 5, which we show in Fig. 4 as an opaque barrier containing an aperture o f arbitrary shape. (Later, we consider an aperture that is a single narrow slit, which produced the diffraction pattern shown in Fig. 1.) C in Fig. 4 is a screen or photographic film that receives the light that passes through or around the diffracting object. We can calculate the pattern of light intensity on screen C by subdividing the wavefront into elementary areas rfA, each o f which becomes a source of an expanding
Figure 3 The diffraction pattern of a razor blade viewed in monochromatic light. Note the fringes near the edges.
Figure 4 Diffraction occurs when coherent wavefronts of light fall on opaque barrier B, which contains an aperture of arbitrary shape. The diffraction pattern can be viewed on screen C.
Huygens wavelet. The light intensity at an arbitrary point
P is found by superimposing the wave disturbances (that is, the E vectors) caused by the wavelets reaching P from all these elementary sources. The wave disturbances reaching P differ in amplitude and in phase because (1) the elementary sources are at varying distances from P, (2) the light leaves the elemen tary sources at various angles to the normal to the wavefront, and (3) some sources are blocked by barrier B\ others are not. Diffraction calculations, which are simple in principle, may become difficult in practice. The calcu lation must be repeated for every point on screen C at which we wish to know the light intensity. We followed
Section 46-1
Diffraction and the Wave Theory o f Light
969
Figure 5 Light from point source S illuminates a slit in the opaque barrier B. The slit extends a long distance above and below the plane of the figure; this distance is much greater than the slit width a. The intensity at point P on screen C depends on the relative phases of the light re ceived from various parts of the slit, (a) If source S and screen C are moved to large distances from the slit, both the incident and emergent light at B consist of nearly parallel rays. (b) Rather than using large distances, the source and the screen can each be placed in the focal plane of a lens; once again, parallel light rays enter and leave the slit, (c) Without the lens, the rays are not parallel.
exactly this program in calculating the double-slit inten sity pattern in Section 45-3. The calculation there was simple because we assumed only two elementary sources, the two narrow slits. Figure 5 shows another representation of Fig. 4, in the form o f ray diagrams. The pattern formed on the screen depends on the separation between the screen C and the aperture B. In general, we can consider three cases: 1. Very small separation. When C is very close to B, the waves travel only a short distance after leaving the aper
ture, and the rays diverge very little. The effects o f diffrac tion are negligible, and the pattern on the screen is the geometric shadow of the aperture. 2. Very large separation. Figure 5a represents the situa tion when the screen is so far from the aperture that we can regard the rays as parallel or, equivalently, the wavefronts as planes. (In this case, we also assume the source to be far from the aperture, so that the incident wavefronts are also planes. The same effect can be achieved by illumi nating the aperture with a laser.) One way of achieving this condition, which is known as Fraunhofer diffraction.
970
Chapter 46
Diffraction
in the laboratory is to use two converging lenses, as in Fig. 5b. The first lens converts the diverging light from the source into a plane wave, and the second lens focuses plane waves leaving the aperture to point P. All rays that reach P leave the aperture parallel to the dashed line Px drawn from P through the center of the second lens. 3. Intermediate separation. In the case shown in Fig. 5c, the screen can be at any distance from the aperture, and the rays entering and leaving the aperture are not parallel. This general case is called Fresnel diffraction. Although Fraunhofer diffraction is a special limiting case o f the more general Fresnel diffraction, it is an im portant case and is easier to handle mathematically. We assumed Fraunhofer diffraction in our analysis of double slit interference in Chapter 45. In this book we deal only with Fraunhofer diffraction.
shown. The ray xP | passes undeflected through the center of the lens and therefore determines 6. Ray r, originates at the top of the slit and ray rj at its center. If 6 is chosen so that the distance bb' in the figure is one-half wavelength, r, and Tj are out of phase and interfere destructively at P ,. The same is true for a ray just below r, and another just below T2 . In fact, for every ray passing through the upper half o f the slit, there is a corresponding ray passing through the lower half, originating at a point a/2 below the first ray, such that the two rays are out o f phase at P ,. Every ray arriving at P, from the upper half o f the slit interferes destructively with one coming from the bottom half of the slit. The intensity at P, is therefore zero, and P, is the first minimum o f the diffraction pattern. Since the distance bb' equals (a/2) sin 6, the condition for the first minimum can be written a . A - sin 0 = - ,
2
2
or
46-2
SIN G LE-SL IT DIFFRACTION
The simplest diffraction pattern to analyze is that pro duced by a long narrow slit. In this section we discuss the locations o f the minima and maxima in the pattern as shown in Fig. 1. In the next section we calculate the inten sity o f the pattern as a function of position on the screen. Figure 6 shows a plane wave falling at normal incidence on a slit o f width a. Let us first consider the central point Pq. Rays that leave the slit parallel to the central horizon tal axis are brought to a focus at Pq. These rays are cer tainly in phase at the plane o f the slit, and they remain in phase as they are brought to a focus by the lens (see, for example. Fig. 17a of Chapter 44). Since all rays arriving at Pqare in phase, they interfere constructively and produce a maximum o f intensity at P q . We now consider another point on the screen. Light rays that reach P, in Fig. 7 leave the slit at the angle 6, as
Incident
a sin 0 = A.
( 1)
Equation 1 shows that the central maximum becomes wider as the slit is made narrower. If the slit width is as small as one wavelength (a = A), the first minimum occurs at 6 = 90° (sin 0 = 1 in Eq. 1), which implies that the central maximum fills the entire forward hemisphere. We assumed a condition approaching this in our discus sion of double-slit interference in Section 45-1. In Fig. 8 the slit is divided into four equal zones, with a ray leaving the top o f each zone. Let 0 be chosen so that the distance bb' is one-half wavelength. Rays r, and Tj then cancel at P^. Rays rj and r^ are also one-half wave length out o f phase and also cancel. Consider four other rays, emerging from the slit a given distance below these four rays. The two rays below r, and rj cancel, as do the two rays below rj and r4 . We can proceed across the entire slit and conclude again that no light reaches P2 ; we have located a second point of zero intensity.
Figure 6 Conditions at the central maximum of the diffraction pattern.
Section 46-2
Single-Slit Diffraction
971
Figure 7 Conditions at the first minimum of the diffraction pattern. The angle 6 is such that the distance bb' is one-half wavelength.
Incident wave
Figure 8 Conditions at the second minimum of the diffraction pattern. The angle 6 is such that the distance bb' is one-half wavelength.
Incident wave
the angle d. By extension o f Eqs. 1 and 2, the general formula for the minima in the diffraction pattern on screen C is
The condition described (see Fig. 8) requires that
a . „ A -s in 0 = - . or a sin 0 = 2A.
(2)
For a given slit width a and wavelength A, Eq. 2 gives the position on the screen of the second minimum in terms o f
a sin 0 = mA
m = 1, 2, 3, . . .
(minima).
(3)
There is a maximum approximately halfway between each adjacent pair of minima. Later in the chapter we
972
Chapter 46
Diffraction
derive a formula for the intensity of the diffracted light, from which the locations o f the maxima can be found exactly. Note that Eq. 3 suggests two minima (and corre sponding maxima) for each m , one at an angle 6 above the central axis and one below (corresponding to m < 0). In deriving Eq. 3, consider how the assumption o f paraUel rays (Fraunhofer diffraction) has simplified the analysis.
Sample Problem 1 A slit of width a is illuminated by white light. For what value of a does the first minimum for red light (A = 650 n m ) f a lla t0 = IS**? Solution At the first minimum, m = 1 in Eq. 3. Solving for a, we then find m k _ (1X650 nm) ^ sin ^ sin 15® = 2510 nm = 2.51 pm . For the incident light to flare out that much (± 15 ®) the slit must be very narrow indeed, amounting to about four times the wave length (and far narrower than a fine human hair, which may only be about 100 pm in diameter).
Sample Problem 2 In Sample Problem 1, what is the wave length k' of the light whose first diffraction maximum (not counting the central maximum) falls at 15®, thus coinciding with the first minimum for red light? Solution
Maxima occur about halfway between minima, so
there is a maximum at 15 ®when the first minimum is at 10®and the second minimum is at 20®. In this case, for the second mini mum, a sin ^ = 2A', or A' = i(2510 nmXsin 20®) = 430 nm. Light of this wavelength is violet. The second maximum for light of wavelength 430 nm always coincides with the first minimum for light of wavelength 650 nm, no matter what the slit width. If the slit is relatively narrow, the angle 6 at which this overlap occurs is relatively large, and conversely.
46-3 INTENSITY IN SINGLE-SLIT DIFFRACTION_________________ In Section 46-2, we located the positions o f the minima of the single-slit diffraction pattern. We now wish to find an expression for the intensity o f the entire pattern as a func tion o f the diffi^ction angle 6. This expression will permit us to find the location and intensity o f the maxima. Figure 9 shows a slit of width a divided into N parallel strips, each o f width Ax. The strips are very narrow, so that each strip can be regarded as a radiator o f Huygens wavelets, and all the li ^ t from a given strip arrives at point P with the same phase. The waves arriving at P
Figure 9 A slit of width a is divided into N strips of width Ax. The inset shows more clearly the conditions at the second strip. In the differential limit, the width dx of each strip becomes infinitesimally small and the number of strips becomes infinitely large. Here and in the next figure we take iV = 18 for clarity.
Section 46-3
Intensity in Single-Slit Diffraction
973
from any pair of adjacent strips have the same (constant) phase difference A
phase difference _ path difference ia)
In or A 0 = ^ Ax sin 0, A
(4)
where Ax sin 0, as the detail of Fig. 9 shows, is the path difference for rays originating from corresponding points of adjacent strips. If the angle 0 is not too large, each strip produces a wave o f the same amplitude A£o at The net effect at P is due to the superposition of N vectors of the same amplitude, each differing in phase from the next by A 0. To find the intensity at P, we must first find the net electric field of the N vectors. In Section 45-3, we introduced a graphical method for adding wave disturbances that enabled us to calculate the intensity in double-slit interference. That method is based on representing each wave disturbance as a phasor (a ro tating vector) and finding the resultant phasor amplitude by vector addition, taking into account the relative phase given by Eq. 4. The resultant electric field varies with 0, because the phase difference A 0 varies with 0. Let us consider some examples of the addition of phasors in single-slit diffraction. We first consider the result ant electric field at point Pq(the center of the diffraction pattern on the screen). In this case 0 = 0, and Eq. 4 gives A 0 = 0 as the phase difference between adjacent strips. According to the method of Section 45-3, we then lay N vectors o f length A^o head to tail and parallel to one another (A 0 = 0). The resultant E qis shown in Fig. lOfl. This is clearly the maximum value that the resultant of these N vectors can take, so we label it E ^ . As we move away from 0 = 0, the phase difference A 0 assumes a definite nonzero value. Again laying the vec tors head to tail, each differing in direction from the previous one by A 0 , we obtain the resultant shown in Fig. 106. Note that E qis smaller than it was in Fig. 10a. Now consider the first minimum of the diffraction pat tern (point P, in Fig. 7). At this point the intensity is zero, so the resultant E q must be zero. This means that the N phasors, laid head to tail, must form a closed loop, as in Fig. 10c. Beyond the first minimum, the phase shift A 0 is still larger, and the chain of vectors coils around through an angle greater than 360°. At a certain angle (corresponding to a certain phase shift, as in Fig \0 d \ the resultant has its greatest length within this loop, corresponding to the first maximum beyond the central one. Note that the intensity o f this maximum is much smaller than the in tensity of the central maximum, represented in Fig. 10a. Eventually, this loop closes on itself, giving a resultant of zero and corresponding to the second minimum. Our goal in finding the intensity of the single-slit dif-
Figure 10 Phasors in single-slit diffraction, showing condi tions at (a) the central maximum, (6) a direction slightly re moved from the central maximum, (c) the first minimum, and (d) the first maximum beyond the central maximum. This figure corresponds to A^= 18 in Fig. 9.
fraction pattern for any 0 is to evaluate the phase shift according to Eq. 4 and find the resultant E q, as in Fig. 106. The square of this resultant then gives the relative inten sity, as in Section 45-3. The light arriving at P from a given strip is in phase only if the strip is infinitesimally small and the number of strips is correspondingly large. The chain of phasors o f Fig. 106 then approaches the arc of a circle, as drawn in Fig. 11. The length of the arc is £■„, while the amplitude we seek for the resultant field is indicated by the chord E q. The angle 0 is the total phase difference between the rays from the top and bottom of the strip; as Fig. 11 shows, 0 is also the angle between the two radii P. From this figure we can write (h
E q= 2R sin y .
974
Chapter 46
Diffraction
From Fig. 11, 0 in radian measure is ^
R
Combining yields E„
.
or
Eg = Er
sm a
(5)
a
in which
a
2
( 6)
•
From Fig. 9, recalling that (f> is the phase difference between rays from the top and the bottom o f the slit and that the path difference for these rays is a sin 6, we have phase difference _ path difference 2n A ’
Figure 11 A construction used to calculate the intensity in single-slit diffraction. The situation corresponds to that of Fig.
10ft.
or
Combining with Eq. 6 yields
1.0
0 -
0.8
-
0.6
-20
-15
-1 0
-5
-20 ( 6)
0.2
0
10
15
20
_ , /s in a V
6 (degrees)
(a)
-15
-10
-5
(7)
Equation 5, with a evaluated according to Eq. 7, gives the amplitude of the wave disturbance for a single-slit diffraction pattern at any angle 6. The intensity Ig for the pattern is proportional to the square of the amplitude, so
-0.4 -
na . .
0
10
15
$ (degrees)
1.0
20
( 8)
Equation 8, combined with Eq. 7, gives the result we seek for the intensity o f the single-slit difiraction pattern at any 6. Figure 12 shows plots o f the relative intensity Ig/Ia for several values of the ratio a/X. Note that the pattern be comes narrower as we increase a/X. (See also Fig. 1 of Chapter 43.) Minima occur in Eq. 8 when
a = mn
m = ±l,±2,±3,. . . .
(9)
Combining with Eq. 7 leads to
as i nd = niX
(c)
m = ± 1, ± 2 , ± 3 , . . .
(minima),
which is the result derived in the preceding section (Eq. 3). In that section, however, we derived only this result, ob taining no quantitative information about the intensity of the difiraction pattern at places in which it was not zero. Here (Eq. 8) we have complete intensity information. 6 (degrees)
Figure 12 The intensity distribution in single-slit diffraction for three different values of the ratio a/X, The wider the sht, the narrower is the central diffraction peak. As indicated in (ft), Ad gives a measure of the width of the central peak.
Sample Problem 3 Calculate, approximately, the relative in tensities of the secondary maxima in the single-sUt Fraunhofer diffraction pattern.
Section 46-4 Solution The secondary maxima lie approximately halfWay between the minima and are roughly given by (see Problem 15) m=l,2, 3 ,...
,
with a similar result for m < 0. Substituting into Eq. 8 yields ^
r sin (m -f \) n Y
' '”L (m+i)7T J ’ which reduces to h
1
(m + i )2 7t^ This yields 7^//n, = 0.045 ( m = l ) , 0.016 (w = 2), 0.0083 (m = 3), and so forth. The successive maxima decrease rapidly in intensity.
Sample Problem 4 Derive the width Ad of the central maxi mum in a single-slit Fraunhofer diffraction (see Fig. \2b). The width can be represented as the angle between the two points in the pattern where the intensity is one-half that at the center of the pattern. Solution Point x in Fig. 12^ is chosen so that /
Diffraction at a Circular Aperture
975
46-4 DIFFRACTION AT A CIRCULAR APERTURE_________ In focusing an image, a lens passes only the light that falls within its circular perimeter. From this point o f view, a lens behaves like a circular aperture in an opaque screen. Such an aperture forms a diffraction pattern analogous to that of a single slit. Diffraction effects often limit the abil ity of telescopes and other optical instruments to form precise images. The image formed by a lens can be distorted by other effects, including chromatic and spherical aberrations. These effects can be substantially reduced or eliminated by suitable shaping of the lens surfaces or by introducing correcting elements into the optical system. However, no amount of clever design can eliminate the effects of dif fraction, which are determined only by the size of the aperture (the diameter of the lens) and the wavelength of the light. In diffraction, nature imposes a fundamental limitation on the precision of our instruments. When we used geometrical optics to analyze lenses, we assumed diffraction not to occur. However, geometrical optics is itself an approximation, being the limit of wave optics. If we were to make a rigorous wave-optical analysis of the formation o f an image by a lens, we would find that diffraction effects arise in a natural way. Figure 13 shows the image of a distant point source of light (a star) formed on a photographic film placed in the focal plane of the converging lens of a telescope. It is not a
( 10)
To solve this on your calculator, enter the “radian” mode. Pick any starting value for say = 1. Plug this value into the right side of Eq. 10 and solve, obtaining 1.19. Equation 10 re quires that this value must then be equal to , which it is clearly not (1 # 1.19). Take 1.19 as the new trial value, and again evalu ate the right-hand side, obtaining 1.31. We still do not have a solution that satisfies Eq. 10(1.19# 1.31), but we are closer than we were on our first try. Continue in this way, using the result of one calculation as the starting point of the next, until the differ ence between the calculated value of the right-hand side of Eq. 10 and the starting value becomes as small as you like. (You can set this up as a program for a calculator or a computer and have it do the repetitions automatically. You can also have it stop when the difference between successive values becomes smaller than a limit you can set.) This method is called the iterative technique for solving equations. After 10 iterations, the result is a ^ = 1.39156, and additional iterations change only the fifth decimal place. Inserting this value into Eq. 7, we obtain 6x = s\n~^ The width of the curve is then found from A6 = 2 e = 10.2^
Figure 13 The diffraction pattern of a circular aperture. The central maximum is sometimes called the Airy disk (after Sir George Airy, who first solved the problem of diffiraction by a circular aperture in 1835). Note the circular secondary maxima.
976
Chapter 46
Diffraction
Figure 14 The images of two distant point sources (stars) formed by a converging lens. The diameter of the lens (which is the diffracting aperture) is 10 cm, so that a/X = 200,000 if the effective wavelength is about 500 nm. In {a) the stars are so close together that their images can scarcely be distinguished, owing to the overlap of their diffraction patterns. In {b) the stars are farther apart and their separation meets Rayleigh’s criterion for resolution of their images. In (c) the stars are still farther apart and their images are well resolved. Computer-generated profiles of the intensities are shown below the images.
point, as the (approxim ate) geom etrical optics treatm en t suggests, b u t a circular disk su rrounded by several progres sively fainter secondary rings. C om parison w ith Fig. 1 leaves little d o u b t th at we are dealing w ith a diffraction phenom enon. The m athem atical analysis o f diffraction by a circular aperture, which is beyond the level o f this text, shows that (under F raun h o fer conditions) the first m in im u m occurs at an angle from the central axis given by
In Fig. 146 the angular separation o f the two point sources is such th at the central m axim um o f the diffrac tion pattern o f one source falls on the first m in im u m of the diffraction pattern o f the other. This is called R a y leigh 5 criterion for resolving images. From Eq. 11, two objects th at are barely resolvable by Rayleigh’s criterion m ust have an angular separation of
sin 6 = 1.22 - , a
Since the angles involved are rather small, we can replace sin 0 R by 0 r , so
( 11)
where d is the d iam eter o f the aperture. T his is to be com pared w ith Eq. 1, sin 0 = - , a which locates the first m in im u m o f a slit o f w idth a. These expressions differ by the factor 1.22 , w hich arises w hen we divide the circular aperture into elem entary Huygens sources an d integrate over the aperture. T he fact th a t lens images are diffraction p atterns is im p o rtan t w hen we wish to distinguish tw o distant point objects whose angular separation is small. Figure 14 shows the visual appearances and the corresponding in tensity patterns for tw o d istant poin t objects (stars, say) w ith sm all angular separations and approxim ately equal central intensities. In Fig. 14^ the objects are not resolved; th a t is, they can n o t be distinguished from a single point object. In Fig. 146 they are barely resolved, and in Fig. 14c they are fully resolved.
0,
. - , / l . 22 A\
j
0r = 1 .2 2 - ,
( 12)
in which 0 r is expressed in radians. If the angular separa tion 9 between the objects is greater th an 0 r , we can resolve the two objects; if it is less, we cannot. T he angle 0 R is the sm allest angular separation for which resolution is possible, using Rayleigh’s criterion. W hen we wish to use a lens to resolve objects o f small angular separation, it is desirable to m ake the central disk o f the diffraction pattern as small as possible. T his can be done (see Eq. 12) by increasing the lens diam eter or by using a shorter wavelength. O ne reason for constructing large telescopes is to produce sharper images so th at we can exam ine astronom ical objects in finer detail. The images are also brighter, not only because the energy is concentrated into a sm aller diffraction disk b u t because the larger lens collects m ore light. T hus fainter objects, for exam ple, distant galaxies, can be seen.
Section 46-5
Double-Slit Interference and Diffraction Combined
977
(^) The linear separation is Ax = /0 R = (0.24 mX2.10 X 10-^ rad) = 5.0 pm, or about 9 wavelengths of the light.
46-5 DOUBLE-SLIT INTERFERENCE AND DIFFRACTION COM BINED Figure 15 An image of a chain of streptococcus bacteria (di ameter 10“ ^ m) obtained with an electron microscope. Note the sharpness of the image, which would not be possible using visible light.
To reduce diffraction effects in microscopes we often use ultraviolet light, which, because of its shorter wave length, permits finer detail to be examined than would be possible if the same microscope used visible light. We shall see in Chapter 50 that beams of electrons behave like waves under some circumstances. In the electron micro scope such beams may have an effective wavelength of 4 X 10"^nm ,ofthe order o f lOHimes shorter than that of visible light. This permits the detailed examination of tiny objects such as bacteria or viruses (Fig. 15). If such a small object were examined with an optical microscope, its structure would be hopelessly concealed by diffraction.
Sample Problem 5 A converging lens 32 mm in diameter has a focal length / of 24 cm. (a) What angular separation must two distant point objects have to satisfy Rayleigh’s criterion? Assume that A = 550 nm. (b) How far apart are the centers of the diffrac tion patterns in the focal plane of the lens? Solution
{a) From Eq. 12
A _ (1.22X550 X IQ-^m) e^= L22- = d 32 X 10-3 m = 2.10 X 10“ 3 rad = 4.3 arc seconds.
In our analysis of double-slit interference (Section 45-1) we assumed that the slits were arbitrarily narrow; that is, that a c X. For such narrow slits, the central part o f the screen on which the light falls is uniformly illuminated by the diffracted waves from each slit. When such waves interfere, they produce interference fringes o f uniform intensity. In practice, for visible light, the condition a c A is usually not met. For such relatively wide slits, the inten sity of the interference fringes formed on the screen is not uniform. Instead, the intensity o f the fringes varies within an envelope due to the diffraction pattern of a single slit. The effect of difiraction on a double-slit interference pattern is illustrated in Fig. 16, which compares the double-slit pattern with the difiraction pattern produced by a single slit o f the same width as each o f the double slits. You can see from Fig. 16a that the difiraction does indeed provide an intensity envelope for the more closely spaced double-slit interference fringes. Let us now analyze the combined interference and dif fraction pattern o f Fig. 16a. The interference pattern for two infinitesimally narrow slits is given by Eq. 11 o f Chap ter 45, or, with a small change in notation. ^9,int
^m,int
A
(13)
where
= ^ sin 6
(14)
in which d is the distance between the center-lines o f the slits.
Figure 16 (a) Interference fringes for a double-slit system in which the slit width is not negligible in comparison with the wavelength, (b) The diffraction pattern of a single slit of the same width. Note that the diffraction pattern modu lates the intensity of the interference fringes, as shown in part (a).
978,
Chapter 46
Diffraction
The intensity for the diffracted wave from either slit is given by Eq. 8, or, again with a small change in notation, /
= /
(15)
where
na .
a = — Sin 6.
( 16)
(a)
$ (degrees)
We find the combined effect by regarding i„, in Eq. 13 as a variable amplitude, given in fact by o f Eq. 15. This assumption, for the combined pattern, leads to
Ie = IJcosfi)-
(sina a V) ’
(17)
in which we have dropped all subscripts referring sepa rately to interference and diffraction. Later in this section we derive this result using phasors. Let.us express this result in words. At any point on the screen the available light intensity from each slit, consid ered separately, is given by the diffraction pattern o f that slit (Eq. 15). The diffraction patterns for the two slits, again considered separately, coincide because parallel rays in Fraunhofer diffraction are focused at the same spot. Because the two diffracted waves are coherent, they interfere. The effect o f interference is to redistribute the available energy over the screen, producing a set o f fringes. In Sec tion 45-1, where we assumed a c X, the available energy was virtually the same at all points on the screen so that the interference fringes had virtually the same intensities (see Fig. 8 o f Chapter 45). If we relax the assumption a c A, the available energy is no/uniform over the screen but is given by the diffraction pattern o f a slit of width a. In this case the interference fringes have intensities that are determined by the intensity o f the diffraction pattern at the location o f a particular fringe. Equation 17 is the math ematical expression of this argument. This is especially clear in Fig. 17, which shows (a) the “interference factor” in Eq. 17 (that is, the factor cos^)?), (b) the “diffraction factor” (sin a/ a f , and (c) their product. Figure 18 is a plot of the relative intensity le/Im given by Eq. 17 for = 50A and for three values of a/X. It shows clearly that for narrow slits {a = A) the fringes are nearly uniform in intensity. As the slits are widened, the intensi ties o f the fringes are markedly modulated by the “diffrac tion factor” in Eq. 17, that is, by the factor (sin a/a)X\ compare with Fig. 12. If we decrease the slit width a, the envelope of the fringe pattern becomes broader, and the central peak spreads out (compare Figs. 18a and 186). As the slit width a ap proaches zero, a ^ 0 and sin a /a —* 1. Thus Eq. 17 re duces to Eq. 13, which describes interference from a pair o f vanishingly narrow slits. If we let the slit separation d approach zero, the two slits coalesce into a single slit of width a. From Eq. 14, as —>0, and Eq. 17 re
( 6)
(c)
(degrees)
$ (degrees)
Figure 17 (a) Interference fringes that would be produced by a double slit of vanishingly narrow widths. (6) The difiiaction pattern for a slit of finite width, (c) The pattern of interference fringes formed by two slits of the same width as that of (b). This pattern is equivalent to the product of the curves shown in (a) and (b). Compare Fig. 16a.
duces to Eq. 15, the diffraction equation for a single slit of width a. If we increase the slit width a, the envelope o f the fringe pattern becomes narrower, and the central peak becomes sharper (compare Figs. 186 and 18c). The separation be tween the fringes, which depends on d/X, does not change. If we increase the slit separation d, the fringes are closer together, but the envelope o f the fringe pattern, which depends on a/A, does not change. If we increase the wavelength o f the incident light, both the diffraction and interference patterns broaden: the dif fraction envelope becomes wider and the fringe separa tion increases. The reverse effect occurs as we decrease the wavelength. Put another way, the relationship between the diffraction envelope and the interference fringes (for example, the number o f fringes in the central peak) de pends on the ratio d/a and is independent o f A. The double-slit pattern illustrated in Fig. 17 combines interference and diffraction in an intimate way. Both are superposition effects that depend on adding wave distur bances at a given point, taking phase differences properly
Section 46-5
Double-Slit Interference and Diffraction Combined
1.0
Figure 18 Interference fnnges for a dou ble slit with slit separation d = 50A. Three different slit widths are shown.
J
-15
-1 0
-5
0
5
979
10
15
B (degrees)
(a)
e (degrees)
f\
-15 (c)
a = lO X
A
-10
10
15
$ (degrees)
into account. If the waves to be combined originate from a
finite (and usually small) number of elementary coherent radiators, as in the double slit, we call the effect interfer ence. If the waves to be combined originate by subdividing a wave into infinitesimal coherent radiators, as in our treatment o f a single slit, we call the effect diffraction. This distinction between interference and diffraction is conve nient and useful. However, it should not cause us to lose sight o f the fact that both are superposition effects and that often both are present simultaneously, as in the double-slit experiment.
distance from the central maximum to the first minimum of the fringe envelope? Solution {a) The intensity pattern is given by Eq. 17, the fnnge spacing being determined by the interference factor cos^ From Sample Problem 2, Chapter 45, we have Ay = - j - . Substituting yields Ay =
(480 X 10-’ mX52 X 10"^ m)
0.12X 10-^m
= 2.1 mm. Sample Problem 6 In a double-slit experiment, the distance D of the screen from the slits is 52 cm, the wavelength Ais 480 nm, the slit separation disOAl mm, and the slit width a is0.025 mm. (fl) What is the spacing between adjacent fnnges? (b) What is the
(b)
The angular position of the first minimum follows from
Eq. 1, or . „ A 480X 10-»m sin 0 = - = ^ ,— = 0.0192. a 25 X 10 ‘ m
980
Chapter 46
Diffraction
This is so small that, with little error, we can put sin 0 ^ \2i n 0 ^ 6 , s o
= Z) tan ^
= (52 X lO'^ mXO.0192)
= 10 mm. You can show that there are about 9 fringes in the central peak of the diffraction envelope.
Sample Problem 7 What requirements must be met for the central maximum of the envelope of the double-slit interference pattern to contain exactly 11 fringes? Solution The required condition will be met if the sixth mini mum of the interference factor (c o s ^ in Eq. 17) coincides with the first minimum of the diffraction factor [(sin a l a f in Eq. 17]. The sixth minimum of the interference factor occurs when /? = ( l l / 2 ) 7 r
in Eq. 17. The first minimum in the diffraction term occurs for
Figure 19 Each slit in a double slit is divided into N strips. In the differential limit, the strips become infinitesimally small and infinitely numerous. Here, as we did in Fig. 9, we show A^= 18.
a = 7T in Eq. 17. Dividing (see Eqs. 14 and 16) yields ^ = ^ = Ii a a 2 * This condition depends only on the ratio of the slit separation d to the slit width a and not at all on the wavelength. For larger A the pattern is broader than for smaller A, but there are always 11 fringes in the central peak of the envelope.
Phaser Derivation of Eq. 17 (Optional) Figure 19 shows the geometry appropriate for the analysis of the double slit using phasors. Each of the two slits is divided into N zones, as was done for the single slit in Fig. 9. The net electric field at P is found from the superposition of the electric field vectors from the upper slit and the N electric field vectors from the lower slit. The phasor method allows us to combine these contributions to the electric field at P, taking into account their relative phases. Figure 20 shows the first N phasors (corresponding to the upper slit of Fig. 19) and their resultant E",, as in Fig. 11. There is a phase difference A 0 = 0/A/^ between each of the iVphasors. To add the second group of N phasors (corresponding to the lower slit) we must find the phase angle between the last phasor from the upper slit and the first phasor from the lower slit. We than draw the N phasors from the lower slit and find their resultant, Ej • The sum of the E, and Ej phasors gives the resultant Eg that characterizes the double slit. From Fig. 20, we see that E^ is the base of an isosceles triangle whose sides have equal lengths E, and E 2, which are given by Eq. 5. From the geometry of Fig. 20, £ « = 2£ , s i n | ,
(18)
Figure 20 Phasor diagram used to calculate the resultant electric field in double-slit interference.
which gives
J = 7 T -((J + 0). Using Eq. 20 to evaluate sin S/2, we find sin - = sin y - ------— j = cos —^
^ + S + ^ + i= n ,
(19)
.
(21)
From the expression path difference _ phase difference
A where J, the apex angle of the triangle, can be found from
(20)
“
271
with the phase difference between the two rays (from the bottom of the upper slit and the top of the lower slit, as shown in Fig. 19) of i and the path difference of (d — a) sin 6, we have
Questions
1 = j ( < / - a ) s i n 0. Combining this with Eq. 7,
981
Inserting this result into Eq. 18 and using Eq. 5 for the magni tude of £■, (or we obtain £ , = 2£ „ — a
cos/?.
(22)
Squaring Eq. 22 gives the intensities as
which is j u s t a c c o r d i n g to Eq. 14. Substituting into Eq. 21, we find
Ie = H c o s f i f { ^ y ,
sin - = cos p.
which is identical to Eq. 17. Note that, as was the case in Eq. 11 of Chapter 45, /„ = 4/ q. ■
QUESTIONS 1. Distinguish between Fresnel and Fraunhofer diffraction. Do different physical principles underlie them? If so, what are they? If the same broad principle underlies them, what is it? 2. In what way are interference and diffraction similar? In what way are they different? 3. Suppose that you hold a single narrow vertical slit in front of the pupil of your eye and look at a distant light source in the form of a long heated filament. Is the diffraction pattern that you see a Fresnel or a Fraunhofer pattern? 4. Do diffraction effects occur for virtual images as well as for real images? Explain. 5. Do diffraction effects occur for images formed by (a) plane mirrors and (b) spherical mirrors? Explain. 6 . Comment on this statement: “Diffraction occurs in all re gions of the electromagnetic spectrum.” Consider the x-ray region and the microwave region, for example, and give arguments for believing the statement to be true or false. 7. We have claimed (correctly) that Maxwell’s equations pre dict all the classical optical phenomena. Yet in Chapter 45 (Interference) and in this chapter (Diffraction), there is little mention of Maxwell’s equations. Is there an inconsistency here? Where is the impact of Maxwell’s equations felt? Dis cuss. 8 . If we were to redo our analysis of the properties of lenses in Chapter 44 by the methods of geometrical optics but with out restricting our consideration to paraxial rays and to “thin” lenses, would diffraction phenomena emerge from the analysis? Discuss. 9. Why is the diffraction of sound waves more evident in daily experience than that of light waves? 10. Sound waves can be diffracted. About what width of a single slit should you use if you wish to broaden the distribution of an incident plane sound wave of frequency 1 kHz? 11. Why do radio waves diffract around buildings, although light waves do not? 12. A loudspeaker horn, used at a rock music concert, has a rectangular aperture 1 m high and 30 cm wide. Will the pattern of sound intensity be broader in the horizontal plane or in the vertical? 13. A particular radar antenna is designed to give accurate measurements of the altitude of an aircraft but less accurate measurements of its direction in a horizontal plane. Must
the height-to-width ratio of the radar antenna be less than, equal to, or greater than unity? 14. Describe what happens to a Fraunhofer single-slit diffrac tion pattern if the whole apparatus is immersed in water. 15. In single-slit diffraction, what is the effect of increasing (a) the wavelength and (b) the slit width? 16. While listening to the car radio, you may have noticed that the AM signal fades, but the FM signal doesn’t, when you drive under a bridge. Could diffraction have anything to do with this? 17. What will the single-slit diffraction pattern look like if X> al 18. What would the pattern on a screen formed by a double slit look like if the slits did not have the same width? Would the location of the fringes be changed? 19. The shadow of a vertical flagpole cast by the Sun has clearly defined edges near its base, but less-well-defined edges near its top end. Why? 20. Sunlight falls on a single slit of width 1/im. Describe qualita tively what the resulting diffraction pattern looks like. 21. In Fig. 8, rays r, and are in phase; so are Tj and r^. Why isn’t there a maximum intensity at Fj rather than a min imum? 22. When we speak of diffraction by a single slit we imply that the width of the slit must be much less than its length. Sup pose that, in fact, the length was equal to twice the width. Make a rough guess at what the diffraction pattern would look like. 23. In Fig. 7 the optical path lengths from the slit to point Pqare all the same. Why? 24. In Fig. lOJ, why is E q, which represents the first maximum beyond the central maximum, not vertical? {Hint: Consider the effects of a slight winding or unwinding of the coil of phasors in this figure.) See Problem 16. 25. Give at least two reasons why the usefulness of large tele scopes increases as we increase the lens diameter. 26. Are diffraction effects associated with reflecting telescopes, such as the Hubble Space Telescope, that use mirrors in stead of lenses? If so, why do we go to the effort of putting such telescopes in space? 27. We have seen that diffraction limits the resolving power of optical telescopes (see Fig. 14). Does it also do so for large radio telescopes?
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Chapter 46
Diffraction
28. Diffraction is more of a nuisance in a telescope than in a camera. Why? 29. The double-slit pattern of Fig. 21a seen with a monochro matic light source is somehow changed to the pattern of Fig. 2 1b. Consider the following possible changes in conditions: (a) the wavelength of the light was decreased; (b) the wave length of the light was increased; (c) the width of each slit was increased; (d) the separation of the slits was increased; (e) the separation of the slits was decreased; ( / ) the width of each slit was decreased. Which selection(s) of the above changes could explain the alteration of the pattern? 30. In double-slit interference patterns such as that of Fig. 16a, we said that the interference fringes were modulated in in-
Figure 21
Question 29.
tensity by the diffraction pattern of a single slit. Could we reverse this statement and say that the diffraction pattern of a single slit is intensity-modulated by the interference fringes? Discuss.
PROBLEMS Section 46~2 Single-Slit Diffraction 1. When monochromatic light is incident on a slit 0.022 mm wide, the first diffraction minimum is observed at an angle of 1.8° from the direction of the incident beam. Find the wavelength of the incident light. 2. Can you demonstrate the wave nature of x rays by diffract ing them through a single slit? Determine the maximum slit width that could be used if a central maximum angular width of 0.12 mrad can just be detected and you guess the wavelength of the x rays to be 0.10 nm. 3. Monochromatic light of wavelength 441 nm falls on a narrow slit. On a screen 2.16 m away, the distance between the second minimum and the central maximum is 1.62 cm. (a) Calculate the angle of diffraction d of the second mini mum. (^) Find the width of the slit. 4. Light of wavelength 633 nm is incident on a narrow slit. The angle between the first minimum on one side of the central maximum and the first minimum on the other side is 1.97 °. Find the width of the slit. 5. A single slit is illuminated by light whose wavelengths are and Xf,, so chosen that the first diffraction minimum of the Afl component coincides with the second minimum of the A^ component, (a) What relationship exists between the two wavelengths? (b) Do any other minima in the two patterns coincide? 6 . A plane wave, with wavelength of 593 nm, falls on a slit of width 420 pm . A thin converging lens, having a focal length of 71.4 cm, is placed behind the slit and focuses the light on a screen. Find the distance on the screen from the center of the pattern to the second minimum. 7. In a single-sht diffraction pattern the distance between the first minimum on the right and the first minimum on the left is 5.20 mm. The screen on which the pattern is displayed is 82.3 cm from the sht and the wavelength is 546 nm. Calculate the slit width. 8 . The distance between the first and fifth minima of a single slit pattern is 0.350 mm with the screen 41.3 cm away from the slit, using light having a wavelength of 546 nm. (a) Cal culate the diffraction angle 0 of the first minimum, (b) Find the width of the slit. 9. A slit 1.16 mm wide is illuminated by light of wavelength 589 nm. The diffraction pattern is seen on a screen 2.94 m
away. Find the distance between the first two diffraction minima on the same side of the central maximum. 10. Manufacturers of wire (and other objects of small dimen sions) sometimes use a laser to continually monitor the thickness of the product. The wire intercepts the laser beam, producing a diffraction pattern like that of a single slit of the same width as the wire diameter, see Fig. 22. Suppose a H e-N e laser, wavelength 632.8 nm, illuminates a wire, the diffraction pattern being projected onto a screen 2.65 m away. If the desired wire diameter is 1.37 mm, what would be the observed distance between the two tenth-order min ima on each side of the central maximum?
Section 46-3 Intensity in Single-Slit Diffraction 11. Monochromatic light with wavelength 538 nm falls on a slit with width 25.2 pm . The distance from the slit to a screen is 3.48 m. Consider a point on the screen 1.13 cm from the central maximum, (a) Calculate 6. {b) Calculate a. (c) Cal culate the ratio of the intensity at this point to the intensity at the central maximum. 12. If you double the width of a single slit, the intensity of the central maximum of the diffraction pattern increases by a factor of four, even though the energy passing through the slit only doubles. Explain this quantitatively. 13. Calculate the width of the central maximum in a single-sht diffraction pattern in which a = 10A. Compare your result with Fig. 12c. See Sample Problem 4. 14. A monochromatic beam of parallel light is incident on a
Problems “collimating” hole of diameter a » A. Point P lies in the geometrical shadow region on a distant screen, as shown in Fig. 23fl. Two obstacles, shown in Fig. 23Z?, are placed in turn over the collimating hole. A is an opaque circle with a hole in it and B is the “photographic negative” of A. Using superposition concepts, show that the intensity at P is iden tical for each of the two diffracting objects A and B (Babinet’s principle). In this connection, it can be shown that the diffraction pattern of a wire is that of a slit of equal width. See “Measuring the Diameter of a Hair by Diffraction,” by S. M. Curry and A. L. Schawlow, American Journal o f Phys ics, May 1974, p. 412.
20
21.
22
23
(a)
T“ a
1_
o
♦
983
Mount Palomar, assuming that this distance is determined by diffraction effects. Assume a wavelength of 565 nm. The wall of a large room is covered with acoustic tile in which small holes are drilled 5.20 mm from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions? Assume the diameter of the pupil of the observer’s eye to be 4.60 mm and the wavelength to be 542 nm. If Superman really had x-ray vision at 0 .12-nm wavelength and a 4.3-mm pupil diameter, at what maximum altitude could he distinguish villains from heroes assuming the min imum detail required was 4.8 cm? A navy cruiser employs radar with a wavelength of 1.57 cm. The circular antenna has a diameter of 2.33 m. At a range of 6.25 km, what is the smallest distance that two speedboats can be from each other and still be resolved as two separate objects by the radar system? The paintings of Georges Seurat consist of closely spaced small dots (« 2 mm in diameter) of pure pigment, as indi cated in Fig. 24. The illusion of color mixing occurs because the pupils of the observer’s eyes diffract light entering them. Calculate the minimum distance an observer must stand from such a painting to achieve the desired blending of color. Take the wavelength of the light to be 475 nm and the diameter of the pupil to be 4.4 mm.
B
ib)
Figure 23
Problem 14.
15. (fl) Show that the values of a at which intensity maxima for single-slit diffraction occur can be found exactly by differen tiating Eq. 8 with respect to a and equating to zero, obtain ing the condition tan a = a. (b) Find the values of a satisfying this relation by plotting graphically the curve j = tan a and the straight line y = a and finding their intersections or by using a pocket calcula tor to find an appropriate value of a by trial and error. (c) Find the (nonintegral) values of m corresponding to suc cessive maxima in the single-slit pattern. Note that the sec ondary maxima do not lie exactly halfway between minima. 16. In Fig. lOd, calculate the angle Eg makes with the vertical; see Question 24 and Problem 15. Section 46~4 Diffraction at a Circular Aperture 17. The two headlights of an approaching automobile are 1.42 m apart. At what (a) angular separation and (b) maxi mum distance will the eye resolve them? Assume a pupil diameter of 5.00 mm and a wavelength of 562 nm. Also assume that diffraction effects alone limit the resolution. 18. An astronaut in a satellite claims to be able to just barely resolve two point sources on the Earth, 163 km below. Cal culate their (a) angular and (b) linear separation, assuming ideal conditions. Take A = 540 nm and the pupil diameter of the astronaut’s eye to be 4.90 mm. 19. Find the separation of two points on the Moon’s surface that can just be resolved by the 200-in. (=5.08-m) telescope at
Figure 24
Problem 23.
24. A “spy in the sky” satellite orbiting at 160 km above the Earth’s surface has a lens with a focal length of 3.6 m. Its resolving power for objects on the ground is 30 cm; it could easily measure the size of an aircraft’s air intake. What is the effective lens diameter, determined by diffraction considera tion alone? Assume A = 550 nm. Far more effective satel lites are reported to be in operation today. 25. (a) A circular diaphragm 60 cm in diameter oscillates at a frequency of 25 kHz in an underwater source of sound used for submarine detection. Far from the source the sound intensity is distributed as a diffraction pattern for a circular hole whose diameter equals that of the diaphragm. Take the speed of sound in water to be 1450 m/s and find the angle between the normal to the diaphragm and the direction of the first minimum, (b) Repeat for a source having an (audi ble) frequency of 1.0 kHz. 26. In June 1985 a laser beam was fired from the Air Force Optical Station on Maui, Hawaii, and reflected back from the shuttle Discovery as it sped by, 354 km overhead. The diameter of the central maximum of the beam at the shuttle position was said to be 9.14 m and the beam wavelength was
984
Chapter 46
Diffraction
500 nm. What is the effective diameter of the laser aperture at the Maui ground station? (Hint: A laser beam spreads because of diffraction; assume a circular exit aperture.) 27. Millimeter-wave radar generates a narrower beam than con ventional microwave radar. This makes it less vulnerable to antiradar missiles, (a) Calculate the angular width, from first minimum to first minimum, of the central “lobe” produced by a 220-GHz radar beam emitted by a 55-cm diameter circular antenna. (The frequency is chosen to coincide with a low-absorption atmospheric “window.”) (b) Calculate the same quantity for the ship’s radar described in Prob lem 22. 28. In a Soviet-French experiment to monitor the Moon’s sur face with a light beam, pulsed radiation from a ruby laser (A = 0.69 pm ) was directed to the Moon through a reflecting telescope with a mirror radius of 1.3 m. A reflector on the Moon behaved like a circular plane mirror with radius 10 cm, reflecting the light directly back toward the telescope on Earth. The reflected light was then detected by a photometer after being brought to a focus by this telescope. What frac tion of the original light energy was picked up by the detec tor? Assume that for each direction of travel all the energy is in the central diffraction circle. 29. It can be shown that, except for ^ = 0, a circular obstacle produces the same diffraction pattern as a circular hole of the same diameter. Furthermore, if there are many such obstacles, such as water droplets located randomly, then the interference effects vanish leaving only the diffraction asso ciated with a single obstacle, (a) Explain why one sees a “ring” around the Moon on a foggy night. The ring is usu ally reddish in color; explain why. (b) Calculate the size of the water droplets in the air if the ring around the Moon appears to have a diameter 1.5 times that of the Moon. The angular diameter of the Moon in the sky is 0.5 ®. (c) At what distance from the Moon might a bluish ring be seen? Some times the rings are white; why? (d) The color arrangement is opposite to that in a rainbow; why should this be so? Section 46-5 Double-Slit Interference and Diffraction Combined 30. (a) Design a double-slit system in which the fourth fringe, not counting the central maximum, is missing, (b) What other fringes, if any, are also missing? 31. Two slits of width a and separation d are illuminated by a coherent beam of light of wavelength A. What is the linear separation of the interference fringes observed on a screen that is a distance D away? 32. Suppose that, as in Sample Problem 7, the envelope of the central peak contains 11 fringes. How many fringes lie between the first and second minima of the envelope? 33. (fl) For d = 2 a 'm Fig. 25, how many interference fringes lie in the central diffraction envelope? (b) If we put ^ = a, the two slits coalesce into a single slit of width la. Show that Eq. 17 reduces to the diffraction pattern for such a slit. 34. (a) How many complete fringes appear between the first minima of the fringe envelope on either side of the central maximum for a double-slit pattern if A= 557 nm, ^ = 0.150 mm, and a = 0.030 mm? (b) What is the ratio of the intensity of the third fringe to the side of the center to that of the central fringe?
Figure 25
Problem 33.
35. Light of wavelength 440 nm passes through a double sht, yielding the diffraction pattern of intensity I versus deflec tion angle 0 shown in Fig. 26. Calculate (a) the slit width and (b) the slit separation, (c) Verify the displayed intensi ties of the m = 1 and m = 2 interference fringes.
Figure 26
Problem 35.
36. An acoustic double-slit system (slit separation d, slit width a) is driven by two loudspeakers as shown in Fig. 27. By use of a variable delay line, the phase of one of the speakers may be varied. Describe in detail what changes occur in the intensity pattern at large distances as this phase difference is varied from zero to In. Take both interference and dif fraction effects into account.
Figure 27
Problem 36.
CHAPTER 47 GRATINGS AND SPECTRA In Chapter 45 we discussed the interference pattern produced when monochromatic light is incident on a double slit: a pattern of bright and dark bands (interferencefringes) is produced. Each of the slits is regarded as an elementary radiator. At first (in Chapter 45) we assumed that the slit width was much smaller than the wavelength of the light, so that light diffractedfrom each slit uniformly covered the observation screen. Later (in Chapter 46) we took the slit width into account and determined the "diffractionfactor" that modulates the interference pattern. In this chapter, we extend our discussion to cases in which the number of elementary radiators or diffraction centers is larger (often much larger) than two. We consider multiple arrays of slits in a plane and also three-dimensional arrays of atoms in a solid (for which we use X rays rather than visible light). In both cases, we must distinguish carefully between the diffracting properties of a single radiator (one slit or atom) and the interference of the waves coherently diffractedfrom the assembly of radiators.
41A
MULTIPLE SLITS
A logical extension of double-slit interference is to in crease the number of slits from two to a larger number N, Figure 1 shows an example with five slits. Such a multi ple-slit arrangement (in which N may be as large as 10^) is called a diffraction grating. As in the case of the double slit, when monochromatic light falls on such a multipleslit arrangement, a pattern of interference fringes is pro duced. For a given wavelength, the spacing of the fringes is determined by the slit separation d, while the relative intensities o f the fringes are determined by diffraction effects associated with the slit width a. In this chapter we analyze the interference patterns for multiple slits. We consider only the Fraunhofer region, in which there is an assumed infinite distance between the light source and the slits as well as between the slits and the screen. Equivalently, parallel light is incident on the slits, and parallel rays emerge from the slits (perhaps to be focused by a lens) to form an image on the screen. Figure 2 shows a portion of the central maximum of the diffraction envelope for = 2 and N = 5, We see that two important changes occur when the number o f slits in creases from two to five: (1) the bright fringes become
Figure 1 An idealized diffraction grating containing five slits. The slit width a is assumed to be much smaller than A, although this condition may not be realized in practice. Also, the focal length / will in practice be much greater than d\ the figure distorts these dimensions for clarity.
narrower, and (2) faint secondary maxima (three in Fig. lb) appear between the bright fringes. Figure 3 shows the results of a theoretical calculation of the intensities (ne-
985
986
Chapter 4 7 Gratings and Spectra
iiiiiiiii Figure 2 The diffraction pattern for a grating with (a) two slits and (b) five slits. Note that, in the case of the five-slit grating, the fringes are sharper (narrower), and secondary maxima of low intensity appear between the bright principal maxima.
glecting the effect of the diffraction envelope), in which the sharpening o f the principal maxima is more apparent. As N increases, the number of secondary maxima in creases and their brightness diminishes, until they be come negligible; correspondingly, the principal maxima become sharper with increasing N. In the following dis cussion we ignore the secondary maxima and consider only the principal maxima. A principal maximum occurs when the path difference between rays from any pair of adjacent slits, which is given by d sin 6, is equal to an integral number o f wave lengths, or
d sm d = mX
m = 0, ± 1 , ± 2 , .
(1)
where m is called the order number. Equation 1 is identi cal with Eq. 1 of Chapter 45 for the maxima of the double slit. Note that if light passing through any pair o f adjacent slits is in phase at a particular point on the screen, then light passing through any pair of slits, even nonadjacent ones, is also in phase at that point. For a given slit separa tion d, the locations o f the principal maxima are deter
mined by the wavelength, and so measuring their loca tions is a means for precise determination o f wavelengths. The locations o f the principal maxima are independent of the number o f slits N, which, as we shall see, determines the width or sharpness o f the principal maxima. The rela tive intensities of the principal maxima within the diffrac tion envelope are determined by the ratio a/A, which does not affect their locations.
Width of the Maxima The sharpening o f the principal maxima as A^is increased can be understood by a graphical argument, using phasors. Figures 4a and Ab show conditions at the central principal maximum for a two-slit and a five-slit grating. The small arrows represent the amplitudes o f the wave disturbances arriving at the screen at the position of the central maximum, for which m = 0, and thus 0 = 0, in Eq. 1. On either side o f the central maximum there is a mini mum of zero intensity, which lies at an angle SOq off the
Relative intensity
V iV = 2
(a) Relative intensity
1r\r\^
.
ib)
N
d h
=
b
k
n
k h d
Figure 3 Calculated intensity patterns for (a) a two-slit and (b) a five-slit grat ing, having the same values of d and k. Note the sharpening of the principal maxima and the appearance of faint secondary maxima in (b)\ compare with Fig. 2. The letters in (b) refer to Fig. 6. This calculation does not include diffrac tion effects due to the slit width; that is, we assume we are near the central region of Fig. 2 where the principal maxima have essentially equal intensities.
Section 47-1
-^>----- D> -------- >
Multiple Slits
987
H> ^5m
='2m
(a)
(c)
( 6)
{d)
Figure 4 {a,b) The conditions at the central maximum for a two-slit and a five-slit grating, respectively. (c,d) The corresponding conditions at a minimum of zero intensity that lies on either side of this central maximum.
central axis, as shown in Fig. 5. Figures Ac and Ad show the phasors at this point. The phase difference between waves from adjacent slits, which is zero at the central principal maximum, must increase by an amount A 0 chosen so that the array of phasors just closes on itself, yielding zero resultant intensity. For N = 2 , A
^
N
( 2)
This increase in phase difference for adjacent waves corresponds to an increase in the path difference AL given by phase difference _ path difference 2^ A ’ or
From Fig. 1, however, the increase in path difference AL at the first minimum is also given by d sin 56q, so that we can write d sm d d Q = — ,
or (4) Since » 1 for actual gratings, sin SOqis ordinarily quite small (that is, the lines are sharp), and to a good approxi mation we may replace sin S9q by 56q, expressed in ra dians, or
Nd
(5)
This equation shows specifically that if we increase A^for a given A and d, then 56qdecreases, which means that the central principal maximum becomes sharper. To obtain the result for any principal maximum, we consider the geometry of Fig. 5, in which the mth princi pal maximum occurs at an angle 6. We move away from the maximum through an angular displacement 56 to
Figure 5 A principal maximum lies at the position given by the angle 6, and the first minimum occurs at the angle SO from that maximum. The angle SO can be taken as a measure of the width or sharpness of the maximum. The width of the central maximum is given by the angle SO q .
arrive at the next minimum; we take this angle to be a measure of the angular width of the maximum. At the maximum, the path difference between rays from adja cent slits is mX (see Eq. 1). At the next minimum, the path difference between rays from adjacent slits is mX + X/N, the additional path length of A/A^beinggiven by Eq. 3. For example, consider the case of A = 10. The additional path length between adjacent slits at the minimum is 0.1 A. The path difference between slits 1 and 6 is therefore 5(mA + 0.1 A) = 5mA + 0.5A; the path lengths differ by a half-inte gral number of wavelengths, so the rays interfere destruc tively. The same is true for slits 2 and 7, slits 3 and 8, and so forth. If the additional path difference is X/N, then rays from the lower N/2 slits undergo pairwise destructive in terference with rays from the upper N/2 slits. At the angle 6 + S6, the path difference between rays from adjacent slits is
d sin {6 + SO) = ^/(sin 0 cos 56 + cos 9 sin 56) « dsin0A-{d cos 6)56, where we assume 59 is small, which allows us to approxi mate cos 59 « 1 and sin 56 « 56. Setting this path differ-
988
Chapter 47
Gratings and Spectra
Figure 6 The figures taken in sequence from {a) to (n) and then from (n) to (a) show conditions as the intensity pattern of a five-slit grating is traversed from the central principal maximum to an adjacent principal maximum. Phase differences between waves from adjacent slits are shown directly; those in going from (n) to (a) are in pa rentheses. Principal maxima occur at (a), secondary maxima at or near (h) and (n), and minima of zero intensity at (d) and (/c). Compare with Fig. 3b.
ence equal to mA + A/A^, its value at the minimum, we obtain d sin 0 + (d cos 6)66 = mA + ~ N or, using Eq. 1, {d cos 6)66 = “ . Solving for 66 gives
central maximum {6 = 0). For a given N, d, and A, the central maximum is the narrowest (cos 0=1); the widths increase as we go to larger 6 (and therefore to larger orders m). Equation 6 shows that 66 becomes smaller (the max ima become sharper) as the product Nd increases. This product (the number of slits times the distance between slits) gives the total width of the grating. Thus the peaks become sharper as the width of the grating increases. The Secondary Maxima (Optional)
66 =
( 6)
Nd cos 6 This result gives the angular width* for the principal max imum that occurs at the angle 6, corresponding to the particular order m. Note that Eq. 6 reduces to Eq. 5 for the * As defined by Eq. 6, the width is the angular interval from the peak to the first minimum. The usual definition of the width of a peak is the full interval covered by the peak at half its maximum height (see, for example, Fig. 12 of Chapter 46). These two measures of the width are roughly equal, and we take Eq. 6 to represent a measure of the width of the peak.
The origin of the secondary maxima that appear for > 2 can also be understood using the phasor method. Figure 6a shows conditions for the central principal maximum for a five-slit grat ing. The phasors are in phase. As we depart from the central maximum, 6 in Fig. 1 increases from zero and the angle between adjacent phasors increases from zero to A0 = (2nlX)(d sin 6). Successive figures show how the resultant wave amplitude E q varies with A
Section 47-2
Diffraction Gratings
989
cx>mpared with Fig. 36, shows that for N = 5 there are three secondary maxima, corresponding to A0 = 110®, 180®, and 250®. Make a similar analysis for = 3 and show that only one secondary maximum occurs. In general, for a grating with N slits, there are —2 secondary maxima. In actual gratings, which commonly contain 10,000 to 50,000 “slits,” the second ary maxima lie so close to the principal maxima or are so re duced in intensity that they cannot be observed experimen tally. ■ Figure 7 A cross section of a blazed grating viewed in re flected light. There is a path difference of d sin 9 between the two rays shown. Sample Problem 1 A certain grating has 10^ slits with a spacing o f d = l . \ /im = 2100 nm. It is illuminated with yellow sodium light (A = 589 nm). Find (a) the angular position of all principal maxima observed and (b) the angular width of the largest-order maximum. Solution
(a) From Eq. 1, we have .
mX
m(589 nm)
which gives e = 16.3® (m = 1), 34.1® (m = 2), and 57.3® (m = 3), with corresponding values at ^ < 0 for m < 0 . For m = 4, sin 1. Thus m = 3 is the highest order observed, which corresponds to a total of seven principal maxima (a central max imum and three on each side of center). (b) For the m = 3 maximum, Eq. 6 gives
se =
A N dcos e
589 nm (10^X2100 nmXcos 57.3®) = 5.2 X 10-5 rad = 0.0030®.
This is an exceedingly narrow principal maximum. Note that Eq. 6, being a dimensionless ratio, gives its result in radian measure. This occurs because we derived Eq. 6 using the approximation sin S 6 ^ S 9 , which is valid only in radian measure.
47-2 DIFFRACTION GRATINGS A typical grating might contain N= 10,000 slits distrib uted over a width of a few centimeters, equivalent to a grating spacing dofa fewmicrometers. As we have seen in Sample Problem 1, when Nd is a few centimeters, the maxima are very narrow, which allows their position to be measured with great precision. Gratings are therefore used to determine wavelengths and to study the structure and intensity of the principal maxima. Any regular periodic structure can serve as a difiiaction grating, for example, the grooves ofa compact disk, which produce a rainbow pattern when light is reflected from the surface of the disk. Gratings can produce their images by transmitted light, as in Fig. 1; there are also reflection gratings, which produce their images in reflected light. In
the grating of Fig. 1, there is a periodic change in the amplitude (and no change in phase) of the light as a func tion of position across the grating. It is also possible to make gratings (of either the reflection or transmission type) that cause a periodic change in the phase (and a negligible change in the amplitude) of the light as a func tion of position across the grating. Most gratings used for visible light, whether of the reflection or transmission type, are phase gratings. Gratings are made by ruling equally spaced parallel grooves in a thin layer ofaluminum or gold deposited on a glass plate, using a diamond cutting point whose motion is automatically controlled by a ruling engine. Once such a master grating has been prepared, replicas can be formed by pouring a liquid plastic on the grating, allowing it to harden, and stripping it off. The stripped plastic, fastened to a flat piece of glass or other backing, forms a good grating. Figure 7 shows a cross section of a common type of reflection phase grating. (Ifthe grating were transparent, it could function as a transmission phase grating, since light passing through different thicknesses will have varying changes in phase.) The angles of the grooves are chosen so that light of a particular order is reflected in a particular direction. In this way the intensity of one particular order can be enhanced over that of other orders. Cutting grat ings in this way is called blazing. Most gratings in use today are blazed gratings. Figure 8 shows a simple grating spectroscope, used for viewing the spectrum of a light source, assumed to emit a number of discrete wavelengths. The light from source S is focused by lens L, on a slit Si placed in the focal plane of lens Lj. The parallel light emerging from collimator C falls on grating G. Parallel rays associated with a particular interference maximum occurring at angle 6 fall on lens L,, being brought to a focus in plane FF'. The image formed in this plane is examined, using a magnifying lens arrangement E (the eyepiece). The entire spectrum can be viewed by rotating telescope T through various angles. Instruments used for scientific research or in industry are more complex than the simple arrangement of Fig. 8. They invariably employ photographic or photoelectric
990
Chapter 4 7 Gratings and Spectra
Figure 8 A simple type of grating spectroscope used to analyze the wavelengths of the light emitted by the source S.
J_
400
In general, gratings may produce several images of spectral lines, corresponding to m = ± 1, ±2, . . . , in Eq. 1, and they can also separate wavelengths that are distril> uted continuously (as in Fig. 10) rather than as sharp spectral lines. The light from a hot, glowing object such as a lamp filament or the Sun gives a continuous spectrum. The Sun’s spectrum also contains sharp spectral lines, which appear as dark lines superimposed on the continu ous spectrum. These lines are caused by absorption of light by atoms of elements in the atmosphere surrounding the Sun. The element helium (from the Greek word helios, meaning the Sun) was discovered from an analysis of these lines. Light can also be analyzed into its component wave lengths if the grating in Fig. 8 is replaced by a prism. In a prism spectrograph each wavelength in the incident beam is deflected through a definite angle 0, determined by the index of refraction of the prism material for that wave length. Curves such as Fig. 4 of Chapter 43, which gives the index of refraction of fused quartz as a function of wavelength, show that the shorter the wavelength, the larger the angle of deflection d. Such curves vary from substance to substance and must be found by measure-
Red
Blue
500
_L
600
700
Wavelength (nm)
Figure 9 Examples of spectra of visible light emitted by gases of sodium (Na) and mercury (Hg).
recording and are called spectrographs. Figure 9 shows examples of spectra of visible light recorded by a spectro graph. Each line in the figure is in effect an image ofthe slit Sx corresponding to one of the many individual wave lengths emitted from the source. For this reason, such images are called spectral lines; a “line” in a spectrum, no matter what technique is used to record the spectrum, is taken to mean a particular wavelength component.
-1
“
“
l l
I
m = -2 I
m= Z
^ n jlK
m= 2
ll":rlTll m=s0 L
_____ L_____ L_____^ _____ ^ ^ _____ I____ 1_____ ^ _____ .1____ L____ 1
-90 -80 -70 -60 -50
-40 -30 -20 -10 d
(degrees)
0
10
I
I
I
20
I
I
30
Figure 10 The spectrum of white light as viewed in a grating spectroscope such as that of Fig. 8. The different orders, identified by the index m, are shown separated vertically for clarity. As actually viewed, they would not be so displaced. The central line in each order corresponds to a wavelength of 550 nm. Diffraction gratings in common use today are designed to concentrate the inten 40 of the light 50 in a 60 80 sity particular70order, and they do not show the ideal symmetrical pat terns illustrated here.
90
Section 4 7-3 Dispersion and Resolving Power
991
ment. Prism instruments are not adequate for accurate
A6 between spectral lines that differ in wavelength by a
absolute measurements of wavelength because the index
small amount AA and (2) the width or sharpness o f the lines. In Sample Problem 2, we calculated the angular separa tion between the closely spaced lines of the yellow sodium doublet, for which AA = 0.59 nm. We found in that case a separation of A0 = 0.014° between the first-order princi pal maxima o f these lines. The angular separation Ad per unit wavelength interval AA is called the dispersion D o f the grating, or
o f refraction o f the prism material at the wavelength in question is usually not known precisely enough. Both prism and grating instruments make accurate compari sons o f wavelength, using a suitable comparison spectrum such as that shown in Fig. 9, in which careful absolute determinations have been made of the wavelengths o f the spectral lines.
Sample Problem 2 A diffraction grating has 10^ rulings uni formly spaced over 25.0 mm. It is illuminated at normal inci dence by yellow light from a sodium vapor lamp. This light contains two closely spaced lines (the well-known sodium dou blet) of wavelengths 589.00 and 589.59 nm. (a) At what angle will the first-order maximum occur for the first of these wave lengths? (b) What is the angular separation between the firstorder maxima for these lines? Solution {a) The grating spacing d is 25(X) nm. The first-order maximum corresponds to m = 1 in Eq. 1. We thus have .
. .J m k \ \ d )
.
/(l)(5 8 9 n m )\ \ 2500 nm /
,,,,
(b) The straightforward way to find the angular separation is to repeat the calculation of part {a) for A = 589.59 nm and to subtract the two angles. A difl&culty, which can best be appre ciated by doing the calculation, is that we must carry a large number of significant figures to obtain a meaningful value for the difference between the angles. To calculate the difference in angular positions directly, let us solve Eq. 1 for sin 6 and differ entiate the result, treating 6 and A as variables: . . mk sm 0 = - 3a
cos 6de = ^dX. d If the wavelengths are close enough together, as in this case, dk can be replaced by AA, the actual wavelength difference; dS then becomes A^, the quantity we seek. This gives m AA dcosO
(1X0.59 nm) (2500 nmXcos 13.6®)
= 2.4X 10-^ rad = 0.014®. Note that although the wavelengths involve five significant fig ures, our calculation, done this way, involves only two or three, with consequent reduction in numerical manipulation.
47-3 DISPERSION AND RESOLVING POW ER__________ The ability o f a grating to produce spectra that permit precise measurement o f wavelengths is determined by two intrinsic properties o f the grating: (1) the separation
For lines of nearly equal wavelengths to appear as widely separated as possible, we would like our grating to have the largest possible dispersion. To see what physical property o f the grating determines its dispersion, we differentiate Eq. 1 {d sin 6 —wA), treat ing 6 and A as variables, which gives
d cos 6 d6 = m dX, or, in terms o f small differences instead o f differentials,
d cos 6 AO = m AX.
(8)
The dispersion D is given by A0/AA, or
D=
m dcosO
(9)
The dispersion increases as the spacing between the slits decreases. We can also increase the dispersion by working at higher order (large m), as Fig. 10 illustrates. Note that the dispersion does not depend on the number o f rulings.
Resolving Power If a grating produces lines o f large width, then the maxima o f spectral lines o f closely spaced wavelengths may over lap, making it difficult to determine whether such lines have one or more components and to measure the wave lengths o f the lines to high precision. We therefore want to select a grating that produces the narrowest possible lines. We obtain a reasonable measure of the ability to resolve nearby lines of different wavelengths by applying Ray leigh’s criterion (see Section 46-4): if the maximum o f one line falls on the first minimum o f its neighbor, we should be able to resolve the lines. In Section 47-1, we defined the width of a spectral line in just that way, as the angular interval SOfrom the maximum to the first minimum. The limit o f resolution of the grating occurs when two lines in the spectrum are separated by a wavelength interval AA such that the difference 80 between their angular posi tions is given by Eq. 6. We define the resolving power R o f the grating as R - ±
00)
If the lines are to be narrow (80 is small), then the cone-
992
Chapter 47
TABLE 1
Gratings and Spectra
PROPERTIES OF THREE GRATINGS^
Grating
N
t/(nm )
e
A B C
5,000 5,000
10,000
10,000
10,000
2.9“ 5.7“ 2.9“
5,000
R 5,000 5,000
10,000
0 .0 0 0 5 7 “
Z)(10 ^rad/nm)
1.0 2.0 1.0
" For X = 500 nm and m = 1.
(a) 0 .0 0 1 1 4 “
Spending wavelength interval AA must be small, and the resolving power must be large. We should therefore choose a grating with the largest R. To find the physical property of the grating that deter mines R , let us solve Eq. 8 for the spacing A 6 between nearby lines and (using Rayleigh’s criterion) set this result equal to the width S6 of the line, given by Eq. 6 as the spacing between the maximum and first minimum. This gives m AX _ A d cos 6 N d cos 6 ’
0 .0 0 0 5 7 “
and solving for R (=A/AA) gives R = Nm.
(c)
(11)
The resolving power, like the dispersion, increases with the order number. Unlike the dispersion, R depends on the number of lines but is independent of their separa tion d. T o maximize the resolving power, we choose a grating with the largest number of lines. For a given slit spacing d, the grating with the greatest total width has the greatest resolving power (that is, it produces the sharpest spectral lines). Dispersion and resolving power measure different aspects o f a diffraction grating’s ability to produce cleanly separated lines. Consider, for example, three gratings A, B, and C whose properties are listed in Table 1. Suppose that the gratings are illuminated with light consisting of a doublet o f lines at 500 nm separated by an interval AA = 0.10 nm. We have chosen the proF>erties of grating^ such that the two lines of the doublet in the first-order maxi mum are just at the limit of resolution; that is, the maxi mum o f one line falls on the minimum of the other, as shown in Fig. 11 a. Grating B has twice the dispersion of A but the same resolving power, and it produces the spec trum shown in Fig. W b .ln effect all angular intervals are scaled by a factor of 2, including the angular width and angular separation of the peaks. If our measurement with grating A had been limited by our ability to determine small angular intervals, changing to grating B would im prove the measurement. Grating C has twice the resolving power of A but the same dispersion. The peaks in Fig. 1 \c appear with the same angular separation as those in Fig. 1 la, but with smaller widths. The maximum of one peak now clearly falls outside the first minimum of the other, and the two lines are more clearly distinguished from one another using grating C.
Figure 11 The intensity pattern of two lines at A = 500 nm separated by AA = 0.10 nm, produced by the three gratings of Table 1. Grating B has the largest dispersion and grating C the largest resolving power.
The total widths of the three gratings, equal to the prod uct N d, are 50 mm for grating^, 25 mm for grating fi, and 100 mm for grating C. Note from Fig. 11 that the peak widths depend inversely on the grating width, as suggested by Eq. 6.
Sample Problem 3 A grating has 9600 lines uniformly spaced over a width IV = 3.00 cm and is illuminated by light from a mercury vapor discharge, (a) What is the expected dispersion, in the third order, in the vicinity of the intense green line (A = 546 nm)? (b) What is the resolving power of this grating in the fifth order? Solution
(a) The grating spacing is given by IT N
3.00X lQ -^ m _ , , , , 9600 3125 nm.
We must find the angle 6 at which the line in question occurs. From Eq. 1, we have 9 = sin
3125 nm
We can now calculate the dispersion. From Eq. 9 D=
m dcos 6
3 (3125 nmXcos 31.6“) = 1.13 X 10“ ^ rad/nm = 0.0646“/nm = 3.87 arc min/nm.
Section 47’4 X-Ray Diffraction (b)
From Eq. 11 R = Nm = (9600X5) = 4.80 X \0*.
993
From Eq. 11, the number of rulings needed to achieve this re solving power (in first order) is R
Thus, near A = 546 nm and in fifth order, a wavelength differ ence given by (see Eq. 10)
9Q8
= — = Z ^ = 998 rulings. m 1 Since the grating has about 12 times as many rulings as this, it can easily resolve the sodium doublet lines, as we have already shown in part (c).
A _ 546m n__ R 4.80 X l( y ® can be resolved. Sample Problem 4 A diffraction grating has 1.20 X 10^ rulings uniformly spaced over a width W = 2.50 cm. It is illuminated at normal incidence by yellow light from a sodium vapor lamp. This light contains two closely spaced lines of wavelengths 589.00 and 589.59 nm. (a) At what angle does the first-order maximum occur for the first of these wavelengths? (b) What is the angular separation between these two lines (in first order)? (c) How close in wavelength can two lines be (in first order) and still be resolved by this grating? (d) How many rulings can a grating have and just resolve the sodium doublet lines? Solution
(a) The grating spacing d is given by ^
W N
2.50 X 10-2 m 1.20 X icy
2083 nm.
The first-order maximum corresponds to m = 1 in Eq. 1. We thus have
. . (mk\ . /(lX 5 8 9 .0 0 n m )\ ,, 6 = sin M — I = sin ' I ---------) = 16.4°. \ d /
\
2083 nm
/
(b) Here the dispersion of the grating comes into play. From Eq. 9, the dispersion is
m
Z) = d cos d
1
(2083 nmXcos 16.4°) = 5.01 X 10“^ rad/nm.
From Eq. 7, the defining equation for dispersion, we have ^0 = D Ak = (5.01 X 10-^ rad/nmX589.59 nm - 589.00 nm) = 2.95 X 10“^ rad = 0.0169° = 1.02 arc min. As long as the grating spacing d remains fixed, this result holds no matter how many lines there are in the grating.
47-4 X-RAY DIFFRACTION__________ X rays are electromagnetic radiation with wavelengths of the order of 0.1 nm (compared with 5(X) nm for a typical wavelength o f visible light). Figure 12 shows how x rays are produced when electrons from a heated filament Fare accelerated by a potential difference V and strike a metal target. For such small wavelengths a standard optical diffrac tion grating, as normally employed, cannot be used. For A = 0.10 nm and d = 3000 nm, for example, Eq. 1 shows that the first-order maximum occurs at 0 = sin"'
(mX\
.
/(IX O .IO nm )\ i 3 X 10^ nm j
= 0.0019” This is too close to the central maximum to be practical. A grating with A is desirable, but, because x-ray wave lengths are about equal to atomic diameters, such gratings cannot be constructed mechanically. In 1912 it occurred to physicist Max von Laue that a crystalline solid, consisting as it does o f a regular array of atoms, might form a natural three-dimensional “diffrac tion grating” for x rays. Figure 13 shows that if a colli mated beam of x rays, continuously distributed in wave length, is allowed to fall on a crystal, such as sodium chloride, intense beams (corresponding to constructive interference from the many diffracting centers o f which
(c) Here the resolving power of the grating comes into play. From Eq. 11, the resolving power is R = Nm = {\.2 0 X 10^X1)= 1 2 0 X 10^. From Eq. 10, the defining equation for resolving power, we have ^
A R
589 nm 1.20 X l( y
....
This grating can easily resolve the two sodium lines, which have a wavelength separation of 0.59 nm. Note that this result de pends only on the number of grating rulings and is independent of d, the spacing between adjacent rulings. {d) From Eq. 10, the defining equation for R, the grating must have a resolving power of AA
0.59 nm
Figure 12 X rays are generated when electrons from heated filament F, accelerated through a potential difference V, strike a metal target T in the evacuated chamber C. Window W is transparent to x rays.
994
Chapter 47
Gratings and Spectra
“0
pattern on the photographic film S.
Figure 14 A Laue x-ray diffraction pattern from a crystal of sodium chloride.
the crystal is made up) appear in certain sharply defined directions. If these beams fall on a photographic film, they form an assembly o f “Laue spots.” Figure 14, which shows an actual example o f these spots, demonstrates that the hypothesis o f Laue is indeed correct. The atomic ar rangements in the crystal can be deduced from a careful study o f the positions and intensities of the Laue spots in much the same way that we might deduce the structure o f an optical grating (that is, the detailed profile of its slits) by a study o f the positions and intensities o f the lines in the interference pattern. Other experimental arrangements have supplanted the Laue technique to a considerable extent today, but the principle remains unchanged (see Question 25). Figure 15 shows how sodium and chlorine atoms (strictly, Na"^ and Cl“ ions) are stacked to form a crystal o f sodium chloride. This pattern, which has cubic sym metry, is one o f the many possible atomic arrangements exhibited by solids. The model represents the unit cell for
Figure 15 A model of a sodium chloride crystal, showing how the sodium ions Na^ (small spheres) and chloride ions C r (large spheres) are stacked in the unit cell, whose edge Oq has the length 0.563 nm.
sodium chloride. This is the smallest unit from which the crystal may be built up by repetition in three dimensions. You should verify that no smaller assembly of atoms pos sesses this property. For sodium chloride the length Uqof the cube edge o f the unit cell is 0.563 nm. Each unit cell in sodium chloride has four sodium ions and four chlorine ions associated with it. In Fig. 15 the sodium ion in the center belongs entirely to the cell shown. Each o f the other twelve sodium ions shown is shared with three adjacent unit cells so that each contrib utes one-fourth of an ion to the cell under consideration. The total number of sodium ions is then 1 + i(12) = 4. By similar reasoning you can show that although there are fourteen chlorine ions in Fig. 15, only four are associated with the unit cell shown. The unit cell is the fundamental repetitive diffracting unit in the crystal, corresponding to the slit (and its adja cent opaque strip) in the optical diffraction grating of Fig. 1. Figure 16a shows a particular plane in a sodium chloride crystal. If each unit cell intersected by this plane is represented by a small cube. Fig. 166 results. You may imagine each of these figures extended indefinitely in three dimensions. Let us treat each small cube in Fig. 166 as an elemen tary diffracting center, corresponding to a slit in an optical grating. The directions (but not the intensities) o f all the diffracted x-ray beams that can emerge from a sodium chloride crystal (for a given x-ray wavelength and a given orientation of the incident beam) are determined by the geometry of this three-dimensional lattice o f difiracting centers. In exactly the same way the directions (but not the intensities) of all the diffracted beams that can emerge from a particular optical grating (for a given wavelength and orientation of the incident beam) are determined only by the geometry of the grating, that is, by the grating spacing d. Representing the unit cell by what is essentially
Section 47-4 X-Ray Diffraction
a point, as in Fig. 166, correst)onds to representing the slits in a diffraction grating by lines, as we did in discussing the double-slit experiment in Section 45-1. The intensities of the lines from an optical diffraction grating depend on the diffracting characteristics o f a single slit, determined in particular by the slit width a; see, for example. Fig. 2 for a set of slits. The characteristics o f actual optical gratings are determined by the profile o f the grating rulings. In exactly the same way the intensities of the diffracted beams emerging from a crystal depend on the diffracting characteristics o f the unit cell. Fundamentally, the x rays are diffracted by electrons, diffraction by nuclei being negligible in most cases. Thus the diffracting characteris tics o f a unit cell depend on how the electrons are distrib uted throughout the volume o f the cell. By studying the directions of diffracted x-ray beams, we can learn the basic symmetry of the crystal. By studying the intensities we can learn how the electrons are distributed in a unit cell. Figure 17 shows an example o f this technique.
|<-ao^
1
ao
995
J
Bragg’s Law Bragg's law predicts the conditions under which dif fracted x-ray beams from a crystal are possible. In deriv ing it, we ignore the structure of the unit cell, which is related only to the intensities of these beams. The dashed sloping lines in Fig. 18a represent the intersection with the
(b) Figure 16 (a) A plane through a crystal of NaCl, showing the Na and Cl ions, (b) The corresponding unit cells in this sec tion. Each cell is represented by a small black square.
W
\
\ -----N— H
N--- C
//
w
N IN
N
\C==N
M
II
I
I I
0
H— N-----
\
\
\u
A H
Scale (nm) 0.1 0 .2 0 .3 0 .4 0.5
(a)
(6)
Figure 17 (a) Electron density contours for phthalocyanine (CjjHjgNg) determined from the intensity distribution of scattered x rays. The dashed curves represent a density of one electron per 0.01 nm^, and each adjacent curve represents an increase of one electron per 0.01 nm^. (b) A structural representation of the molecule. Note that the greatest electron density occurs in (a) near the N atoms, which have the largest number of electrons (7). Note also that the H atoms, which contain only a single electron, are not prominent in (a).
996
Chapter 4 7 Gratings and Spectra
\
\ \
\
^ \
\
\
rays from adjacent planes {abc in Fig. 18Z?) must be an integral number of wavelengths or
\
ld s \v iO = m k
\
X\
\
X,
X
\
H W X V 'H X H '
X \ X \ \ X \ X
X
\
X^
\
\ \
\
\
\
X
\
\
m = 1, 2, 3, . . . .
This relation is called Bragg's law after W. L. Bragg who first derived it. The quantity d in this equation (the inter planar spacing) is the perpendicular distance between the planes. For the planes of Fig. 18a we see that d is related to the unit cell dimension a^ by d = ^
s
\
(12)
V 5‘
(13)
x' (a)
Incident wave
If an incident monochromatic x-ray beam falls at an arbitrary angle d on a particular set o f atomic planes, a diffracted beam will not result because Eq. 12 will not, in general, be satisfied. If the incident x rays are continuous in wavelength, diffracted beams will result when wave lengths given by 2 i/sin 0 A= m
Figure 18 (a) A section through the NaCl lattice of Fig. 16. The dashed lines represent an arbitrary set of parallel planes connecting unit cells. The interplanar spacing is d. (b) An in cident beam falls on a set of planes. A strong diffracted beam will be observed if Bragg’s law is satisfied.
plane o f the figure o f an arbitrary set of planes passing through the elementary diffracting centers. The perpen dicular distance between adjacent planes is d. Many other such families o f planes, with different interplanar spac ings, can be defined. Figure 1ib shows an incident wave striking the family o f planes, the incident rays making an angle 6 with the plane.* For a single plane, mirror-like “reflection” occurs for any value o f 0. To have a constructive interference in the beam diffracted from the entire family o f planes in the direction 6, the rays from the separate planes must rein force each other. This means that the path difference for
* In x-ray diffraction it is customary to specify the direction of a wave by giving the angle between the ray and the plane (the glancing angle) rather than the angle between the ray and the normal.
w = 1, 2, 3, . . .
are present in the incident beam (see Eq. 12). X-ray diffraction is a powerful tool for studying both x-ray spectra and the arrangement of atoms in crystals. To study the spectrum of an x-ray source, a particular set of crystal planes, having a known spacing d, is chosen. Dif fraction from these planes locates different wavelengths at different angles. A detector that can discriminate one angle from another can be used to determine the wave length of radiation reaching it. On the other hand, we can study the crystal itself, using a monochromatic x-ray beam to determine not only the spacings of various crystal planes but also the structure of the unit cell. The DNA molecule and many other equally complex structures have been mapped by x-ray diffraction methods.
Sample Problem 5 At what angles must an x-ray beam with A= 0.110 nm fall on the family of planes represented in Fig. 1 if a diffracted beam is to exist? Assume the material to be sodium chloride (Uq = 0.563 nm). Solution The interplanar spacing d for these planes is given by Eq. 13, or , ( - ^ - 2 : 2 5 ^ - 0 . 2 5 2 ; nm. V5 >/5 Equation 12 gives ^
/(m X 0.110nm )\ (.(2X0.252 . „ . ) )■
Diffracted beams are possible for 0 = 12.6® (/w = 1), ^ = 25.9® (m = 2), ^ = 40.9® (m = 3), and ^ = 60.9® (m = 4). Higher order beams cannot exist because they require that sin ^ > 1. Actually, the unit cell in cubic crystals such as NaCl has sym
Section 47-5 metry properties that require the intensity of diffracted x-ray beams corresponding to odd values of m to be zero. (See Prob lem 42.) Thus the only beams that are expected are ^ = 25.9® (m = 2) and 6 = 60.9® (m = 4).
47-5 HOLOGRAPHY (Optional) The light emitted by an object contains the complete informa tion on the size and shape of the object. We can consider that information to be stored in the wavefronts of the light from the object, specifically in the variation of intensity and phase of the electromagnetic fields. If we could record this information, we could reproduce a complete three-dimensional image of the ob ject. However, photographic films record only the intensity vari ations; the films are not sensitive to phase variations. It is there-
Figure 19 Apparatus for producing holograms. A portion of the beam from a source of coherent light (a laser, for instance) illuminates an object. The light diffracted by the object inter feres on the film with a portion of the original beam, which serves as the reference.
Figure 20 To view a hologram, it is illuminated with light identical to the reference beam. A three-dimensional virtual image can be seen, at the location of the original object.
Holography
(Optional)
997
fore not possible to use a photographic negative to reconstruct a three-dimensional image. One exception to this restriction occurs in the case of x-ray diffraction from a crystal. Because of the regular spacing of the atoms of a crystal, we can easily deduce the relative phases of the diffracted waves reaching the film from different atoms. This possibility was realized by W. L. Bragg, who illuminated a photo graphic negative of an x-ray diffraction pattern and so recon structed the image of a crystal. In this “double diffraction” method, diffraction of radiation from a diffraction pattern gives an image of the original object. For objects whose atoms are not arranged in such a periodic array, this simple method of image reconstruction does not work. A scheme for recording the intensity and phase of the waves from objects was developed in 1948 by Dennis Gabor, who was awarded the 1971 Nobel Prize in physics for this discovery. This type of image formation is called holography, from the Greek words meaning “entire picture,” and the image is called a holo gram. The procedure is illustrated in Fig. 19. A wave diffracted from an object interferes on the photographic film with a refer ence wave. The interference between the two waves serves as the means for storing on the film information on the phase of the wave from the object. When the photographic image is viewed using light identical with the reference beam, a three-dimen sional virtual image of the original object is reconstructed (Fig. 20). A second image (a real image), not shown in Fig. 20, is also produced by the hologram. Because the film is illuminated uniformly by the diffracted light from the object and the reference beam, every piece of the film contains the information necessary to reproduce the threedimensional image. The hologram itself (Fig. 21) shows only the interference fringes; in general, it is necessary to use a suitable monochromatic and coherent beam to reconstruct the image. For this reason, active development of holography did not occur until the early 1960s, when lasers became commonly available. Some holograms can be viewed in white light. White-light holograms use a thick photographic emulsion, in which light is reflected by successive layers of grains in the film. Constructive interference occurs in the reflected light for the wavelength of the original reference beam, and destructive interference occurs for
Figure 21 A close-up view of a hologram, showing the inter ference pattern.
998
Chapter 47
Gratings and Spectra
Figure 22 Two different views of the same hologram, taken from different directions. Note the relative movement of the objects in the images.
349
404
Figure 23 A holographic interference pattern of a top violin plate vibrating at different fre quencies. The frequencies (in Hz) are shown above the plates.
other wavelengths. By using reference beams of several different colors, a full-color image can be produced.* The hologram reconstructs a true three-dimensional image; for example, nearby objects appear “in front o f” more distant objects, and by moving your head from side to side you can change the relative spatial orientation of the objects. Figure 22 shows two different views of the same hologram, illustrating the parallax effect of viewing the hologram from two different direc tions. Holograms have a variety of applications in basic and applied * See “White-Light Holograms,” by Emmett N. Leith, Scientific American, October 1976, p. 80.
science. For example, in producing holograms the object must be kept absolutely still while the film is exposed; any small move ment would change the relative phase between the diffracted and reference beams and thereby change the interference pattern stored on the film. If a hologram is made by superimposing on the film two successive exposures of a vibrating object, such as the top or bottom plate of a violin, locations on the object that moved between the exposures by an integral number of wave lengths will show constructive interference, while parts of the object that moved by a half-integral number of wavelengths (A/2, 3A/2,. . . ) will show destructive interference. Figure 23 shows an example of the use of this technique, called holographic interfer ometry. ■
QUESTIONS 1. Discuss this statement: “A diffraction grating can just as well be called an interference grating.” 2. How would the spectrum of an enclosed source that is formed by a diffraction grating on a screen change (if at all) when the source, grating, and screen are all submerged in water? 3. {a) For what kind of waves could a long picket fence be
considered a useful grating? (b) Can you make a diffraction grating out of parallel rows of fine wire strung closely to gether? 4. Could you construct a diffraction grating for sound? If so, what grating spacing is suitable for a wavelength of 0.5 m? 5. A crossed diffraction grating is ruled in two directions, at right angles to each other. Predict the pattern of light inten-
Questions sity on the screen if light is sent through such a grating. Is there any practical value to such a grating?
6 . Suppose that, instead of a slit, a small circular aperture were placed in the focal plane of the collimating lens in the tele scope of a spectrometer. What would be seen when the tele scope is illuminated by sodium light? Why then do we usu ally call spectra line spectra? 7. In a grating spectrograph, several lines having different wavelengths and formed in different orders might appear near a certain angle. How could you distinguish between their orders? 8 . You are given a photograph of a spectrum on which the angular positions and the wavelengths of the spectrum lines are marked, (a) How can you tell whether the spectrum was taken with a prism or a grating instrument? (b) What infor mation could you gather about either the prism or the grat ing from studying such a spectrum? 9. A glass prism can form a spectrum. Explain how. How many “orders” of spectra will a prism produce?
17.
18.
19. 20.
21.
10. For the simple spectroscope of Fig. 8 show (a) that 6 in creases with Afor a grating and (b) that 6 decreases with Afor a prism. 11. According to Eq. 6 the principal maxima become wider (that is, S6 increases) the higher the order m (that is, the larger 6 becomes). According to Eq. 11 the resolving power becomes greater the higher the order m. Explain this appar ent paradox. 12. Explain in your own words why increasing the number of slits in a diffraction grating sharpens the maxima. Why does decreasing the wavelength do so? Why does increasing the grating spacing d do so? 13. How much information can you discover about the struc ture of a diffraction grating by analyzing the spectrum it forms of a monochromatic light source? Let A = 589 nm, for example. 14. Assume that the limits of the visible spectrum are 430 and 680 nm. How would you design a grating, assuming that the incident light falls normally on it, such that the first-order spectrum barely overlaps the second-order spectrum? 15. (a) Why does a diffraction grating have closely spaced rul ings? (b) Why does it have a large number of rulings? 16. Two light beams of nearly equal wavelengths are incident on a grating of rulings and are not quite resolvable. However, they become resolved if the number of rulings is increased. Formulas aside, is the explanation of this that: (a) more light can get through the grating? (b) the principal maxima be-
22. 23. 24.
25.
999
come more intense and hence resolvable? (c) the diffraction pattern is spread more and hence the wavelengths become resolved? (d) there is a large number of orders? or (e) the principal maxima become narrower and hence resolvable? The relation R = Nm suggests that the resolving power of a given grating can be made as large as desired by choosing an arbitrarily high order of diffraction. Discuss this possibility. Show that at a given wavelength and a given angle of diffrac tion the resolving power of a grating depends only on its width IV {= Nd). How would you experimentally measure (a) the dispersion D and (b) the resolving power R o fa grating spectrograph? For a given family of planes in a crystal, can the wavelength of incident x rays be (a) too large or (b) too small to form a diffracted beam? If a parallel beam of x rays of wavelength Ais allowed to fall on a randomly oriented crystal of any material, generally no intense diffracted beams will occur. Such beams appear if (a) the x-ray beam consists of a continuous distribution of wavelengths rather than a single wavelength or (b) the speci men is not a single crystal but a finely divided powder. Explain each case. Does an x-ray beam undergo refraction as it enters and leaves a crystal? Explain your answer. Why cannot a simple cube of edge Uq/ I in Fig. 15 be used as a unit cell for sodium chloride? In some respects Bragg reflection differs from plane grating diffraction. Of the following statements, which one is true for Bragg reflection but not true for grating diffraction? (a) Two different wavelengths may be superposed, (b) Radia tion of a given wavelength may be sent in more than one direction, (c) Long waves are deviated more than short waves, (d) There is only one grating spacing, (e) Diffraction maxima of a given wavelength occur only for particular angles of incidence. In Fig. 24a we show schematically the Debye-Scherrer ex perimental arrangement and in Fig. 24b a corresponding x-ray diffraction pattern, (a) Keeping in mind that the Laue method uses a large single crystal and an x-ray beam contin uously distributed in wavelength, explain the origin of the spots in Fig. 14. (Hint: Each spot corresponds to the direc tion of scattering from a family of planes.) (b) Keeping in mind that the Debye - Scherrer method uses a large number of small single crystals randomly oriented and a monochro matic beam of x rays, explain the origin of the rings. (Hint: Because the small crystals are randomly oriented, all possi ble angles of incidence are obtained.)
Debye-Scherrer
Figure 24 Question 25.
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Chapter 4 7 Gratings and Spectra
PROBLEMS Section 47~1 Multiple Slits 1. A diffraction grating 21.5 mm wide has 6140 rulings. {a) Calculate the distance d between adjacent rulings. {b) At what angles will maximum-intensity beams occur if the incident radiation has a wavelength of 589 nm? 2. A diffraction grating 2.86 cm wide produces a deviation of 33.2 ®in the second order with light of wavelength 612 nm. Find the total number of rulings on the grating. 3. With light from a gaseous discharge tube incident normally on a grating with a distance 1.73 pm between adjacent slit centers, a green line appears with sharp maxima at measured transmission angles 0 = ± 17.6®, 37.3®, —37.1®, 65.2®, and —65.0®. Compute the wavelength of the green line that best fits the data. 4. A narrow beam of monochromatic light strikes a grating at normal incidence and produces sharp maxima at the follow ing angles from the normal: 6 ®40', 13® 30', 20® 20', 35® 40'. No other maxima appear at any angle between 0® and 35 ®40'. The separation between adjacent ruling centers in the grating is 5040 nm. Find the wavelength of light used. 5. Light of wavelength 600 nm is incident normally on a dif fraction grating. Two adjacent principal maxima occur at sin 6 = 0.20 and sin 6 = 0.30. The fourth order is missing. (a) What is the separation between adjacent slits? (b) What is the smallest possible individual slit width? (c) Name all orders actually appearing on the screen with the values de rived in (a) and (b). 6 . A diffraction grating is made up of slits of width 310 nm with a 930-nm separation between centers. The grating is illumi nated by monochromatic plane waves, A = 615 nm, the angle of incidence being zero, (a) How many diffraction maxima are there? (b) Find the width of the spectral lines observed in first order if the grating has 1120 slits. 7. Derive this expression for the intensity pattern for a threeslit “grating” :
10. A three-slit grating has separation d between adjacent slits. If the middle slit is covered up, will the halfwidth of the intensity maxima become broader or narrower and by what factor? See Problem 8. 11. A diffraction grating has a large number N of slits, each of width a. Let denote the intensity at some principal maximum, and let I,, denote the intensity of the ^th adjacent secondary maximum, (a) If /c A^, show from the phasor diagram that, approximately, = l/(^ + (Compare this with the single-slit formula.) {b) For those secondary maxima that lie roughly midway between two adjacent principal maxima, show that roughly IJImax = X/N^. (c) Consider the central principal maximum and those adjacent secondary maxima for which k ^ N. Show that this part of the diffraction pattern quantitatively resem bles that for one single slit of width Na. Section 47~2 Diffraction Gratings 12. A diffraction grating has 200 rulings/mm and a principal maximum is noted at 0 = 28®. (a) What are the possible wavelengths of the incident visible light? (b) What colors are they? 13. A grating has 315 rulings/mm. For what wavelengths in the visible spectrum can fifth-order diffraction be observed? 14. Show that in a grating with alternately transparent and opaque strips of equal width, all the even orders (except m = 0) are absent. 15. Given a grating with 4(X) rulings/mm, how many orders of the entire visible spectrum (400- 7(X) nm) can be produced? 16. Assume that light is incident on a grating at an angle ip as shown in Fig. 25. Show that the condition for a diffraction maximum is d(sin ^ + sin ^) = wA
m = 0 , 1, 2, . . . .
Only the special case ^ = 0 has been treated in this chapter (compare with Eq. 1).
I = i/m (l H- 4 cos 0 + 4 cos^ 0), where ,
0=
2nd sin 6
------ 5 ------ .
Assume that a c X and be guided by the derivation of the corresponding double-slit formula (Eq. 17 of Chapter 46). 8 . (a) Using the result of Problem 7, show that the halfwidth of the fringes for a three-slit diffraction pattern, assuming 6 small enough so that sin 6 ^ 6, is AO^
A 3 .2 d '
(b) Compare this with the expression derived for the two-slit pattern in Problem 25, Chapter 45 and show that these results support the conclusion that for a fixed slit spacing the interference maxima become sharper as the number of slits is increased. (a) Using the result of Problem 7, show that a three-slit “grating” has only one secondary maximum. Find (b) its location and (c) its relative intensity.
Figure 25
Problem 16.
17. A transmission grating with d = 1.50 pm is illuminated at various angles of incidence by light of wavelength 600 nm. Plot as a function of angle of incidence (0 to 90 ®) the angular deviation of the first-order diffracted beam from the inci dent direction. See Problem 16.
Problems 18. Assume that the limits of the visible spectrum are arbitrarily chosen as 430 and 680 nm. Calculate the number of rulings per mm of a grating that will spread the first-order spectrum through an angular range of 20.0 ®. 19. White light (400 nm < A < 7(X) nm) is incident on a grating. Show that, no matter what the value of the grating spacing d, the second- and third-order spectra overlap. 20. A grating has 350 rulings/mm and is illuminated at normal incidence by white light. A spectrum is formed on a screen 30 cm from the grating. If a 10-mm square hole is cut in the screen, its inner edge being 50 mm from the central maxi mum and parallel to it, what range of wavelengths passes through the hole?
Section 47~3 Dispersion and Resolving Power 21. The “sodium doublet” in the spectrum of sodium is a pair of lines with wavelengths 589.0 and 589.6 nm. Calculate the minimum number of rulings in a grating needed to resolve this doublet in the second-order spectrum. 22. A grating has 620 rulings/mm and is 5.05 mm wide. (a) What is the smallest wavelength interval that can be resolved in the third order at A = 481 nm? (b) How many higher orders can be seen? 23. A source containing a mixture of hydrogen and deuterium atoms emits light containing two closely spaced red colors at A = 656.3 nm whose separation is 0.180 nm. Find the mini mum number of rulings needed in a diffraction grating that can resolve these lines in the first order. 24. (a) How many rulings must a 4.15-cm-wide diffraction grat ing have to resolve the wavelengths 415.496 nm and 415.487 nm in the second order? (b) At what angle are the maxima found? 25. In a particular grating the sodium doublet (see Problem 21) is viewed in third order at 10.2 ®to the normal and is barely resolved. Find (a) the ruling spacing and (b) the total width of the grating. 26. Show that the dispersion of a grating can be written D
~ ^ . A
27. A grating has 40,000 rulings spread over 76 mm. (a) What is its expected dispersion D in ®/nm for sodium light (A = 589 nm) in the first three orders? (b) What is its resolving power in these orders? 28. Light containing a mixture of two wavelengths, 500 nm and 600 nm, is incident normally on a diffraction grating. It is desired ( 1) that the first and second principal maxima for each wavelength appear at ^ ^ 30®, (2) that the dispersion be as high as possible, and (3) that the third order for 600 nm be a missing order, (a) What should be the separation be tween adjacent slits? (b) What is the smallest possible in dividual slit width? (c) Name all orders for 600 nm that actually appear on the screen with the values derived in (a) and (b). 29. A diffraction grating has a resolving power R = A/AA = Nm. (a) Show that the corresponding frequency range Av that can just be resolved is given by Av = c/NmA.
1001
(b) From Fig. 1, show that the “times of flight” of the two extreme rays differ by an amount At = (Nd/c) sin 6. (c) Show that (A v)( A/) = 1, this relation being independent of the various grating parameters. Assume N Section 47-4 X-Ray Diffraction 30. X rays of wavelength 0.122 nm are found to reflect in the second order from the face of a lithium fluoride crystal at a Bragg angle of 28.1®. Calculate the distance between adja cent crystal planes. 31. A beam of x rays of wavelength 29.3 pm is incident on a calcite crystal of lattice spacing 0.313 nm. Find the smallest angle between the crystal planes and the beam that will result in constructive reflection of the x rays. 32. Monochromatic high-energy x rays are incident on a crystal. If first-order reflection is observed at Bragg angle 3.40®, at what angle would second-order reflection be expected? 33. An x-ray beam, containing radiation of two distinct wave lengths, is scattered from a crystal, yielding the intensity spectrum shown in Fig. 26. The interplanar spacing of the scattering planes is 0.94 nm. Determine the wavelengths of the X rays present in the beam.
Diffraction angle
Figure 26
Problem 33.
34. In comparing the wavelengths of two monochromatic x-ray lines, it is noted that line A gives a first-order reflection maximum at a glancing angle of 23.2® to the face of a crys tal. Line B, known to have a wavelength of 96.7 pm, gives a third-order reflection maximum at an angle of 58.0® from the same face of the same crystal, {a) Calculate the inter planar spacing, (b) Find the wavelength of line A. 35. Monochromatic x rays are incident on a set of NaCl crystal planes whose interplanar spacing is 39.8 pm. When the beam is rotated 51.3® from the normal, first-order Bragg reflection is observed. Find the wavelength of the x rays. 36. Show that, in Bragg diffraction by a monochromatic beam of X rays, no intense maxima will be obtained if the wave length of the X rays is greater than twice the largest crystal plane separation. See Question 20. 37. Prove that it is not possible to determine both wavelength of radiation and spacing of Bragg reflecting planes in a crystal by measuring the angles for Bragg reflection in several orders. 38. Assume that the incident x-ray beam in Fig. 27 is not mono chromatic but contains wavelengths in a band from 95.0 to 139 pm. Will diffracted beams, associated with the planes
1002
Chapter 47
Gratings and Spectra
Incident beam
j_ d \
Figure 27
Problems 38 and 40.
shown, occur? If so, what wavelengths are diffracted? As sume d = 275 pm. 39. First-order Bragg scattering from a certain crystal occurs at an angle of incidence of 63.8®; see Fig. 28. The wavelength of the X rays is 0.261 nm. Assuming that the scattering is from the dashed planes shown, find the unit cell size 40. Monochromatic x rays (A = 0.125 nm) fall on a crystal of sodium chloride, making an angle of 42.2® with the refer ence line shown in Fig. 27. The planes shown are those of Fig. 18fl, for which J = 0.252 nm. Through what angles must the crystal be turned to give a diffracted beam asso ciated with the planes shown? Assume that the crystal is turned about an axis that is perpendicular to the plane of the page. 41. Consider an infinite two-dimensional square lattice as in
Figure 28
\
\
\
\
\
Problem 39.
Fig. \6b. One interplanar spacing is obviously a^ itself. {a) Calculate the next five smaller interplanar spacings by sketching figures similar to Fig. \Sa. (b) Show that the gen eral formula is d = ao/yfP -in?, where h and k are both relatively prime integers that have no common factors other than unity. 42. In Sample Problem 5 the m = 1beam, permitted by interfer ence considerations, has zero intensity because of the dif fracting properties of the unit cell for this geometry of beams and crystal. Prove this. (Hint: Show that the “reflection" from an atomic plane through the top of a layer of unit cells is canceled by a “reflection” from a plane through the mid dle of this layer of cells. All odd-order beams prove to have zero intensity.)
CHAPTER 48 POLARIZATION ♦
In Chapter 41, we showed electromagnetic waves traveling such that the electric field vector E and magnetic field vector B are perpendicular to each other and to the direction ofpropagation o f the wave. That is, electromagnetic waves are transverse waves. This prediction follows from M axwell’s equations. In many o f the experiments we have described so far, light waves do not reveal their transverse nature. For example, reflection, refraction, interference, and diffraction can occur fo r longitudinal waves (such as sound) as well as for transverse waves. Thomas Young (whom we also remember for the double-slit experiment) in 1817 provided the experimental basis fo r believing that light waves are transverse, building on experiments by his contemporaries Arago and Fresnel on the phenomenon we now call double refraction (see Section 48-4). In this chapter, we consider the polarization o f light and other electromagnetic waves. The direction o f polarization refers to the direction o f the E vector o f the wave. We discuss different types o f polarization, including linear and circular, and we consider the experimental techniques for producing and detecting polarized light.
48-1 POLARIZATION Consider the experimental arrangement shown in Fig. 1. A microwave transmitter on the left is connected to a dipole antenna. Charges surging up and down in the an
tenna produce an electromagnetic wave whose E vector is (at large distances from the dipole) parallel to its axis. When this wave is incident on the antenna of the microwave receiver at the right, the E vector o f the wave causes charges to move up and down in the antenna. These mov ing charges produce a signal in the receiver.
Microwave transmitter Antenna
Figure 1 The electromagnetic wave generated by the transmitter is polarized in the plane of the page, its E vector being parallel to the axis of the transmitting antenna. The receiving an tenna can detect this wave with maximum effectiveness if its antenna also lies in the plane and parallel to E. If the receiving antenna were rotated through 90® about the direction of propagation, no signal would be detected.
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1004
Chapter 48
Polarization
Plane of polarization
X
If the transmitter were rotated by 90° about the direc tion o f propagation o f the wave, the signal in the receiver would drop to zero. In this case, the E vector o f the wave would be at right angles to the axis o f the receiving an tenna; the wave would produce no movement o f chaige along the antenna and thus no signal in the receiver. A similar result would be obtained if the receiver were ro tated instead o f the transmitter. Figure 2 represents an electromagnetic wave such as that of Fig. 1. As is always the case, the E and B vectors are perpendicular to one another and to the direction o f projv agation o f the wave, which is the basic picture o f a trans verse wave. By convention, we define the direction of polarization o f the wave to be the direction of the E vector (the y direction in Fig. 2). The plane determined by the E vector and the direction o f propagation o f the wave (the xy plane in Fig. 2) is called the plane ofpolarization o f the wave. Note that specifying two directions of an electro magnetic wave (the direction o f propagation and the direc tion o f E) completely specifies the wave, because the direc tion o f B is fixed by these two directions.* • Recall the Poynting vector, S = (E X B)///,,, discussed in Sec tion 41-4, where S is in the direction of propagation of the wave. Given S and E, we can find the magnitude and direction of B.
Figure 2 An instantaneous snapshot of a traveling electromagnetic wave showing the E and g vectors. The wave is linearly polarized, in this case in the y direction. The plane o f polarization is defined to be the plane containing the E vector and the direction of propagation; in this case, the plane of polarization is the x y plane.
The wave illustrated in Fig. 2 is said to be linearlypolar ized (also called plane polarized). This means that the E field remains in a fixed direction (the y direction in Fig. 2) as the wave propagates. As in the experiment shown in Fig. 1, linearly polarized electromagnetic waves in the microwave or radio regions can be produced by orienting the axis of a dipole trans mitting antenna in a certain direction. For example, waves used to transmit television signals in the United States are polarized in a horizontal plane; for that reason. TV receiving antennas are mounted on the roofs of houses in a horizontal plane. (In England, TV signals are transmitted with a vertical plane o f polarization, and so antennas are mounted in a vertical plane.) The motions o f the electrons in the microwave antenna of Fig. 1 are coherent', they act in unison to transmit a polarized electromagnetic wave (see Fig. 3a). In ordinary sources o f light, such as an incandescent bulb or the Sun. the atoms behave independently and emit waves whose planes of polarization are randomly oriented about the direction of propagation (Fig. 3Z>). This light is transverse but unpolarized', that is, there is no preferred plane of polarization. The symmetry about the direction o f propa gation conceals the true transverse nature o f the waves. Laser light, on the other hand, is coherent and polarized.
ly
V
{a) Figure 3 (a) A linearly polarized wave, such as that of Fig. 2, viewed from along the direction of propagation. The wave is moving out of the plane of the page. Only the di rection of the E vector is shown, (b) An unpolarized wave, which can be considered to be a random superposition of many polarized waves, (c) An equivalent way of showing the unpolarized wave, as two waves linearly polarized at right angles to one another and with a random phase difference between them. The orientation of the y and z axes about the direction of propagation is completely arbitrary.
Section 48-2
Figure 3c shows an alternative and useful way to represent an unpolarized wave. The random E vectors are repre sented by components on any two perpendicular axes (here y and z). For unpolarized waves, the components have equal amplitudes, and the phase difference between them varies randomly with time.
48-2 POLARIZING SHEETS Figure 4 shows unpolarized light falling on a sheet o f commercial polarizing material called Polaroid.^ There exists in the sheet a certain characteristic polarizing direc tion, shown by the parallel lines. The sheet transmits only those wavetrain components whose electric field vectors vibrate parallel to this direction and absorbs those that vibrate at right angles to this direction. The light emerging from the sheet is linearly polarized. The polarizing direc tion o f the sheet is established during the manufacturing process by embedding certain long-chain molecules in a flexible plastic sheet and then stretching the sheet so that the molecules are aligned parallel to each other. Radiation with its E vector parallel to the long molecules is strongly absorbed, while radiation with its E vector perpendicular to that direction passes through the sheet. In Fig. 5 the polarizing sheet or polarizer lies in the plane o f the page, and the direction of propagation is out o f the page. The vector E shows the plane of vibration of a randomly selected wavetrain falling on the sheet. Two vector components, E^ (of magnitude E sin 6) and E^ (of magnitude E cos 6), can replace E, one parallel to the polarizing direction and one at right angles to it. Only the component E^ is transmitted; the component E^ is ab sorbed within the sheet. When unpolarized light is incident on an ideal polariz ing sheet, the intensity of the polarized light transmitted through the sheet is half the incident intensity, no matter what the orientation o f the sheet. We can see this from the representation o f the incident unpolarized light given in Fig. 3c, in which each of the components has, on the average, half the intensity of the incident light. Because the orientation of the axes in Fig. 3c is arbitrary, we are free to choose one of them to be along the direction of transmission o f the polarizing sheet on which it is inci dent. Since this component o f the light would be com pletely transmitted and the other completely absorbed, the sheet transmits 50% of the incident light. We can reach the same conclusion from Fig. 5, in which a wave t There are other ways of producing polarized light without using this well-known commercial product. We mention some of them later. Also see “The Amateur Scientist,” by Jearl Walker, Scientific American, December 1977, p. 172, for ways of making polarizing sheets and quarter-wave and half-wave plates and for various experiments that can be done with them.
Polarizing Sheets
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Polarizing sheet P i
Figure 4 Unpolarized light is linearly polarized (and reduced in intensity by half) after passing through a single polarizing sheet. The parallel lines, which are not actually visible on the sheet, suggest its polarization direction. l:v
Figure 5 Another view of the action of a polarizing sheet. A linearly polarized wave (perhaps one of those shown in Fig. 3b) oriented in a random direction 6 falls on the sheet. The y component of E is transmitted, and the z component is ab sorbed.
polarized in an arbitrary direction is incident on a polariz ing sheet. The component Ey (= E cos 9) is transmitted, so the transmitted intensity is proportional to Ej = E^ cos^ 9, If the incident light is unpolarized, we find the total transmitted intensity by averaging this expression over all possible orientations of the plane of polarization of the incident light, that is, over all possible values o f 9. The average value o f cos^ 9 is i, so we again conclude that half the incident light is transmitted. Owing to reflection and partial absorption of the light along the polarizing direction, real polarizing sheets may transmit only 40% o f the incident intensity. In our discussions, we assume ideal polarizers. Let us place a second polarizing sheet P2 (usually called, when so used, an analyzer) as in Fig. 6. If P2 is rotated
Polarizer
Figure 6 Unpolarized light is not transmitted by two polariz ing sheets whose polarizing directions are perpendicular to one another.
1006
Chapter 48
Polarization
Figure 7 Two sheets of polarizing mate rial are placed over an illustration from a book. In (a) the polarization directions of the two sheets are parallel, so that light passes through; in {b) the polarization di rections are perpendicular, so that no light passes through. (The illustration shows the Luxembourg Palace in Paris. Malus discov ered the phenomenon of polarization by reflection, using a calcite crystal to view sunlight reflected from the windows of this building.)
about the direction of propagation, there are two posi tions, 180° apart, at which the transmitted light intensity falls to zero; these are the positions in which the polarizing directions of Pi and Pi are at right angles. If the amplitude of the linearly polarized light incident on Pi is £■„, the amplitude o f the light that emerges is £■„ cos 6, where 6 is the angle between the p)olarizing directions o f P, and Pi. Recalling that the intensity of the light beam is proportional to the square o f the amplitude, we see that the transmitted intensity I varies with 6 ac cording to l = I^cos^0, (1) in which the maximum value of the transmitted inten sity, occurs when the polarizing directions of P, and Pi are parallel, that is, when 0 = 0 or 180°. Figure la, in which two overlapping polarizing sheets are in the parallel posi tion (0 = 0 or 180° in Eq. 1), shows that the intensity of the light transmitted through the region of overlap has its maximum value. In Fig. lb one or the other of the sheets has been rotated through 90° so that 0 in Eq. 1 has the value 90° or 270°; the intensity of the light transmitted through the region of overlap is now a minimum. Equa tion 1, called the law of Malus, was discovered by Etienne Louis Malus (1775-1812) experimentally in 1809, using polarizing techniques other than those so far described (see Section 48-3). Historically, polarization studies were made (by Young and by Malus, for example) to investigate the nature o f light. Today we reverse the procedure and deduce some thing about the nature o f an object from the polarization state o f the light emitted by, or scattered from, that object. It has been possible to deduce, from studies o f the polariza tion of light reflected from them, that the grains o f cosmic dust present in our galaxy have been oriented in the weak galactic magnetic field (about 10“* T) so that their long dimension is parallel to this field. Polarization studies
have suggested that Saturn’s rings consist of ice crystals. The size and shape o f virus particles can be determined by the polarization of ultraviolet light scattered from them. Information about the structure o f atoms and nuclei is obtained from polarization studies o f their emitted radia tions in all parts of the electromagnetic spectrum. Thus we have a useful research technique for structures ranging in size from a galaxy (10"^^^ m) to a nucleus (10"'“*m). Polarized light also has many applications in industry and in engineering. Figure 8 shows a piece of plastic that
Figure 8 A piece of plastic is viewed between crossed polar izing sheets. The light and dark patterns show regions of stress in the structure.
Section 48-3
Polarization by Reflection
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48-3 POLARIZATION BY REFLECTION
Figure 9 A portable computer with a liquid crystal display.
has been stressed and placed between polarizing sheets. The stress pattern is revealed, allowing engineers to refine their designs to reduce stress at critical locations in the structure.* Figure 9 shows a common liquid crystal dis play, which uses polarized light to form letters and num bers, such as on watches and calculator displays. The liq uid crystal is a material with stretched molecules like polarizing sheets; however, the long direction can be made to follow an applied electric field. The liquid crystal is arranged so that it normally transmits light through the polarizer and analyzer. When the electric field (from a battery) is applied to certain regions, the molecules line up in such a way that no light is transmitted through those regions, which form the dark patterns of the display.
Sample Problem 1 Two polarizing sheets have their polarizing directions parallel so that the intensity /„ of the transmitted light is a maximum. Through what angle must either sheet be turned if the intensity is to drop by one-half? Solution
From Eq. 1, since I =
Malus discovered in 1809 that light can be partially or completely polarized by reflection. Anyone who has watched the Sun’s reflection in water, while wearing a pair of sunglasses made of polarizing material, has probably noticed the effect. It is necessary only to tilt the head from side to side, thus rotating the polarizing sheets, to observe that the intensity o f the reflected sunlight passes through a minimum. Figure 10 shows an unpolarized beam falling on a glass surface. The E vectors are resolved into two components (as in Fig. 3c), one perpendicular to the plane o f incidence (the plane o f Fig. 10) and one parallel to this plane. On the average, for completely unpolarized incident light, these two components are of equal amplitude. For glass or other dielectric materials, there is a particu lar angle of incidence, called the polarizing angle dp (also known as Brewster’s angle), at which the reflection coeffi cient for the polarization component in the plane o f Fig. 10 is zero. This means that the beam reflected from the glass, although of low intensity, is linearly polarized, with its plane of polarization perpendicular to the plane of incidence. This polarization of the reflected beam can easily be verified by analyzing it with a polarizing sheet. When light is incident at the polarizing angle, the com ponent with polarization parallel to the plane o f incidence is entirely refracted, while the perpendicular component is partially reflected and partially refracted. Thus the re fracted beam, which is of high intensity, is partially polar ized. If this refracted beam passed out o f the glass into the air and were then incident on a second glass surface (again at angle 0p), the perpendicular component would be re flected, and the refracted beam would have a slightly
Incident unpolarized
Reflected
we have
\Im = ^n. COS^ e, or 0 = cos-‘ ^ ± ^ ^ = ± 45°, ± 1 3 5 '. The same effect is obtained no matter which sheet is rotated or in which direction.
* For examples of how such models are used to study classical architecture, see “The Architecture of Christopher Wren,” by Harold Dorn and Robert Mark, Scientific American, July 1981, p. 160, and “Gothic Structural Experimentation,” by Robert Mark and William W. Clark, Scientific American, November 1984, p. 176.
Figure 10 For a particular angle of incidence 6^, the re flected light is completely polarized. The refracted light is par tially polarized. The dots indicate polarization components perpendicular to the plane of the page, and the double arrows indicate polarization components parallel to the plane of the page.
1008
Chapter 48
Polarization
Incident
Figure 11 Polarization of light by a stack of glass plates. Unpolarized light is incident at the angle 6^. All reflected waves are polar ized perpendicular to the plane of the page. After passing through several layers, the transmitted wave no longer contains any ap preciable component polarized perpendicu lar to the page.
Light almost polarized in plane of page
greater polarization. By using a stack o f glass plates, we obtain reflections from successive surfaces, and we can increase the intensity o f the emerging reflected beam (see Fig. 11). The perpendicular components are progressively removed from the transmitted beam, making it more completely polarized in the plane of Fig. 11. At the polarizing angle it is found experimentally that the reflected and the refracted beams are at right angles, or (Fig. 10)
Solution
From Eq. 3
(9p = tan-‘ 1.50 = 56.3^ The angle of refraction follows from Snell’s law: sin 0p = n sin 6^, or sin
=
= 0.555
or
0, = 33.7°.
0p + (?, = 9O°. From Snell’s law, «, sin 0p =
«2
48-4 DOUBLE REFRACTION________
sin 6^.
Combining these equations leads to n, sin 0p =
«2
sin (90° —0p) =
«2
cos 6p,
or ( 2)
where the incident beam is in medium 1 and the refracted beam in medium 2. If medium 1 is air (Hi = 1), this be comes tan 0p = (3) where n is the index of refraction of the medium on which the light is incident. Equation 2 is known as Brewster's law after Sir David Brewster (1781-1868), who deduced it empirically in 1812. It is possible to prove this law rigorously from Maxwell’s equations (see also Question 14). Sample Problem 2 We wish to use a plate of glass (n = 1.50) in air as a polarizer. Find the polarizing angle and the angle of refraction.
In earlier chapters we assumed that the speed o f light, and thus the index of refraction, is independent o f the direc tion o f propagation in the medium and o f the state of polarization of the light. Liquids, amorphous solids such as glass, and crystalline solids having cubic symmetry normally show this behavior and are said to be optically isotropic. Many other crystalline solids are optically anisotropic (that is, not isotropic).*** Optical anisotropy is re sponsible for the stress pattern illustrated in Fig. 8, al though in this case the material is not crystalline. Figure 12, in which a polished crystal of calcite (CaCOj) is laid over a printed pattern, shows the optical anisotropy of this material; the image appears double. Furthermore,*
* Solids may be anisotropic in many properties: mechanical (mica cleaves readily in one plane only), electric (a cube of crys talline graphite does not have the same electric resistance be tween all pairs of opposite faces), magnetic (a cube of crystalline nickel magnetizes more readily in certain directions than in others), and so forth.
Section 48-4
1 I-------- 1 II-------CAlCfTE I I CALCITE i
EED 1"“'"I
Figure 12 A view through a birefringent crystal, showing the two images that result from the two different indices of refrac tion. The double images can be seen where there is no strip of polarizing material. The polarization axis of each strip is par allel to its long direction. Note that the two images have per pendicular polarizations.
Double Refraction
1009
birefringence. This phenomenon was studied by Huy gens, who described it in his Treatise on L ight published in 1678. If the two emerging beams in Fig. 13 are analyzed with a polarizing sheet, they are found to be linearly polarized with their planes of vibration at right angles to each other. Figure 12 shows that each of the two crossed polarizers transmits only one of the two images (but not the other). Some doubly refracting materials are strongly absorb ing for one polarization component, while the other passes through with little absorption. Such materials are called dichroic. Polarizing sheets are examples o f dichroic material. If experiments are carried out at various angles of inci dence, one of the beams in Fig. 13 (represented by the ordinary ray, or o-ray) is found to obey Snell’s law of refraction at the crystal surface, just like a ray passing from one isotropic medium into another. The second beam (represented by the extraordinary ray, or e-ray) does not obey Snell’s law. In Fig. 13, for example, the angle of incidence for the incident light is zero but the angle o f refraction of the e-ray, contrary to the prediction of Snell’s law, is nonzero. In general, the e-ray does not even lie in the plane of incidence. This difference between the waves represented by the o- and e-rays with respect to Snell’s law can be explained in these terms:
1. The o-wave travels in the crystal with the same speed
Voin all directions. In other words, the crystal has, for this wave, a single index of refraction solid.
Figure 13 Unpolarized light falling on a birefringent mate rial (such as a calcite crystal) splits into two components, the o-ray (which follows Snell’s law of refraction) and the ^-ray (which does not follow Snell’s law). The two refracted rays have perpendicular polarizations, as shown.
2. The e-wave travels in the crystal with a speed that varies with direction from to v^. In other words, the index of refraction, defined as c/v, varies with direction from n^ to n^.
the two images show perpendicular polarizations, as indi cated in Fig. 13, which shows a beam of unpolarized light falling on a calcite crystal at right angles to one o f its faces. The single beam splits into two at the crystal surface. The “double-bending” o f a beam transmitted through calcite, exhibited in Figs. 12 and 13, is called double refraction or
TABLE 1
The quantities n^ and n^ are called the principle indices o f refraction for the crystal. Problem 19 suggests how to measure them. Table 1 shows these indices for six doubly refracting cyrstals. For three of them the ^-wave is slower; for the other three it is faster than the o-wave. Some dou bly refracting crystals (such as mica and topaz) are more complex optically than calcite and require three principal
PRINCIPAL INDICES OF REFRACTTION OF SEVERAL DOUBLY REFRACTING CRYSTALS^
Crystal
Formula
Ice Quartz Wurzite Calcite Dolomite Siderite
HjO SiOj ZnS CaCOj C aO M gO -2C O j FeO C O j
' For sodium light, A= 589 nm.
no 1.309 1.544 2.356 1.658 1.681 1.875
just like an isotropic
ne 1.313 1.553 2.378 1.486 1.500 1.635
n e -n o +0.004 +0.009 + 0.022 -0 .1 7 2 -0.181 -0 .2 4 0
1010
Chapter 48
Polarization
o-wave surface e-wave surface
Figure 14 Huygens wave surfaces produced by a point source S imbedded in calcite. The polarization states of three o-rays and three e-rays are shown by the dots and arrows, re spectively. In the general case (see ray Sb, for instance), the polarization direction is not perpendicular to the ray.
indices o f refraction for a complete description o f their optical properties. Crystals whose basic structure is cubic (such as NaCl; see Fig. 15 of Chapter 47) are optically isotropic, requiring only one index o f refraction. The behavior for the speeds o f the two waves traveling in calcite is summarized by Fig. 14, which shows two wave surfaces spreading out from an imaginary point light source S imbedded in the crystal. The characteristic direc
Figure 15 Unpolarized light falls at nor mal incidence on a slab cut from a calcite crystal. Huygens wavelets are shown, as in Fig. 14. (a) No double refraction or speed difference occurs, (b) No double re fraction occurs, but there is a speed differ ence. (c) Both double refraction and a speed difference occur, (d) Same as (c), but showing the polarization states and the emerging rays.
Successive o- and ewave fronts
e-ray o-ray (c)
tion in the crystal in which v„ = is called the optic axis. The optic axis is a property o f the crystal itself and is independent of the polarization or direction o f propaga tion o f the light. The o-wave surface in Fig. 14 is a sphere, because the medium is isotropic for o-waves. The e-wave surface can not be spherical, because the speed o f the e-wave varies with direction relative to the optic axis. The e-wave sur face is an ellipsoid o f revolution about the optic axis. The two wave surfaces represent light having two different polarization states. If we consider for the present only rays lying in the plane of Fig. 14, then (1) the plane o f polariza tion for the o-rays is perpendicular to the figure, as sug gested by the dots, and (2) that for the e-rays coincides with the plane of the figure, as suggested by the double arrows. We describe the polarization states more fully at the end of this section. We can use Huygens’ principle to study the propaga tion o f light waves in doubly refracting crystals. The most general situation may be quite complicated, with the ewave emerging in a different plane than the o-wave. How ever, we may orient the crystal so the propagation direc tions for the incident wave, the o-wave, and the o-wave are all in the same plane. In the following discussion, we assume this has been done. Figure 15a shows the special case in which unpolarized light falls at normal incidence on a calcite slab cut from a crystal in such a way that the optic axis is normal to the
id)
Section 48-4
surface. Consider a wavefront that, at time t = 0, coin cides with the crystal surface. Following Huygens, we may let any point on this surface serve as a radiating center for a double set o f Huygens wavelets, such as those in Fig. 14. The plane o f tangency to these wavelets represents the new position of this wavefront at a later time t. The inci dent beam in Fig. 15a is propagated through the crystal without deviation at speed The beam emerging from the slab has the same polarization character as the inci dent beam. The calcite slab, in these special circum stances only, behaves like an isotropic material, and no distinction can be made between the o- and the ^-waves. They both travel parallel to the optic axis and so have the same speed. Figure \5b shows two views of another special case, namely, unpolarized incident light falling at right angles on a slab cut so that the optic axis is parallel to its surface. In this case too the incident beam is propagated without deviation. However, the propagation direction is perpen dicular to the optic axis, and those waves that are polar ized perpendicular to the axis have a different speed than those that are polarized along the axis. The first are owaves and their speed is the second are ^-waves and their speed is There will be a phase difference between the o-waves and ^-waves as they emerge from the bottom of the slab. Figure 15c shows unpolarized light falling at normal incidence on a calcite slab cut so that its optic axis makes an arbitrary angle with the crystal surface. Two spatially separated beams are produced, as in Fig. 13. They travel through the crystal at different speeds, that for the ^?-wave being and that for the c-wave being intermediate be tween Vo and Vg. Note that ray xa represents the shortest optical path for the transfer o f light energy from point x to the c-wavefront. Energy transferred along any other ray, in particular along ray xb, would have a longer transit time, a consequence of the fact that the speed of c-waves varies with direction. Figure 15d represents the same case as Fig. 15c. It shows the rays emerging from the slab, as in Fig. 13, and makes clear that the emerging beams are polarized at right angles to each other; that is, they are
Double Refraction
1011
for all directions of displacement of the electrons from their equilibrium positions. In doubly refracting crystals, however, k varies with direction. For electron displacements that lie in a plane at right angles to the optic axis, k has the constant value /c^, no matter how the displacement is oriented in this plane. For displacements parallel to the optic axis, k has the larger value (for calcite) k^. Note carefully that the speed of a wave in a crystal is determined by the direction in which the E vectors vibrate and not by the direction of propagation. It is the transverse E-vector vibrations that call the restoring forces into play and thus deter mine the wave speed. Note too that the stronger the restoring force, that is, the larger k, the faster the wave. For waves traveling along a stretched cord, for example, the restoring force for the transverse displacements is determined by the tension F in the cord. Equation 18 of Chapter \9 (v = 4F/p) shows that an in crease in F means an increase in the wave speed v. Figure 16, a long weighted “tire chain” supported at its upper end, provides a one-dimensional mechanical analogy for double refraction. It applies specifically to o- and e-waves traveling at right angles to the optic axis, as in Fig. 15b. If the supporting block is made to oscillate, as in Fig. 16a, a transverse wave travels along the chain with a certain speed. If the block oscillates lengthwise, as in Fig. 16b, another transverse wave is also propa gated. The restoring force for the second wave is greater than for the first, the chain being more rigid for vibrations in its plane (Fig. 16^) than perpendicular to the plane (Fig. 16a). Thus the second wave travels along the chain with a greater speed. In the language of optics we would say that the speed of a transverse wave in the chain depends on the orientation of the plane of vibration of the wave. If we oscillate the top of the chain in a random way, the wave disturbance at a point along the chain
cross-polarized.
A Mechanical Analogy (Optional) We now seek to understand, in terms of the atomic structure of optically anisotropic crystals, how cross-polarized light waves with different speeds can exist. Light is propagated through a crystal by the action of the vibrating E vectors of the wave on the electrons in the crystal. These electrons, which experience elec trostatic restoring forces if they are moved from their equilib rium positions, are set into forced periodic oscillation about these positions and pass along the transverse wave disturbance that constitutes the light wave. The strength of the restoring forces may be measured by a force constant /c, as for the simple harmonic oscillator discussed in Chapter 15 (for which F = —kx). In optically isotropic materials the force constant k is the same
( 6)
Figure 16 A one-dimensional mechanical model for double refraction, (a) Vibration perpendicular to the plane of the chain, (b) Vibration in the plane of the chain.
1012
Chapter 48
Polarization
can be described as the sum of two waves, polarized at right angles and traveling with different speeds. This corresponds ex actly to the optical situation of Fig. \5b. For waves traveling parallel to the optic axis, as in Fig. 15a, or for waves in optically isotropic materials, the appropriate me chanical analogy is a single weighted hanging chain. Here there is only one speed of propagation, no matter how the upper end oscillates. The restoring forces are the same for all orientations of the plane of polarization of waves traveling along such a chain. These considerations allow us to understand more clearly the polarization states of the light represented by the double-wave surface of Fig. 14. For the (spherical) o-wave surface, the E-vector vibrations must be everywhere at right angles to the optic axis. If this is so, the same force constant always applies, and the o-waves travel with the same speed in all directions. More specifically, if we draw a ray in Fig. 14 from S to the o-wave surface, considered three-dimensionally (that is, as a sphere), the E-vector vibrations are always at right angles to the plane de fined by this ray and the optic axis. Thus these vibrations are always at right angles to the optic axis. For the (ellipsoidal) ^-wave surface, the E-vector vibrations in general have a component parallel to the optic axis. For rays such as Sa in Fig. 14 or for the ^-rays of Fig. 15b, the vibrations are completely parallel to this axis. Thus a relatively strong force constant (in calcite) is operative, and the wave speed is relatively high. For ^-rays such as Sb in Fig. 14, the parallel component of the E-vector vibrations is less than 1(X)%, so that the corresponding wave speed is less than v^. For ray Sc in Fig. 14, the parallel component is zero, and the distinction between o- and ^-rays disappears. ■
vibration is at 45 ° to the optic axis, they have equal ampli tudes. Since the waves travel through the crystal at differ ent speeds, there is a phase difference
Ey =
sin (ot
and
E^ = E^ cos 0 )t,
where y and z represent arbitrary perpendicular axes for a wave propagating in the x direction. These equations represent the equivalence between a circularly polarized wave and two linearly polarized waves with equal ampli tudes and a 90° phase difference. The resultant inten sity in the incident circularly polarized wave is propor when the tional to E^ = Ej + E], which equals components of the electric field are given by 4. Hence
48-5 g R C U L A R POLARIZATION Let linearly polarized light of angular frequency o (= 2nv) fall at normal incidence on a slab of calcite cut so that the optic axis is parallel to the face o f the slab, as in Fig. 17. The two waves that emerge are linearly polarized at right angles to each other, and, if the incident plane o f
(4)
(5) Let the polarizing direction of the sheet make an arbi trary angle 6 with the y axis as shown in Fig. 19. The instantaneous amplitude o f the linearly polarized wave transmitted by the sheet is £ = Fj sin 0 -I- Ey cos 6 = £■„ cos o)t sin 6
£ „ sin (ot cos 6
= £ „ sin (cot + 6).
(6)
The intensity of the wave transmitted by the sheet is pro portional to E^, or I
E i,s in ^ ((o t + 6).
(7)
The eye and other measuring instruments respond only to the average intensity /, which is found by replacing sin^ (cot + 6) by its average value over one or more cycles (= i), so ( 8) Figure 17 Linearly polarized light falls on a doubly refract ing slab cut with its optic axis parallel to the surface. The plane of polarization makes an angle of 45® with the optic axis.
Comparison with Eq. 5 shows that inserting the tx)larizing sheet reduces the intensity by one-half. The orientation of the sheet makes no difference, since 6 does not appear in this equation; this is to be expected if circularly polarized
Section 48-5
Circular Polarization
1013
Figure 18 (a) Two waves of equal ampli tude and linearly polarized in perpendicu lar directions move in the x direction. Only the E vectors are shown. The waves differ in phase by 90°, such that one reaches its maximum when the other is zero, (b) The resultant amplitude of the approaching wave as seen by observers at the numbered positions shown on the x axis. Note that, as the wave propagates, each observer will see at later times what the previous ob server has seen. For instance, one-quarter cycle after the instant of this snapshot, the condition shown here for observer 7 will occur for observer 8. The resultant E vector thus appears to each observer to rotate clockwise with time.
Figure 19 Circularly polarized light falls on a polarizing sheet. E^ and E^ are instantaneous values of the two compo nents, which have maximum values E„.
light is represented by a rotating vector, because all orien tations about the propagation direction are equivalent. When unpolarized light is incident on a polarizing sheet, the intensity o f the transmitted light is also reduced by i, independent of the orientation of the sheet, as we dis cussed in Section 48-2. A simple polarizing sheet there fore cannot be used to distinguish between unpolarized and circularly polarized light. To distinguish between circularly polarized and unpo larized light, we can use a quarter-wave plate. Suppose circularly polarized light is incident on a quarter-wave plate whose optic axis has an arbitrary orientation. Com
ponents o f the incident light along and perpendicular to the direction of the optic axis differ in phase by 90°. After passing through the quarter-wave plate, an additional phase difference of 90° is introduced, which will either add to or subtract from the previous phase difference, depending on the orientation of the axis o f the quarterwave plate. The resulting phase difference is either 0° or 180°; that is, the polarization components along two per pendicular axes reach their maximum values at the same instant. The total E field is the sum of these two vectors and makes an angle of 45 ° with the two components. The emerging light is therefore linearly polarized in a direction at an angle of ± 4 5 ° with the optic axis, which we could demonstrate by placing a polarizing sheet in the path o f the light and rotating the sheet to show the extinction of the intensity. This experiment is in effect the reverse of Fig. 17, in which circularly polarized light emerges when linearly polarized light is incident on a quarter-wave plate. Here we have linearly polarized light emerging when circularly polarized light is incident. This is an example of timereversal symmetry in nature; if we reverse all motions in a physical situation, the result must also be an allowed phys ical situation. While certain very weak forces between elementary particles may not follow this symmetry, all other known forces, including electromagnetism and gravity, strictly follow the time-reversal symmetry.
Sample Problem 3 A quartz quarter-wave plate is to be used with sodium light (A = 589 nm). What is the minimum thick ness of such a plate?
1014
Chapter 48
Polarization
Solution Two waves travel through the slab at speeds corre sponding to the two principal indices of refraction given in Table 1 (Hg = 1.553 and = 1.544). If the crystal thickness is x, the number of wavelengths of the first wave contained in the crystal is ^
Xg
A ’
where Xg is the wavelength of the ^-wave in the crystal and Ais the wavelength in air. For the second wave the number of wave lengths is ^
A,
A ’
where A^ is the wavelength of the o-wave in the crystal. The difference Ng — must be w + i, where m = 0 , 1, 2, . . . . The minimum thickness corresponds to m = 0, in which case
The wave component whose vibrations are at right angles to the optic axis (the o-wave) can be represented as E^ = (E q sin 45®) sin (cot —90®) = — ]= E q v2 = —£■„ cos cot,
c o s
cot
the 90® phase shift representing the action of the quarter-wave plate. Note that E^ reaches its maximum value one-fourth of a cycle later than Ey does, for, in calcite, wave E^ (the o-wave) travels slower than wave Ey (the ^-wave). To decide the direction of rotation, let us locate the tip of the rotating electric vector at two instants of time, (Fig. 20a) / = 0 and (Fig. 20^) a short time t^ later chosen so that cu/j is a small angle. At r = 0 the coordinates of the tip of the rotating vector (see Fig. 20a) are Ey = Q and
E^ = —E ^.
At t = ti these coordinates become, approximately, Ey =
This equation yields x=
A 4 (rig -n ,)
589 nm = 0.016 mm. (4X 1.553- 1.544)
This plate is rather thin. Most quarter-wave plates are made from mica; the sheet is split to the correct thickness by trial and error.
Sample Problem 4 A linearly polarized light wave of ampli tude E q falls on a calcite quarter-wave plate with its plane of polarization at 45 ° to the optic axis of the plate, which is taken as the y axis; see Fig. 20. The emerging light will be circularly polarized. In what direction will the electric vector appear to rotate? The direction of propagation is out of the page. Solution The wave component whose vibrations are parallel to the optic axis (the ^-wave) can be represented as it emerges from the plate as Ey = {E q
c o s
45®) sin cot = - ^ E q sin cot = V2
sin cot.
sin coti ^ E^coti
E , = - E ^ coscur, Figure 20b shows that the vector representing the emerging cir cularly polarized light is rotating counterclockwise; by conven tion such light is called left-circularly polarized, the observer always being considered to face the light source. You should verify that if the plane of vibration of the incident light in Fig. 20 is rotated through ±90®, the emerging light will be right-circularly polarized.
48-6 SCATTERING OF LIGHT A light wave, falling on a tran sp aren t solid, causes the electrons in the solid to oscillate periodically in response to the tim e-varying electric vector o f the incident wave. T he wave th at travels through the m edium is the resultant o f the incident wave and the radiations from the oscilla-
Figure 20 Sample Problem 4. Linearly po larized light falls (from behind the page) on a quarter-wave plate. The incident light is po larized at 45 ®with the y and z axes, (a) At a particular time t = 0, the emerging E vector points in the —z direction, (b) A short inter val of time r, later the vector has rotated to a new position. In this case the E vector ro tates counterclockwise as seen by an observer on the X axis facing the light source.
Section 48-6
ting electrons. The resultant wave has a maximum inten sity in the direction o f the incident beam, falling off rap idly on either side. The lack o f sideways scattering, which would be essentially complete in a large “perfect” crystal, comes about because the oscillating chaiges in the me dium act cooperatively or coherently. When light passes through a liquid or a gas, we find much more sideways scattering. The oscillating electrons in this case, being separated by relatively large distances and not being bound together in a rigid structure, act independently rather than cooperatively. Thus a rigid cancellation o f wave disturbances that are not in the for ward direction is less likely to occur; there is more side ways scattering. Light scattered sideways from a gas can be wholly or partially polarized, even though the incident light is un polarized. Figure 21 shows an unpolarized beam moving upward on the page and striking a gas atom at O. The electrons at O oscillate in response to the electric compo nents o f the incident wave, their motion being equivalent to two oscillating dipoles whose axes are in the y and z directions at O. For transverse electromagnetic waves, an oscillating dipole does not radiate along its own axis. Thus an observer at O' would receive no radiation from the dipole at O oscillating in the z direction. The radiation reaching O' would come entirely from the dipole at O oscillating in the y direction and would be linearly polar ized in the y direction. As observer O' moves off the z axis, the radiation be comes less than fully polarized, because the dipole at O oscillating along the z axis can radiate somewhat in these directions. At points along the x axis, the transmitted (x > 0) or backscattered (x: < 0) radiation is unpolarized.
Scattering o f Light
1015
because both dipoles can radiate equally well in the x direction. A familiar example of this effect is the scattering o f sunlight by the molecules of the Earth’s atmosphere. If the atmosphere were not present, the sky would appear black except in the direction of the Sun, as observed by astro nauts orbiting above the atmosphere. We can easily check with a polarizer that the light from the cloudless sky is at least partially polarized. This fact is used in polar explora tion in the so/ar compass. In this device we establish direc tion by noting the nature o f the polarization o f the scat tered sunlight. As is well known, magnetic compasses are not useful in these regions. It has been learned* that bees orient themselves in their flights between their hive and the pollen sources by means of polarization o f the light from the sky; bees’ eyes contain built-in polarization sensing devices. It still remains to be explained why the light scattered from the sky is predominantly blue and why the light received directly from the Sun— particularly at sunset when the length of the atmosphere that it must traverse is greatest— is red. The cross section of an atom or molecule for light scattering depends on the wavelength, blue light being scattered more effectively than red light. Since the blue light is more strongly scattered, the transmitted light has the color of normal sunlight with the blues largely removed; it is therefore more reddish in appearance. The conclusion that the scattering cross section for blue light is higher than that for red light can be made reason-
* See “Polarized-Light Navigation by Insects,” by Rudiger Wehner, Scientific American, July 1976, p. 106.
Figure 21 An unpolarized incident wave is scattered by an atom at O. The wave scattered toward O' on the z axis is linearly polarized.
1016
Chapter 48
Polarization
able with a mechanical analogy. An electron in an atom or molecule is bound there by strong restoring forces. It has a definite natural frequency, like a small mass suspended in space by an assembly o f springs. The natural frequency for electrons in atoms and molecules is usually in a region corresponding to violet or ultraviolet light. When light is allowed to fall on such bound electrons, it sets up forced oscillations at the frequency of the incident light beam. In mechanical resonant systems it is possible to “drive” the system most effectively if we impress on it an external force whose frequency is as close as possible to the natural resonant frequency. In the case of light, the frequency o f blue light is closer to the natural resonant frequency o f the bound electron than is that of red light. We would expect the blue light to be more effective in causing the electron to oscillate, and it is more effectively scattered. Double Scattering
(Optional)
Experiments similar to that shown in Fig. 2 1 can demonstrate that electromagnetic waves must be transverse; that is, there can be no component of the E vector parallel to the direction of propagation. Suppose there were such a component along the direction of the incident wave (the x direction in Fig. 21). Then the electrons at O would oscillate in all three directions, and the scattered wave directed toward O' would show all three possible polarization directions (two transverse and one longitudinal). This radiation would thus be unpolarized. If the incident radia tion is only transverse, as in Fig. 21, the radiation propagated to O' is linearly polarized. The question as to the transverse nature
of the radiation is thus equivalent to determining whether the radiation traveling to O' is polarized or unpolarized. There is another way to make this determination. Let us place a second scatterer at O'. A dipole at O' will oscillate in response to the incident (polarized) wave in only one direction (the y' direction, that of the incident E vector, as shown in Fig. 22). Radiation scattered by that dipole can travel in the ± x ' direc tions, but (for transverse radiation) not in they' direction. Thus a detector D measuring the intensity of the radiation should see a maximum in the ± x ' directions and a mimimum of zero inten sity in the y' direction. Such an experiment, as illustrated in Fig. 22, is called a double scattering experiment. Note that the polar ization of the radiation scattered by the first target is determined through the intensity of the radiation scattered by the second target. If the radiation traveling to O' were not polarized (and not purely transverse), then the detector D would record the same intensity in all directions. We can establish the transverse nature of electromagnetic radiation either by measuring the polarization of the radiation scattered from the first target (as shown in Fig. 21) or the inten sity distribution of radiation scattered from the second target (as shown in Fig. 22). For some radiations (such as light), polariza tion measurements are relatively easy to make, and the double scattering method provides no great advantage. For other radia tions (such as Xrays or gamma rays), double scattering is usually the preferred method. Indeed, following the discovery of x rays in 1898, there was speculation whether they were waves or parti cles. A double scattering experiment, performed in 1906 by Charles Barkla, established that x rays, like visible light, were transverse in nature and helped to confirm that x rays are part of the electromagnetic spectrum. ■
Figure 22 The polarized radiation scattered at O can be scattered by an other atom at O'. A detector D mea sures the intensity of the radiation scattered by O' at various locations 6 in the x 'y ' plane.
-z, z
/
7 ^
k ZL
Section 48-7
In this chapter, we have described such properties of elec tromagnetic waves as polarization and scattering based on analysis in terms of the wave picture. As an alternative and complementary explanation, we can consider the quantum picture, in which the properties o f the radiation are associated not with the fields but with individual quanta o f radiation (photons). As an example, we review the linear momentum carried by a monochromatic light wave. In Section 41-5, we showed that the absorption by an object of energy U frDm a light wave is accompanied by the transfer of mo mentum P to the object, where U and P are related by (9) where c is the speed of light. In contrast to the wave pic ture, we can regard the light as a stream o f photons, each o f which carries an energy E. The photon is a massless particle, for which Eq. 32 o f Chapter 21 gives E = pc, so the momentum p carried by each particle (photon) is given by (10)
Comparison o f Eqs. 9 and 10 indicates the relationship between the photon and the wave pictures, or equiva lently between the quantum and classical (nonquantum) domains. The absorption o f energy U from a light wave is accomplished by the absorption of many individual pho tons of energy E by the atoms o f the object. Similarly, the momentum ^delivered to the object by the light wave can be analyzed in terms o f the momentum p delivered to individual atoms by photons in the beam. The absorption o f a circularly polarized light wave can, in an analogous way, deliver angular momentum to an object. Classical electromagnetism gives the relationship between the energy U and the angular momentum L as (O
( 11)
where (o is the angular frequency of the wave. According to quantum mechanics, the energy £ o f a photon can be written (see Eq. 38 of Chapter 8, AE = hv)
E - h v - ^ o ,.
1017
momentum, which we write here as /. Hence Eq. 12 can be written
48-7 TO THE QUANTUM LIM IT
o - f .
To the Quantum Lim it
( 12)
where h is the Planck constant. In Section 13-6, we showed that h /ln is the basic quantum unit o f angular
1 -^ .
(13)
CD
In the quantum picture, when an atom absorbs a photon of energy E, its angular momentum changes by a definite amount /. Comparing Eqs. 11 and 13, we see the corre spondence between the classical and quantum descrip tions. The total angular momentum L absorbed by the object can be regarded as the net effect of the quanta o f angular momentum / absorbed by individual atoms. Classical physics, including the wave description o f elec tromagnetic radiation, works perfectly well in analyzing a wide class of phenomena, including diffraction, polariza tion, and scattering. It is not necessary to invoke the quantum theory to explain these effects (although they can often be equally well explained based on quantum effects, as we have discussed in this section). For example, the Barkla x-ray double scattering experiment, discussed in the previous section, can also be interpreted in the quantum picture if we assign to each photon an intrinsic angular momentum (“spin”) and demand that individual photons must have their spins aligned parallel or antipar allel to their direction of propagation. This is in fact the behavior that quantum theory predicts for photons. This competition between particle and wave descrip tions o f phenomena associated with electromagnetic waves dates from the time o f Newton, who sought to explain refraction based on a particle theory o f light. Ulti mately, it is interference and diffraction experiments, such as we discussed in Chapters 45 and 46, that lead us to favor the wave interpretation. Beginning in the early 20th century, a new class of experiments was done that upset the conventional view of electromagnetic waves. The photoelectric effect (in which a metal surface irradiated with light emits electrons) and Compton scattering (in which the wavelength o f the radia tion scattered in the geometry of Fig. 21 is found to differ from the incident wavelength) cannot be accommodated in the wave picture. Further difficulties with classical phys ics arose when particles such as electrons were found to exhibit wavelike behavior under certain circumstances. The quantum theory, developed in the 1920s, offers an alternative explanation for all o f these failures o f classical physics and stresses the complementary roles of the wave and particle pictures. Chapters 4 9 -5 6 in the extended version o f this text present an introduction to the quan tum theory and some o f its many applications, ranging from the quark structure o f elementary particles to the origin and evolution of the universe.
1018
Chapter 48
Polarization
QUESTIONS 1. It is said that light from ordinary sources is unpolarized. Can you think of any common sources that emit polarized light? 2. Light from a laboratory gas discharge tube is unpolarized. How can this be made consistent with the fact that atoms and molecules radiate as electric dipoles whose radiation is linearly polarized? 3. Polarizing sheets contain long hydrocarbon chains that are made to line up in a parallel array during the production process. Explain how a polarizing sheet is able to polarize light. (Hint: Electrons are relatively free to move along these chains.) 4. As we normally experience them, radio waves are almost always polarized and visible light is almost always unpolar ized. Why is this so? 5. What determines the desirable length and orientation of the rabbit ears on a portable TV set? 6. Why are sound waves unpolarized? 7. Suppose that each slit in Fig. 4 of Chapter 45 is covered with a polarizing sheet, the polarizing directions of the two sheets being at right angles. What is the pattern of light intensity on screen C? (The incident light is unpolarized.) 8. Why do sunglasses made of polarizing materials have a marked advantage over those that simply depend on ab sorption effects? What disadvantages might they have? 9. Unpolarized light falls on two polarizing sheets so oriented that no light is transmitted. If a third polarizing sheet is placed between them, can light be transmitted? If so, explain how. 10. Sample Problem 1 shows that, when the angle between the two polarizing directions is turned from 0® to 45 ®, the inten sity of the transmitted beam drops to one-half its initial value. What happens to this “missing” energy? 11. You are given a number of polarizing sheets. Explain how you would use them to rotate the plane of polarization of a linearly polarized wave through any given angle. How could you do it with the least energy loss? 12. In the early 1950s, 3-D movies were very popular. Viewers wore polarizing glasses and a polarizing sheet was placed in front of each of the two projectors needed. Explain how the system worked. Can you suggest any problems that may have led to the early abandonment of the system? 13. A wire grid, consisting of an array of wires arranged parallel to one another, can polarize an incident unpolarized beam of electromagnetic waves that pass through it. Explain the facts that (a) the diameter of the wires and the spacing be tween them must be much less than the incident wavelength to obtain effective polarization and (b) the transmitted com ponent is the one whose electric vector oscillates in a direc tion perpendicular to the wires. 14. Brewster’s law, Eq. 2, determines the polarizing angle on reflection from a dielectric material such as glass; see Fig. 10. A plausible interpretation for zero reflection of the parallel component at that angle is that the charges in the dielectric are caused to oscillate parallel to the reflected ray by this component and produce no radiation in this direction. Ex plain this and comment on the plausibility. 15. Explain how polarization by reflection could occur if the
light is incident on the interface from the side with the higher index of refraction (glass to air, for example). 16. Find a way to identify the polarizing direction of a polariz ing sheet. No marks appear on the sheet. 17. Is the optic axis of a doubly refracting crystal simply a line or a direction in space? Has it a direction sense, like an arrow? What about the characteristic direction of a polarizing sheet? 18. If ice is doubly refracting (see Table 1), why don’t we see two images of objects viewed through an ice cube? 19. Is it possible to produce interference effects between the o-beam and the ^-beam, which are separated by the calcite crystal from the incident unpolarized beam in Fig. 13, by recombining them? Explain your answer. 20. From Table 1, would you expect a quarter-wave plate made from calcite to be thicker than one made from quartz? 21. Does the ^-wave in doubly refracting crystals always travel at a speed given by c/n^? 22. In Figs. 15a and 15b describe qualitatively what happens if the incident beam falls on the crystal with an angle of inci dence that is not zero. Assume in each case that the incident beam remains in the plane of the figure. 23. Devise a way to identify the direction of the optic axis in a quarter-wave plate. 24. If linearly polarized light falls on a quarter-wave plate with its plane of vibration making an angle of (a) 0® or (b) 90® with the axis of the plate, describe the transmitted light, (c) If this angle is arbitrarily chosen, the transmitted light is called elliptically polarized; describe such light. 25. You are given an object that may be (a) a disk of grey glass, (b) a polarizing sheet, (c) a quarter-wave plate, or (^/) a half wave plate (see Problem 21). How could you identify it? 26. Can a linearly polarized light beam be represented as a sum of two circularly polarized light beams of opposite rotation? What effect has changing the phase of one of the circular components on the resultant beam? 27. Could a radar beam be circularly polarized? 28. How can a right-circularly polarized light beam be trans formed into a left-circularly polarized beam? 29. A beam of light is said to be unpolarized, linearly polarized, or circularly polarized. How could you choose among them experimentally? 30. A parallel beam of light is absorbed by an object placed in its path. Under what circumstances will (a) linear momentum and (b) angular momentum be transferred to the object? 31. When observing a clear sky through a polarizing sheet, you find that the intensity varies on rotating the sheet. This does not happen when viewing a cloud through the sheet. Why? 32. In 1949 it was discovered that light from distant stars in our galaxy is slightly linearly polarized, with the preferred plane of vibration being parallel to the plane of the galaxy. This is probably due to nonisotropic scattering of the starlight by elongated and slightly aligned interstellar grains (see Prob lem 31 in Chapter 24). If the grains are oriented with their long axes parallel to the interstellar magnetic field lines, as discussed in Section 48-2, and they absorb and radiate elec-
Problems tromagnetic waves like the oscillating electrons in a radio antenna, how must the magnetic field be oriented with re spect to the galactic plane?
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33. Verify that Eq. 11 is dimensionally correct. 34. Is polarization or interference a better test for identifying waves? Do they give the same information?
PROBLEMS Section 48~1 Polarization 1. The magnetic field equations for an electromagnetic wave in free space are B^ = B sin (ky + (ot), By = B^ = 0. (a) What is the direction of propagation? (b) Write the elec tric field equations, (c) Is the wave polarized? If so, in what direction? 2. Prove that two linearly polarized light waves of equal ampli tude, their planes of vibration being at right angles to each other, cannot produce interference effects. (Hint: Prove that the intensity of the resultant light wave, averaged over one or more cycles of oscillation, is the same no matter what phase difference exists between the two waves.) Section 48~2 Polarizing Sheets 3. A beam ofunpolarized light of intensity 12.2 mW/m^ falls at normal incidence upon a polarizing sheet, (a) Find the max imum value of the electric field of the transmitted beam. (b) Calculate the radiation pressure exerted on the polariz ing sheet. 4. Unpolarized light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the transmitted light is one-third the intensity of the incident beam? Assume that each polarizing sheet is ideal, that is, that it reduces the intensity of unpolarized light by exactly 50%. 5. Three polarizing plates are stacked. The first and third are crossed; the one between has its axis at 45 ®to the axes of the other two. What fraction of the intensity of an incident unpolarized beam is transmitted by the stack? 6. A beam of linearly polarized light strikes two polarizing sheets. The characteristic direction of the second is 90“ with respect to the incident light. The characteristic direction of the first is at angle 0 with respect to the incident light. Find angle 6 for a transmitted beam intensity that is 0.100 times the incident beam intensity. 7. A beam of unpolarized light is incident on a stack of four polarizing sheets that are lined up so that the characteristic direction of each is rotated by 30“ clockwise with respect to the preceding sheet. What fraction of the incident intensity is transmitted? 8. A beam of light is linearly polarized in the vertical direction. The beam falls at normal incidence on a polarizing sheet with its polarizing direction at 58.8“ to the vertical. The transmitted beam falls, also at normal incidence, on a sec ond polarizing sheet with its polarizing direction horizontal. The intensity of the original beam is 43.3 W/m^. Find the intensity of the beam transmitted by the second sheet. 9. Suppose that in Problem 8 the incident beam was unpolar ized. What now is the intensity of the beam transmitted by the second sheet?
10. A beam of light is a mixture of polarized light and unpolar ized light. When it is sent through a Polaroid sheet, we find that the transmitted intensity can be varied by a factor of five depending on the orientation of the Polaroid. Find the rela tive intensities of these two components of the incident beam. 11. At a particular beach on a particular day near sundown the horizontal component of the electric field vector is 2.3 times the vertical component. A standing sunbather puts on Polar oid sunglasses; the glasses suppress the horizontal field com ponent. (a) What fraction of the light energy received before the glasses were put on now reaches the eyes? (b) The sunbather, still wearing the glasses, lies on his side. What frac tion of the light energy received before the glasses were put on reaches the eyes now? 12. It is desired to rotate the plane of vibration of a beam of polarized light by 90“. (a) How might this be done using only polarizing sheets? (b) How many sheets are required in order for the total intensity loss to be less than 5.0%? Section 48-3 Polarization by Reflection 13. (a) At what angle of incidence will the light reflected from water be completely polarized? (b) Does this angle depend on the wavelength of the light? 14. Light traveling in water of index of refraction 1.33 is inci dent on a plate of glass of index of refraction 1.53. At what angle of incidence is the reflected light completely linearly polarized? 15. Calculate the range of polarizing angles for white light inci dent on fused quartz. Assume that the wavelength limits are 400 and 700 nm and use the dispersion curve of Fig. 4, Chapter 43. 16. When red light in vacuum is incident at the polarizing angle on a certain glass slab, the angle of refraction is 31.8 “. What are (a) the index of refraction of the glass and (b) the polariz ing angle? Section 48-4 Double Refraction 17. Linearly polarized light of wavelength 525 nm strikes, at normal incidence, a wurzite crystal, cut with its faces paral lel to the optic axis. What is the smallest possible thickness of the crystal if the emergent o- and ^-rays combine to form linearly polarized light? See Table 1. 18. A narrow beam of unpolarized light falls on a calcite crystal cut with its optic axis as shown in Fig. 23. (a) For t = 1.12 cm and for = 38.8 “, calculate the perpendicular dis tance between the two emerging rays x and y. (b) Which is the o-ray and which the ^-ray? (c) What are the states of polarization of the emerging rays? (d) Describe what hap pens if a polarizer is placed in the incident beam and rotated. (Hint: Inside the crystal the E-vector vibrations for one ray
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Chapter 48
Polarization
Figure 24
Figure 23
Problem 18.
are always perpendicular to the optic axis and for the other ray they are always parallel. The two rays are described by the indices and in this plane each ray obeys Snell’s law.) 19. A prism is cut from calcite so that the optic axis is parallel to the prism edge as shown in Fig. 24. Describe how such a prism might be used to measure the two principal indices of refraction for calcite. (Hint: See hint in Problem 18; see also Sample Problem 3, Chapter 43.) Section 48~5 Circular Polarization 20. Find the greatest number of quarter-wave plates, to be used with light of wavelength 488 nm, that could be cut from a dolomite crystal 0.250 mm thick. 21. What would be the action of a halfwave plate (that is, a plate
Problem 19.
twice as thick as a quarter-wave plate) on (a) linearly polar ized light (assume the plane of vibration to be at 45® to the optic axis of the plate), (b) circularly polarized light, and (c) unpolarized light? 22. A polarizing sheet and a quarter-wave plate are glued to gether in such a way that, if the combination is placed with face A against a shiny coin, the face of the coin can be seen when illuminated with light of appropriate wavelength. When the combination is placed with face A away from the coin, the coin cannot be seen, (a) Which component is on face A and (b) what is the relative orientation of the compo nents? Section 48-7 To the Quant tm Limit 23. Assume that a parallel beam of circularly polarized light whose power is 106 W is absorbed by an object, (a) At what rate is angular momentum transferred to the object? (b) If the object is a flat disk of diameter 5.20 mm and mass 9.45 mg, after how long a time (assuming it is free to rotate about its axis) would it attain an angular speed of 1.50 rev/s? Assume a wavelength of 516 nm.
CHAPTER 49 LIGHT AND QUANTUM PHYSICS Thus fa r have studied radiation— including not only light but all o f the electromagnetic spectrum— through the phenomena o f reflection, refraction, interference, diffraction, and polarization, all o f which can be understood by treating radiation as a wave. The evidence in support o f this wave behavior is overwhelming. We now move o ff in a new direction and consider experiments that can be understood only by making quite a different assumption about electromagnetic radiation, namely, that it behaves like a stream o f particles. The concepts o f wave and particle are so different that it is hard to understand how light (and other radiation) can be both. In a wave, for example, the energy and momentum are distributed smoothly over the wavefront, while they are concentrated in bundles in a stream ofparticles. We delay a discussion o f this dual nature until Chapter 50. In the meantime, we ask that you not worry about this puzzle and that you consider the compelling experimental evidence that radiation has this particlelike nature. This begins our study o f quantum physics, which leads eventually to our understanding o f the fundamental structure o f matter.
49-1 THERM AL RADIATION________ We see most objects by the light that is reflected from them. At high enough temperatures, however, bodies be come self-luminous, and we can see them glow in the dark. Incandescent lamp filaments and bonfires (see Fig. 1) are familiar examples. Although we see such objects by the visible light that they emit, we do not have to linger too long near a bonfire to believe that it also emits copiously in the infrared region of the spectrum. It is a curious fact that quantum physics, which dominates our modem view of the world around us, arose from the study— under controlled laboratory conditions— of the radiations emitted by hot objects. Radiation given off by a body because of its tempera ture is called thermal radiation. All bodies not only emit such radiation but also absorb it from their surroundings. If a body is hotter than its surroundings it emits more radiation than it absorbs and tends to cool. Normally, it will come to thermal equilibrium with its surroundings, a condition in which its rates of absorption and emission of radiation are equal.
The spectmm of the thermal radiation from a hot solid body is continuous, its details depending strongly on the temperature. If we were steadily to raise the temperature of such a body, we would notice two things: (1) the higher the temperature, the more thermal radiation is emitted— at first the body appears dim, then it glows brightly; and (2) the higher the temperature, the shorter is the wave length of that part of the spectmm radiating most intensely— the predominant color of the hot body shifts from dull red through bright yellow-orange to bluish “white heat.” Since the characteristics of its spectmm depend on the temperature, we can estimate the tempera ture of a hot body, such as a glowing steel ingot or a star, from the radiation it emits. The eye sees chiefly the color corresponding to the most intense emission in the visible range. The radiation emitted by a hot body depends not only on the temperature but also on the material of which the body is made, its shape, and the nature of its surface. For example, at 2000 K a polished flat tungsten surface emits radiation at a rate of 23.5 W/cm^; for molybdenum, how ever, the corresponding rate is 19.2 W/cm^. In each case the rate increases somewhat if the surface is roughened.
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Light and Quantum Physics
in any way by the material o f the cavity, its shape, or its size. Cavity radiation (radiation in a box) helps us to un derstand the nature of thermal radiation, just as the ideal gas (matter in a box) helped us to understand matter in its gaseous form. Figure 2 shows a cavity radiator made o f a thin-walled cylindrical tungsten tube about 1 mm in diameter and heated to incandescence by passing a current through it. A small hole has been drilled in its wall. It is clear from the figure that the radiation emerging from this hole is much more intense than that from the outer wall o f the cavity, even though the temperatures of the outer and inner walls are more or less equal. There are three interrelated properties o f cavity radiation— all well verified in the laboratory— that any theory o f cavity radiation must explain. 1. The Stefan - Boltzmann law. The total radiated power per unit area o f the cavity aperture, summed over all wavelengths, is called its radiant intensity I(T) and is re lated to the temperature by
I(T) = a T \
Figure 1 Students contemplating thermal radiation. The study of such radiation, under controlled laboratory condi tions, laid the foundations for modem quantum mechanics.
Other differences appear if we measure the distribution in wavelength o f the emitted radiation. Such details make it hard to understand thermal radiation in terms o f simpler physical ideas; it reminds us o f the complications that arise in trying to understand the properties of real gases in terms o f a simple atomic model. The “gas problem” was managed by introducing the notion o f an ideal gas. In much the same spirit, the “radiation problem” can be made manageable by introducing an “ideal radiator” for which the spectrum o f the emitted thermal radiation de pends only on the temperature o f the radiator and not on the material, the nature of the surface, or other factors. We can make such an ideal radiator by forming a cavity within a body, the walls o f the cavity being held at a uniform temperature. We must pierce a small hole through the wall so that a sample of the radiation inside the cavity can escape into the laboratory to be examined. It turns out that such thermal radiation, called cavity radi ation, *has a very simple spectrum whose nature is indeed determined only by the temperature of the walls and not
• Also known as black-body radiation, because an ideal black body (one that absorbs all radiation incident on it) would emit the same type of radiation. We assume that the dimensions of the cavity are much greater than the wavelength of the radiation.
(1)
in which cr(= 5.670 X 10“ * W/m^-K"*) is a universal con stant, called the Stefan-Boltzmann constant. Ordinaiy hot objects always radiate less efficiently than do cavity radiators. We express this by generalizing Eq. 1 to / ( ^ ) = €a7’^
(2)
in which e, a dimensionless quantity, is called the emissivity o f the surface material. For a cavity radiator, c = 1, but for the surfaces of ordinary objects, the emissivity is always less than unity and is almost always a function of temperature. 2. The spectral radiancy. The spectral radiancy R(X) tells us how the intensity of the cavity radiation varies with
Figure 2 An incandescent tungsten tube with a small hole drilled in its wall. The radiation emerging from the hole is cavity radiation.
Section 49-1
wavelength for a given temperature. It is defined so that the product /?(A) dX gives the radiated power per unit area that lies in the wavelength band that extends from X to X + dX. R{X) is a statistical distribution function o f the same type we considered in Chapter 24. We can find the radiant intensity I(T) for any temperature by adding up (that is, by integrating) the spectral radiancy over the complete range o f wavelengths. Thus
I{T )= \ R{X)d}.
(fixed n
(3)
Figure 3 shows the spectral radiancy for cavity radia tion at four selected temperatures. Equation 3 shows that we can interpret the radiant intensity I(T) as the area under the appropriate spectral radiancy curve. We see from the figure that, as the temperature increases, so does this area and thus the radiant intensity, as Eq. 1 predicts. 3. The Wien displacement law. We can see from the spectral radiancy curves o f Fig. 3 that the wave length at which the spectral radiancy is a maximum, de creases as the temperature increases. Wilhelm Wien (Ger man, 1864-1928) deduced that varies as \/T and that the product ^ ^ T is a universal constant. Its mea sured value is A „ ^ r = 2 8 9 8 //m -K . (4) This relationship is called the Wien displacement law; Wien was awarded the 1911 Nobel prize in physics for his research into thermal radiation.
Thermal Radiation
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Sample Problem 1 How hot is a star? The “surfaces” of stars are not sharp boundaries like the surface of the Earth. Most of the radiation that a star emits is in thermal equilibrium with the hot gases that make up the star’s outer layers. Without too much error, then, we can treat starlight as cavity radiation. Here are the wavelengths at which the spectral radiancies of three stars have their maximum values: Star
Afnox 240 nm 500 nm 850 nm
Sirius Sun Betelgeuse
Appearance Blue-white Yellow Red
(a) What are the surface temperatures of these stars? {b) What are the radiant intensities of these three stars? (c) The radius r of the Sun is 7.0 X 10* m and that of Betelgeuse is over 500 times larger, or 4.0 X 10" m. What is the total radiated power output (that is, the luminosity L) of these stars? Solution
(a) From Eq. 4 we find, for Sirius, 2898//m»K / 2898/im \ 240 nm
The temperatures for the Sun and for Betelgeuse work out in the same way to be 5800 K and 3400 K, respectively. At 5800 K, most of the radiation from the Sun’s surface lies within the visible region of the spectrum. This suggests that over ages of
X (/tm) Figure 3 Spectral radiancy curves for cavity radiation at four selected temperatures. Note that as the temperature increases, the wavelength of the maximum spectral radiancy shifts to lower values.
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Chapter 49
Light and Quantum Physics
evolution, eyes have adapted to the Sun to become most sensi tive to those wavelengths that it radiates most intensely. (b) For Sirius we have, from the Stefan-Boltzmann law (Eq. 1) /= = (5.67 X 10-«W/m2*K^X 12,000 K)^ = 1 .2 X 10’ W/m2.
The radiant intensities for the Sun and for Betelgeuse work out to be 6.4 X 10^ W/m^ and 7.7 X 10^ W /m^ respectively. (c) We find the luminosity of a star by multiplying its radiant intensity by its surface area. Thus, for the Sun, L = I{4nr^) = (6.4 X 10^ W/m2X47rX7.0 X 10* m)^ = 3.9 X 1026 W. For Betelgeuse the luminosity works out to be 1.5 X 10^' W, about 38,000 times larger. The enormous size of Betelgeuse, which is classified as a “red giant,” much more than makes up for the relatively low radiant intensity associated with its low surface temperature. The colors of stars are not strikingly apparent to the average observer because the retinal cones, which are responsible for color vision, do not function well in dim light. If this were not so, the night sky would be spangled with color.
49-2 PLANCK’S RADIATION LAW Is there a simple formula, derivable from basic principles, that fits the experimental radiancy curves of Fig. 3? In September 1900 there were two suggested formulas, nei ther o f which could fit the curves over the entire range o f wavelengths. The first, due originally to Lord Rayleigh but later de rived independently by Einstein and modified by James Jeans, was developed rigorously from its classical base. Unfortunately, it completely fails to fit the curves, not even passing through a maximum. However, the Rayleigh - Jeans formula, as it is called, does fit the curves quite well in the limit of very long wavelengths. Figure 4 shows the spectral radiancy curve for cavity radiation at 2000 K, along with the Rayleigh-Jeans prediction. The good fit we sp>eak of occurs for wavelengths much greater than 50 ^m, far beyond the scale of that figure. The Rayleigh-Jeans formula, unsatisfactory though it may be, is the best that classical physics has to offer. Wilhelm Wien also derived a theoretical expression for the spectral radiancy. His formula (see also Fig. 4) is much better. It fits the curves quite well at short wavelengths, passes through a maximum, but departs noticeably at the long-wavelength end o f the scale. However, Wien’s for mula was not based on classical radiation theory but in stead on a conjecture— it has been called a “guess” — that there is an analogy between the spectral radiancy curves and the Maxwell speed distribution curves for the molecules o f an ideal gas.
Figure 4 The solid curve shows the experimental spectral ra diancy for radiation from a cavity at 2000 K. The predictions of the classical Rayleigh-Jeans law and Wien’s law are shown as dashed lines. The shaded vertical bar represents the range of visible wavelengths.
Thus we have two formulas, one agreeing with experi ment at long wavelengths and the other at short wave lengths. Max Planck,* seeking to reconcile these two radi ation laws, made an inspired interpolation between them that turned out to fit the data at all wavelengths. Planck’s radiation formula, announced to the Berlin Physical Soci ety on October 19, 19(X), is 1 /?(/) = ^5
1
(5)
in which a and b are empirical constants, chosen to give the best fit of Eq. 5 to the experimental data. Figure 5 shows how good the agreement is. Even though correct, Planck’s formula was originally only empirical and did not constitute a true theory. Planck set to work at once to derive his formula from simple assumptions and, in 2 months, he succeeded. In the process he recast his formula slightly, presenting the two arbitrary constants it contained in a different form. In this new notation, Planck’s radiation law becomes
* Max Planck (1858-1947) was a German theoretical physicist whose specialization in thermodynamics led him to the study of thermal radiation and the discovery of the quantization of en ergy, for which he was awarded the 1918 Nobel prize in physics. Under his leadership, theoretical physics flourished in Germany in the 1920s; young physicists trained by Planck and his col leagues produced a complete mathematical formulation of the quantum theory. In his later life, Planck wrote extensively on religious and philosophical issues.
Section 49-3
The Quantization o f Energy
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49-3 THE QUANTIZATION OF ENERGY_______________________
Figure 5 Planck’s radiation law fitted to experimental data for a cavity radiator at 1595 K.
/?(A) =
2nc^h
1
A5
—1■
( 6)
The two adjustable constants a and b in Eq. 5 are here replaced by quantities involving two different constants, the Boltzmann constant k (see Section 23-1) and a new constant, now called the Planck constant h; the quantity c is the speed o f light. By fitting Eq. 6 to the experimental data, Planck could find values for k and h. His values were within a percent or so o f their presently accepted values, which are A: = 1.381 X 10-2^ J/K and
We turn now to the assumptions made by Planck in deriv ing his radiation law and to the significance o f the con stant h that appears in it. These assumptions and their consequences were not immediately clear to Planck’s con temporaries or for that matter (as he confirmed later) to Planck himself. In what follows we describe the situation as it appeared some 6 or 7 years after Planck first ad vanced his theory. It seems to be true that the basic prem ise underlying Planck’s radiation law— the quantization of energy— was not understood at any earlier date. Planck derived his radiation law by analyzing the inter play between the radiation in the cavity volume and the atoms that make up the cavity walls. He assumed that these atoms behave like tiny oscillators, each with a char acteristic frequency of oscillation. These oscillators radi ate energy into the cavity and absorb energy from it. It should be possible to deduce the characteristics of the cavity radiation from the characteristics of the oscillators that generate it. Classically, the energy of these tiny oscillators is a smoothly continuous variable. We certainly assume this for large-scale oscillators such as pendulums or m assspring systems. It turns out, however, that in order to derive Planck’s radiation law it is necessary to make a radical assumption; namely, atomic oscillators may not
emit or absorb any energy E but only energies chosenfrom a discrete set, defined by
h = 6.626 X 1 0 -^ J -s.
E = nhv, Sample Problem 2 Figure 4 suggests that Planck’s radiation law (Eq. 6) approaches the classical Rayleigh-Jeans law at long wavelengths. To what expression does Planck’s law reduce as A^ 0 0 ? Solution form
For algebraic convenience, we can write Eq. 6 in the
A*
e ^ -1 ’
in which x = hc/kkT. As A—►oo, we see that x —►0. Recalling that .
2!
.
.
3!
(see Appendix H) allows us to make the approximation \ ^ X.
Thus we have
R{X)
_ Inc'^h 1 _ Inc'^h / ^JcT\ _ In ckT A" ■
Note that the Planck constant h, a sure identifier of a quantum formula, conveniently cancels out as we approach the classical long-wave limit. The above result, in fact, is precisely the classi cal Rayleigh-Jeans expression for the spectral radiancy.
« = 1, 2, 3, . . . ,
(7)
in which v is the oscillator frequency. Here the Planck constant h is introduced into physics for the first time. We say that the energy of an atomic oscillator is quantized and that the integer « is a quantum number. Equation 7 tells us that the oscillator energy levels are evenly spaced, the interval between adjacent levels being hv\ see Fig. 6. The assumption of energy quantization is indeed a radi cal one, and Planck himself resisted accepting it for many years. In his words, “My futile attempts to fit the elemen tary quantum of action [that is, h] somehow into the classical theory continued for a number of years, and they cost me a great deal of effort.’’ Max von Laue, the 1914 Nobel laureate in physics and a student of Planck’s, has written: “After 1900 Planck strove for many years to bridge, if not to close, the gap between the older and the quantum physics. The effort failed, but it had value in that it provided the most convincing proof that the two could not be joined.” Let us look at energy quantization in the context of a large-scale oscillator such as a swinging pendulum. Our experience suggests that a pendulum can oscillate with any reasonable total energy and not only with certain selected energies. As friction causes the pendulum ampli-
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Light and Quantum Physics
Figure 6 The energy levels for atomic oscillators at three se lected frequencies. The quantum numbers of some of the levels are indicated. On the right is shown the energy k T for a classical oscillator at 2000 K.
■20
3h
> S ^ 2 O ) c
15
•10 :kT
UJ
1-
10
15
20
25
Frequency (10^^ Hz)
tude to decay, it seems that the energy is dissipated in a perfectly continuous way and not in “jumps” or “quanta.” However, because the Planck constant is so small, there is no basis in such everyday experience to dismiss energy quantization as a violation of “common sense.” The “jumps” are there; they are just far too small for us to detect. We would not apply quantum theory to a pendulum because classical theory works perfectly well in that case. We now see classical theory as a limiting case of quantum theory, the two being connected by the correspondence principle, which states that:
As the amplitude of the oscillations dies away by friction, quan tum theory predicts that the energy E will fall in “jumps” whose size is A £ = /zv = (6.63X 10-^J*sX 0.50s-') = 3.3X 10-^ J. Thus E '
3.3 X 10-^ J = 2.2 X 0.015 J
Energy measurements of such precision simply cannot be made. The quantum jumps of this oscillator are too small to be de tected, and we are quite safe in treating the problem by the methods of classical physics. (b) From the quantization relation, Eq. 7, we have
Quantum theory must agree with classical theory in the limit in which classical theory is known to agree with experiment. Another way of stating the correspondence principle is:
an enormous number! It is not surprising that we cannot detect energy quantization in the operation of an oscillator, we cannot measure changes of one unit out of 4.6 X 10’‘.
Quantum theory must agree with classical theory in the limit of large quantum numbers. It is in this way that the swinging pendulum and the oscil lating atom relate to each other. The classical limit is illustrated in the following sample problem.
Sample Problem 3 A 300-g body, connected to a spring whose force constant /c is 3.0 N/m, is oscillating with an amplitude A of 10 cm. Treat this system as a quantum oscillator and find {a) the energy interval between adjacent energy levels and {b) the quan tum number that describes the oscillations. Solution
{a) The frequency of oscillation is found from 1 [k v = — W— 2nym
1 /3.0 N/m / V ,— = 0.50 s '. 2 n y 0.3 kg
The total mechanical energy E of the oscillating system is E = {loP = K3.0 N/mXO. 10 m)^ = 0.015 J.
The quantization of energy simply does not show up for large-scale oscillators. The smallness o f the Planck con stant h makes the graininess in the energy too fine to detect. In much the same way, we cannot tell by waving our hand through air that it is made up of molecules. The Planck constant might as well be zero as far as classical systems are concerned and, indeed, one way to reduce quantum formulas to their limiting classical coun terparts is to let /»—» 0 in those formulas. In a similar way, we reduce relativistic formulas to their limiting classical counterparts by letting c where c is the speed of light. This leaves the question: “Why should letting the wave length increase mean that we are approaching a realm in which classical physics holds?” The answer is that as the wavelength increases, the frequency decreases and thus the basic energy quantum (= hv) becomes smaller. To tell whether we are in a classical or a quantum situation we
Section 49-4
must compare this quantity with kT, which is a (classical) measure of the mean translational energy of a particle at temperature T. If hv kT, the “graininess” of the en ergy of the atomic oscillators (which is measured by hv) will not be noticed and we are in the classical realm. To sum up then, we approach classical situations as v —»•0, as A-♦ 00, or (for that matter) asT—*
49-4 THE HEAT CAPACITY OF SOLIDS _____________________ Energy quantization was slow to be accepted, not an un usual fate for a radically new idea. It is not hard to see why. The systems whose energies were first quantized were the hypothetical “oscillators” that Planck assumed to form the walls of a cavity radiator. In fact, there are no such simple, one-dimensional, harmonic oscillators. The atoms that make up the wall are far more complex. Energy quantization started to come into its own only after 1907, when Einstein showed that the same ideas that had worked so well for the cavity radiation problem could be used to solve another problem, that of the heat capaci ties of solids. Here, as we shall see, the systems whose energies are to be quantized are real and familiar atoms. If you transfer heat Q to 1 mole of a solid and if a temperature rise AT results, the molar heat capacity is defined from (constant volume). C = -^ (8) AT We have chosen to transfer heat under conditions of con stant volume, so that the distances between atoms remain constant and any added energy appears entirely as energy of oscillation of the atoms about their fixed lattice sites. We take the amount of the substance to be 1mole, so that comparisons from element to element can be made on the basis of the same number of atoms. See Section 25-3 for more details on molar heat capacities and for the relation ship between Cy, which is easier to calculate, and Cp, which is easier to measure. Table 1 shows the molar heat capacities of some ele mental solids at or near room temperature. Aglance at the table shows a regularity known as the Dulong and Petit rule, after the investigators who first pointed it out in 1819. It asserts simply that, with a few exceptions, all solids have the same molar heat capacity, namely, about 25 J/mol •K. Values that were substantially less than this were called “anomalous” in those early days. Figure 7, which shows the molar heat capacity of lead, aluminum, and carbon as a function of temperature, clar ifies the situation. We see that Cy for all three elements approaches the same limiting value at high temperatures. That carbon appears “anomalous” in Table 1 simply re
The Heat Capacity o f Solids
1027
TABLE 1 MOLAR HEAT CAPACITIES OF SOME SOLIDS" Solid
Cy (J/mol K)
Aluminum Beryllium Bismuth Boron Cadmium Carbon (diamond) Copper Gold Lead Platinum Silver Tungsten
23 11 25 13 25 6 24 25 25 25 24 24
" All measurements were made at room temperature; three ''anomalous*' values have been offset for emphasis.
fleets the fact that, for this element, room temperature is not a very high temperature. What does classical physics predict for the molar heat capacity of a solid? The atoms in a solid are arranged in a threedimensional lattice. Each atom, bound to its lattice site by electromagnetic forces, oscillates about that site with an amplitude that increases as the temperature increases. Each atom behaves like a tiny oscillator with three inde pendent degrees of freedom, corresponding to three inde pendent directional axes along which the atom is free to move. The classical equipartition of energy theorem asso ciates an energy of {kT with each degree of freedom. The three-dimensional oscillator has six degrees of freedom, two for each direction (corresponding to the kinetic and potential energies for motion of the oscillation in that direction). The internal energy per mole of a solid is then E,,, = 6({kT)N^ = 3RT,
(9)
Figure 7 The molar heat capacities of three solids as a func tion of temperature.
1028
Chapter 49
Light and Quantum Physics
in which is the Avogadro constant and R is the univer sal gas constant. If the solid is held at a constant volume, we can replace Q in Eq. 8 (the heat transferred per mole) by the change in internal energy per mole. Doing so yields Cy = AEiaJAT, which becomes d_Eu V “
( 10)
dT
in the differential limit. Substituting from Eq. 9 yields finally Cy = -^ (3 R T ) = 3R.
(11)
Classical theory predicts the molar heat capacity to be constant, the same for all substances, and independent o f temperature. Substituting the value of R (= 8.31 J/mol • K) yields Cy = 24.9 J/mol • K. This agrees very well with the limiting value of Cy at high temperatures, as Fig. 7 and Table 1 show. However, there is no indication from this classical theory of the variation at lower temper atures that is shown in Fig. 7.
Quantum Theory o f Heat Capacity We turn now to the prediction o f quantum theory. Ein stein assumed that the energies of the atomic oscillators in the solid were quantized according to Eq. 7, and he as signed to each oscillator an average energy per direction, not of kT as in the classical case, but o f
hv E = ^hvIkT_ I *
( 12)
This is the same expression used by Planck for the average energy o f the oscillators in the cavity radiation problem. In Eq. 12, v is the natural vibrational frequency of the oscillating atom, which Einstein left as an adjustable con stant. Multiplying Eq. 12 by the Avogadro constant and also by a factor o f 3 to take account of the three directions, we obtain the internal energy per mole:
F
3N.hv
= ^hvIkT ^ J
(13)
Differentiating, as in Eq. 10, gives eventually C y = ^ = 3 /? ( /,v W ( ^ J , , - l ) ,
(14)
as Einstein’s prediction for the molar heat capacity. There is only one adjustable parameter in Eq. 14, the oscillator frequency v. Commonly, this frequency v is expressed in terms o f a characteristic Einstein temperature = hv/k. This temperature can be chosen so that Einstein’s equa tion fits the data rather well, although there are small deviations at low temperatures, deviations that had not yet been experimentally established when Einstein pro posed his theory. The failure to agree with experiment at low tempera ture can be traced to the fact that Einstein— perhaps deliberately— made an overly simple assumption, namely, that the oscillations of a particular atom are not influenced by those of its neighbors. In 1912 the Dutch physicist Peter Debye refined Einstein’s theory by taking the interaction of the atomic oscillators with neighboring atoms into account. Figure 8 shows the excellent agree ment of the Debye theory with experiment for a number of solids. The temperature scale in that figure is dimen sionless, being a constant that has a different value for each material. When these characteristic Debye tempera tures, as they are called, are properly assigned, we see how nicely all the experimental points fall on the same theoreti cal curve. This agreement is a major triumph for quan tum theory! Figures 7 and 8 immediately suggest the explanation for the “anomalous” values of Table 1. For these sub stances, Td is much greater than room temperature, so that the heat capacity has not yet reached its limiting value. At high temperatures, you can show (see Problem 22) that Einstein’s expression for the heat capacity (Eq. 14) reduces to the classical result (Eq. 11). This occurs for the same reasons that we discussed at the end o f Sample
Figure 8 The quantum theory result for the heat ca pacity of solids is in excellent agreement with the ex perimental results. The horizontal scale is the dimen sionless ratio of the temperature T to the Debye temperature the latter having a characteristic value for each substance.
T!T^
Section 49-5
Problem 3. In a solid, the frequency v o f the atomic oscil lators was assumed by Einstein to have a single constant value, characteristic o f the substance. Thus as T ^ we approach the condition in which hv kT. This, as we have seen, means that the energy interval between adja cent levels for the atomic oscillators (= hv) is much less than the mean translational energy of the atoms (meas ured classically by kT). Under these conditions, the en ergy quantization o f the atomic oscillators is not appar ent, and classical conditions hold.
The Photoelectric Effect Vacuum
♦ Slifjing contact
1029
. Quartz window
+
r-VNAAAAAAA/WWWW\n Sample Problem 4 In terms of the Einstein temperature Te , find the temperature at which the heat capacity of a substance has half its classical value. Solution The classical value is 5R, so we seek the temperature at which Cy in Eq. 14 has the value 3f?/2, or 3/?
(f;)’
ghv/kT
^
2R
Figure 9 An apparatus for studying the photoelectric effect. The arrows show the direction of the current in the external circuit, which is opposite to the direction of motion of the electrons. The voltmeter V measures the externally applied voltage V„,.
Letting x = h v/kT = T’e/T’, we can write this as , e^ _ 1 ^ ( e ^ - 1)2 2 ■ There is no analytic technique for solving this equation. A nu merical solution can be found on a calculator by trial and error or on a computer by calculating and displaying a table of values of the function on the left-hand side and noting the value of ;c when the function has the value The result is x = 2.98. Since x = T^/T, we have T/T^ = x " ' = 0.336, or r = o .3 3 6 rE .
49-5 TH E PHOTOELECTRIC EFFECT________________________ We were led to the idea o f energy quantization by looking at the interplay between matter and radiation at the walls of a cavity radiator. Here we consider another example of a radiation-matter interaction, the photoelectric effect. It involves the Planck constant in a central way and extends the idea o f quantization to the very nature o f radiation itself. Figure 9 shows a typical apparatus used to study the photoelectric effect. Light of frequency v falls on a metal surface (emitter E) and, if the frequency is high enough, the light will eject electrons out o f the surface. If we set up a suitable potential difference Fbetween E and the collec tor C, we can collect these photoelectrons, as we call them, and measure them as a photoelectric current i. The potential difference Vthat acts between the emitter and the collector is not the same as the potential differ-
ence supplied by the external battery and read on the voltmeter in Fig. 9. There is also a second em f— a hidden battery, if you will— associated with the fact that the emitter and the collector are almost always made of different materials. If suitable precautions are taken, this contact potential difference remains constant throughout the experiment. The potential difference V that the electrons “see” is the algebraic sum o f these two quantities, or (15) In all that follows we shall assume that this contact poten tial difference has been measured and taken into account, and we shall express all our results in terms of Fas defined by Eq. 15. Figure 10 (curve a) shows the photoelectric current as a function of the potential difference V. We see that if V is positive and large enough, the photoelectric current reaches a constant saturation value, at which all photo electrons ejected from E are collected by C. If we reduce Vto zero and then reverse it, the photoelec tric current does not immediately drop to zero because the electrons emerge from emitter E with nonzero speeds. Some will reach the collector even though the potential difference opposes their motion. However, if we make the reversed potential difference large enough, we reach a value Fq, called the stopping potential, at which the photo electric current does indeed drop to zero. This potential difference, multiplied by the electronic charge e, gives us the kinetic energy of the most energetic o f the emit ted photoelectrons: K„,^ = eV^. (16) The stopping potential Fq, and thus K ^ , is indepen dent o f the intensity o f the incident light. Curve b in Fig.
1030
Chapter 49
Light and Quantum Physics increase as the light beam is m ade m ore intense. However, Fig. 10 shows th at (= ^ ^o) is independent of the light intensity', this has been tested over a range o f intensities o f 10^. 2. Thefrequency problem. A ccording to the wave theory, the photoelectric effect should occur for any frequency of the light, provided only th a t the light is intense enough to supply the energy needed to eject the photoelectrons. However, Fig. 11 shows th at there exists, for each surface, a characteristic cutoff frequency Vq. For frequencies less
than Vq, the photoelectric effect does not occur, no matter how intense the illumination. 3. The time delay problem. If the energy acquired by a ratus of Fig. 9. The intensity of the incident light is twice as great for curve b as for curve a.
10, in w hich the light intensity has been doubled, shows this. A lthough the saturation cu rren t is also doubled, the stopping potential rem ains unchanged. Figure 11 is a plot o fth e stopping potential as a function o f the frequency o f the incident light. W e see by extrapola tion th at there is a sharp cutofffrequency Vqcorresponding to a stopping potential o f zero. F or light o f a lower fre quency th an this, no photoelectrons at all are em itted. T here sim ply is no photoelectric effect. T hree m ajor features o f the photoelectric effect cannot be explained in term s o f the classical wave theory o f light. As for the cavity radiation an d the heat capacity prob lems, the failure o f classical wave theory in these cases is n o t a m atter o f a sm all num erical disagreem ent. T he fail ure is total an d indisputable. H ere are the three problem s:
1. The intensity problem. W ave theory requires th a t the oscillating electric vector E o f the light wave increases in am plitude as the intensity o f the light beam is increased. Since the force applied to the electron is eE, this suggests th a t the kinetic energy o f the photoelectrons should also
photoelectron is absorbed directly from the wave incident on the m etal plate, the “ effective target area” for an elec tron in the m etal is lim ited and probably n o t m uch m ore th an th at o f a circle o f diam eter roughly equal to th at o f an atom . In the classical theory, the light energy is uniform ly distributed over the wavefront. T hus, if the light is feeble enough, there should be a m easurable tim e lag, which we shall estim ate in Sam ple Problem 5, betw een the im ping ing o f the light on the surface an d the ejection o f the photoelectron. D uring this interval the electron should be absorbing energy from the beam until it had accum ulated enough to escape. However, no detectable time lag has
ever been measured. In the next section we see how q u an tu m theory solves these problem s in providing the correct interpretation of the photoelectric effect.
Sample Problem 5 A potassium foil is placed a distance r (= 0.5 m) from a light source whose output power Pq is 1.0 W. How long would it take for the foil to soak up enough energ> (= 1.8 eV) from the beam to eject an electron? Assume that the ejected photoelectron collected its energy from a circular area of the foil whose radius equals the radius of a potassium atom (1.3 X 10-‘Om). Solution If the source radiates uniformly in all directions, the intensity / of the light at a distance r is given by LOW /= ^ = . : = 0.32 W/m2. 47rr^ 47t(0.5 m)^ The target area .4 is 7t( 1.3 X lO” '®m )^or5.3X 10"^° m^ so that the rate at which energy falls on the target is given by P = 7/4 = (0.32 W/m2X5.3 X 10”
m^)
= 1.7X 10-20 J/s. If all this incoming energy is absorbed, the time required to accumulate enough energy for the electron to escape is
Figure 11 The stopping potential as a function of frequency for a sodium surface. The data come from Millikan’s meas urements in 1916.
_ / 1.8 eV \ / 1 . 6 X 10->’ J \ ' \1 .7 X 10-“ J / s / \ lev ) Our selection of a radius for the effective target area was some what arbitrary, but no matter what reasonable area we choose.
Section 49-6 we would still calculate a “soak-up time” within the range of easy measurement. However, no time delay has ever been observed under any circumstances, the early experiments setting an upper limit of about 10“ ’ s for such delays.
49-6 EINSTEIN’S PH O TO N THEORY_______________________
Einstein's Photon Theory
1031
tation o f the photoelectric effect. As for the first objection (“ the intensity problem ’’), there is com plete agreem ent o f the p hoton theory w ith experim ent. If we double the light intensity, we double the n u m b er o f photons an d thus double the photoelectric current; we do not change the energy o f the individual photons or the nature o f the indi vidual photoelectric processes described by Eq. 18. T he second objection (“ the frequency problem ” ) is m et by Eq. 18. If equals zero, we have hvo = 4>,
In 1905 Einstein m ade a rem arkable assum ption about the natu re o f light; nam ely, that, u n d er som e circum stances, it behaves as if its energy is concentrated into localized bundles, later called photons. T he energy £* o f a single p h oton is given by E = hv,
(17)
where v is the frequency o f the light. T his n otion th at a light beam behaves like a stream o f particles is in sharp contrast to the notion th at it behaves like a wave. In the wave theory o f light, the energy is not co ncentrated into bundles b u t is spread o u t uniform ly over the wavefronts. W hen Planck, in 1900, derived his radiation law and first introduced the q u an tity h in to physics, he m ade use o f the relation E = hv. H e applied it, however, not to the radiation w ithin the cavity b u t to the atom ic oscillators th at m ade up its walls. Planck treated the cavity radiation on the basis o f wave theory, b u t Einstein was later able to derive Planck’s radiation law on the basis o f his photon concept. H is m ethod was both clear an d sim ple and avoided m any o f the special assum ptions th a t Planck had found it necessary to m ake in his pioneering effort. If we apply E instein’s p h oton concept to the photoelec tric effect, we can write hv = (f)
(18)
where hv is the energy o f the photon. E quation 18 says th at a single pho to n carries an energy hv into the surface where it is absorbed by a single electron. P art o f this en ergy (0 , called the work fu n ctio n o f the em itting surface) is used in causing the electron to escape from the m etal surface. T he excess energy {hv — (f>) becom es the electron kinetic energy; if the electron does n o t lose energy by internal collisions as it escapes from the m etal, it will still have this m uch kinetic energy after it emerges. T hus represents the m ax im u m kinetic energy th a t the p h oto electron can have outside the surface.* C onsider how E instein’s p h o to n hypothesis m eets the three objections raised against the w ave-theory interpre
* The work function represents the energy needed to remove the least tightly bound electrons from the surface. More tightly bound electrons require a larger energy and (for a fixed photon energy) emerge with a kinetic energy smaller than .
w hich asserts th a t the photon has ju st enough energy to eject the photoelectrons a n d none extra to app ear as ki netic energy. If v is reduced below Vq, hv will be sm aller th an 0 an d the individual photons, no m atter how m any o f them there are (that is, no m atter how intense the illu m ination), will not have enough energy to eject photoelec trons. T he th ird objection (“the tim e delay problem ” ) follows from the photon theory because the required energy is supplied in a concentrated bundle. It is not spread u n i form ly over the beam cross section as in the wave theory. Let us rewrite Einstein’s photoelectric equation (Eq. 18) by substituting for from Eq. 16. T his yields, after rearrangem ent, V^ = ih/e)v-{
(19)
T hus Einstein’s theory predicts a linear relationship be tween Vq an d v, in com plete agreem ent with experim ent; see Fig. 11. T he slope o f the experim ental curve in this figure should be h/e, so h
ab
e
be
2.30 V - 0.68 V (10 X 10'“ - 6 . 0 X 10'“) H z = 4.1 X 10-'* V -s.
W e can find h by m ultiplying this ratio by the electron charge e, A = (4.1 X 10-'5 V -s)(1 .6 X 1 0 - '’ C )
= 6.6X 10-3“ J-s. From a m ore careful analysis o f these and o th er data, including data taken w ith lithium surfaces, M illikan found the value A = 6.57 X 10"^“ J -s w ith a n a c c u r a c y o f ab out 0.5%. This agreem ent with the value o f h derived from Planck’s radiation form ula is a striking confirm a tion o f Einstein’s photon concept. W hen Einstein first advanced his p h oton theory o f light, the facts o f photoelectricity were not nearly as well established experim entally as we have described. Precise photoelectric m easurem ents are difficult, and it was not until 1916 th at M illikan successfully subjected Einstein’s photoelectric equation to rigorous exjjerim ental test. Al though M illikan showed th a t this equation agreed with experim ent in every detail, he him self rem ained u n co n vinced th a t Einstein’s light particles were real. H e w rote o f E instein’s “ bold, not to say reckless, hypothesis” and
1032
Chapter 49
Light and Quantum Physics
w rote furth er th at E instein’s pho to n concept “ seems at present to be wholly u n ten ab le.” Planck, the very originator o f the con stan t A, did not at once accept E instein’s photo n s either. In recom m ending Einstein for m em bership in the Royal Prussian A cadem y o f Sciences in 1913, he wrote: “ th at he m ay som etim es have m issed the target in his speculations, as for exam ple in his theory o f light q u an ta, can n o t really be held against h im .” It is n o t unusual for truly novel ideas to be accepted only slowly, even by leading scientists such as M illikan an d Planck. It was, incidentally, for his pho to n theory as applied to the photoelectric effect th a t Einstein received the N obel prize in physics for 1921.
Sample Problem 6 Find the work function of sodium from the data plotted in Fig. 11. Solution The intercept of the straight line in Fig. 11 on the frequency axis is the cutoff frequency Vq. Putting Fq = 0 and V= Vq in Eq. 19 yields 0 = /zvo = (6.63X 10-^J*s)(4.39X 10'^ Hz) = 2.91 X 10-*’ J = 1.82 eV. We note from Eq. 19 that a determination of the Planck constant h involves only the slope of the straight line in Fig. 11 and a determination of the work function 0 involves only the intercept. Convince yourself that in the first case you need not take the contact potential difference into account but in the second case you must do so.
Sample Problem 7 At what rate per unit area do photons strike the metal plate in Sample Problem 5? Assume a wavelength of 589 nm (yellow sodium light). Solution Recall our previous definition (see Section 41-4) of the intensity of light: energy per unit time per unit area (the area being taken as perpendicular to the direction of propagation of the light). Here we consider the intensity (for monochromatic light) in terms of photons as the energy per photon times the rate per unit area at which the photons strike a surface perpendicular to their motion. The two interpretations of intensity are equiva lent. The intensity of the light falling on the plate is, from Sample Problem 5,
7 2.0X 10'«eV/m2*s ^ ^ ^ , r = —= — . , . ---------- = 9.5 X 10'^photons/m^-s. E 2.1eV/photon ^ Even at this modest light intensity the photon rate is very great, with about 10'^ photons falling on 1 mm^ each second.
49-7 THE COM PTON EFFECT Cavity radiation, which involved largely the infrared part of the spectrum, was our first example of the interaction of radiation with matter. The photoelectric effect, our second example, involved visible and ultraviolet light. Here we describe the Compton* effect, in which the key experiments occur in the x-ray and the gamma-ray re gions of the electromagnetic spectrum. The Compton effect, which involves the scattering of radiation from atoms, can readily be understood in terms of billiard-ball-like collisions between photons and elec trons. In the explanation we must take into account not only the energy of the photons but also their linear mo mentum, a property that we have not needed to introduce so far. We have seen that Einstein’s analysis o f the heat capacity of a solid in quantum terms went far to convince people to accept the notion of energy quantization. In the same way, Compton’s analysis of the effect that bears his name went far to convince people of the reality of pho tons. In Compton’s experiment, a beam of x rays with sharply defined wavelength X falls on a graphite target T. as in Fig. 12. For various angles of scattering 0 , the inten sity of the scattered x rays is measured as a function of their wavelength. Figure 13 shows Compton’s experimen-
* Arthur H. Compton ( 1892- 1962) discovered in 1923 that the wavelengths of x rays change after they are scattered from elec trons. He received the 1927 Nobel prize in physics for this discov ery. Later he became the director of the laboratory at the Univer sity of Chicago where the first nuclear reactor was built.
Detector
7 = (0.32 J/m2-sKl eV/1.6X 10"'^J)
= 2.0X 10'*eV/m2-s. Each photon has an energy given by ^
he _ (6.63 X 10"^ J •SX3.00 X 10« m/s) X 5.89 X 10-^ m slits
The rate per unit area r at which photons strike the plate is then the intensity divided by the energy per photon, or
Figure 12 The experimental setup for observing the Comp ton effect. The detector can be moved to different angles 0.
Section 49-7
The Compton Effect
1033
Compton explained his experimental results by postu lating that the incoming x-ray beam behaved not as a wave but as an assembly of photons of energy E (= hv) and that these photons experienced billiard-ball-like collisions with the free electrons in the scattering target. In this view, the scattered radiation consists of the recoiling photons emerging from the target. Since the incident photon transfers some of its energy to the electron with which it collides, the scattered photon must have a lower energy £■'. It must therefore have a lower frequency v' { = E f h \ which implies a larger wavelength A' (= c/v'). This point of view accounts, at least qualitatively, for the wavelength shift AA. Note how different this particle model o f x-ray scattering is from that based on the wave picture. Now let us analyze a single photon-electron collision quantitatively. Figure 14 shows a collision between a pho ton and an electron. The electron is assumed to be ini tially at rest and essentially free, that is, not bound to the atoms of the scatterer. (This approximation holds for the loosely bound outer electrons, whose binding energy is much less than the energy of the x-ray photon.) Let us apply the law of conservation of energy to this collision. Since the recoil electrons may have a speed v that is com parable with that of light, we must use the relativistic expression for the kinetic energy of the electron. From the relativistic expression for the conservation of energy (see Section 21-9) we may write
E=E, or
hv + mc'^ = hv' + mc'^ + K, 70
75 Wavelength (pm)
( 20)
80
Figure 13 Compton’s experimental results for four different values of the scattering angle
tal results. We see that although the incident beam con sists essentially of a single wavelength A, the scattered x rays have intensity peaks at two wavelengths; one of them is the same as the incident wavelength, but the other (A') is larger by an amount AA. This Compton shift AA varies with the angle at which the scattered x rays are observed. The presence of a scattered wave of wavelength A' can not be understood if the incident x rays are regarded as an electromagnetic wave. In the wave picture, the incident wave of frequency v causes electrons in the scattering target to oscillate at that same frequency. These oscilla ting electrons, like charges surging back and forth in a small radio transmitting antenna, radiate electromag netic waves that again have this same frequency v. Thus, according to this interpretation, the scattered wave should have the same frequency and the same wavelength as the incident wave. This conclusion disagrees with the experi mental evidence (Fig. 13), which shows a variation in the wavelength o f the scattered wave.
Electron v = 0
Before
Figure 14 A photon of wavelength A strikes an electron at rest. The photon is scattered at an angle 0 with an increased wavelength A'. The electron moves off with speed v at the angle 6.
1034
Chapter 49
Light and Quantum Physics
where mc^ is the rest energy o f the struck electron and K is its (relativistic) kinetic energy. Substituting c/X for v (and c/X' for v') and using Eq. 25 of Chapter 7 for the relativistic kinetic energy, we have
he
he ,
,
( 21) \^ \-{ v /e Y
/
Now let us apply the (vector) law o f conservation o f linear momentum to the collision of Fig. 14. We first need an expression for the momentum of a photon. In Section 41 -5 we saw that if an object completely absorbs an energy U from a parallel light beam that falls on it, the light beam, according to the wave theory of light, simultaneously transfers to the object a linear momentum f//c. In Section 48-7, we showed that we could consider this situation from the standpoint o f a beam of photons of energy E, each delivering momentum p = Eje to the absorbing ob ject. In this case,
^
E e
hv e
h X'
( 22)
For the electron, the relativistic expression for the lin ear momentum is given by Eq. 22 of Chapter 9, m s
P=
■J1 — {v/cf
We can then write for the conservation of the x compo nent o f linear momentum
h X
h X
, ,
- = — cos (t>+ ,
mv
: COS 6
V1 — (v/eY
(2 3 )
and for the y component
^
h .
mv
sin 6. 0 = TT sm 0 A' ^l-(v/ef
sion in Fig. 14, the incident photon being scarcely de flected) to I h /m c (for 0 = 180% corresponding to a “head-on” collision, the incident photon being reversed in direction). Remember that the Compton shift AA is a purely quan tum effect, not expected to occur on the basis o f classical physics. As in cavity radiation and the photoelectric ef fect, the presence of the Planck constant h in the expres sion for the Compton shift (Eq. 25) indicates a quantum phenomenon. Equation 25 shows that AA —> 0 as A —> 0. The method of letting the Planck constant approach zero is a formal way of testing quantum equations to see whether they predict what would happen if the laws of classical physics applied not only to large objects but also to atoms and electrons. It remains to explain the presence o f the peak in Fig. 13 for which the wavelength does not change on scattering. This peak results from collisions between photons and electrons that, instead of being nearly free, are tightly bound in an ionic core in the scattering target. During photon collisions the bound electrons behave like very heavy free electrons. This is because the ionic core as a whole recoils during the collision. Thus the effective mass M for a carbon scatterer is approximately the mass o f a carbon nucleus. Since this nucleus contains six protons and six neutrons, we have approximately, M = 12 X 1 840m = 22,000m. If we replace m by Af in Eq. 25, we see that the Compton shift for collisions with tightly bound electrons is immeasurably small.
(24)
Our aim is to find AX{=X' —X), the wavelength shift o f the scattered photons, so that we may compare it with the experimental results o f Fig. 13. Compton’s experiment did not involve observations of the recoil electron in the scattering block. O f the five variables (A, A', v, 0 , and 6) that appear in the three equations (21,23, and 24) we may eliminate two. We choose to eliminate rand 6, which deal only with the electron, thereby reducing the three equa tions to a single relation among the variables. Carrying out the necessary algebraic steps (see Problem 64) leads to this simple result for the change in wavelength o f the scattered photons:
Sample Problem 8 X rays with A = 100 pm are scattered from a carbon target. The scattered radiation is viewed at 90° to the incident beam, {a) What is the Compton shift AA? (A) What kinetic energy is imparted to the recoiling electron? Solution (a) Putting 0 = 90° in Eq. 25, we have, for the Compton shift. AA = — (1 —cos 0) me 6.63 X 1 0 -^J* s (1 —cos 90°) (9.11 X 10"^' kgX3.00 X 10* m/s) = 2.43X 10-‘2m = 2.43 pm. (A) Using V= c/A, we can write Eq. 20 as
Substituting A' = A -I- AA and solving for K, we obtain
A/. = A '- A = — ( 1 - C O
me
S 0 ).
(25)
The Compton shift AX depends only on the scattering angle (f>and not on the initial wavelength A. Equation 25 predicts within experimental error the observed Compton shifts o f Fig. 13. Note from the equation that AA varies from zero (for 0 = 0, corresponding to a “grazing” colli
he AX A(A + AA) _ (6.63 X 10-^ J-SX3.00 X 10* m/sX2.43 X 10" m) (100 X 10"'2 mXlOO pm + 2.43 pmX10"'2 m/pm) = 4.72X 10"'M = 295 eV. You can show that the initial photon energy E in this case
Section 49-8 (=hv = hc/X) is 12.4 keV so that the photon lost about 2.4% of its energy in this collision. A photon whose energy was ten times as large (= 124 keV) can be shown to lose 20% of its energy in a similar collision. This is consistent with the fact that AX does not depend on the intial wavelength. More energetic x rays, which have smaller wavelengths, will experience a larger percent in crease in wavelength and thus a larger percent loss in energy.
49-8 LINE SPECTRA________________ Experimental results from the photoelectric effect and the Compton effect give indisputable evidence for the exis tence o f the photon or the particlelike nature of electro magnetic radiation. Historically, however, the photon concept emerged from the study of thermal radiation, which has a continuous spectrum of energies. The discrete (quantized) nature of the photon energies in this case is hidden in this broad distribution of energies. A more di rect verification of the quantized nature of the radiation would result from detecting individual photons and mea suring their energies. The complication in the case of thermal radiation occurs because the atoms in the cavity walls behave coop eratively and must be analyzed using statistical considera tions. If we analyze absorbing or emitting systems that are isolated from one another, we find that the radiation spec trum is not continuous but discrete, consisting of individ ual wavelengths separated by gaps where no radiation occurs. In the case of visible light, spectra are often dis played and analyzed using spectroscopes with prisms or diffraction gratings, such as that of Fig. 8 of Chapter 47, which give spectra such as Fig. 9 of Chapter 47. These spectra are called line spectra, and the individual components are called spectral lines. Line spectra can
TABLE 2
Entity
15a I5b 15c \5d I5e 15/ _15£___
H j molecule N H j molecule H Q molecule Fe atom Mo atom *’®Hg nucleus Proton
Wavelength Region (m) 40 1 X 10-2 3 X 10-‘ 3 X 10-’ 6 X 10-" 3 X 10-'2 4 X 10-'*
1035
result from the emission or absorption o f radiation by any isolated system, including molecules, atoms, nuclei, or subnuclear particles. (Line spectra of atoms and mole cules were available in Planck’s time, but they were not interpreted in terms of energy quantization until after Planck and Einstein developed the photon concept.) Analysis of radiation as photons of a definite energy strongly suggests that the system emitting or absorbing the radiation has discrete energy states, such that the differ ence in energy between the states equals the photon en ergy, as indicated in Figs. 18 and 19 of Chapter 8. (Here we are neglecting the small “recoil” energy needed to conserve linear momentum in the absorption or emission process.) In effect, this is a consequence of Einstein’s in terpretation of the spectrum o f cavity radiation: quantiza tion of the radiation implies quantization of the sources of the radiation. By studying line spectra, we can learn about the energy states of the atoms or other systems that emit the radiation. In the next chapter we discuss quantum mechanics, a mathematical procedure for calculating the energies of these discrete states, based on assuming a par ticular force to act between the components of the system (for example, the electrostatic force between the electron and the nucleus in an atom). The calculated energies of the states can then be compared with those deduced from experiment to see if the assumption about the force that acts in the system is reasonable. Figure 15 shows examples of line spectra from mole cules, atoms, nuclei, and particles, which have contrib uted to our understanding of their internal structure. These spectra, whose origins are listed in Table 2, indicate the great variety of line spectra that can be measured in the laboratory and the corresponding range o f wave lengths. In Chapter 51, we discuss the line spectrum of atomic hydrogen, which provided the insight that led to the quantum theory of atomic structure.
SOME SELECTED LINE SPECTRA
Figure
Line Spectra
Spectrum Region
Mode
Radio Microwave Infrared Ultraviolet X ray Gamma ray Gamma ray
Absorption Absorption Absorption Emission Emission Emission Absorption
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Chapter 49
0.160
Light and Quantum Physics
0.165
0.170
Magnetic field (tesia)
(a)
20 (e)
40
60
80
Wavelength (pm)
( 6)
Wavelength (pm) if)
Wavelength ifim) (c)
Wavelength (nm)
340
i d)
345
ig)
Figure 15 Some selected line spectra. See Table 2 for their identifications.
Photon energy (GeV)
100
Questions
1037
QUESTIONS 1. “Pockets” formed by the coals in a coal fire seem brighter than the coals themselves. Is the temperature in such pock ets appreciably higher than the surface temperature of an exposed glowing coal? Explain this common observation. 2. The relation /( T) = aT^ (Eq. 1) is exact for true cavities and holds for all temperatures. Why don’t we use this relation as the basis of a definition of temperature at, say, 100®C? 3. Do all incandescent solids obey the fourth-power law of temperature, as Eq. 2 seems to suggest? 4. A hole in the wall of a cavity radiator is sometimes called a black body. Why? 5. If we look into a cavity whose walls are maintained at a constant temperature, no details of the interior are visible. Explain. 6. By simply looking at the sky at night, can you pick out stars that are hotter than the Sun? Cooler than the Sun? What do you look for? Is the star’s brightness a clue? 7. Betelgeuse, the prominent red star in the constellation Orion, has a surface temperature that is much lower than that of the Sun, yet it radiates energy into space at a much higher rate than the Sun does. How can that be? 8. Less than a few percent of the energy supplied to a 100-W lamp appears in the form of visible light. What happens to the rest of it? What could be done to increase this percent age? Why hasn’t it already been done? 9. Your skin temperature is about 300 K. In what region of the electromagnetic spectrum do you emit thermal radiation most intensely? 10. Spectral radiancy curves for cavity radiators at different tem peratures do not intersect; see Fig. 3. Suppose, however, that they did. Can you show that this would violate the second law of thermodynamics? 11. We claim that all objects radiate energy by virtue of their temperature and yet we cannot see all objects in the dark. Why not? 12. Is energy quantized in classical physics? 13. Show that the Planck constant has the dimensions of angu lar momentum. Does this necessarily mean that angular momentum is a quantized quantity? 14. For quantum effects to be “everyday” phenomena in our lives, what order of magnitude value would h need to have? [SeeG. Gamow, Mr. Tompkins in Wonderland (Cambridge University Press, Cambridge, 1957), for a delightful popular ization of a world in which the physical constants c, G, and h make themselves obvious.] 15. For you to be able to detect energy quantization by watching a swinging pendulum, what order of magnitude would the Planck constant have to be? (Hint: See Sample Problem 3.) 16. Explain in your own words just why assuming that the en ergy of the atomic oscillators in a solid is quantized leads to a reduction in the heat capacity at low temperatures. 17. The classical law of equipartition of energy (see Section 23-6) leads to the Rayleigh-Jeans radiation law when ap plied to cavity radiation and to the Dulong-Petit law when applied to the heat capacities of solids. In both cases there is serious disagreement with experiment. Can you relate these
18.
19.
20.
21.
two failures of the equipartition law and explain why energy quantization leads, in each case, to theories that do agree with experiment? “For the cavity radiation problem and the heat capacity of solids problem, the disagreements between experiment and classical theory in certain ranges of the variables are not small but are total, and beyond all dispute.” Can you iden tify, in each case, the specific disagreements to which this statement refers? Explain why a tube used to examine photoelectric emission is (a) evacuated and (b) fitted with a window made of quartz rather than glass. Determine whether or not relativistic mechanics is needed to verify the photoelectric equation (Eq. 18) with a 1% un certainty. Note that typical stopping potentials are a few volts. In Fig. 10, why doesn’t the photoelectric current rise verti cally to its maximum (saturation) value when the applied potential difference is slightly more positive than Vq?
22. In the photoelectric effect, why does the existence of a cutoff frequency speak in favor of the photon theory and against the wave theory? 23. Why are photoelectric measurements so sensitive to the na ture of the photoelectric surface? 24. An insulated metal plate yields photoelectrons when first illuminated by ultraviolet light but then doesn’t give up any more. Explain. 25. Why is it that even for incident radiation that is monochro matic the photoelectrons are emitted with a spread of veloci ties? 26. We claim that all the energy of an absorbed photon is given to an emitted photoelectron. Why can we neglect the energy taken by the lattice? 27. Is the Compton effect more supportive of the photon theory of light than is the photoelectric effect? Explain your answer. 28. Consider the following procedures: (a) bombard a metal with electrons; (b) place a strong electric field near a metal; (c) illuminate a metal with light; (d) heat a metal to a high temperature. Which of the above procedures can result in the emission of electrons? 29. A certain metal plate is illuminated by light of a definite frequency. Whether or not photoelectrons are emitted as a result depends on which of the following features: (a) inten sity of illumination; (b) length of time of exposure to the light; (c) thermal conductivity of the plate; (d) area of the plate; (e) material of the plate? 30. Does Einstein’s theory of photoelectricity, in which light is postulated to be a stream of photons, invalidate the double slit interference experiment in which light is postulated to be a wave? 31. Explain the statement that one’s eyes could not detect faint starlight if light were not particlelike. 32. How can a photon energy be given b y E = bv when the very presence of the frequency v in the formula implies that light is a wave?
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33. Distinguish between the Planck relation E = nhv (Eq. 7) and the Einstein relation £ = /zv (Eq. 17). 34. A photon has no rest mass since it can never be at rest with respect to any observer. If energy equals how can a photon have any energy? 35. The momentum p of a photon is given by p = h/L Why is it that c, the speed of light, does not appear in this expression? 36. In discussing the propagation of light we sometimes use straight rays, sometimes waves, and still other times discrete photons. To what extent, if at all, are these views compatible with one another? Are there cases in which one view is clearly superior to the others? 37. Given that E = hv for a photon, the Doppler shift in fre quency of radiation from a receding light source would seem to indicate a reduced energy for the emitted photons. Is this in fact true? If so, what happened to the conservation of energy principle? (See “Questions Students Ask,” The Phys ics Teacher, December 1983, p. 616.) 38. Photon A has twice the energy of photon B. What is the ratio of the momentum of A to that of B1 39. How does a photon differ from a material particle?
40. What is the direction of a Compton scattered electron with maximum kinetic energy compared with the direction of the incident monochromatic photon beam? 41. Why, in the Compton scattering picture (Fig. 14), would you expect A Ato be independent of the materials of which the scatterer is composed? 42. Why don’t we observe a Compton effect with visible light? 43. Light from distant stars is Compton scattered many times by free electrons in outer space before reaching us. This shifts the light toward the red. How can this shift be distin guished from the Doppler red shift due to the motion of receding stars? 44. In both the photoelectric effect and the Compton effect there is an incident photon and an ejected electron. What is the difference between these two effects? 45. List and discuss the assumptions made by Planck in con nection with the cavity radiation problem, by Einstein in connection with the photoelectric effect, and by Compton in connection with the Compton effect. 46. Describe several experimental methods that can be used to determine the value of the Planck constant h.
PROBLEMS Section 49~I Thermal Radiation 1. In 1983 the Infrared Astronomical Satellite (IRAS) detected a cloud of solid particles surrounding the star Vega, radiat ing maximally at a wavelength of 32 pm . What is the temper ature of this cloud of particles? Assume an emissivity of unity. 2. Low-temperature physicists would not consider a tempera ture of 2.0 mK (0.0020 K) to be particularly low. At what wavelength is the spectral radiancy of a cavity at this temper ature a maximum? To what region of the electromagnetic spectrum does this radiation belong? What are some of the practical difficulties of operating a cavity radiator at such a low temperature? 3. Calculate the wavelength of maximum spectral radiancy and identify the region of the electromagnetic spectrum to which it belongs for each of the following: (a) The 2.7-K cosmic background radiation, a remnant of the primordial fireball, {b) Your body, assuming a skin temperature of 34 ®C. (c) A tungsten lamp filament at 18(X) K. {d) The Sun, at an assumed surface temperature of 5800 K. {e) An ex ploding thermonuclear device, at an assumed fireball tem perature of 10^ K. ( / ) The universe immediately after the Big Bang, at an assumed temperature of 10^* K. Assume cavity radiation conditions throughout. 4. (a) The effective surface temperature of the Sun is 5800 K. At what wavelength would you expect the Sun to radiate most strongly? In what region of the spectrum is this? Why then does the Sun appear yellow? (b) At what temperature is cavity radiation most visible to the human eye? See Fig. 1 in Chapter 42. 5. A cavity whose walls are held at 1900 K has a small hole, 1.00 mm in diameter, drilled in its wall. At what rate does energy escape through this hole from the cavity interior?
6. Calculate the thermal power radiated from a fireplace as suming an emissivity of 0.90, an effective radiating surface of 0.50 m^, and a radiating temperature of 5(X)®C. Does your answer seem reasonable? 7. (a) Show that a human body of area 1.80 m^ emissivity € = 1.0, and temperature 34 ®Cemits radiation at the rate of 910 W. (b) Why, then, do people not glow in the dark? 8. A cavity at absolute temperature T, radiates energy at a power level of 12.0 m W. At what power level does the same cavity radiate at temperature 2T^*> 9. A cavity radiator has its maximum spectral radiancy at a wavelength of 25.0 pm, in the infrared region of the spec trum. The temperature of the body is now increased so that the radiant intensity /( T) of the body is doubled, (a) What is this new temperature? (b) At what wavelength will the spec tral radiancy now have its maximum value? 10. A 100-W incandescent lamp has a coiled tungsten filament whose diameter is 0.42 mm and whose extended length is 33 cm. The effective emissivity under operating conditions is 0.22. Find the operating temperature of the filament. 11. An oven with an inside temperature = 215®C is in a room with a temperature of T^ = 26.2®C. There is a small opening of area A = 5.20 cm^ in one side of the oven. How much net power is transferred from the oven to the room? (Hint: Consider both oven and room as cavities with € = 1.) 12. A thermograph is a medical instrument used to measure radiation from the skin. For example, normal skin radiates at a temperature of about 34 X and the skin over a tumor radiates at a slightly higher temperature, (a) Derive an ap proximate expression for the fractional difference A /// in the radiant intensity between adjacent areas of the skin that are at slightly different temperatures T and T-\- AT. (b) Evaluate this expression for a temperature difference
Problems of 1.3 C®. Assume that the skin radiates with a constant emissivity. 13. A convex lens 3.8 cm in diameter and of focal length 26 cm produces an image of the Sun on a thin black screen the same size as the image. Find the highest temperature to which the screen can be raised. The effective temperature of the Sun is 5800 K. 14. The filament of a particular 100-W light bulb is a cylindrical wire of tungsten 0.280 mm in diameter and 1.80 cm long. See Appendix D for needed data on tungsten. Assume an emissivity of unity and ignore absorption of energy by the filament from the surroundings, (a) Calculate the operating temperature of the filament, (b) How long does it take for the filament to cool by 500 C® after the bulb is switched off? 15. Consider a planet, with radius R, revolving about the Sun in a circular orbit of radius r. Suppose that the planet has no atmosphere (and therefore no “greenhouse effect” on its surface temperature), (a) Show that the surface temperature T of the planet is given from the relation where Fsun is the radiant power output of the Sun. (b) Evalu ate the temperature numerically for the Earth. Section 49~2 Planck*s Radiation Law
21.
22.
23.
24.
25.
16. Show that the wavelength A^ax at which Planck’s spectral radiation law, Eq. 6, has its maximum is given by Eq. 4: A ^ = (2898//m -K )/r. (Hint: Set dR!dk = 0\ an equation will be encountered whose numerical solution is 4.965.) 17. (a) By integrating the Planck radiation law, Eq. 6, over all wavelengths, show that the power radiated per square meter of a cavity surface is given by
(Hint: Make a change in variables, letting x = hc/XkT, The definite integral
/:
x^dx ex- 1
will be encountered, which has the value 7tV15.) (b) Verify that the numerical value of the constant a is 5.67 X 10-« W/(m2-K^). 18. (a) An ideal radiator has a spectral radiancy at 400 nm that is 3.50 times its spectral radiancy at 200 nm. What is its temperature? (b) What would be its temperature if its spec tral radiancy at 200 nm were 3.50 times its spectral radiancy at 400 nm? Section 49~4 The Heat Capacity o f Solids 19. In terms of the Einstein temperature Tg, at what tempera ture will the molar internal energy of a solid achieve one-half its classical value of 3/?T? 20. (a) Show that the molar internal energy of a solid can be written, according to Einstein’s theory of heat capacities, as E.nx
3RTf
in which x = T eIT, where
is the Einstein temperature
26.
1039
hv/k. (b) Verify that E'in, approaches its classical value of 3 R T as r —►«>. In terms of Einstein’s theory of heat capacity, (a) what is the molar heat capacity at constant volume of a solid at its Einstein temperature? Express your answer as a percentage of its classical value of 3R. (b) What is the molar internal energy at the Einstein temperature? Express your answer as a percentage of its classical value of 3RT. Show that, at high enough temperatures, Einstein’s expres sion for the heat capacity of a solid, Eq. 14, reduces to the classical formula, Eq. 11. The Einstein temperatures of lead, aluminum, and beryl lium may be taken as 68 K, 290 K, and 690 K, respectively. For each of these elements, find (a) the frequency v of its atomic oscillators, (b) the spacing A E between adjacent oscillator levels, and (c) the effective spring constant k. The Einstein temperature of aluminum may be taken as 290 K. According to Einstein’s theory of heat capacity, what are (a) its molar internal energy (see Problem 20) at 150 K and (b) its molar heat capacity, under constant-volume con ditions, at 150 K? A 12.0-g block of aluminum is heated from 80 K up to 180 K, under constant-volume conditions. How much heat is required according to (a) the classical theory of heat capac ity and (b) Einstein’s quantum theory of heat capacity? The Einstein temperature for aluminum may be taken to be 290 K. Assume that 25.0 g of aluminum at 80.0 K are mixed thor oughly with 12.0 g of aluminum at 200 K in an insulated container. What is the final temperature of the mixture? Assume that Einstein’s theory of heat capacities is valid and that, at these relatively low temperatures, the differences between the heat capacity at constant volume and that at constant pressure may be neglected. Assume further that there are no energy exchanges between the two aluminum specimens and the container. The Einstein temperature of aluminum may be taken to be 290 K.
Section 49-6 Einstein's Photon Theory 27. (a) By using the “best” values of the fundamental constants, as found in Appendix B, show that the energy £ of a photon is related to its wavelength A by _
1240eV*nm
^ --------3----- ■ This result can be useful in solving many problems, (b) The orange-colored light from a highway sodium lamp has a wavelength o f589 nm. How much energy is possessed by an individual photon from such a lamp? 28. Consider monochromatic light falling on a photographic film. The incident photons will be recorded if they have enough energy to dissociate a AgBr molecule in the film. The minimum energy required to do this is about 0.60 eV. Find the cutoff wavelength greater than which the light will not be recorded. In what region of the spectrum does this wavelength fall? 29. An atom absorbs a photon having a wavelength of 375 nm and immediately emits another photon having a wavelength
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Light and Quantum Physics
o f580 nm. What was the net energy absorbed by the atom in this process? 30. {a) A spectral emission line of hydrogen, important in radioastronomy, has a wavelength of 21.11 cm. What is its corresponding photon energy? (b) At one time the meter was defined as 1,650,763.73 wavelengths of the orange light emitted by a light source containing krypton-86 atoms. What is the corresponding photon energy of this radiation? 31. Most gaseous ionization processes require energy changes of 1.0 X 1 0 " to 1.0 X 10" J. What region then of the Sun’s electromagnetic spectrum is chiefly responsible for creating the ionosphere in the Earth’s atmosphere?
Wavelength (nm) 433.9 404.7 365.0 312.5 253.5 Stopping potential (V) 0.55 0.73 1.09 1.67 2.57
42.
43.
32. Under ideal conditions the normal human eye will record a visual sensation at 540 nm if incident photons are absorbed at a rate as low as 100 s"‘. To what power level does this correspond? 33. You wish to pick a substance for a photocell operable with visible light. Which of the following will do (work function in parentheses): tantalum (4.2 eV), tungsten (4.5 eV), alu minum (4.2 eV), barium (2.5 eV), lithium (2.3 eV), cesium (1.9 eV)?
44.
34. Satellites and spacecraft in orbit about the Earth can become charged due, in part, to the loss of electrons caused by the photoelectric effect induced by sunlight on the space vehi cle’s outer surface. Suppose that a satellite is coated with platinum, a metal with one of the largest work functions: (f) = 5.32 eV. Find the smallest-frequency photon that can eject a photoelectron from platinum. (Satellites must be designed to minimize such charging.)
45.
35. (a) The energy needed to remove an electron from metallic sodium is 2.28 eV. Does sodium show a photoelectric effect for red light, with A = 678 nm? (b) What is the cutoff wave length for photoelectric emission from sodium and to what color does this wavelength correspond?
46.
47.
36. Find the maximum kinetic energy in eV of photoelectrons if the work function of the material is 2.33 eV and the fre quency of the radiation is 3.19 X 10'^ Hz. 37. Incident photons strike a sodium surface having a work function of 2.28 eV, causing photoelectric emission. When a stopping potential of 4.92 V is imposed, there is no photo current. Find the wavelength of the incident photons. 38. Light of wavelength 200 nm falls on an aluminum surface. In aluminum, 4.2 eV is required to remove an electron. What is the kinetic energy of (a) the fastest and (b) the slowest emitted photoelectrons? (c) Find the stopping po tential. (d) Calculate the cutoff wavelength for aluminum.
48.
49.
39. (a) If the work function for a metal is 1.85 eV, what would be the stopping potential for light having a wavelength of 410 nm? (b) What would be the maximum speed of the emitted photoelectrons at the metal’s surface? 40. The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 491 nm is 710 mV. When the incident wavelength is changed to a new value, the stopping potential is found to be 1.43 V. (a) What is this new wavelength? (b) What is the work function for the surface? 41. Millikan’s photoelectric data for lithium are:
50.
Make a plot like Fig. 11, which is for sodium, and find (a) the Planck constant and (b) the work function for lithium. A lithium surface for which the work function is 2.49 eV is irradiated with light of frequency 6.33 X 10*^ Hz. The loss of electrons causes the metal to acquire a positive potential. What must this potential have become by the time its value prevents further loss of electrons from the surface? A satellite in Earth orbit maintains a panel of solar cells at right angles to the direction of the Sun’s rays. Assume that the solar radiation is monochromatic with a wavelength of 550 nm and arrives at the rate of 1.38 kW/m^. What must be the area of the panels in order that “one mole of photons’’ arrives each minute? In the photon picture of radiation, show that if two parallel beams of light of different wavelengths are to have the same intensity, then the rates per unit area at which photons pass through any cross section of the beams are in the same ratio as the wavelengths. An ultraviolet light bulb, emitting at 4(X) nm, and an infra red light bulb, emitting at 700 nm, each are rated at 130 W. {a) Which bulb radiates photons at the greater rate? (b) How many more photons does it generate per second than does the other bulb? To remove an inner, most tightly bound, electron from an atom of molybdenum requires an energy of 20 keV. If this is to be done by allowing a photon to strike the atom, (a) what must be the associated wavelength of the photon? (b) In what region of the spectrum does the photon lie? (c) Could this process be called a photoelectric effect? Discuss your answers. X rays with a wavelength of 71.0 pm eject photoelectrons from a gold foil, the electrons originating from deep within the gold atoms. The ejected electrons move in circular paths of radius r in a region of uniform magnetic field B. Experi ment shows that rB = 188 p T -m . Find (a) the maximum kinetic energy of the photoelectrons and (b) the work done in removing the electrons from the gold atoms that make up the foil. A special kind of light bulb emits monochromatic light at a wavelength of 630 nm. It is rated at 70.0 W and is 93.2% efficient in converting electrical energy to light. How many photons will the bulb emit over its 730-h lifetime? Assume that a 100-W sodium-vapor lamp radiates its en ergy uniformly in all directions in the form of photons with an associated wavelength of 589 nm. (a) At what rate are photons emitted from the lamp? (b) At what distance from the lamp will the average flux of photons be 1.00 photon/(cm^ • s)? (c) At what distance from the lamp will the average density of photons be 1.00 photon/cm^? (d) Calcu late the photon flux and the photon density 2.00 m from the lamp. Show, by analyzing a collision between a photon and a free electron (using relativistic mechanics), that it is impossible for a photon to give all its energy to the free electron. In other words, the photoelectric effect cannot occur for completely
Problems free electrons; the electrons must be lx)und in a solid or in an atom. Section 49-7 The Compton Effect 51. A particular x-ray photon has a wavelength of 41.6 pm. Calculate the photon’s (a) energy, (b) frequency, and (c) momentum. 52. Find (a) the frequency, (b) the wavelength, and (c) the mo mentum of a photon whose energy equals the rest energy of the electron. 53. By how much does a sodium atom slow down upon absorb ing a photon of wavelength 589 nm with which it collides head-on? 54. The quantity h/mc in Eq. 25 is often called the Compton wavelength, Ac, of the scattering particle and that equation is written A A = Ac(l —cos 0).
{a) Calculate the Compton wavelength of an electron. Of a proton, (b) What is the energy of a photon whose wavelength is equal to the Compton wavelength of the electron? Of the proton? (c) Show that in general the energy of a photon whose wavelength is equal to the Compton wavelength of a particle is just the rest energy of that particle. 55. Photons of wavelength 2.17 pm are incident on free elec trons. (a) Find the wavelength of a photon that is scattered 35.0® from the incident direction, (b) Do the same if the scattering angle is 115®. 56. A 511 -keV gamma-ray photon is Compton-scattered from a free electron in an aluminum block, (a) What is the wave length of the incident photon? (b) What is the wavelength of the scattered photon? (c) What is the energy of the scattered photon? Assume a scattering angle of 72.0®. 57. Show that A£'/£’, the fractional loss of energy of a photon during a Compton collision, is given by E
hv' = ---- T (1 - cos 0). mc^
1041
58. What fractional increase in wavelength leads to a 75% loss of photon energy in a Compton collision with a free electron? 59. Find the maximum wavelength shift for a Compton colli sion between a photon and a free proton. 60. A 6.2-keV x-ray photon falling on a carbon block is scat tered by a Compton collision and its frequency is shifted by 0.010%. (a) Through what angle is the photon scattered? (b) How much kinetic energy is imparted to the electron? 61. An x-ray photon of wavelength A= 9.77 pm is backscattered by an electron ( 0 = 180®). Determine {a) the change in wavelength of the photon, (b) the change in energy of the photon, and (c) the final kinetic energy of the electron. 62. Calculate the fractional change in photon energy for a Compton collision with 0 in Fig. 14 equal to 90® for radia tion in (a) the microwave range, with A = 3.00 cm, (b) the visible range, with A= 500 nm, (c) the x-ray range, with A= 0.10nm , and (d) the gamma-ray range, with A = 1.30 pm. What are your conclusions about the imf)ortance of the Compton effect in these various regions of the electro magnetic spectrum, judged solely by the criterion of energy loss in a single Compton encounter? 63. Through what angle must a 215-keV photon be scattered by a free electron so that it loses 10.0% of its energy? 64. Carry out the necessary algebra to eliminate v and 6 from Eqs. 21, 23, and 24 to obtain the Compton shift relation, Eq. 25. 65. (a) Show that when a photon of energy E scatters from a free electron, the maximum recoil kinetic energy of the electron is given by £2 E + m cyi • (b) Find the maximum kinetic energy of the Comptonscattered electrons knocked out of a thin copper foil by an incident beam of 17.5-keV x rays.
CHAPTER 50 THE
W
NATURE OF MATTER
M
E
Physicists have rarely gone wrong in relying on the underlying symmetries o f nature. For example, after learning that a changing magnetic field produces an electric field, it is a good bet to guess (and it turns out to be true) that a changing electric field produces a magnetic field. The electron was known to have an antiparticle (a particle o f the same mass but opposite charge), and one might guess that the proton also has an antiparticle. To confirm this guess, a proton accelerator o f the proper energy (see Sample Problem 9 o f Chapter 21) was built, and the antiproton was discovered. In the previous chapter, we discussed the particlelike properties o f light and other radiation, which we traditionally analyze as waves. On the basis o f symmetry, we are led to ask the following question: Does matter, which we traditionally analyze as particles, also have wavelike properties? In this chapter we show that this hypothesis turns out to be correct, and we discuss the mechanics o f this wavelike behavior, which is known as quantum mechanics. As we shall see in the remaining chapters, quantum mechanics provides the means to understand the fundamental behavior o f physical systems from solids to quarks.
50-1 THE WAVE BEHAVIOR OF PARTICLES____________________ Before we discuss the analysis of the wave behavior of particles, let us try to persuade you that they really do show this kind o f behavior. As we have seen, electromag netic radiation can under most circumstances be treated as waves; its particlelike properties are directly revealed only in a few special experiments. We have also had a great deal o f success at treating matter as composed of entities with only particlelike properties; for example, there is no need to take the wave nature into account in a game o f billiards or in the construction of a building. However, there are a number of experiments that can be understood only if the entities that we have ordinarily analyzed as particles behave as waves. In Chapters 45 and 46, we discussed interference and diffraction experiments, and we pointed out that the ap pearance of an interference or diffraction pattern is a defi nite signal o f wave-type behavior. If we are going to search for direct evidence o f the wave-type behavior o f particles, interference and diffraction experiments are a logical place to begin.
Figure 1a shows a beam of electrons incident on a dou ble slit. The electrons are accelerated to a chosen energy in a potential difference V, and after passing through the double slit they strike a fluorescent screen (like a TV screen). The resulting pattern on the screen leaves an image on the photographic film. The experimental setup of Fig. la looks very much like that for double-slit inter ference with light waves. The results of this experiment are shown in Fig. \b. There should be no doubt that we are observing an inter ference pattern. If the electrons had no wavelike behavior, we should expect to see bright regions on the film only in front o f the two slits; clearly there is more to this outcome, and the wavelike behavior of the electrons is responsible for it. We can replace the double slit with a circular aperture, which produces the diffraction pattern shown in Fig. 2. Once again, there is clearly more to this result than the passage of particles through an aperture. Figure 3 compares the results of diffraction at a straight edge for beams of electrons and light. The comparison is convincing evidence that electrons have wavelike behav ior. Is there something unique about electrons that causes
1043
1044
Chapter 50
The Wave Nature o f Matter Fluorescent screen •Photographic film
— ( I Electrons
50 kV
(a)
Figure 1 (a) Apparatus for producing double-slit interference with electrons. A filament F produces a spray of electrons, which are accelerated through 50 kV, pass through the single slit, and strike the double slit. They produce a visible pattern when they strike a fluorescent screen, which can be photo graphed. (b) The resulting electron double-slit interference pattern, showing the interference fringes.
Figure 2 An electron diffraction pattern using a circular ap erture of diameter 30 pm and 100-keV electrons. Compare with the optical pattern (Fig. 2 of Chapter 46).
Figure 3 Diffraction of (a) light and (b) electrons from a straight edge.
this behavior? Let us instead try the experiment with neu trons, which differ from electrons in several respects: neu trons are more massive (by a factor of about 2000), they are uncharged, and they are composite particles (as op posed to electrons, which are fundamental “point” parti cles). If neutrons also show wavelike behavior, we should suspect that this behavior has nothing to do with the spe cial properties of electrons but instead may be character istic of particles in general. To show interference with neutrons, a wire made of material (boron, for example) that is highly absorbing of neutrons is placed in a gap (of width slightly larger than the wire) in a similarly absorptive material, creating in effect a double slit. A detector scans across the transmitted neutron beam and measures the intensity as a function of location. The results are shown in Fig. 4. Once again, if neutrons behaved like traditional particles, we should ex pect to find peaks in the transmitted intensity only di rectly in front of the slits. Here, on the other hand, we see definite evidence of interference effects. Figure 5 shows the results of an experiment in which a beam of helium atoms was passed through a double slit. Although the results are not as dramatic as those for the electrons and neutrons, there is again evidence for inter ference fringes similar to those obtained with light. These experiments use very different types o f particles and different types of slit systems and detector systems, yet they all have one feature in common: the particles
Section 50-2
The De Broglie Wavelength
1045
Figure 4 Intensity pattern of neutrons passing through a double slit.
seem to be undergoing some sort of interference. Such experiments provide direct evidence for the wave nature o f particles. At this point, you are probably wondering how a parti cle can produce an interference pattern. Our analysis of the double-slit experiment in terms of waves in Chapter 45 was based on parts of a single wavefront passing through each slit and then recombining on the screen. Is it possible for parts of an electron or a neutron or a helium atom to pass through each slit and then to recombine? This is a difficult question, but one that is essential to the understanding of quantum behavior. The answer is yes, and we discuss it further in the final section of this chap ter.
Sample Problem 1 In the data shown in Fig. 5, the slit separa tion d was 8 //m and the detector was a distance Z) = 64 cm from the slits. From the observed spacing between the fringes, find the wavelength of the helium atoms. Solution In Sample Problem 2 of Chapter 45, we found the spacing between adjacent interference fringes to be
Figure 5 Intensity pattern of helium atoms passing through a double slit.
A
^
From Fig. 5, we estimate about 8 //m for the spacing between the minima (which is the same as the spacing between the maxima), and so D
0.64 m
m.
50-2 THE DE BROGLIE WAVELENGTH The experiments discussed in the previous section were done in recent years, when the precise apparatus needed to produce narrow slits or stable beams was available. However, the proposal that particles have a wavelike na ture was made long before these results were obtained, on the basis of indirect arguments based partly on the sym metry of nature. In 1924 Louis de Broglie, a physicist and a member o f a distinguished French aristocratic family, puzzled over the fact that radiation seemed to have a dual wave-particle aspect, but matter (at that time) seemed entirely particle like. On the other hand, matter and radiation had other aspects in common: both are forms of energy, each can be transformed into the other, and both are governed by the spacetime symmetries of the theory of relativity. De Bro glie began to think that matter should also have a dual character and that particles such as electrons should have wavelike properties. Equation 22 of Chapter 49 provided a connection be tween a wavelike property of radiation, the wavelength, and a particlelike property, the momentum: p = A/A, where h is Planck’s constant. De Broglie suggested that
1046
Chapter 50
The Wave Nature o f Matter
this same relationship connects the particlelike and wave like properties of matter. That is, associated with a free particle moving with linear momentum p there is a sinu soidal wave having a wavelength A given by
P
(1)
The wavelength of a particle computed according to Eq. 1 is called its de Broglie wavelength. Note that Planck’s constant provides the connecting link between the wave and particle natures of both matter and radiation. Equation 1 immediately shows why we don’t observe the wave behavior of ordinary objects. Planck’s constant is so small 10“^^J • s) that the wavelengths of ordinary objects are many orders of magnitude smaller than the size of a nucleus! No double slit could possibly be con structed on this scale to reveal the wave nature. In the atomic or subatomic realm, however, the momentum p can be sufficiently small to bring the de Broglie wave length into the range in which wave properties can be observed, as we illustrated in the previous section. De Broglie’s relationship provides us with a means to calculate the wavelength associated with the wave behav ior o f matter. It does not indicate anything about the amplitude of the wave, nor does it suggest the physical variable that is oscillating as the wave travels. We deal with these questions in Section 50-6.
Sample Problem 2 Calculate the de Broglie wavelength of (a) a virus particle of mass 1.0 X 10“ '^ kg moving at a speed of 2.0 mm/s, and (b) an electron whose kinetic energy is 120 eV. Using Eq. 1, we find p
50-3 TESTING DE BROGLIE’S HY POTHESIS__________________ If you want to prove that you are dealing with a wave, a convincing thing to do is to measure the wavelength. In 1801, for example, Thomas Young made a strong case for the wave nature of visible light when he measured its wavelength, using double-slit interference. To measure a wavelength using the double-slit (or any similar) method, we need two or more diffracting centers (slits) separated by a distance that is of the order o f magni tude of the wavelength itself Sample Problem 2a shows at once that it is hopeless to try to measure the wavelength of even so small a particle as a virus; we would need two “slits” separated by 10“ m. That is why our daily experi ences with large moving objects gives no clues to the wave nature of matter. Sample Problem 2b, however, suggests that we should be able to measure the wavelength o f a moving electron. We now describe two ways to do so. 1. The Davisson - Germ er experim ent. Sample Problem 2b suggests that we should be able to use a crystal as a
diffraction grating to measure the de Broglie wavelengths of electrons with kinetic energies of a few hundred elec tron volts. Figure 6 shows the apparatus used for this purpose by C. J. Davisson and L. H. Germer of what is now the AT&T Bell Laboratories. In 1937 Davisson shared the Nobel prize for this work. In the Davisson-Germer apparatus of Fig. 6, electrons are accelerated from a heated filament F by an adjustable potential difference V. The beam, made up o f electrons whose kinetic energy is eV, is then allowed to fall on a crystal C, which, in their experiment, was of nickel. They set detector D at an arbitrary angle
_ h _ 6.6 X 1 0 -^J* s mv (1.0 X 10-'^ kgX2.0 X 10'^ m/s)
ib) ^ _____________ 6.6X 1 0 -^J» s_____________ V2(9.11 X 10"^' kgX120 eV)(1.6X lO -'^J/eV) = 1.1 X lO-'o m. Even for so small an object as a virus particle moving slowly, the de Broglie wavelength is too small for observation (smaller than an atomic nucleus). For larger objects, the wave behavior is entirely unobservable. For the electron in part (b \ however, the de Broglie wavelength is about the same size as an atom, and (as we shall see) by using atoms as diffracting objects for electrons we can verify that the de Broglie wavelength does indeed charac terize the wave behavior of electrons.
Figure 6 The apparatus used in the Davisson-Germer ex periment. Electrons are emitted from the filament F and ac celerated by the adjustable potential difference V. After reflec tion from the crystal C, they are recorded by the detector D, which can be moved to various angular positions 0.
Section 50-3
Testing De Broglie's Hypothesis
1047
Figure 7 The results obtained by Davis son and Germer for five different acceler ating voltages, shown as polar plots of current / as a function of the angle . A strong diffraction peak is observed in (c) at 0 = 50" for F = 5 4 V.
difference V, Figure 7 shows the results of five such runs. We see that there is a strong diffracted beam for 0 = 50® and F = 54 V. If either the angle or the accelerating po tential are changed, the intensity of the diffracted beam drops. Figure 8 is a simplified representation of the nickel crystal C of Fig. 6. Because this low-energy electron beam does not penetrate very far into the crystal, it is sufficient to consider the diffraction to take place in the plane of atoms on the surface. The situation is very similar to light reflected from a diffraction grating. In this case the grating lines are the parallel rows of atoms lying on the crystal surface, and the grating spacing is the interval D in Fig. 8. The principal maxima for such a grating must satisfy Eq. 1 o f Chapter 47,
mk = D sin
(m = 1, 2, 3,
).
( 2)
For their crystal Davisson and Germer knew that D = 215 pm. For m = 1, which corresponds to a first-order diffraction peak, Eq. 2 leads to A = Z) sin 0 = (215 pm)(sin 50®) = 165 pm.
-9 ®
^
^
Figure 8 The crystal surface acts like a diffraction grating with spacing D.
The expected de Broglie wavelength for a 54-eV electron, calculated as in Sample Problem 2b, is 167 pm, in good agreement with the measured value. De Broglie’s predic tion is confirmed. 2. G. P. Thomson's experiment. In 1927 George P. Thomson, working at the University of Aberdeen in Scot land, independently confirmed de Broglie’s equation, using a somewhat different method. As in Fig. 9a, he directed a monoenergetic beam of 15-keV electrons through a thin metal target foil. The target was specifically not a single crystal (as in the Davisson-Germer experi ment) but was made up of a large number of tiny, ran domly oriented crystallites. If a photographic film is placed parallel to the target, as shown in Fig. 9a, the central beam spot will be surrounded by diffraction rings. Figure 9b shows this pat tern for an x-ray beam incident on an aluminum target. Figure 9c shows the pattern for aluminum when an elec tron beam of the same wavelength replaces the x-ray beam. A simple glance at these two diffraction patterns leaves no doubt that both originate in the same way. Nu merical analysis of the patterns confirms de Broglie’s hy pothesis in every detail. Thomson shared the 1937 Nobel prize with Davisson for his electron diffraction experiments. George P. Thom son was the son of J. J. Thomson, who won the Nobel prize in 1906 for his discovery of the electron and for his measurement of its charge-to-mass ratio. It has been said that “Thomson, the father, was awarded the Nobel prize for having shown that the electron is a particle, and Thomson, the son, for having shown that the electron is a wave.” Today the wave nature of matter is taken for granted, and diffraction studies by beams of electrons or neutrons are used routinely to study the atomic structures of solids or liquids. Matter waves serve as a valuable supplement to X rays as tools for structure analysis. Electrons, for exam ple, are less penetrating than x rays and so are particularly useful for studying surfaces. X rays interact largely with the electrons in a target, and for that reason it is not easy to
1048
Chapter 50
The Wave Nature o f Matter
Circular diffraction ring
Photographic film
Figure 9 (a) An arrangement for producing a diffraction pattern using a powdered or crys talline target, (b) The x-ray diffraction pattern of a powdered aluminum target, (c) The elec tron diffraction pattern of the same target. The electron energy has been chosen so that the de Broglie wavelength is the same as the x-ray wavelength used in (/?).
use them to locate light atoms— particularly hydrogen— which have few electrons. Neutrons, on the other hand, interact largely with the nucleus o f the atom and can be used to fill this gap. Figure 10, for example, shows the structure o f solid benzene as deduced from neutron dif fraction studies.
atoms and compare with the wavelength estimated in Sample Problem 1. Solution
From Eq. 3,
A„ = -
6.63X 10-^J* s
V5(4.0026 uKl.66 X 10”
kg/uXl.38 X 1 0 " J/KX83 K)
= 1.07 X lO-'o m. Sample Problem 3 Suppose a beam of atoms emerges from an oven at a temperature T. The beam has a Maxwellian distribu tion of speeds (see Section 24-3). Based on this distribution, it can be shown (see Problem 20) that the most probable value of the de Broglie wavelength of the atoms in the beam is ^
yISmkT
(3)
The data of Fig. 5 were obtained for helium atoms (m = 4.0026 u) emerging from an oven at a temperature r = 83 K. Find the most probable de Broglie wavelength for these helium
Figure 10 The atomic structure of solid benzene as deduced from neutron diffraction. The solid circles show the location of the six carbon atoms that form the familiar benzene ring. The dotted circles show the locations of the hydrogen atoms.
This value agrees very well with the estimate obtained in Sample Problem 1, verifying once again that the de Broglie wavelength characterizes the wave behavior of material particles. Sample Problem 4 Nuclear reactors are often designed so that a beam of low-energy neutrons emerges after passing through a graphite cylinder in the shielding wall (see Fig. 11). After many collisions with the carbon atoms, the neutrons are in thermal equilibrium with them at room temperature (293 K). Such neu trons are called thermal neutrons, (a) Find the most probable de Broglie wavelength in a beam of thermal neutrons, (b) Let a beam of these neutrons be incident on a crystal C in which the spacing between the Bragg planes is ^ = 0.304 nm. An intense first-order Bragg diffraction is observed for neutrons of wave length Ap when the Bragg scattering angle 0 is as shown in Fig. 11. Find the angle 6.
Figure 11 Sample Problem 4. An arrangement for observing neutron diffraction. The neutrons emerging from the graphite thermalizing column have a distribution of energies. After Bragg reflection from the crystal C, the beam at the angle 6 is monoenergetic.
Section 50-4 Solution
(a) Using Eq. 3, we have V5(1.67 X 10-27 kgXl.38 X
J/KK293 K)
= 1.14 X 10“ '° m = 0.114 nm. {b) The Bragg formula for x-ray diffraction was given as Eq. 12 of Chapter 47, 2 Jsin 6 = mk
(m = 1, 2, 3, . . . ).
The same formula can be applied to the diffraction of particles, if we use the de Broglie wavelength A. Solving for the angle 6, we obtain n
•
x (m k ^ \
•
, /(lK 0 .1 1 4 n m )\
By diffracting the neutrons in this way, a monoenergetic beam can be obtained. This beam can then be diffracted by other materials to study their structure, as was done to obtain Fig. 10.
50-4 WAVES, WAVE PACKETS, AND PARTICLES As we have just seen, the evidence that matter is wavelike is very strong. Still, we cannot forget that the evidence that matter is particlelike is just as strong. The basic difference between these two points of view is that the position of a particle can be localized in both space and time but a wave can not, being spread out in both of these dimensions. Let us begin to reconcile these two approaches by seeing whether we can put together an assembly of waves in such a way that we end up with something that reminds us of a particle. What we will have to say will hold for all kinds of waves, whether they be water waves, sound waves, electro
Ajc = 00
V
—
1. Localizing a wave in space. Figure 12a is a “snapshot” of a wave taken at an arbitrary time, say, t = 0. The wave extends from jc = — t o x = + « ) and has a sharply defined wavelength Xqand a correspondingly sharply de fined wave number /cq(= In lko ), as Fig. 12b shows. How ever, there is nothing about this wave that suggests the localization in space that we associate with the word “par ticle.” Put another way, if the wave of Fig. 12a is to repre sent a particle, the uncertainty Ax of its position along the X axis is infinite: it could be anywhere along the axis. We recall that it is possible (see Section 19-7) to create almost any wave shape we want by adding sine waves with properly chosen wave numbers, amplitudes, and phases. Figure 13a shows a wave packet that can be put together in this way. This collection of infinite waves adds to make a sine wave over a certain region of width Ax and, by de structive interference, adds to zero everywhere else. We now have a localization in space, measured by Ax, the length of the packet. The price we have paid is the sacrifice of the “purity” of our original wave because our packet now no longer contains a single wave number k^ but rather a spread of wave numbers centered about k^\ see Fig. 13Z?. Let A/c in Fig. 136 be a rough measure of the spread of wave numbers that forms the packet of Fig. 13a. It is reasonable to believe that the sharper (that is, the more particlelike) we wish the wave packet to be, the broader the range of wave numbers that we must use to build it up. In Fig. 12a, for example, the “packet” was not sharp at all (Ax 00) but, on the other hand, we needed only a single wave number to “build it up” (A/c = 0). At the other ex treme, we could build a very sharp wave packet (Ax ^ 0), but we would need to combine waves with a very large
/V
^
AAAAAAA/\AAAAAAA
v v v v v V
/ v / v
\ / 1/ \ / \ / \ / \ / \ / '
(a)
^
\)
^k = Q
(
6)
O.Skr,
1049
magnetic waves, or de Broglie waves. We discuss, in se quence, localizing a wave in space and in time.
6.63 X 10-5^ J-s ^P =
Waves, Wave Packets, and Particles
\.0kn
l.bkr,
2.0k^
Figure 12 (a) A harmonic wave viewed at / = 0. (6) The distribution of wave num bers, shown as a plot of the amplitude of the harmonic component as a function of its wave number. In this plot, all waves with k4^k^ have an amplitude of zero.
1050
Chapter 50
The Wave Nature o f Matter Figure 13 (a) \ wave packet of length Ax, viewed at / = 0. (b) The relative amplitudes of the various harmonic components that combine to make up the packet. The central peak has a width
( 6)
spread in wave number (A/: °o) to do so. In general, as Ax decreases, A/c increases, and conversely. The relation ship between them proves to be very simple; namely.
A k -A x - 1.
(4)
The symbol ~ in Eq. 4 should be taken to mean “is of the order of,” because so far we have not defined Ax or Ak very precisely.* 2. Localizing a wave in time. A particle is localized in time as well as in space. If we replaced the space variable x in Fig. 12a by the time variable t (and the wavelength Xqby the period Tq), that figure would then show how our wave would vary with time as it passes a particular fixed point, say, X = 0. As before, there is nothing at all about this wave that suggests the localization in time that we asso ciate with the word “particle,” because a particle would pass our observation point at a particular time, rather than spread over an infinite time interval. We can build up a wave packet in time as well as in space. Figure 13a can illustrate this, provided we replace the space variables by the corresponding time variables, as above, and also replace the wave number k^ by the angu lar frequency cUq. By analogy with Eq. 4, the duration At o f our new wave packet is related to the spread Acu of angular frequencies needed to make up the wave packet by Acu*A/~l. (5)
This equation has many practical applications. For ex ample, much of our information in today’s society, in cluding telephone communication, radar, and computer data storage, is sent from point to point in the form of pulses. The electronic amplifiers through which these pulses pass should be sensitive over the full range o f fre quencies included in the pulses they are designed to han dle. Equation 5 tells us that the shorter the time duration of the pulse, the greater must be the frequency bandwidth (as it is called) of the amplifier, and conversely.
Sample Problem 5 A radar transmitter emits pulses 0.15 //s long at a wavelength of 1.2 cm. (a) To what central frequency should the radar receiver be tuned? (b) What is the length of the wave packet? (c) What should be the frequency bandwidth of the receiver? That is, to what range of frequencies should it be able to respond? Solution
(a) The central frequency Vq (= (Oq/ I ti) is given by
(6) The length of the wave packet is Ax = cA / = (3.00X 10® m/sX0.15X 1 0 -^ ) = 45 m. (c) The receiver’s bandwidth is given approximately by Eq. 5, or Av =
Aco In
1
2nAt
1 2n{0.\5 X 10“^ s)
= 1.1 X 10^ H z = 1.1 MHz. * The estimate given by Eq. 4 represents the best we can do in constructing a wave packet. It is possible to do much worse (A^ • Ax » 1), but it is not possible to do much better (A/c •Ax can never be c 1).
If the receiver cannot respond to frequencies throughout this range, it will not be able to reproduce faithfully the shape of the transmitted radar pulse.
Section 50-5
50-5 HEISENBERG’S UNCERTAINTY RELATIONSHIPS Equation 4 applies to all kinds of waves. Let us apply it to de Broglie waves. We write, for the quantity Ak that ap pears in that equation.
Here we have identified A with the de Broglie wavelength o f the particle and substituted h/p^ for it. The subscript on the momentum reminds us that we are dealing with mo tion along the X axis only. Substituting this result into Eq. 4 leads to A/:* Ax = ^ Ap;,-Ax' 1
h
or Ap;p*Ax~ h/ln. Taking into account the fact that momentum is a vec tor, we can generalize this relationship to A p j,* A x ~ / z/27T, t^Py-ts.y - h lln , A p ^ -A z ^ hl2n.
( 6)
Equations 6 are the Heisenberg uncertainty relationships, first derived by Werner Heisenberg in 1927. They can be regarded as the mathematical formulation of Heisen
berg's uncertainty principle: It is not possible to determine both the position and the momentum of a particle with unlimited precision. It is our goal in quantum mechanics to represent a particle by a wave packet that has large amplitude where the particle is likely to be found and small amplitude elsewhere. The width Ax of the wave packet indicates something about the probable location of the particle. However, as we have seen, construction of such a wave packet requires the superposition of waves with a range Ak in wave number or, equivalently, a range Ap^ in mo mentum. Hence, another way of stating the uncertainty principle is: a particle cannot be described by a wave
packet in which both the position and the momentum have arbitrarily small ranges. As you make the range of one of them smaller, the range of the other becomes larger, ac cording to Eq. 6. Even though an individual measurement of the mo mentum o f a particle can yield an arbitrarily precise value, that value can be anywhere in a range Ap^ about the “true” value p^,. (In effect, quantum mechanics tells us that we cannot determine the “true” value p^^ except to within a range Ap^.) If we repeat the measurement a large number of times on identically prepared systems, our re
Heisenberg's Uncertainty Relationships
1051
sults will cluster about p^ with a statistical distribution characterized by the width Ap^. Similarly, measurements of position will cluster about the position x with a statisti cal distribution characterized by the width Ax. These limitations have nothing whatever to do with the practical problems o f measurement. Equations 6, in fact, assume ideal instruments. In practice, you will always do worse. Sometimes these relationships are written with a “ > ” symbol replacing the “ ” symbol, reminding us of this fact. When we use the word “particle” to describe objects such as electrons, it conjures up in our mind the image o f a tiny dot moving along a path, its position and velocity being well-defined at every moment. This way o f thinking is a natural extension of familiar experiences with objects like baseballs and pebbles that we can see and touch. We must, however, accept the fact that this picture simply does not hold up experimentally beyond the limits set by the uncertainty principle. The quantum world is a world beyond our direct experience, and we must be prepared for new ways of thinking. In Sample Problem 6 we shall see that the uncertainty principle does not limit our precision of measurement when we are dealing with large objects such as golf balls. Here ordinary instrumental errors overwhelm the funda mental limits set by this principle. When we deal with electrons and other elementary particles, however, the situation is quite different, as Sample Problem 6 shows.
Sample Problem 6 {a) A free 10-eV electron moves in the x direction with a speed of 1.88 X 10^ m/s. Assume that you can measure this speed to a precision of 1%. With what precision can you simultaneously measure its position? (b) A golf ball has a mass of 45 g and a speed of 40 m/s, which you can measure with a precision of 1%. What limits does the uncertainty principle place on your ability to measure its position? Solution
(a) The electron’s momentum is
p^ = mv^ = (9.W X 10-^‘ kgX1.88X 10^ m/s) = 1.71 X 10"2^kg-m/s. The uncertainty Ap^ in momentum is 1% of this or 1.71 X 10“ ^^ kg-m/s. The uncertainty in position is then, from Eq. 5, A x~
1.06 X 10-^ J-s h / l n ___________________ = 6.2 nm, ~ A ^ ~ 1.71 X 10-2<^kg*m/s
which is about 30 atomic diameters. Given your measurement of the electron’s momentum, there is simply no way that you can simultaneously pin down its position to a better precision than this. (b) This example is exactly like part (a) except that the golf ball is much more massive and much slower than the electron. The same calculation yields, in this case, Ax ~ 6 X 10“ ^' m. This is a very small distance, about 10’^ times smaller than the diameter of a typical atomic nucleus. Where large objects are concerned the uncertainty principle sets no meaningful limit to
1052
Chapter 50
The Wave Nature o f Matter
the precision of measurement. You could never have discovered this principle by studying flying golf balls!
The Uncertainty Principle and Single-Slit Diffraction Here we learn more about the uncertainty principle by seeing how it works in a particular case. Consider a beam o f electrons o f speed Vq, moving upward as in Fig. 14. We set ourselves this task: measure simultaneously and with unlimited precision the horizontal position x and the ve locity component for an electron in this beam. As we shall see, this task (which violates the uncertainty princi ple) cannot be accomplished. To measure x let us block the beam with a screen A in which we put a slit of width A x If an electron passes through this slit, we can claim to know its horizontal position to this precision. By narrowing the slit we can improve the precision o f this measurement as much as we wish. So far so good. However, something happens that per haps we hadn’t counted on. The electron beam— being wavelike— flares out by diffraction as it passes through the slit. If we put a suitably sensitive screen B in Fig. 14, a typical single-slit diffraction pattern shows up. Electrons that form the left half of this pattern must have been moving to the left (some faster, some slower) as they emerged from the slit. Those that form the right half must have been moving to the right. Even though— as the sym-
metry of the arrangement requires— the average ya\\it of for the emerging electrons is zero, individual electrons can have nonzero values. There is a particular value of that will cause the electron to land at the first minimum o f the diffraction pattern, identified by the angle 0, in Fig. 14. We take this value of Vx— somewhat arbitrarily— as a rough measure of the uncertainty of our knowledge of and we call it The location o f the first minimum of the diffraction pattern is given by Eq. 1 o f Chapter 46 (sin 0, = A/Ax). If we assume that is small enough, we can replace sin d, by 0 ,, obtaining « A/Ax. To reach the first minimum it must be true that 0, « A u^/P q.
Combining these two relations leads to ^ x ^ kVQ.
Now A, the de Broglie wavelength of the electron, is equal to h/p or h/mvQ; putting this into the above and rearrang ing, we find
Apj,-Ax^ h. This is certainly consistent with Eq. 6; minor differences (the factor 2n) result from the arbitrary way we have de fined Ax and Ap^,. See how the uncertainty principle operates in this case. If we want to pin down the horizontal position o f the electron, we must narrow the slit. This, however, broad ens the diffraction pattern so that Ap^ increases. On the other hand, if we want to pin down the horizontal mo mentum component of the electron, we must somehow reduce the angular width o f the diffraction pattern. The only way to do this is to widen the slit but that, in turn, means that we no longer know the horizontal position of the electron as precisely as we did. As we try to increase our knowledge about one variable, we simultaneously re duce our knowledge of the other. The uncertainty princi ple is not a statement about electrons (or other particles); it is a statement about our ability simultaneously to deter mine certain properties of those particles.
The Energy-Tim e Uncertainty Relationship Thus far we have considered only the wavelengths o f mat ter waves and have said nothing about their frequencies. By analogy with Einstein’s photon equation (E = hv), the uncertainty in the frequency o f a matter wave is re lated to the uncertainty in the energy E o f the correspond ing particle by Av = A E /h . Substituting this into Eq. 5 yields, with Aco = I n Av, Figure 14 An incident beam of electrons is diffracted at the slit in screen A. If the slit is made narrower, the diffraction pattern becomes wider.
A E -A t-h /ln ,
(7)
which is the mathematical relationship o f the uncertainty
Section 50-6
principle expressed in terms of different parameters. In words, it says: It is not possible to determine both the energy and the time coordinate of a particle with unlimited precision. All energy measurements carry an inherent uncertainty unless you have an infinite time available for the measure ment. In an atom, for example, the lowest energy state (the so-called ground state) has a well-defined energy be cause the atom normally exists indefinitely in that state. The energies of all states of higher energy (the excited states) are less precisely defined because the atom— sooner or later—will move spontaneously to a state of lower energy. On average, you have only a certain time A/ available so that your energy measurement will be uncer tain by an amount A£* given by (hl2n)/At. Sample Problem 7 In 1974 an important new particle, more than three times as massive as the proton, was discovered simul taneously and independently by two groups of physicists, using the high-energy accelerators at the Brookhaven National Labora tory and at Stanford University. The rest energy of this particle was measured to be 3097 MeV, the uncertainty of the measure ment being only 0.063 MeV. Such a massive particle was ex pected to decay extremely rapidly into particles of smaller mass. What is the mean time interval between production and decay for these short-lived particles? Solution The answer follows from the uncertainty principle, in the form of Eq. 7. Solving for At yields A t-
h /ln _ 1.06X 1 0 -^ J -s A E ~ (0.063 MeVK1.60 X lO"'^ J/MeV) = 1.1 X 10"20s.
This time interval can be identified with the lifetime of the parti cle. By ordinary standards this seems to be a short lifetime but, in fact, the experimenters were astonished by how long it was or— equivalently— by how sharply the rest energy of the particle was defined. Theory had predicted that the decay of this massive particle (called ij/) would be very much faster. One observer remarked that the observed slowing down of the decay was as if Cleopatra, floating on the Nile in her Royal Barge, had dropped a pebble over the side and the falling pebble, as of today, had not yet hit the water! This new particle proved to be so significant that the leaders of the two groups. Burton Richter and Samuel Ting, were awarded the Nobel prize in 1976 for its discovery. The uncertainty principle is used in this way to deduce the lifetimes of the excited states of molecules, atoms, and unstable fundamental particles of all kinds.
50-6 TH E W AVE FU N C TIO N ________
Bythis time you should be comfortable with the fact that a moving particle can be viewed as a wave, and you should know how to measure its wavelength. It remains to ask:
The Wave Function
1053
“What is the quantity whose variation in time and space makes up this wave?” To put it loosely: “What is wav ing?” For a wave in a string we can represent the wave distur bance by the transverse displacement y. For sound waves we use the differential pressure Ap and for electromag netic waves the electric field vector E. For waves represent ing particles, we introduce the wavefunction *F. The prob lem at hand may be that of a proton traveling along the axis of an evacuated tube in a particle accelerator, a con duction electron moving through a copper wire, or an electron moving about the nucleus of a hydrogen atom. Whatever the case may be, if we know the wave function (x, y, z, t) for every point of space and for every instant of time, we know all that can be known about the behav ior of the particle. Before we look into the physical meaning of the wave function, let us consider a problem that involves radiation rather than matter: a plane electromagnetic wave travel ing through free space. We can think of such a wave (fol lowing Maxwell) as an arrangement of electric and mag netic fields that varies in space and time or (following Einstein) as a beam of photons, each moving with the speed of light. On the first picture, the rate per unit area at which energy is transported by the wave (see Section 41-4) is proportional to E^, where E is the amplitude of the electric field vector. On the second picture, this rate is proportional to the average number of photons per unit volume of the beam, each photon having an energy hv. We see here a connection between the wave and particle pictures of radiation, namely, the notion—first advanced by Einstein—that the square of the electric field intensity is a direct measure of the average density of photons. Max Bom proposed that the wave function 'F for a beam of particles be interpreted in this same way, namely, that its square is a direct measure of the average density of particles in the beam. In many problems, however, such as the stmcture of the hydrogen atom, there is only a single electron present. What can it then mean to speak of the “average density of particles”? Bom proposed that in such cases we should interpret the square of the wave function at any point as the probability (per unit volume) that the particle will be at that point.* Specifically, if dVis a volume element located at a point whose coordinates are x, y, z, then the probability that the particle will be found in that volume element at time t is proportional to 4'^ dV. Perhaps by analogy with ordinary mass density (a mass per unit volume), we call the square of the wave
* The wave function y is usually a complex quantity; that is, it involves which we represent by the symbol i. By (more properly written as |*Fp) we mean the square of the absolute value of the wave function. This is always a real quantity. We give a physical interpretation only to the square of the wave function, not to the wave function itself
1054
Chapter 50
The Wave Nature o f Matter
function a probability density, that is, a probability per unit volume. Note that the relationship between the wave function and its associated particle is statistical, involving only the probability that the particle will find itself within a speci fied volume element. In classical physics we also deal with particles on a statistical basis (see Chapters 23 and 24), but in those cases statistical methods are just a handy way of dealing with large numbers of particles. In quantum me chanics, however, the statistical nature is inherent and is dictated by the uncertainty principle, which, as we have seen, sets limits to the meaning that we can attach to the word “particle.” The probability that our particle will be somewhere must be equal to unity (corresponding to a 100%chance to find it) so that we have
/
dV = \
(normalization condition),
(8)
the integration being taken over all space. To normalizes wave function means to multiply it by a constant factor, chosen so that Eq. 8 is satisfied. We save for last an obvious question: In any given problem, how do we know what the wave function is? Waves on strings and sound waves are governed by New ton’s laws of mechanics. Electromagnetic waves are pre dicted and described by Maxwell’s equations. From where do the wave functions come? In 1926 Erwin Schrodinger, inspired by de Broglie’s concept, thought along these lines: geometrical optics deals with rays and with the motion of light in a straight line; it turned out to be a special case of a much more
n = 15
general wave optics. Newtonian mechanics also has “rays” (trajectories) and straight-line motion (of free par ticles). Could it turn out to be a special case of a much more general—but as yet undiscovered—wave mechan ics? Schrodinger derived a remarkably successful theory based on this analogy. Its central feature is a differential equation, now known as Schrodinger's equation, which governs the variation in space and time of the wave func tion 4^ for a wide range of problems. We obtain solutions to problems in classical mechanics by manipulating New ton’s laws of motion; we obtain solutions to problems in electromagnetism by manipulating Maxwell’s equations; in exactly the same spirit, we obtain solutions to atomic problems by manipulating Schrodinger’s equation. In the next section we shall study an important problem from the point of view of wave mechanics, that of a parti cle trapped in a region from which it can never escape (or can it?).
50-7 TR A PPED PARTICLES A N D PROBABILITY D E N SIT IE S
Before we consider the situation for matter waves, let us review two analogous examples involving mechanical waves and electromagnetic waves. In Sections 19-9 and 19-10, we considered the standing waves that can occur in a string of length L that is fixed at both ends. The fixed ends of the string are constrained by the supports to be vibrational nodes, that is, locations where the amplitude is zero at all times. Only a limited set of wavelengths can occur for these standing waves. As shown in Section 19-10, these allowed wavelengths can be written («= 1,2,3, . . . ) n or, in terms of wave number, , 2n nn A
( « = 1 , 2 , 3 , . . . ).
(9)
(10)
At any point along the string, the amplitude of vibration is y«(-«) =
sin k„x =
sin — ( « = 1 ,2 ,3 , . . . ),
Figure 15 Four standing wave patterns for a stretched string of length L, clamped rigidly at each end. These patterns are determined from Eq. 11.
(11)
where is the maximum displacement of the string. Figure 15 shows examples of the vibrational patterns of these standing waves, which might characterize some of the lower vibrational modes of a guitar string or a violin string. In electromagnetism, a similar situation results when a plane electromagnetic wave oscillates back and forth (in one dimension) between two perfectly reflecting surfaces
Section 50-7
(mirrors, for instance) separated by a distance L. Such a situation might occur for light waves in a laser. Just as in the mechanical case, a standing wave is established in the cavity. This standing electromagnetic wave can be consid ered as the superposition of two similar waves traveling in opposite directions. At the ends of the cavity, where the reflection occurs from a conducting material such as the silvering on a mirror, the electric field must drop to zero (which is true for all ideal conductors under conditions of electrostatics). Imposing these conditions that £ = 0 at X= 0 and x = L, we find that only certain wavelengths are permitted for the standing wave; the permitted wave lengths are given by Eq. 9, and the amplitude of the elec tric field oscillations can be written Enix) =
sin ^
(/I = 1, 2, 3, . . . ). (12)
Figure 16a shows a plot of E„{x) as a function of x for the lowest modes of oscillation (n = 1and 2). Note the simi larity between this figure and Fig. 15, which showed the mechanical standing wave on a string. Figure 16 b shows a plot of E l ( x ) , which is proportional to the energy density of the wave, as we discussed in Sec tion 41-4. In terms of the photon picture, we can regard the standing wave as a collection of photons, and Fig. 16 b represents the density of photons as a function of x. That is, in the mode of oscillation with n = 1 you would find the greatest density ofphotons at x = L/2 and the smallest density next to the walls. In the « = 2 mode you would find a minimum in density at x = L/2 and maxima at X = L/4 and 3L/4. Suppose we reduce the intensity of light in the cavity until it contains only a single photon. Figure \6b would still apply, but we must change its interpretation slightly, since it is no longer appropriate to speak of the density of photons. Instead, we use a related statistical concept: the square ofthe electricfieldamplitude at aparticular coordi nate gives the probability tofind the photon at that loca tion. Figure \6b shows locations where the probability to
n=2
n=2
n=\
n= 1
(6)
Trapped Particles and Probability Densities
1055
find the photon is large and where it is zero. Note that we do not consider the actual location of the photon, but instead its probability to be found in a certain location. These characteristics of mechanical and electromag netic standing waves in one dimension can be directly carried over to matter waves. Consider a particle confined to move between two perfectly reflecting walls a distance L apart. Figure 17 shows a device that might be used to trap an electron. Although this is a large-scale device, it is possible to construct microscopic devices that accomplish the same result. For example, “quantum well” structures are built from a few atomic layers of semiconducting ma terial surrounded by insulating material; such devices are used for optical communication and logic gates. In the apparatus shown in Fig. 17a, the electron can move freely in the central section, where no forces act on it. When it reaches either end, it encounters a region in which the potential changes rapidly from 0 (which we take to be that of the central section) to —Kq, the potential associated with either battery. Equivalently, the potential energy of the electron is 0 in the central section and Uq {=eVo) in the outer sections (Fig. \lb). If the kinetic en ergy of the electron in the central region is less than Uq, then classically it has insufficient energy to escape from the well and it oscillates back and forth between the walls. We seek a description of the motion of the electron in the potential well using the language of wave mechanics. While this may seem like a trivial problem far removed from, say, the structure of atoms, it turns out to demon strate the important features of wave-mechanical behav ior in a way that avoids the mathematical complexity of more complicated systems. In describing mechanical and electromagnetic standing waves, we used functions y{x) and E„ (x), which lack the time dependence that must be present to describe a wave. However, as we showed above, for our analysis we were more interested in the variations of amplitude with posi tion, and so it was not necessary to consider the time dependence. We shall do the same in the case of matter
Figure 16 (a) The electric field for two elec tromagnetic standing wave patterns in a cavity of length L. These patterns are determined from Eq. 12. (b) The density of photons in the cavity can be found from the square of the fields.
1056
Chapter 50
The Wave Nature o f Matter
Uix)
l/n
( 6)
Figure 17 (a) An arrangement that can be used to confine an electron to a region of length L along the x axis, (b) The potential energy of the electron. In any real device, the potential energy would not change immediately from 0 to f/o; the graph of the potential energy would have rounded comers and nonvertical sides.
waves. Instead o f seeking the general wave function 4'(x, t), we shall consider only the spatial part, which we write as yA^x). We begin by assuming the walls to be perfectly reflect ing for all particles; that is, we consider an infinitely deep well, such that The Schrodinger equation in this case turns out to be identical with the wave equations that describe mechanical or electromagnetic waves. Its solution is
y/„{x) = As\nk„x = A sin ( « = 1 , 2 , 3 , . . . ),
(13)
where the allowed set of wave numbers or wavelengths is given by Eqs. 9 and 10. The constant A must be deter mined by the normalization condition (see Sample Prob lem 10). The function y/„{x) has no physical interpretation. However, the square o f the wave function does have a physical meaning— it gives the probability density P„(x): Pnix) = lv'„(x)p =
The (kinetic) energy is then
2m
‘
( « = 1 , 2, 3, . . . ),
(16)
where m is the mass of the particle. When we write the energy in this way, the index n is called a quantum num ber. The allowed energies are plotted in Fig. 19. The elec tron is permitted to occupy only those states o f motion corresponding to this set of energies; no other energies are permitted for the particle. Once we have found the permitted energies and wave functions, we have solved the problem o f the trapped particle using the techniques of quantum mechanics. The quantum solution shows a number of unexpected fea tures that are not part of the classical solution for a trapped particle. Let us consider some of these.
n = 15
sin^
( « = 1 , 2 , 3,
).
(14)
Just as was the case for the square o f the electric field in Fig. 16, the square of the wave function at a particular location indicates the probability to find the electron at that location. Some o f these probability densities are plot ted in Fig. 18. The energy o f the particle (which is entirely kinetic inside the well, since C/ = 0) is restricted to a certain set of values; we say that the energy is quantized. Let us see how this comes about. The allowed de Broglie wavelengths of the particle are given by Eq. 9, and so the magnitude of its momentum is restricted to the values
_ h _nh 2L
. _ , T ,
).
(15)
Figure 18 The probability density PJ^x), computed accord ing to Eq. 14, for four different values of the quantum num ber n. The horizontal lines show the classical expectations for the probability density.
Section 50-7 -E^ = 16(/i2/8mL2)
n = 4n = 3-
= 9{h^lSmL^)
n = 2-
.E 2 = 4(/i2/8mL2)
n = 1-
- E j = h^lSmL^
Figure 19 The allowed energy levels, calculated from Eq. 16, for an electron trapped in a one-dimensional region.
1. The electron cannot be at rest in the well. The lowest energy state, called the ground state, corresponds to « = 1 in Eq. 16 and Fig. 18. This lowest energy is not zero. We cannot reduce the energy of the particle (that is, its kinetic energy) to zero. The minimum energy, given by E ,=
h^ im L ^ ’
(17)
is called the zero-point energy for the infinite well. In other quantum systems, the zero-point energy may take differ ent forms, but the phenomenon exists for all quantum systems. Even at the absolute zero of temperature, where a particle is in the lowest possible energy state, it still has motion and energy. In effect, the zero-point motion occurs as a result of confining the particle to a region of space. Let us see how the uncertainty principle helps us understand this effect. If we confine a particle in the well, we know its position to within an uncertainty of approximately L. The corre sponding uncertainty in its momentum is, from Eq. 6, ^P x-
2nL
*
The smaller the region in which we confine the particle, the larger is the uncertainty in its momentum. If we know the kinetic energy of the particle, we know the magnitude of its momentum precisely, but we don’t know the direction. An uncertainty in momentum of ^p^ suggests that the particle may be moving to the right with momentum p^ = \b^Px = h /A nL or to the left with mo mentum Px = — = —h/A nL, This is comparable to the momentum of this state given by Eq. 15; that is, p, = h / l L , the difference of a factor of 2;: having to do with the arbitrary way we have defined the uncertainties. The esti mate does indicate that the zero-point motion is consist ent with the uncertainty relationship, and that the motion is a result o f our confining the particle to a region of space. 2. The electron spends more time in certain parts of the trap than in others. For this one-dimensional problem a “volume” element becomes a line element so that the product Pn{x) dx is the probability that the electron will be found in the interval x\ox-\- dx.k glance at the probabil ity density curves of Fig. 18 shows that in the ground state {n = 1), the electron is much more likely to be found near the center o f the trap than near its ends. Once again we
Trapped Particles and Probability Densities
1057
have a prediction in sharp contrast with the prediction of classical mechanics. According to classical theory all posi tions between the walls of the trap are equally likely, as the horizontal lines in Fig. 18 suggest. For both the quantum and the classical curves, the area under the curve is unity, as the normalization condition (Eq. 8) requires. For states of higher quantum number— and thus of higher energy— the distribution of the electron probabil ity density across the trap becomes more uniform and the quantum prediction merges with the classical prediction. This agreement between classical and quantum physics for large quantum numbers is called the correspondence principle and is discussed in Section 50-9. 3. The electron can escape from its trap. So far we have dealt with a well of infinite depth. A major quantum surprise awaits us if we relax this requirement and deal with the more realistic case of a well of finite depth. In Fig. 20 we compare two wells of the same width (= 2 X 10” *®m, about the size of a large atom), one o f the wells being infinitely deep and the other having a depth of only 20 eV. To find the allowed energies and the corresponding probability densities for a finite well, we need the full power of the Schrodinger equation. Here we simply give the results, without proof.* We consider the ground state only. Figure 20 shows that the ground state energy for the finite well (=4.45 eV) is substantially less than the ground state energy for the infinite well (=9.41 eV, calculated from Eq. 16). We can tell that this should be the case simply by inspecting the probability density curves of Fig. 20. For the infinite well, half a de Broglie wavelength fits neatly and exactly between the rigid walls of the well. For the finite well, however, the de Broglie wavelength is too large to fit in this way; it spills over beyond the walls. Now if the de Broglie wavelength is larger for the finite well, the momentum (=A/A) must be smaller, which means the energy must also be smaller, just as we observe. Thus the reduced energy for the finite well is consistent with the form of its probability density curve. The spilling over of the exponential tails of the probabil ity curve (Fig. 20b) beyond the walls means that there is a finite probability that the electron will be found outside the well! This curve has been normalized (see Eq. 8) so that the area under the curve is unity. The area under the two exponential tails is 0.074, which means that, if you measured the position of the trapped electron, you would find it outside the well 7.4% of the time. How can an electron whose energy is only 4.45 eV escape from a well that is 20 eV deep? It is clearly a classi cal impossibility. It is as if you put a jelly bean into a closed box and sometimes (but not always) the jelly bean
* See, for example, Robert Eisberg and Robert Resnick, Quan tum Physics o f Atoms, Molecules, Solids, Nuclei, and Particles, 2nd edition (Wiley, 1985), Appendices G and H.
1058
Chapter 50
The Wave Nature o f Matter
40
Figure 20 A potential well of {a) infinite depth and (b) finite depth (20 eV) are compared. The wells have the same width (0.2 nm). The ener gies £ , and the probability densities P,(x) are compared.
30
> -Uq= 20
20
10-
eV
E i ( = 9 .4 1 eV) - E j (= 4 .4 5 eV)
0*- L -
^— L—^
Infinite well
Finite well
materialized outside. This is so unlikely for jelly beans that we can safely use the word “impossible.” However, electrons are not jelly beans. They are governed by quan tum laws and not at all by classical Newtonian laws. How are we to understand this behavior of the electron? Once again the uncertainty principle provides the an swer. Recall that, when we applied the uncertainty princi ple to an electron trapped in an infinite well, we assumed that Ax « L, the width of the well, and Ap^,« Ip ^. For the finite well the electron’s momentum, as we have just seen, is sm aller than it is for the infinite well. Therefore, for our finite well, the uncertainty in position in the ground state must be larger than it is for the infinite well. Thus, for the finite well, the uncertainty in position is larger than the width o f the well! We should not then be surprised to find the electron outside the well from time to time.
Sample Problem 8 Consider an electron confined by electrical forces to an infinitely deep potential well whose length L is 1(X) pm, which is roughly one atomic diameter. What are the ener gies of its three lowest allowed states and of the state with « = 15? Solution
Sample Problem 9 Consider a 1-/zg speck of dust moving back and forth between two rigid walls separated by 0.1 mm. It moves so slowly that it takes 100 s for the particle to cross this gap. What quantum number describes this motion? Solution
The energy of the particle is
E (=K) = = K1 X 10-’ kgXl X lO”'' m /sY = 5X 10-22 J. Solving Eq. 16 for « yields L rr-r;
1 X 10-^ ’m V(8X10-’ kgX5X lO-^M) 6.63X 1 0 -^J* s
« 3 X 10'^ This is a very large number. It is experimentally impossible to distinguish between « = 3 X 10'^ and n = 3 X 10'^ + 1, so that the quantized nature of this motion would never reveal itself If you compare this example with the previous one, you will see that, although its mass and its kinetic energy are both extremely small, our speck of dust is still a gross macroscopic object when compared to an electron. Quantum mechanics gives the correct answers but, since these answers coincide in this case with the answers given by classical physics, the complications of the quantum calculations are not needed.
From Eq. 16, with n = 1, we have
h^ _ (6.63 X 10-^J»s)"_______ ^1 = 8mL2 (8X9.11 X 10"^' kgXlOO X 1 0 " m)^ = 6.03X 10-'* J = 37.7 eV. The energies of the remaining states (n = 2, 3, and 15) are 22 X 37.7 eV, 3^ X 37.7 eV, and 15^ X 37.7 eV or 151 eV, 339 eV, and 8480 eV, respectively.
Sample Problem 10 Evaluate the normalization constant A in Eq. 13, which gives the probability density for a particle trapped in an infinitely deep well of width L. Solution For this one-dimensional problem, the “volume” ele ment is a length element and the normalization equation (Eq. 8) becomes
Section 50~8
/:
P n M d x = \.
in which L is the width of the well. Substituting for P„(x) from Eq. 14 yields
.
2
nTtx ,
I sin^ —p—ax =
Jo
^
This integral is carried out most easily by introducing a new variable 6, defined from
mix
e=-
L
'
With this change, the integral becomes — f"* sin^ 0 nn Jo
= — ( 1 0 - 2 sin nn \ 2 4
/|o
= 1.
Evaluating and solving for A lead eventually to (18) Note that the normalization constant A does not involve the quantum number n and is thus the same for all states of the system. Sample Problem 11 An electron is trapped in an infinitely deep well of width L. If the electron is in its ground state, what fraction of its time does it spend in the central third of the well? Solution In the preceding example we showed that the normal ization constant A that appears in Eq. 13 is V2/L, so that the probability density for the ground state, which corresponds to /I = 1, is given from Eq. 14 by P^{x) = - sin^ — .
Barrier Tunneling
1059
We pointed out that it was as if you put a jelly bean in a closed box and found it outside the box a certain fraction of the times that you checked. Things like this don’t hap pen to massive objects like jelly beans, but they do happen to electrons and to other light particles. Here we discuss a related quantum phenomenon, the penetration of classically impenetrable barriers. In this case it is as if you tossed a jelly bean at a window pane and— to your surprise— it materialized on the other side with the glass unbroken. Again, don’t expect this to hap pen for jelly beans. Barrier tunneling, as it is called, cer tainly does happen for electrons and is, as we shall see, a phenomenon of great practical importance. Figure 2 Ifl shows such a barrier, of height U and thick ness L. An electron of total energy E approaches the barrier from the left. Classically, because E < U , the elec tron would be reflected at the barrier and would move back in the direction from which it came. In wave me chanics, however, there is a finite chance that the electron will penetrate the barrier and continue its motion to the right. We can describe the situation by assigning a reflection coefficient R and a transmission coefficient T, the sum of these two quantities being unity. Thus, for example, if T = 0.05, 5 of every 100 electrons fired at the barrier will, on the average, get through and 95 will be reflected. Figure 2\b shows the probability density for the situa tion. To the left of the barrier the reflected matter wave has a smaller amplitude than the incident wave so that, although there is interference, there are no points at which the cancellation is total. Within the barrier the wave decays exponentially, just as it did outside the potential well in Fig. 20^?. On the far side of the barrier we have a
The integral of this quantity over the entire well is unity, and the fraction that we seek is given by f2L/3 f
J lp
2 P ,{x)dx = y
T J^/3
nx s in ^ ^ d x . l
Evaluating this integral as in the previous sample problem leads to / = 0.61. Thus the electron in its ground state spends 61 % of its time in the central third of its trap, and about 19.5% in each of the outer two-thirds (0.195 + 0.61 +0.195 = 1). If the electron obeyed the laws of classical physics, it would spend exactly one-third of its time in each of these regions of its trap. The probability density curve for the ground state, displayed in Fig. 18, supports graphically the calculation that we have made in this example.
(a)
(b)
50-8 BARRIER TUNNELING_________ In Section 50-7 we saw that an electron trapped in a well from which— classically— it could not escape has never theless a finite probability of being found outside the well.
Figure 21 A particle of total energy E is incident from the left on a barrier of height U. I represents the incident beam of particles, R the reflected beam, and T the beam transmitted through the barrier, {b) The probability density for the wave describing this particle. The incident and reflected beams combine to produce standing waves to the left of the barrier.
1060
Chapter 50
The Wave Nature o f Matter
traveling matter wave of reduced amplitude, which gives a uniform probability density. From the Schrddinger equation we can show that the transmission coefficient T is given by* T =^ eo-2kL
(19)
in which
in^m{U —E) - i
~h^
*
This formula is an approximation that holds only for barriers that are either high enough and/or thick enough so that the transmission coefficient T is small { T 1). Nevertheless, Eq. 19 displays well enough the main fea tures o f the barrier-tunneling phenomenon. The value of the transmission coefficient is very sensi tive to the thickness of the barrier L and to the factor /c, which, in turn, depends on the mass m of the particle and the height U o f the barrier. Equation 19 shows us that the transmission coefficient decreases if we increase either the thickness L or the height U o f the barrier. This is just what we expect from the correspondence principle. The trans mission coefficient also decreases as the mass of the parti cle increases, becoming vanishingly small very rapidly indeed as we proceed from electrons to jelly beans. Again, this is just what we expect from the correspondence princi ple. Sample Problem 12 shows some numerical predic tions of Eq. 19.
Barriers and Waves The penetration of barriers by waves of all kinds is not uncommon in classical physics. It is only when the wave is a matter wave and when, in addition, we choose to focus our attention on its associated particle, that nonclassical behavior presents itself. Consider Fig. 22a, which represents an incident electro magnetic wave (visible light) falling on a glass-air inter face at an angle of incidence such that total internal reflec tion occurs. When we treated this subject in Section 43-6, we assumed that there was no penetration of the incident ray into the air space beyond the interface. However, that treatment was based on geometrical optics, which, as we know, is always an approximation, being a limiting case o f the more general wave optics. In much the same way, Newtonian mechanics (with its raylike trajectories) is a limiting case of the more general wave mechanics. If we analyze total internal reflection from the wave optics point of view, we learn that there is a penetration of the wave, for a distance of the order of a few wavelengths, beyond the interface. Speaking very loosely, we can say that such a penetration is necessary because the incident
* See, for example, Robert Eisberg and Robert Resnick, Quan tum Physics o f Atoms, Molecules, Solids, Nuclei, and Particles 2nd edition (Wiley, 1985), Section 6-5.
wave must “feel out” the situation locally before it can “know for sure” that an interface is present. In Fig. 22b, we place the face of a second glass prism parallel to the interface, the gap between them being no more than a few wavelengths. The incident wave can then “tunnel” through this narrow “barrier” and generate a transmitted wave T. The energy in the transmitted wave comes at the expense of the reflected wave R, which is now reduced in intensity. The comparison with the barrier tunneling of a matter wave is direct. In one case we deal with an electromagnetic wave (governed by Maxwell’s equations) and in the other with a matter wave (governed by Schrodinger’s equation). You can check out the phenomenon shown in Fig. 22b using a glass of water. Look down into the glass at the side wall, at such an angle that the light entering your eye has been totally reflected from the wall. The wall will look silvery when this condition holds. Then press your (moist ened) fingertip against the outside of the glass. You will be able to see the ridges of your fingerprint because, at those points, you have interfered with the total reflection pro cess, as in Fig. 22b. The valleys between the ridges of your prints are still far enough away from the glass surface that the reflection here remains total, and you see simply a silvery whorl. It is also possible to demonstrate the phenomenon of Fig. 22b on a large scale by using incident microwaves and large paraffin prisms. In this case the wavelength may be a few centimeters so that the gap between the prisms can also be of this order of magnitude.
Barrier Tunneling: Some Examples The barrier tunneling of matter waves is an important phenomenon in the natural world and has many practical applications. For a simple example consider a bare copper wire that has been cut and the two ends twisted together. It still conducts electricity readily, in spite of the fact that the wires are coated with a thin layer of copper oxide, an
(a)
Figure 22 (a) An incident light beam I undergoes total inter nal reflection at the glass-air interface, (b) The beam tunnels through the narrow air gap, and as a result there is a transmit ted beam T in the second glass. This condition is called frus trated total internal reflection. In these drawings, the width of the beam represents its intensity.
Section 50-8 Scanning
Figure 23 A needle is scanned over the surface of a sample in a scanning tunneling microscope.
insulating material. How do the electrons get through this (extremely thin) oxide layer? By barrier tunneling. For a more exotic example, consider the core of the Sun, where the Sun’s energy is being generated by thermo nuclear fusion processes. Such processes involve the fus ing together o f light nuclei to form heavier ones, with the release of energy. Suppose that two protons are rushing together at high speed. They must get extremely close before their strong attractive nuclear forces can take effect and cause them to fuse. Meanwhile, they are slowed down by the repulsive Coulomb force that tends to drive them apart. They are, we can say, separated by a Coulomb barrier. The likelihood of fusion depends critically on the ability of the protons to tunnel through this mutual barrier. Without barrier tunneling the solar furnace would shut down and the Sun would collapse into itself. The emission of (positively charged) alpha particles by radioactive nuclei and the spontaneous fission of heavy nuclei into two large fragments are among other natural processes in which tunneling plays a role. Among practical applications we may list the tunnel diode, in which the flow of electrons (by tunneling) through a device can be rapidly turned on or off by con trolling the height of the barrier (by varying an externally applied voltage, for instance). This can be done with a very short response time (of the order of 10“ *‘ s or 10 ps)
Barrier Tunneling
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so that the device is suitable for applications where speed of response is critical. The 1973 Nobel prize was shared by three “tunnelers,” Leo Esaki (tunneling in semiconduc tors), Ivar Giaver (tunneling in superconductors), and Brian Josephson (the Josephson junction, a quantum switching device based on tunneling). In a scanning tunneling microscope, a fine needle tip is scanned mechanically (in a TV-like raster pattern) over the surface of the sample being investigated, as in Fig. 23. Electrons from the sample tunnel through the gap be tween the sample and the needle and are recorded as a “tunnel current.” Normally, this tunnel current would vary widely as the gap between the sample and the needle changes during the scan. However, a mechanism is provided that automatically moves the needle up or down during the scan, so as to keep the tunnel current— and thus the gap— constant. The needle’s vertical position can then be displayed on a screen as a function of its location, producing a threedimensional plot of the surface. The 1986 Nobel prize was awarded to Gerd Binnig and Heinrich Rohrer for the development of the scanning tunneling microscope.* Figure 24 shows the result of a scan over a graphite surface. The “bumps” suggest individual carbon atoms. Features as small as 1/100 of an atomic diameter can be resolved with this remarkable device.
Sample Problem 12 Consider an electron whose total energy E is 5.0 eV approaching a barrier whose height V is 6.0 eV, as in Fig. 2 la. Let the barrier thickness L be 0.70 nm. (a) What is the de Broglie wavelength of the incident electron? (b) What trans mission coefficient follows from Eq. 19? (c) What would be the transmission coefficient if the barrier thickness were reduced to
* See “The Scanning Tunneling Microscope,” by Gerd Binnig and Heinrich Rohrer, Scientific American, August 1985, p. 50.
Figure 24 The regular arrangement of carbon atoms on the surface of graphite is revealed in this image made with a scanning tunneling mi croscope.
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0.30 nm? If its height were increased to 7.0 eV? If the incident particle were a proton? Solution (a) Before the electron reaches the barrier, its total energy E is entirely kinetic, the potential energy in that region being zero. Proceeding as in Sample Problem 2Z?, we find A= 0.55 nm. Thus the barrier is about 0.70 nm/0.55 nm or about 1.3 de Broglie wavelengths thick. (b) We have
:-yf
% n^m (U -E) h^
l%n\9.\ \ X 10-3' kgX6.0 eV - 5.0 eVK 1.60 X 10"‘’ J/eV) (6.63X 10-^J*s)2 = 5.12X 10’ m -‘. The quantity kL is then (5.12 X 10^ m “ ‘)(700 X 10“ 3.58, and the transmission coefficient is T=
m) =
= 7.7 X 10“^
Of every 100,(XX) electrons that strike the barrier, only 77 will tunnel through it. (c) Making the appropriate changes in the solution to part (b), we find: L = 0.30nm : 7 = 0 .1 0
U=7.0eV: w =
1836we:
T = 5 . 9 X 10“ ^ T = 10“ ' ^ .
It is easier for the electron to penetrate the thinner barrier, but more difficult to penetrate the higher one. The more massive proton penetrates hardly at all. (Imagine how small T would be for a jelly bean!)
50-9 THE CORRESPONDENCE PRINCIPLE____________________ In several cases in this chapter and the previous one, we have tried to make comparisons between classical and quantum behaviors. For example, in Sample Problem 3 o f Chapter 49, we showed that the quantized behavior of an oscillator of ordinary size is too small to be observable; we are therefore safe in treating that oscillator using classi cal (nonquantum) techniques and in regarding the energy of the oscillator as a continuous (rather than a quantized) variable. In this chapter, we showed in Sample Problem 2 that the de Broglie wavelength of a virus particle is unobservably small; in Sample Problem 6b that the uncertainty principle should not affect your golf game; and in Sample Problem 9 that the energy quantization of a trapped dust particle cannot be observed. We seem to have two sets of rules for analyzing me chanical behavior: we use quantum mechanics for “small” particles and classical mechanics for “large” par ticles. Clearly a golf ball is a large particle, but even dust particles and viruses can be regarded as “large.” Perhaps
you are wondering where to draw the line between classi cal particles and quantum particles. Niels Bohr was similarly puzzled when he attempted (before de Broglie’s bold hypothesis led to the develop ment of quantum theory) to work out the structure of atoms based on the nuclear model of electrons orbiting a central nucleus. Bohr’s model was based on discrete en ergy levels (and discrete transitions as the electron jumped from one level to another) for the atom, but he knew that a “classical” atom would be characterized by a continuous spectrum of radiation as the electron spiraled in toward the nucleus. As in the case of the quantum system, Bohr faced the dilemma of one set of rules for systems on one scale and a different set o f rules on another scale. Bohr resolved his problem by proposing the correspond dence principle, which can be stated in general terms as:
Quantum theory must agree with classical theory in the limit of large quantum numbers. This avoids the problem of having to find a boundary between the two different systems. The predictions of quantum mechanics must be identical to those o f classical mechanics as the quantum system grows to classical di mensions. For example, consider the probability densities for a particle trapped in a well (Fig. 18). The behavior for n = \ or n = 2 differs markedly from the classical behav ior of a uniform probability density in the well. However, for « = 15, the probability density has become much more uniform. As n increases, the oscillations o f P are packed closer and closer together, so that if we examine the probability in an interval of length Ax greater than L/«, we find no change as the interval is moved through out the well. Here we are approaching the classical situa tion of a uniform probability density for large quantum numbers. The correspondence principle tells us that we do not need to draw a line between classical and quantum behav iors. If we are in doubt whether to apply classical or quan tum laws to a virus or a dust particle, we now know that we are safe in applying the quantum laws, the results of which must duplicate the classical laws if we are in a region where classical behavior is expected. Indeed, by using the probability densities calculated from the Schrodinger equation and doing the appropriate averag ing, it can be shown that the average force on a particle in the quantum_regime equals the mass times the average acceleration: F = md. In the limit of large quantum num bers, the fluctuations from the average become negligible, and F = ma becomes exact. Even though the Schrodinger equation looks very different from Newton’s second law, its outcomes reduce to Newton’s second law in the limit of large systems. This justifies our use of Newton’s second law, which is easier to apply in the large limit, to bodies composed of atoms that are individually governed by the Schrodinger equation.
Section 50-10
50-10 WAVES AND PARTICLES On several earlier occasions we have promised to address the question of how an electron (or a photon) can be wavelike under some circumstances and particlelike under others. We now keep that promise. First, we re mind you in Table 1 of the clear experimental evidence that both matter and radiation do indeed have this dual character. Our mental images of “wave” and “particle” are drawn from our familiarity with large-scale objects such as ocean waves and tennis balls. In a way it is fortunate that we are able to extend these concepts into the atomic domain and to apply them to entities such as the electron, which we can neither see nor touch. We say at once, however, that no single concrete mental image, combining the features of both wave and particle, is possible in the quantum world. As Paul Davies, physicist and science writer, has written: “It is impossible to visualize a wave-particle, so don’t try.” What then are we to do? Niels Bohr, who not only played a major role in the development o f quantum mechanics but also served as its major philosopher and interpreter, has shown the way with his principle of complementarity, which states:
The wave and the particle aspects of a quantum entity are both necessary for a complete description. However, the two aspects cannot be revealed simultaneously in a single experiment. The aspect that is revealed is determined by the nature of the experiment being done. Consider a beam of light, perhaps from a laser, that passes across a laboratory table. What is the nature of the light beam? Is it a wave or a stream of particles? You cannot answer this question unless you interact with the beam in some way. If you put a diffraction grat ing in the path of the beam you reveal it as a wave. If you interpose a photoelectric apparatus (Section 49-5), you will need to regard the beam as a stream o f particles (pho tons) if you are to interpret your measurements in a satis factory way. Try as you will, there is no single experiment that you can carry out with the beam that will require you to interpret it as a wave and as a particle at the same time.
TABLE 1
Wave nature Particle nature
Waves and Particles
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Complementarity: A Case Study Let us see how complementarity works by trying to set up an experiment that will force nature to reveal both the wave and the particle aspects of electrons at the same time. In Fig. 25 a beam of electrons falls on a double-slit arrangement in screen A and sets up a pattern o f interfer ence fringes on screen B, This is convincing proof of the wave nature of the incident electron beam. Suppose now that we replace screen B with a small electron detector, designed to generate and record a “click” every time an electron hits it. We find that such clicks do indeed occur. If we move the detector up and down in Fig. 25, we can, by plotting the click rate against the detector position, trace out the pattern of interference fringes. Have we not succeeded in demonstrating both wave and particle? We see the fringes (wave) and we hear the clicks (particle). We have not. The “click” shows that the electron is localized (like a particle) at the detector, but it does not indicate how it got there. The concept of “particle” in volves the concept of “trajectory” and a mental image o f a dot following a path. As a minimum, we want to be able to know which of the two slits in screen A the electron passed through on its way to generating a click in the detector. Can we find out? We can, in principle, by putting a very thin detector in front of each slit, designed so that, if an electron passes through it, it will generate an electronic signal. We can then try to correlate each click, or “screen arrival signal,” with a “slit passage signal,” thus identifying the path of the electron involved. If we succeed in modifying the apparatus to do this, we find a surprising thing. The interference fringes have dis appeared! In passing through the slit detectors, the elec trons were affected in ways that destroyed the interference pattern. Although we have now shown the particle nature of the electron, the evidence for its wave nature has van ished. The converse to our thought experiment is also true. If we start with an experiment that shows that electrons are particles and if we tinker with it to bring out the wave aspect, we will always find that the evidence for particles has vanished. Also, our experiment would work in pre cisely the same way if we substituted a light beam for the incident electron beam in Fig. 25.
SELECTED EXPERIMENTS SHOWING THE DUAL WAVE-PARTICLE NATURE OF MATTER AND OF RADIATION
Matter
Radiation
Davisson-Germer electron diffraction experiments (Section 50-3) J. J. Thomson’s measurement of e/m for the electron (Section 34-2)
Young’s double-slit interference experiment (Section 45-1) The Compton effect (Section 49-7)
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The Wave Nature of Matter
Figure 25 An electron beam falls on a double slit in screen A and produces interference fringes on screen B. Screen B can be replaced by an electron detector, which can be moved along the location occupied by the screen.
A Quantum Puzzle Resolved In Section 50-1, we asked how it is possible for particles to undergo double-slit interference. A particle, after all, m ust travel a definite trajectory. W hat is the source o f the interference?
Figure 26 The buildup of interference fringes as electrons fall on screen B of Fig. 25. In (^7), about 30 electrons have landed on the screen, in (b) about 1000, and in (c) about 10,000. The probability density of the wave describing the electron determines where the electrons will land on the screen.
As the beginning o f an answ er we look again at the thought experim ent o f Fig. 25, in which the pattern of fringes on screen B is neatly accounted for by the alternat ing constructive and destructive interference o f m atter wavelets radiating from each o f the two slits in screen A. The connection o f these waves with the particle is that the square o f their associated wave function at any point gives the probability (per unit volum e) th at the particle will be found at that point. Thus, on screen B, electrons will pile up at those places where this probability am plitude is large, and they will be found in lesser abundance at those places where it is small. Figure 26, a co m puter sim ulation, shows how the fringes build up with tim e for a weak inci dent beam . These considerations apply even if the incident beam is deliberately m ade so weak that, by calculation, there should be — on average— only a single electron in the apparatus at any given tim e. You m ight think that, be cause the single electron that happens to be in the appa ratus m ust go through one slit or the other, the fringes m ust vanish; after all, you may reason, the electron can not interfere with itself and there is nothing else for it to interfere with. However, experim ent shows that the fringes will still be form ed, built up slowly as electron after electron falls on screen B. Even under these conditions the associated wave always passes through both slits, and it is w hat determ ines where the electrons are likely to fall on screen B. To get a better idea o f the role o f the wave in the m otion o f a particle, consider Fig. 27, in which a particle (an electron, say) is generated at point 7 and detected at point F. How does it travel this straight-line path? The q u an tu m answer is that the wave explores all possi ble paths, as the figure suggests, assigning an equal proba bility to each. However, only for the straight line connect ing the two points do the waves add constructively, yielding a high probability that the particle will be found there if sought. For points not near this straight line, the waves cancel each other by destructive interference, the cancellation being m ore severe the m ore massive the par ticle. It is in this way that the trajectories o f particles in N ew tonian m echanics are related to their associated waves. It is not hard to im agine the effect o f inserting a double slit in Fig. 27 between I and F. In the process o f exploring
Figure 27 An electron moves from I to F. The waves that describe its journey interfere constructively along the straight path and destructively along all other paths. The wave ex plores all possible paths between I and F.
Questions
the possible paths, only those waves passing through the slits survive. The particle is more likely to be found in
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regions on the screen of high probability density, and in this way the interference pattern is formed.
QUESTIONS 1. How can the wavelength of an electron be given by A = h/pl Doesn’t the very presence of the momentum p in this for mula imply that the electron is a particle? 2. In a repetition of Thomson’s experiment for measuring e/m for the electron (see Section 34-2), a beam of electrons is collimated by passage through a slit. Why is the beamlike character of the emerging electrons not destroyed by diffrac tion of the electron wave at this slit? 3. Why is the wave nature of matter not more apparent in our daily observations? 4. Considering the wave behavior of electrons, we should ex pect to be able to construct an “electron microscope” using short-wavelength electrons to provide high resolution. This, indeed, has been done, (a) How might an electron beam be focused? (b) What advantages might an electron microscope have over a light microscope? (c) Why not make a proton microscope? A neutron microscope? 5. How many experiments can you recall that support the wave theory of light? The particle theory of light? The wave theory of matter? The particle theory of matter? 6. Is an electron a particle? Is it a wave? Explain your answer, citing relevant experimental evidence. 7. If the particles listed below all have the same energy, which has the shortest wavelength: electron; a particle; neutron; proton? 8. What common expression can be used for the momentum of either a photon or a particle? 9. Discuss the analogy between (a) wave optics and geometri cal optics and (b) wave mechanics and classical mechanics. 10. Does a photon have a de Broglie wavelength? Explain. 11. Discuss similarities and differences between a matter wave and an electromagnetic wave. 12. Can the de Broglie wavelength associated with a particle be smaller than the size of the particle? Larger? Is there any relation necessarily between such quantities? 13. If, in the de Broglie formula A = h/mv, we let m —►<» , do we get the classical result for particles of matter? 14. Considering electrons and photons as particles, how are they different from each other? 15. Is Eq. 1 for the de Broglie wavelength, A = h/p, valid for a relativistic particle? Justify your answer. 16. How could Davisson and Germer be sure that the “54-V” peak of Fig. 7 was a first-order diffraction peak, that is, that m = 1 in Eq. 2? 17. Do electron diffraction experiments give different informa tion about crystals than can be obtained from x-ray diffrac tion experiments? From neutron diffraction experiments? Give examples. 18. Why are the hydrogen atoms clearly visible in Fig. 10 but not in Fig. 17 of Chapter 47?
19. In Fig. 9b (made with x rays) the diffraction circles are speckled, but in Fig. 9c (made with electrons) they are smooth. Can you explain why? 20. Electromagnetic waves will penetrate seawater to a certain extent if their frequency is low enough. This is the basis of one plan to communicate with submerged submarines. A difficulty with this plan is that the lower the frequency, the longer the time it takes to transmit a message (in Morse code pulses, say). Can you explain why this should be? 21. Why is the Heisenberg uncertainty principle not more read ily apparent in our daily observations? 22. (a) Give examples of how the process of measurement dis turbs the system being measured, {b) Can the disturbances be taken into account ahead of time by suitable calcula tions? 23. You measure the pressure in a tire, using a pressure gauge. The gauge, however, bleeds a little air from the tire in the process, so that the act of measuring changes the property that you are trying to measure. Is this an example of the Heisenberg uncertainty principle? Explain. 24. “The energy of the ground state of an atomic system can be precisely known, but the energies of its excited states are always subject to some uncertainty.” Can you explain this statement on the basis of the uncertainty principle? 25. “If an electron is localized in space, its momentum becomes uncertain. If it is localized in time, its energy becomes un certain.” Explain this statement. 26. The quantity ^(x), the amplitude of a matter wave, is called a wavefunction. What is the relationship between this quan tity and the particles that form the matter wave? 27. In Section 50-7 we solved the wave mechanical problem of a particle trapped in an infinitely deep well without ever using (or even writing down) Schrodinger’s equation. How were we able to do that? 28. A standing wave can be viewed as the superposition of two traveling waves. Can you apply this view to the problem of a particle confined between rigid walls, giving an interpreta tion in terms of the motion of the particle? 29. The allowed energies for a particle confined between rigid walls are given by Eq. 16. First, convince yourself that, as n increases, the energy levels become farther apart. How can this possibly be? The correspondence principle would seem to require that they move closer together as n increases, approaching a continuum. 30. How can the predictions of wave mechanics be so exact if the only information we have about the positions of the electrons in atoms is statistical? 31. In the « = 1 state, for a particle confined between rigid walls, what is the probability that the particle will be found in a small-length element at the surface of either wall?
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The Wave Nature o f Matter
32. What are the dimensions of P^ix) in Fig. 18? What is the value of the classically expected probability density, repre sented by the horizontal lines? What value do the areas under the curves have? How does the area under any curve compare with the area under the horizontal line? All these questions can be answered by inspection of the figure. 33. In Fig. 18 what do you imagine the curve for F„(x) for /2 = 100 looks like? Convince yourself that these curves ap proach classical expectations as « —► . 34. We have seen that barrier tunneling works for matter waves and for electromagnetic waves. Do you think that it also works for water waves? For sound waves? 35. Comment on the statement: “A particle can’t be detected while tunneling through a barrier, so that it doesn’t make sense to say that such a thing actually happens.” 36. List examples of barrier tunneling occurring in nature and in manufactured devices. 37. A proton and a deuteron, each having 3 MeV of energy, attempt to penetrate a rectangular potential barrier of height 10 MeV. Which particle has the higher probability of suc ceeding? Explain in qualitative terms. 38. A laser projects a beam of light across a laboratory table. If you put a diffraction grating in the path of the beam and observe the spectrum, you declare the beam to be a wave. If instead you put a clean metal surface in the path of the beam and observe the ejected photoelectrons, you declare this same beam to be a stream of particles (photons). What can you say about the beam if you don’t put anything in its path?
39. State and discuss (a) the correspondence principle, {b) the uncertainty principle, and (c) the complementarity prin ciple. 40. In Fig. 25, why would you expect the electrons from each slit to arrive at the screen over a range of positions? Shouldn’t they all arrive at the same place? How does your answer relate to the complementarity principle? 41. Several groups of experimenters are trying to detect gravity waves, perhaps coming from our galactic center, by measur ing small distortions in a massive object through which the hypothesized waves pass. They seek to measure displace ments as small as 10“^' m. (The radius of a proton is - 1 0 “ *^ m, a million times larger!) Does the uncertainty principle put any restriction on the precision with which this measurement can be carried out? 42. Figure 18 shows that for « = 3 the probability function P„(x) for a particle confined between rigid walls is zero at two points between the walls. How can the particle ever move across these positions? (Hint: Consider the implications of the uncertainty principle.) 43. In Sample Problem 8, the electron’s energy is determined exactlyhy the size of the box. How do you reconcile this with the fact that the uncertainty in the location of the electron cannot exceed 1(X) pm and, if the uncertainty principle is to be obeyed, the electron’s momentum must be correspond ingly uncertain?
PROBLEMS Section S0~2 The De Broglie Wavelength 1. A bullet of mass 41 g travels at 960 m/s. (a) What wavelength can we associate with it? (Jb) Why does the wave nature of the bullet not reveal itself through diffraction effects? 2. Using the classical relation between momentum and kinetic energy, show that the de Broglie wavelength of an electron can be written {a) as 1.226 nm
A= -
4k
in which K is the kinetic energy in electron volts, or (b) as
where Ais in nm, and V is the accelerating potential in volts. (Use the best values of the needed constants as found in Appendix B.) 3. Calculate the wavelength of a 1.00-keV {a) electron, (b) pho ton, and (c) neutron. 4. The wavelength of the yellow spectral emission line of so dium is 589 nm. At what kinetic energy would an electron have the same de Broglie wavelength? 5. If the de Broglie wavelength of a proton is 0.113 pm, (a) what is the speed of the proton and {b) through what electric potential would the proton have to be accelerated from rest to acquire this speed?
6. Singly charged sodium ions are accelerated through a po tential difference of 325 V. (a) What is the momentum acquired by the ions? (^) Calculate their de Broglie wave length. 7. The existence of the atomic nucleus was discovered in 1911 by Ernest Rutherford, who properly interpreted some ex periments in which a beam of alpha particles was scattered from a foil of atoms such as gold, (a) If the alpha particles had a kinetic energy of 7.5 MeV, what was their de Broglie wavelength? (b) Should the wave nature of the incident alpha particles have been taken into account in interpreting these experiments? The distance of closest approach of the alpha particle to the nucleus in these experiments was about 30 fm. (The wave nature of matter was not postulated until more than a decade after these crucial experiments were first performed.) 8. The highest achievable resolving power of a microscope is limited only by the wavelength used; that is, the smallest detail that can be separated is about equal to the wavelength. Suppose one wishes to “see” inside an atom. Assuming the atom to have a diameter of 1(X) pm this means that we wish to resolve detail of separation about 10 pm. (a) If an electron microscope is used, what minimum energy of electrons is needed? (b) If a light microscope is used, what minimum energy of photons is needed? (c) Which microscope seems more practical for this purpose? Why?
Problems 9. The 32-GeV electron accelerator at Stanford provides an electron beam of small wavelength, suitable for probing the fine details of nuclear structure by scattering experiments. What is this wavelength and how does it compare with the size of an average nucleus? (At these energies it is sufficient to use the extreme relativistic relationship between momen tum and energy; namely, p = E/c. This is the same relation ship used for light and is justified when the kinetic energy of a particle is much greater than its rest energy, as in this case. The radius of a middle-mass nucleus is about 5.0 fm.) 10. Consider a balloon filled with (monatomic) helium gas at 18X and 1.0 atm pressure. Calculate (a) the average de Broglie wavelength of the helium atoms and (b) the average distance between the atoms. Can the atoms be treated as particles under these conditions? 11. A nonrelativistic particle is moving three times as fast as an electron. The ratio of their de Broglie wavelengths, particle to electron, is 1.813 X lO”"*. By calculating its mass, identify the particle. See Appendix B. 12. (a) A photon in free space has an energy of 1.5 eV and an electron, also in free space, has a kinetic energy of that same amount. What are their wavelengths? (b) Repeat for an en ergy of 1.5 GeV. 13. In an ordinary color television set, electrons are accelerated through a potential difference of 25.0 kV. Find the de Bro glie wavelength of such electrons (a) using the classical ex pression for momentum and (b) taking relativity into ac count. 14. What accelerating voltage would be required for electrons in an electron microscope to obtain the same ultimate resolv ing power as that which could be obtained from a gammaray microscope using 136-keV gamma rays? (Hlnl: See Problem 8.) Section 50~3 Testing De Broglie*s Hypothesis 15. A neutron crystal spectrometer utilizes crystal planes of spacing d — lZ .l pm in a beryllium crystal. What must be the Bragg angle B so that only neutrons of energy K = 4.2 eV are reflected? Consider only first-order reflections. 16. A beam of thermal neutrons from a nuclear reactor falls on a crystal of calcium fluoride, the beam direction making an angle Q with the surface of the crystal. The atomic planes parallel to the crystal surface have an interplanar spacing of 54.64 pm. The de Broglie wavelength of neutrons in the incident beam is 11 .(X) pm. For what values of ^ will the first three orders of Bragg-reflected neutron beams occur? {Hint. Neutrons, which carry no charge and are thus not subject to electrical forces, are not refracted as they pass through a crystal surface. Thus neutron diffraction can be treated in strict analogy with x-ray diffraction.) 17. In the experiment of Davisson and Germer (a) at what angles would the second- and third-order diffracted beams corresponding to a strong maximum in Fig. 7 occur, pro vided they are present? (b) At what angle would the firstorder diffracted beam occur if the accelerating potential were changed from 54 to 60 V? 18. A potassium chloride (KCI) crystal is cut so that the layers of atomic planes parallel to its surface have a spacing of 314 pm between adjacent lines of atoms. A beam of 380-eV elec
1067
trons is incident normally on the crystal surface. Calculate the angles 0 at which the detector must be positioned to record strongly diffracted beams of all orders present. 19. A beam of low-energy neutrons emerges from a reactor and is diffracted from a crystal. The kinetic energies of the neu trons are contained in a band of width AK centered on kinetic energy K. Show that the angles for a given order of diffraction are spread over a range A6 given in degrees by
where 6 is the diffraction angle for a neutron with kinetic energy K. 20. A beam of atoms emerges from an oven that is at a tempera ture T. The distribution of the speeds of the atoms in the beam is proportional to (see Section 24-3). (a) Show that the distribution of de Broglie wavelengths of the atoms is proportional to and (b) that the most probable de Broglie wavelength is y/SmkT Section 50~4 Waves, Wave Packets, and Particles 21. Using a rotating shutter arrangement, you listen to a 540-Hz standard tuning fork for 0.23 s. What approximate spread of frequencies is contained in this acoustic pulse? 22. The signal from a television station contains pulses of full width Ar 10 ns. Is it feasible to transmit television in the AM broadcasting band, which runs from about 500 to 1600 kHz? Section SOS Heisenberg*s Uncertainty Relationships 23. A nucleus in an excited state will return to its ground state, emitting a gamma ray in the process. If its mean lifetime is 8.7 ps in a particular excited state of energy 1.32 MeV, find the uncertainty in the energy of the corresponding emitted gamma-ray photon. 24. An atom in an excited state has a lifetime of 12 ns; in a second excited state the lifetime is 23 ns. What is the uncer tainty in energy for a photon emitted when an electron makes a transition between these two states? 25. A microscope using photons is employed to locate an elec tron in an atom to within a distance of 12 pm. What is the minimum uncertainty in the momentum of the electron located in this way? 26. Imagine playing baseball in a universe where Planck’s con stant was 0.60 J*s. What would be the uncertainty in the position of a 0.50-kg baseball moving at 20 m/s with an uncertainty in velocity of 1.2 m/s? Why would it be hard to catch such a ball? 27. Find the uncertainty in the location of a particle, in terms of its de Broglie wavelength A, so that the uncertainty in its velocity is equal to its velocity. Section 50-7 Trapped Particles and Probability Densities 28. What must be the width of an infinite well such that a trapped electron in the « = 3 state has an energy of 4.70 eV?
1068
Chapter 50
The Wave Nature o f Matter
29. (a) Calculate the smallest allowed energy of an electron con fined to an infinitely deep well with a width equal to the diameter of an atomic nucleus (about 1.4 X 10“ m). (b) Repeat for a neutron, (c) Compare these results with the binding energy (several MeV) of protons and neutrons in side the nucleus. On this basis, should we expect to find electrons inside nuclei? 30. The ground-state energy of an electron in an infinite well is 2.6 eV. What will the ground-state energy be if the width of the well is doubled? 31. An electron, trapped in an infinite well of width 253 pm, is in the ground (n = \) state. How much energy must it ab sorb to jum p up to the third excited (n = 4) state? 32. (a) Calculate the fractional difference between two adjacent energy levels of a particle confined in a one-dimensional well of infinite depth, (b) Discuss the result in terms of the correspondence principle. 33. (a) Calculate the separation in energy between the lowest two energy levels for a container 20 cm on a side containing argon atoms, (b) Find the ratio with the thermal energy of the argon atoms at 300 K. (c) At what temperature does the thermal energy equal the spacing between these two energy levels? Assume, for simplicity, that the argon atoms are trapped in a one-dimensional well 20 cm wide. The molar mass of argon is 39.9 g/mol. 34. Consider a conduction electron in a cubical crystal of a conducting material. Such an electron is free to move throughout the volume of the crystal but cannot escape to the outside. It is trapped in a three-dimensional infinite well. The electron can move in three dimensions, so that its total energy is given by (compare with Eq. 16), E=
h^
in which «2, each take on the values 1, 2, . . . . Calculate the energies of the lowest five distinct states for a conduction electron moving in a cubical crystal of edge length L = 250 nm. 35. Consider an electron trapped in an infinite well whose width is 98.5 pm. If it is in a state with « = 15, what are {a) its energy? {b) The uncertainty in its momentum? (c) The un certainty in its position? 36. Repeat Sample Problem 11, but assume now that the elec tron is in the « = 2 state.
37. Where are the points of (a) maximum and (b) minimum probability for a particle trapped in an infinitely deep well of length L if the particle is in the state nl 38. A particle is confined between rigid walls separated by a distance L. (a) Show that the probability P that it will be found within a distance L/3 from one wall is given by j_ / 3\
sin(27T/2/3)\ Inn/Z / •
Evaluate the probability for {b )n = \,(c )n = 2, (d) n = 3. and (e) under the assumption of classical physics. 39. A particle is confined between rigid walls located at x = 0 and X = L. For the « = 4 energy state, (a) sketch the proba bility density curve for the particle’s location. Calculate the approximate probabilities of finding the particle within a region Ax = 0.(XX)3L when (b) A x is located at x = L/8 and (c) at X = 3 L /16. Refer to your figure to see whether or not your results seem reasonable. (Hint: No integration is neces sary.) Section 50-8 Barrier Tunneling 40. In Sample Problem 12, suppose that you can vary the thick ness L of the barrier. To what value should the thickness be adjusted so that 1 electron out of 100 striking the barrier will tunnel through it? 41. (a) A proton and (b) a deuteron (which has the same charge as a proton but twice the mass) are incident on a barrier of thickness 10 fm and height 10 MeV. Each particle has a kinetic energy of 3.0 MeV. Find the transmission probabili ties for them. 42. Consider a barrier such as that of Fig. 21, but whose height U is 6.(X) eV and whose thickness L is 700 pm. Calculate the energy of an incident electron such that its transmission probability is 1 in 1000. 43. Suppose that an incident beam of 5.0-eV protons fell on a barrier of height 6.0 eV and thickness 0.70 nm, and at a rate equivalent to a current of 1.0 kA. How long would you have to wait— on the average— for one proton to be transmit ted? 44. Consider the barrier tunneling situation defined by Sample Problem 12. What fractional change in the transmission coefficient occurs for a 1% increase in (a) the barrier height, (b) the barrier thickness, and (c) the incident energy of the electron?
CHAPTER 51 THE STRUCTURE OF ATOMIC HYDROGEN Ever since it has been known that matter is made up o f atoms, the fundamental question has been: ''What is an atom like?" Our aim in this chapter is to answer this question from the point o f view o f wave mechanics. Understanding the structure o f atoms is essential i f we hope to understand how atoms join to form molecules and solids. Chemistry and solid-state physics both depend on knowledge o f atomic structure acquired from wave mechanics. We start in this chapter with hydrogen, which is both the simplest atom and the most abundant atom in the universe. Understanding how the principles o f wave mechanics account for the structure o f hydrogen leads us to apply similar considerations to explaining the structure o f more complex atoms, which we do in the next chapter. Because o f its simplicity, hydrogen has the advantage that its properties can be calculated exactly and without approximation, which has permitted comparison between prediction and experiment for a variety o f physical theories from quantum mechanics in the 1920s to quantum electrodynamics in the 1940s and 1950s.
51-1 THE BOHR THEORY__________ Most of our knowledge about atoms, molecules, and nu clei comes from studying the radiation emitted or ab sorbed by them, as we illustrated by the line spectra in Fig. 15 o f Chapter 49. This also is the case with atomic hydro gen. A spectrum of atomic hydrogen in the visible region is illustrated in Fig. 1. This spectrum, which might be obtained with a prism or diffraction grating in a spectro graph such as that of Fig. 8 of Chapter 47, had been meas ured with great precision in the late 1800s, and its inter pretation was puzzling for scientists of that era. The initial approach in analyzing this spectrum was to find an empir ical formula that fit the data. It took another 30 years for a theory to be developed that could explain the formula. The spectrum in Fig. 1 shows several regularities. The spacing o f the lines decreases as we go to shorter wave lengths, while the wavelengths themselves approach a limit called the series limit. An empirical formula for the wavelengths o f the lines of atomic hydrogen was devel oped in 1885 by Johannes Balmer, a Swiss high school teacher. Balmer’s formula for the wavelength A in nano meters is
A = 364.6 72 _ 4 ’
« = 3, 4, 5,
( 1)
This series of lines of hydrogen in the visible region is called the Balmer series. In 1890, J. J. Rydberg modified Balmer’s formula and wrote it as "-3,4,5,....
( 2)
where R, called the Rydberg constant, has a value of 1.097 X 10^ m“ Recognizing that 4 can be written as 2^, Rydberg rewrote the formula in a more general form as (3)
n = m + 1, m-I-2, w + 3, . . . , where w = 2 for the Balmer series. The obvious question that occurred was whether there were other series o f lines, corresponding to other fixed values o f m. Soon searchers turned up a series in the infrared corresponding to m = 3 and another in the ultraviolet with m = 1. All these series could be fit by Eq. 3 (called the Balmer-Rydberg for mula) with a given value of m and a series of values of n
1069
1070
Chapter 51
Designation of line
The Structure o f Atomic Hydrogen
^
Hy
®
H,d
H < H,f
CsJ X(nm)
VO
Red
Blue
00
VD
Color
6
viT O )
o>
CO
CO
00 00
ro
CO
Near ultraviolet
Violet
Figure 1 A photograph of the spectral lines of the Balmer series in hydrogen.
starting with m + 1 and ending with the series limit as « —» oo. Figure 2 shows the series for m = 1 (called the Lyman series), m = 2 (the Balmer series), and w = 3 (the Paschen series). The wavelengths of some of these lines are listed in Table 1. The key to understanding this empirical formula was provided by the Danish physicist Niels Bohr in 1913. After completing his Ph.D., Bohr went to England, where he worked first with J. J. Thomson and then with Ernest Rutherford (see the discussions of the Thomson and Rutherford models of the atom in Section 29-7). Bohr immediately recognized the importance o f the Ruther ford nuclear atom in understanding the structure of atoms. He was led to propose a model in which the elec tron circulates about the nucleus like a planet about the Sun (Fig. 3). However, he recognized that such a model would violate one of the predictions of classical physics, namely, that an accelerated electron (even centripetally accelerated) would emit a continuous spectrum of radia tion as it loses energy and spirals into the nucleus. Clearly this does not happen; if Bohr’s planetary structure of the atom is correct, the classical physics o f Newton and Max well must be suspended! (Keep in mind that Bohr’s work occurred 10 years before de Broglie’s bold hypothesis of matter waves.) Realizing that classical physics had come to a dead end on the hydrogen atom problem, Bohr put forward two bold postulates. Both turned out to be enduring features that carry over in full force to our modem point of view.
Lyman series
II III 1
Balmer series
III 1 200
.......L .. ... 400
1. The postulate of stationary states. Bohr assumed that the hydrogen atom can exist for a long time without ra diating in any one of a number of stationary states of well-defined energy. This assumption contradicts classi cal theory, but Bohr’s attitude was: “Let’s assume it any way and see what happens.’’ Note that this postulate says nothing at all about what these states look like. There is, for example, no mention o f orbits. 2. The frequency postulate. Bohr assumed that the hy drogen atom can emit or absorb radiation only when the atom changes from one of its stationary states to another. The energy of the emitted (or absorbed) photon is equal to the difference in energy between these two states. Thus if an atom changes from an initial state o f energy E„ to a final state of (lower) energy E„ , the energy o f the emitted photon is given by
hv„„ = E„ —E„ (Bohr’s frequency postulate). (4) This postulate ties together two new ideas (the photon hypothesis and energy quantization) with one familiar old idea (the conservation of energy). Bohr now sought to interpret the empirical BalmerRydberg formula in terms o f his postulates. We start by
Paschen series
1 600
Moreover, both turned out to be quite general, applying not only to the hydrogen atom but to atomic, molecular, and nuclear systems of all kinds. We discuss each postu late in turn.
■111' 1 B||l , 1 .1: 800
..J ..... .. i 1000 1200 Wavelength (nm)
1
^ ..... i ..... 1400
j ....1.... 1600
i 1800
! 2000
Figure 2 The Lyman, Balmer, and Paschen series of atomic hydrogen. The series limit is at the short wavelength (left) end of each series.
Section 5 1-1
The Bohr Theory
1071
TABLE 1 THE HYDROGEN SPECTRUM (SOME SELECTED LINES) Quantum Number
Name o f Series
n (Upper State)
m (Lower State)
Lyman
Balmer
Paschen
1 1 1 1
2 3 4
2 2 2 2
3 4 5
3 3 3 3
4 5 6
/
121.6 102.6 97.0 91.2
00 (series limit)
656.3 486.1 434.1 364.6
00 (series limit)
1875.1 1281.8 1093.8 822.0
00 (series limit)
recasting this formula (Eq. 3) in the general format of Bohr’s frequency postulate (Eq. 4). If we multiply each side o f Eq. 3 by he and if we replace c/A by v„„, we can write .
Wavelength (nm)
hcR\
(
hcR\
Equation 5 is not yet the end o f the path that Bohr followed. The value of the Rydberg constant in that formula— at this stage— can be found only from experi ment. What is needed is a way o f expressing this constant in terms of other, known physical constants. This, as we shall see, is precisely what Bohr did next.
A term by term comparison with Eq. 4 allows us to infer 17
hcR n
, - -
for the energies of the stationary states of the hydrogen atom. The energy is negative because the atom is in a bound state; that is, work must be done by some external agent to pull it apart. (The potential energy, which is zero at infinite separation of the proton and electron, is nega tive and larger in magnitude than the kinetic energy.) In the same way, the Earth-Sun system is a bound state; work must be done by an external agent to tear this system apart against the gravitational force that holds it together. Figure 4 shows an energy level diagram for the hydro gen atom, the energies being calculated from Eq. 5. Each level is marked with its quantum number n. A downward pointing arrow connecting two levels represents the emis sion of a photon, in accord with Bohr’s frequency postu late (Eq. 4). Table 1 displays the wavelengths of some of the lines shown in this figure.
Sample Problem 1 Calculate the binding energy of the hydro gen atom, that is, the energy that must be added to the atom to remove the electron from its lowest energy state. Solution The energy of the atom when the electron has been removed from it, found by letting « —►» in Eq. 5, is zero. The Series limit
\
Series limit
\
Series limit
\
O i- -
Paschen series
Balmer series
-6
>
i
h
5
-8h
-1 0 -
-
12-
11 14
Figure 3 In the Bohr model of the hydrogen atom, the elec tron moves in a circular orbit about the central proton.
Lyman series
Figure 4 Energy levels and transitions in the spectrum of atomic hydrogen. Compare with the spectral lines represented in Fig. 2.
1072
Chapter 5 1
The Structure o f Atomic Hydrogen
binding energy Ey, is therefore numerically equal to the energy of the atom in its lowest energy state, found by putting n = \ in Eq. 5. That is,
= (6.63X 10-^ J*s)(3.00X 10« m /s)( 1.097 X 10^ m "') = 2.18X 10-'« J = 13.6 eV. This calculated value agrees with the experimentally observed binding energy for the hydrogen atom.
Sample Problem 2 (a) What is the wavelength of the least energetic photon in the Balmer spectrum? (b) What is the wave length of the series limit for the Balmer series? Solution (a) We identify the Balmer series (see Fig. 4) by put ting m = 2 in Eq. 3. From the relation E = hv, the least energetic photon has the smallest frequency and thus the greatest wave length. This means that we must put n = 3 (the smallest possible value) in Eq. 3; any higher value of n would yield a smaller wavelength. With these substitutions we have
3. For generality, we take the central charge to be Ze rather than e, where Z is the atomic number, Z = 1 iden tifying hydrogen. We assume further that M :> m^, where M is the nuclear mass and m* is the mass o f the orbiting electron. Combining Coulomb’s force law with Newton’s second law gives 1 (Zete) 2 = '” . 7 . 47T€o
(6)
in which v is the sjjeed of the electron in its orbit. Solving for Vyields I Z e^ y A neom ^r’’
v(r)
(7)
which tells us the orbital speed if we know the orbit radius. From this result we can write an expression for the frequency of revolution o f the electron in its orbit: Kr) = _ L =
2nr
[.
Z e^
V 116;r^€o
( 8)
'
The kinetic energy follows from
= (1.097 X 10’
(9)
1-524 X 10‘ m - ‘
The potential energy is given by
or A= 6.563 X 10 ^ m = 656.3 nm. (b) Again we put w = 2 in Eq. 3. To find the series limit (see Fig. 4) we let « Equation 3 then becomes j = (1.097 X 10’ m -')
743 X 10^ m - ‘
or A= 3.646 X 10“ ^ m = 364.6 nm.
Z e^ U(r) = - - ^
(10)
so that the total mechanical energy E(r) follows from
E{r) = K(r) A- U(r) = -
Z e^ 87t€or
( 11)
Finally, the angular momentum follows directly from Eq. 7:
Note that both of these numerical results appear in Table 1.
Derivation of the Bohr Theory So far everything we have done has been empirical, that is, based on measured values rather than on derived values. Our goal now is to derive an expression that gives the Rydberg constant or, equivalently, the energy levels (Eq. 5). We shall do this following Bohr’s calculation by invoking the correspondence principle (see Section 50-9): the classical theory (which holds for macroscopic orbits) and the quantum theory must agree where they overlap in the region o f large quantum numbers (large values o f n). We begin by analyzing the properties of an atom, such as that o f Fig. 3, using classical principles. We shall then compare the results with those o f the quantum calcula tion in the limit o f large n. Let us apply Newton’s second law {F = m^a) to the motion o f the electron in the classical orbit shown in Fig.
Thus, if we knew the orbit radius, we could find the orbital linear sp>eed, the frequency of revolution, the ki netic energy, the potential energy, the total mechanical energy, and the angular momentum. We see from their interconnections that if any one o f these quantities turns out to be quantized, all of them will be. There is, however, no quantization o f anything in these purely classical cal culations. We continue by eliminating the radius r between Eqs. 8 and 11 to find a relation between the frequency and the energy: _ / i
3 2 e lE ^ Y ^ Z X ^V
’
in which we have added the subscript cm to remind us that this expression is derived on the basis o f classical mechanics.
Section 51-1
Substituting for E from Eq. 5 gives an expression for the frequency calculated from classical mechanics in the region o f large quantum numbers: =
\
Z'^m ^e^
(14)
)
In classical physics, this frequency of revolution is also the frequency o f the emitted radiation. We turn now to the quantum point of view. In quan tum terms (that is, now using Eq. 4, the second quantum postulate), the frequency Vq„ that corresponds to the clas sical frequency we have just calculated is the lowest emit ted frequency, which is associated with a transition from a state with quantum number n to the next lower state, whose quantum number is n — 1. Putting m = n — \ in Eq. 3 gives 1
1\
A
« (2 'J - 1)
nV
^
This expression should agree with the classical expres sion in the limit o f large quantum numbers. When n » 1, Eq. 15 can be written
2cR
.
for«:»l,
(16)
which is the relationship we seek. We are ready at last to apply the correspondence princi ple. This principle tells us that, in the limit o f large quan tum numbers, the frequency calculated from Eq. 16 (a quantum expression) must equal the frequency v^n, calculated from Eq. 14 (a classical expression). Table 2 shows this principle in action. Equating Eqs. 14 and 16 and solving for the Rydberg constant R, we find
R=
n tfZ ^e * S elh ^c ’
(17)
a theoretically predicted value for the Rydberg constant in
TABLE 2
Quantum Number n 2 5 10 50 100 1,000 10,000 25,000 100,000
Frequency o f Transition to Next Lowest State Vo™(Hz)
8.22 X 5.26 X 6.58 X 5.26 X 6.580 X 6.5797 X 6.5797 X 4.2110 X 6.5798
24.7 X 7.40 X 7.72 X 5.43 X 6.680 X 6.5896 X 6.5807 X 4.2113 X 6.5799
10'“ 10’’ 10'2 10'® 10’ 10‘ 10^ 10^
lO” 10*’ 10'" 10'® 10’ 10‘ 10’ 10"
1073
terms of other fundamental constants: the charge e and the mass of the electron, the speed c of light, and the Planck constant h. Bohr, using data available in his time for these constants, obtained good agreement with the experimentally determined value of R, the agreement today being within extremely narrow limits of experimen tal error. We can now regard the constant R as theoretically de termined and, by substituting Eq. 17 into Eq. 5, obtain E = —
m ^Z ^e* 1
(18)
a purely quantum expression for the energies o f the sta tionary states of the hydrogen atom. This expression is Bohr’s triumph. Everything that he has done so far, in cluding the postulate of stationary states, the frequency postulate, the correspondence principle, and Eq. 18, the expression for the energy of the hydrogen atom states, carries over unchanged into modem quantum mechan ics. By eliminating the energy E between the classical (Eq. 11) and the quantum (Eq. 18) expressions, we can find the radii of the quantized Bohr orbits. They are given by = Z _ fo ^ V
Ze^nm^
« = 1 , 2 , 3 , -------
(19)
The quantity Oq* called the Bohr radius, has the value Oo =
€oh^ = 5.292 X 10“ " m = 52.92 pm. Z e^nm ^
In a formal sense, Oq is the radius of the Bohr orbit corresponding to n = 1, which defines the ground state of the hydrogen atom in Bohr’s semiclassical planetary model o f the one-electron atom, where we visualize the electron moving in planetary orbits. Today we do not believe in such orbits but, based on experiment, we do have some notion o f the size o f the atoms. They are all
THE CORRESPONDENCE PRINCIPLE AND THE HYDROGEN ATOM Frequency o f Revolution in Orbit Vent (Hz)
The Bohr Theory
Difference (%) 67 29 15 3.1 1.5 0.15 0.015 0.007 0.0007
1074
Chapter 51
The Structure o f Atomic Hydrogen
roughly of the order of magnitude of the Bohr radius! It is amazing that, although Bohr put no specific assumption into his theory concerning the size of atoms, it neverthe less generated a number that gave just about the right size. Today we use the Bohr radius as a convenient unit in which to measure lengths on the scale of atomic dimen sions. The fact that the energy (see Eq. 18) and the radius (see Eq. 19) o f the Bohr semiclassical atom are quantized means that other mechanical properties are also quan tized in his planetary model. The quantization of the an gular momentum of the orbiting electron turns out to be particularly simple. It is (see Problem 23)
L„ =
= nh,
n = 1,2,3,
,
(20)
in which we have written h (pronounced “h-bar”) as a convenient shorthand for h/ln. In 1924, 11 years after Bohr presented his theory, de Broglie gave a satisfying physical interpretation of the Bohr rule for the quantization of angular momentum. If we represent the circulating electron in terms o f its de Broglie wave, then the stationary states are those in which the electron’s de Broglie wave joins onto itself with the same phase after each revolution; otherwise, the wave would destroy itself by destructive interference. Put an other way, the de Broglie wavelength must fit around the circumference o f the orbit an integral number of times, or
nX = 2nr,
= 1, 2, 3, . . . ,
as suggested by Fig. 5. Substituting h/p for the de Broglie wavelength in this expression leads directly to Eq. 20. Like the Bohr model itself. Fig. 5 is not consistent with modem quantum theory. Although the quantization of angular momentum plays a central role, it differs some what from Eq. 20. For the ground state of the hydrogen atom, for example, this equation predicts L = h \ modem quantum theory, on the other hand, predicts L = 0, in agreement with experiment. It is to Bohr’s credit that he foresaw the cmcial importance of angular momentum quantization and, indeed, proposed Eq. 20 as an alterna tive basic hypothesis from which his theory could be devel oped; see Problem 27.
Figure 5 A Bohr orbit with the electron represented as a de Broglie wave.
51-2 TH E HYDROGEN ATOM AND SCHRODINGER’S EQUATION____________________ The Bohr theory was surprisingly successful in analyzing the radiations emitted by hydrogen, but it is a very incom plete theory. For example, it doesn’t provide any basis to calculate which among the many permitted radiations are more likely to be emitted, nor does it provide us with the information we need to understand how hydrogen forms molecular bonds with other atoms. To obtain a complete analysis, we must use methods of wave mechanics. As we discussed in Chapter 50, the proper treatment of the electron in any particular dynamical situation must take into account its wave nature. In the case o f an elec tron confined to a region in which no force acts on it (Section 50-7), we saw that the wave behavior was similar to that of a classical standing wave on a string. In the case of an electron subject to a force, especially a force that varies with location, the wave behavior is more compli cated, as the classical string would be if the tension varied with location along its length. To analyze the wave behavior of the electron, we re quire a mathematical procedure in which we can specify the interaction of the electron with its environment and then solve for its motion. This is o f course just what we did in classical physics using Newton’s laws, in which the interactions were described in terms o f forces. The wave-mechanical procedure for studying the be havior o f electrons (and other particles) is based on an equation proposed by the Austrian physicist Erwin Schrodinger (1887-1961) in 1926, just 2 years after de Broglie’s hypothesis concerning matter waves. Schrodinger’s equation, which we shall not present in detail, is for wave mechanics what Newton’s second law is for classical mechanics. We begin by specifying the inter action o f the particle with its environment, which we do in terms o f potential energy rather than force. (The two de scriptions are of course equivalent, as suggested by the one-dimensional expression F = —dU/dx; we can find the force from the potential energy or the potential energy from the force.) We then carry out the mathematical pro cedure specified by Schrddinger’s equation, and the re sults include the wave functions that describe the particle, the quantized energy levels that the particle is permitted to occupy, and a set of quantum numbers that specify the allowed states of motion of the particle. Figure 6 repre sents this procedure symbolically. The box in Fig. 6 might in fact represent a computer, for we currently solve most quantum-mechanical problems of practical interest on computers using numerical methods. The wave functions corresponding to the allowed states of motion encompass all the information about the behav ior o f the particle that can be known. Using those wave
Section 51-2 t Wave functions
I Quantum numbers
Output I
Energies
The Hydrogen Atom and Schrodinger’s Equation
1075
Figure 6 A schematic representation of SchrSdinger’s wave equation as a “machine” in which the potential energy func tion must be supplied as input, and the output consists of the wave functions, quantum numbers, and energy levels that characterize the quantum behavior of the system.
Potential energy function
functions, we can calculate anything we can know about the particle. In the case o f the hydrogen atom, we can use the wave functions resulting from Schrodinger’s equation to find the mean radius of the atom, the probability to find the electron at any specified location, the probability for the electron to make a transition from any specified initial state to any specified final state (emitting or absorbing a photon in the process), the magnetic moment of the atom, and so on. By combining two wave functions, we can even study bonds formed between the two atoms in molecular hydrogen, H 2 . The potential energy that serves as our starting point results from the Coulomb force between the electron and the proton: t/(r) = -
1 4 tco r ■
(21)
Figure 7 is a plot of this familiar potential energy function on a scale appropriate for an atom of hydrogen. We spec ify the distance between the electron and proton in terms o f the Bohr radius Cq defined in Eq. 19. We shall not discuss the mathematical procedure for finding the wave functions,* examples of which are given in Sections 51-7 and 51-8. The energy levels that result from this proce dure are _
m ,e *
1
« = 1, 2, 3, . .
Bohr model (which has the electron moving in a fixed orbit at a unique distance from the nucleus) gives an in complete interpretation. As we shall see in Section 51-7, the electron can be found anywhere from r = 0 to r = » , but r = Uo is its most probable location. The hydrogen atom is a three-dimensional system, and the Schrodinger equation must be solved in three dimen sions. Because of the form of the potential energy (Eq. 21), it is most convenient to solve this problem in spherical coordinates, using as coordinates the radius r and two angles 8 and (f>to fix the direction. When we solve the Schrodinger equation for this system, we find that three quantum numbers are necessary to describe the states of the electron. These quantum numbers are de fined and displayed in Table 3. We discuss these quantum numbers, along with a fourth one based on the electron spin (a relativistic effect that is not predicted by the Schrodinger equation, which is nonrelativistic), later in this chapter. The Bohr theory is of only limited usefulness in under standing the structure of atomic hydrogen and ions with a Radial distance, t / cq
( 22)
which are exactly those obtained from the Bohr model (Eq. 18). This agreement should not surprise us because we have seen that the Bohr theory provided a perfect match with the observed wavelengths of the hydrogen spectral lines. From the Schrodinger wave functions, we can calculate the most probable distance between the electron and pro ton. This turns out to be /j^Oo■That is, in the lowest energy state, the most likely place to find the electron is at a distance o f one Bohr radius from the proton. Here the
* For a full treatment, see Robert Eisberg and Robert Resnick, Quantum Physics o f Atoms. Molecules, Solids, Nuclei, and Par ticles, 2nd edition (Wiley, 1985), Chapter 7.
Figure 7 The potential energy function U{r) for the hydro gen atom. The radial distance between the electron and pro ton is measured in terms of the Bohr radius do.
1076
Chapter 5 1
TABLE 3 Symbol
mi
The Structure o f Atomic Hydrogen
THE HYDROGEN ATOM QUANTUM NUMBERS^ Name
Associated with
Principal quantum number Orbital quantum number Magnetic quantum number
Energy, mean radius
1 ,2,3, . . .
Magnitude of orbital angular momentum Direction of orbital angular momentum
0, 1,2, . . . , « - 1
Allowed Values
0 ,± 1 ,± 2 , . . . , ± /
A fourth quantum number, associated with the spin, will be introduced later.
single electron (He"^, and so forth), and it is of no help at all in understanding details beyond the wave lengths of spectral lines. It provides only a very limited basis for understanding atoms more complex than hydro gen, which can be studied in detail with the Schrodinger equation and the exclusion principle, which is explained in the next chapter. The Bohr theory does not show how to calculate the properties o f systems more complex than a single atom, such as a molecule or a solid. Today we regard the Bohr theory as an important and ingenious step toward understanding the atom, and we should re member that two principles developed by Bohr to make his theory work (the correspondence principle and the existence o f stationary states) are essential parts of the complete quantum theory.
ple, a state with n = 2 can have / = 0 or / = 1. These two states of the hydrogen atom share the same principal quantum number n and have the same energy, even though they represent very different states o f motion.
The Direction of L Let us choose a direction in space, which we arbitrarily label as the z axis, and let us determine the direction o f L with respect to this axis. It turns out that the angular momentum vector L can not take any position with respect to the z axis, but only those positions that have a component along the z axis given by
L^ = m,h.
(25)
in which m,, the magnetic quantum number, may have only the values
51-3 ANGULAR M O M ENTUM
w , = 0, ± 1 , ± 2 , . . . , ± / .
(26)
The energy o f a state is a scalar and, in the hydrogen atom, it is specified by a single quantum number n. The angular momentum of a state, however, is a vector and we see from Table 3 that it takes two quantum numbers, / and m/, to describe it. The angular momentum is doubly quantized, in both magnitude and direction. We discuss each in turn.
This restriction on the direction of L is called space quan
The Magnitude of L
(= 2ft in this case) is less than the magnitude o f L (= 2 .4 5 ft). This will always be the case; the angular mo mentum vector L can never be fully lined up with the z axis. Figure 8 shows the allowed values o f for / = 1,2, and 10. In the latter case we begin to approach the classical situation, in which space quantization has faded away and any orientation o f the angular momentum vector is al lowed. Sample Problem 3 gives further details o f how the correspondence principle operates in this case. The quantization of means that the angle 6 between L and the z axis (see Fig. 8) is quantized, its values being restricted to
In solving the Schrodinger equation, we learn that the angular momentum is quantized. Its allowed values are
L = >//(/+ 1) h
(23)
in which / is the orbital quantum number. For conve nience, we have again introduced the symbol h (pro nounced “h-bar”) as an abbreviation for h/ln. The values that / can have in Eq. 23 depend on the value of the principal quantum number n and are given by
1= 0, 1, 2,
« —
1.
(24)
For example, the ground state o f the hydrogen atom, which has « = 1, must have / = 0 (and thus L = 0), no other value being permitted by Eq. 24. For another exam
tization. We see from Eqs. 2 3 - 2 6 that, for a hydrogen atom state with 1= 2, the magnitude o f L is >/2(2 -I- l)ft or 2.45ft. L^, the component of L along the z axis, may have the values 0, ± 1ft, and ± 2 ft, five components in all. No other
orientations of the angular momentum vector with respect to the z axis are allowed. Note that the maximum value of
6 = cos"' ^ = cos"' i
, W TT)
(27)
Section 51-3 Angular Momentum
1077
Figure 8 The allowed values of for / = 1, 2, and 10. The numbers on the z axis show values of m/. The figures are drawn to different scales.
/=
where we have used Eq. 23 for L and Eq. 25 for L^. The minimum value of 6 occurs when m/ has its greatest value, which is /. For / = 2, for example, you can easily show that this minimum value is ^min = COS
[2H2{2 + 1)] = cos-' 0.817 = 35.3°
For hydrogen atom states with / = 2 it is simply not possi ble for the (orbital) angular momentum vector L to make any smaller angle with the z axis. Once we have selected an axis and determined the com ponent o f L along that axis, the components of L on all other axes are completely uncertain. That is, we can have an exact knowledge o f only one selected component o f L (which we arbitrarily assume to be the z component). Figure 9 suggests a classical vector model that helps in visualizing the space quantization o f L. It shows the vec tor precessing about the z direction, like a spinning top or a gyroscope precessing about a vertical axis in the Earth’s gravitational field. The component remains constant
10
as the motion proceeds, but the x and y components o f L do not have definite values. Heisenberg’s uncertainty principle helps us to under stand the space quantization of the angular momentum vector. In its angular form (compare Eq. 6 o f Chapter 50) this principle is ALj • A 0 ~ h!2n (z component).
(28)
in which
Sample Problem 3 Find the minimum value of 6 in Fig. 9 for / = 1, 10^ 10’, 10^ and 10’. Solution The minimum value of 0 occurs when we put nti = I in Eq. 27. Doing so and rearranging lead to / ^ m in =
Figure 9 A vector representation of the space quantization of the angular momentum L and the magnetic dipole mo ment fi.
COS
'
V 7(7T T )
[-r
We see by inspection that if we let / —► then 6 cos" ‘ 1 = 0 . This is just what we expect from the correspondence principle.
1078
Chapter 51
The Structure o f Atomic Hydrogen
Substituting for / in this equation leads to these results: / 1 1Q2 10^ 10^ 10’
45.0" 5.7" 1.8" 0.57" 0.0018"
momentum, in fact, we almost always mean this maximum projected value. The magnitude of the angular momentum rarely enters quantum calculations and is seldom given. (e) The smallest angle that the angular momentum vector can make with the z axis follows from Eq. 27, with aW/ = /. We then have e = cos-* [//V/(/+ 1)] = cos-' [3/V3(3+ 1)] = c o s - '0.866 = 30°.
For a macroscopic object like a spinning top or a phonograph record, / would be enormously larger than 10’ and would be so close to zero that the difference would be beyond the possibil ity of measurement. Thus as the angular momentum of a spin ning object gets larger and larger, the space quantization of wave mechanics merges gently into the continuous distribution of classical mechanics. We see once again how the correspondence principle works. Computational note: If you use the above formula to calculate ^min for I 10’, your calculator will probably overflow. Take advantage of the fact that (1//) 1 and develop an approxi mate formula. You will need to use both the binomial expansion and the series expansion for cos 6\ see Appendix H.
Sample Problem 4 {a) For « = 4, what is the largest allowed value of /? {b) What is the magnitude of the corresponding angu lar momentum? (c) How many different components on the z axis may this angular momentum vector have? {d) What is the magnitude of the largest projected component? (e) What is the smallest angle that the angular momentum vector can make with the z axis? Solution (a) From Eq. 24, the largest allowed value of / is — 1, so / = 3. {b) From Eq. 23 we have
The angular momentum vector can make no smaller angle with the z axis than this.
Orbital Angular Momentum and Magnetism The Bohr model also suggests that the orbiting electron— a tiny current loop— should have an (orbital) magnetic dipole moment associated with it. Both the angular mo mentum L and the magnetic dipole moment p are vectors and share a common axis. Because the electron has a negative charge, however, these vectors point in opposite directions along this axis. In Section 37-2 we showed that these two vectors are related by
the minus sign showing the opposite directions o f L and fi. Although Eq. 29 was derived on a semiclassical basis, it remains true in wave mechanics. Consider a state in which the z component o f the angu lar momentum is h/ln. Substituting this value for L in Eq. 29 yields _ eh
4 n m ,'
L = V/(/+ 1) (hllit) = '/3 (3 + 1) (6.63 X 10-^ J - s)/(2 t:) = 3.66X lO-’^J-s. In practice, atomic angular momenta are rarely reported in SI units. It would be more customary to report the magnitude of the angular momentum in this case as simply f \ 2 h or 3.46^. See part (d). (c) The number of components that the angular momentum vector may have on the z axis is equal to the number of allowed values of the magnetic quantum number m/. From Eq. 26 this number is 2/ + 1 or 2 X 3 + 1 = 7. {d) The largest projected component is found from Eq. 25, in which the magnetic quantum number W/ is given its largest possible value. From Eq. 26 this largest value is just /, so we have L , = l{h/2n) = (3X6.63 X 10-^J*s)/(27r) = 3.17X 10-^ J-s. Note from part (b) that this is smaller than the magnitude of the angular momentum vector, as it must be. As we remarked in part (b), the maximum projected angular momentum compo nent would be reported as simply 3 h . When we refer to angular
(29)
— L, 2We
This quantity is called the Bohr magneton value
and has the
_ = 9.274 X 10-2^ J/T = 5.788 X IQ-^ eV/T.
(30)
The Bohr magneton is a convenient unit in which to measure atomic magnetic moments, much as we took the Bohr radius a^ as a convenient unit in which to measure atomic distances. Bohr theory predicts that the magnetic dipole moment of the hydrogen atom in its ground state will be one Bohr magneton. The theory is simply not correct on this point. Experiment shows that, in accordance with the predic tions of wave mechanics, both the (orbital) angular mo mentum and the (orbital) magnetic dipole moment of the hydrogen atom in its ground state are zero. This failure of Bohr theory, however, does not stop us from using the Bohr magneton as a convenient unit of measure. If the angular momentum is quantized, the magnetic dipole moment must also be quantized, and in the same
Section 51-3 Angular Momentum
fashion. Combining Eq. 29 (z components only) and Eq. 25 allows us to write eh liz =
47im^
From Eq. 30, we see that the z component of the magnetic dipole moment is given by (31) in which is the Bohr magneton. As shown by Fig. 9, the classical vector model accounts for the space quantization o f fi as well as L. Both vectors precess about the z direc tion, and both are characterized by their z components. The magnetic dipole moment of the atom — much like a compass needle— can respond to an external magnetic field. This gives the atom a convenient “handle” by means o f which we can explore its inner workings by probing from the outside. Because the magnetic dipole moment is rigidly coupled to the angular momentum, keeping track of the former automatically keeps track of the latter as well. We do not have to look far to find evidence that atoms can be the carriers of magnetism. An ordinary iron bar shows no external magnetic properties because its elemen tary atomic magnets are randomly arranged, their effects canceling at all external points. When these elementary atomic magnets are lined up, however, as they are in a bar magnet, their combined magnetic strength is there for all to see. When the magnetic dipole moments of an assembly of atoms are lined up, their angular momenta— to which they are rigidly coupled— must also be lined up. In 1915
1079
Einstein, working with W. J. de Haas (the son-in-law of the great Dutch physicist H. A. Lorentz), carried out an experiment to explore this phenomenon. If an iron bar is suddenly magnetized, perhaps by switching on a current in a solenoid as in Fig. 10, the angular momenta of its atoms suddenly become lined up. Because angular mo mentum must be conserved, the bar as a whole must start to rotate in the opposite sense. This E in s te in -d e H aas effect, as it is called, is small and the measurements are difficult. Bear in mind that in 1915, when this experiment was carried out, wave mechanics had not been discovered, the Bohr theory was only 2 years old, and the intrinsic spin of the electron had yet to be discovered. It turned out later that the Einstein - de Haas effect (and also, for that matter, the ferromagnetism of a bar magnet) is due largely to the intrinsic angular momentum (spin) of the electrons rather than to their orbital angular momen tum. This, however, does not alter the fact that this exper iment demonstrates, in a macroscopic way, that atoms can be the carriers of both magnetism and angular mo mentum.
Sample Problem 5 An unmagnetized iron cylinder, whose radius R '\s5 mm, hangs from a frictionless bearing so that it can rotate freely about its axis; see Fig. 10. A magnetic field is sud denly applied parallel to this axis, causing the magnetic dipole moments of the atoms to align themselves parallel to the field. The atomic angular momentum vectors, which are coupled back-to-back with the magnetic dipole moment vectors, also become aligned and the cylinder will start to rotate in the oppo site sense. Find T, the period of rotation of the cylinder. Assume that each iron atom has an angular momentum of h /ln. The molar mass A /of iron is 55.8 g/mol (=0.0558 kg/mol). Solution The angular momentum of the rotating cylinder (= Lcyi) must be equal in magnitude (though opposite in direc tion) to the angular momentum associated with the aligned atoms (= Latoms )• If H is the number of atoms in the cylinder, is the Avogadro constant, and m is the mass of the cylinder, we can write =
Nmn)
=
(N,,mlM\hl2n).
For the rotating cylinder we have
in which / is the rotational inertia of the cylinder about its rota tional axis and to is its angular speed. Equating these two expressions and solving for T yields
(a)
(6)
Figure 10 The Einstein-de Haas effect, (a) The atomic an gular momentum vectors in an iron cylinder are randomly oriented, (d) When an axial magnetic field is applied, the atomic angular momenta line up as shown and the cylinder as a whole starts to rotate in the opposite sense.
In^R^M N^h (2jt^X5 X 10-’ m)2(0.0558 kg/mol) (6.02 X 10“ mol-'X6.63 X 10” ^ J-s) = 6.90X 10^ s = 19.2 h. This is indeed a slowly rotating cylinder! Actually, Einstein and de Haas suspended their cylinder from a torsion fiber and used
1080
Chapter 51
The Structure o f Atomic Hydrogen
more refined techniques in their experiment dealing with this effect.
51-4 THE STERN -G ERLA CH EXPERIMENT__________________ Space quantizatio n , th a t is, the notion th at an atom ic angular m o m en tu m vector L o r an atom ic m agnetic di pole m o m en t vector // can have only a certain discrete set o f projections on a selected axis, is n ot an easy concept for the classically oriented m ind to accept. Nevertheless, it was predicted theoretically (by W olfgang Pauli) and veri fied experim entally (in 1922, by O tto S tem and W alther G erlach) several years before the developm ent o f wave m echanics. Figure 11 shows the ap paratus o f S tem and G erlach. Silver is vaporized in an electrically heated “ oven,” and silver atom s spray into the external vacuum o f the appa ratus from a sm all hole in the oven wall. T he atom s (which are electrically neutral b u t which have a m agnetic dipole m o m en t) are form ed in to a narrow beam as they pass through a slit in an interposed screen. T he collim ated beam then passes betw een the poles o f an electrom agnet and, finally, deposits itself on a glass d etector plate. O ften in laboratory experim ents we w ant the m agnetic field to be uniform , b u t in this case the pole faces are shaped to m ake the field as nonuniform as possible. T he atom ic beam passes very close to the sharp V-shaped ridge in the u pper pole piece, w here the n onuniform ity o f the m agnetic field is greatest.
A Dipole in a Nonuniform Field Figure 12a shows a dipole o f m agnetic m o m en t //, m aking an angle 6 w ith a uniform m agnetic field. W e can im agine the dipole to be a tiny bar m agnet, w ith the m agnetic dipole m o m en t vector // pointing (by convention) from its south pole to its n o rth pole. W e m ay im agine the forces to be concentrated at the poles as show n in the figure. We
see that, for a uniform field, there is no net force on the dipole. T he upw ard and dow nw ard forces on the poles are o f the sam e m agnitude and they cancel, no m atter w hat the orientation o f the dipole. Figures \2 b and 1 2 c show the situation in a n o n u n i form field. H ere the upw ard and dow nw ard forces do not have the sam e m agnitude because the two poles are im m ersed in fields o f different strengths. In this case there is a net force, both its m agnitude and direction depending on the orientation o f the dipole, th at is, on the value o f 6. In Fig. \2 b this net force is up and in Fig. 1 2 c it is down. T hus the silver atom s in the beam o f Fig. 11, as they pass through the electrom agnet, are deflected up o r dow n, and in greater or lesser am ounts, depending on the orientation o f their m agnetic m o m ent dipole vectors with respect to the m agnetic field. Now let us calculate the deflecting force quantitatively. T he m agnetic potential energy o f a dipole in a m agnetic field B is given by Eq. 38 o f C hapter 34, t/(0) = —// • B = —(// cos 6)B, In ou r m in d ’s eye let us follow the silver atom s in the beam as they m ove through the electrom agnet o f Fig. 11 parallel to the sharp edge. F rom sym m etry (see also Fig. 1 2b, c ), the m agnetic field at this central position has no x or y com ponents. T hus 5 = B ecausep cos 0 = //^, we can write the potential energy as U(e) = - p , B , . T he net force
(32)
on the atom is —(d U /dz) or, from Eq. 32, IT dz •
(33)
N ote th at the deflecting force is determ ined by the deriv ative o f the m agnetic field and does not depend on the m agnitude o f the field itself. In Fig. \2b, c, B^ increases as z increases so th at the derivative is positive. T hus the sign o f the deflecting force in Eq. 33 depends on the sign o f If P 2 is positive (as in Fig. \2b) the atom is deflected upw ard; if it is negative (as in Fig. 12c) the deflection is dow nw ard. O ne troublesom e point rem ains. If the individual
Figure 11
Glass (detector plate
The apparatus of Stem and Gerlach.
Section 51-4
(6 )
The Stern-Gerlach Experiment
1081
(c)
Figure 12 A magnetic dipole, represented as a small bar magnet with two poles, in {a) a uniform field and (b,c) a nonuniform field. The net force acting on the magnet is zero in points up in {b\ and points down in (c).
atoms in the beam behave like tiny bar magnets, why don’t they all simply line up with the magnetic field? Why should any of them point, even partially, in the opposite direction? The answer is that the atoms not only have a magnetic dipole moment; they also have angular mo mentum. The result is that they precess around the field direction (see Fig. 9) rather than line up with it. In the same way a top that is not spinning will simply fall over if you place it at an angle with the Earth’s gravitational field. We saw in Section 13-5, however, that if the top is spin ning, it will precess about this direction. It is the angular momentum that does it!
quantization exists! Stem and Gerlach ended the pub lished report of their work with the words: “We view these results as direct experimental verification of space quanti zation in a magnetic field.” Physicists everywhere agree.
Sample Problem 6 In an experiment of the Stem-Gerlach type, the magnetic field gradient dB^tdz at the beam position was 1.4 T/mm and the length h of the beam path through the magnet was 3.5 cm. The temperature of the “oven” in which the silver was evaporated was adjusted so that the most probable speed V for the atoms in the beam was 750 m/s. Find the separa-
The Experimental Results When the electromagnet in Fig. 11 is turned off (or is operating at very low power), there will be no deflections o f the atoms and the beam will form a narrow line on the detecting plate. When the electromagnet is turned on, however, strong deflecting forces come into play. Then there are two possi bilities, depending on whether space quantization exists or not. (D on’t forget that the entire object of this experi ment is to find out!) If there is no space quantization, the atomic magnetic dipole moment vectors have a continu ous distribution of values, some positive and some nega tive; the beam will simply broaden. On the other hand, if space quantization does exist, there is only a discrete set of values of This means that there is only a discrete set of values for the deflecting force in Eq. 33, and the beam splits up into a number of discrete components. Figure 13 shows what happens. The beam does not broaden but splits cleanly into two subbeams. Space
Figure 13 The results of the Stem-Gerlach experiment, showing the silver deposit on the glass detector plate of Fig. 11, with the magnetic field (a) turned off and (b) turned on. The beam has been split into two subbeams by the magnetic field. The vertical bar at the right in (b) represents 1 mm.
1082
Chapter 51
The Structure o f Atomic Hydrogen
tion d between the two deflected subbeams as they emerge from the magnet; see Fig. 13Z?. The mass m of a silver atom is 1.8 X 10” ^^ kg, and its magnetic moment is 1 Bohr magneton (=9.28 X lO-^M/T). Solution The acceleration of a silver atom as it passes through the electromagnet is given (see Eq. 33) by ^ _ F , _ pA d B Jd z) m m The vertical deflection Az of either of the subbeams as it clears the magnet is m The separation d of the two beams is 2Az, or j
fiz(dBJdz)h^
^
^ ----_ (9.28 X 10-^“ J/TX1.4 X 10^ T/mX3.5 X lO'" m)^ (1.8 X 10-25 kgX750 m/s)2
on its axis. There are many parallels between spin and orbital angular momentum. The spin quantum numbers is analogous to the orbital quantum number /; however, unlike /, the value of s does not change with the electron’s state of motion. All electrons, no matter what their state of motion, have s = i . In fact, we usually consider 5 to be a fundamental property of a particle, along with its mass and electric charge. The spin of the electron can be represented by a vector S o f magnitude (compare Eq. 23)
S=M s+l)h.
(34)
The component o f this vector in the z direction can be written (compare Eq. 25)
S, = m ,h.
(35)
Just like the components of L, the permitted components of S differ by one unit o f h . We therefore see that the permitted values of are
= 1.6 X 10“^ m = 0.16 mm. This is the order of magnitude of the separation displayed in Fig. 13b', note the scale in that figure.
51-5 THE SPINNING ELECTRON The Stem - Gerlach experiment clearly demonstrates that the magnetic moment vector o f an atom can have only a finite number o f discrete directions in space, as opposed to the infinite number allowed by classical physics. How ever, there is a curious feature of the experiment. Figure 13 shows the beam of silver atoms splitting into two com ponents, corresponding to two different orientations of the magnetic moment vector of the atom (or, equiva lently, o f its angular momentum vector, since the two vectors are related by Eq. 29). Yet a glance at Table 3 or Fig. 8 shows that there is always an odd number o f possi ble orientations of the L vector. Put another way, the number o f possible orientations of L is 2/ + 1, and for this to equal 2 we must have / = i; however, this contradicts the restriction that / takes only integer values. The solution to this dilemma was proposed in 1924 by the Austrian-bom physicist Wolfgang Pauli (see Fig. 14). He suggested that there is yet another quantum number that describes the state o f an electron in an atom, and that this quantum number can take the values + i or — In the following year two Dutch graduate students, Samuel Goudsmit and George Uhlenbeck, proposed the notion of electron spin as the physical interpretation of Pauli’s pro posed new quantum number. Spin is often called intrinsic angular momentum, and it is often useful (although strictly not correct) to visualize the spin as the angular momentum of a particle rotating
nis = ± i .
(36)
Associated with the spin angular momentum there is a magnetic moment, which is given by (compare Eq. 29)
m.
(37)
Note the difference in Eqs. 37 and 29 by a factor o f 2. This suggests that the spin angular momentum is twice as ef fective as the orbital angular momentum in producing magnetic effects. For further details on orbital and spin magnetic moments, see Section 37-2. The quantum numbers for the orbital angular momen tum / and its magnetic projection m, arise in a natural way from solving the Schrodinger equation for the hydrogen atom. The spin angular momentum and its magnetic pro jection seem to be introduced arbitrarily with no theoreti cal justification. The English mathematical physicist Paul A. M. Dirac developed a relativistic wave equation similar to the nonrelativistic Schrtxlinger equation, and Dirac showed that solutions to his equation for the hydrogen atom gave the electron spin as a fourth quantum number. To obtain the complete solution for the hydrogen atom, we must replace the Schrodinger machine o f Fig. 6 by a Dirac machine! This is another great triumph for relativ ity theory, without which we would have no theoretical basis for understanding this fundamental part o f the structure of atoms. We can now explain the appearance of two beams in the Stem-Gerlach experiment. The electron in a silver atom happens to occupy a state in which / = 0, so the total angular momentum of the electron is due only to its spin. This spin vector has only two possible orientations rela tive to the magnetic field, hence the two components of the beam.
Section 51-5
The Spinning Electron
1083
Figure 14 Wolfgang Pauli (left) and Niels Bohr watching a “tippy top,” a top that spins for a while on one end and then turns upside down. They are waiting for the “spin flip.”
Every fundamental particle has its characteristic spin and magnetic moment. The proton and neutron, like the electron, have a spin of i . Their magnetic moments are discussed in Section 37-2. In Section 51-8 we consider other observable consequences of the existence of elec tron spin. Consequences of proton and neutron spin have proved to be of great practical value through the phenomenon of nuclear magnetic resonance (see Section 13-6 and Sample Problem 1 of Chapter 37). When a proton is placed in a magnetic field B, an energy change of occurs when the spin changes direction or “flips.” This spin flip can be caused by subjecting the protons to an electromagnetic wave whose frequency is selected such that hv = The field B consists of an external field ^cxt (perhaps due to an electromagnet) and an internal field B^^t (due to the chemical environment in which the proton is found). For example, in a molecule of ethanol, whose formula we may write as CH 3 — CH 2 — OH, each hydrogen nucleus expe riences a different internal field because of its different location in the molecule. By keeping fixed and vary ing the frequency v, we can find several frequencies at which the spin flips occur, each corresponding to a partic ular environment of a hydrogen nucleus in an ethanol molecule. Equivalently, as in Fig. 15, we can keep v fixed and vary B^^i. Either way, we get a unique signature that identifies ethanol. In this way nuclear magnetic resonance
proves to be an important analytical tool in organic chem istry. Other applications include measuring Bi„t in various molecular or solid environments and measuring nuclear magnetic dipole moments.
Figure 15 A nuclear magnetic resonance spectrum of eth anol. All the lines are due to absorption of the incident radia tion when the proton spin flips. The groups of lines corre spond to different groupings of hydrogen within the molecule. The entire horizontal scale is considerably less than lO”'* T.
1084
Chapter 51
The Structure o f Atomic Hydrogen
51-6 COUNTING THE HYDROGEN ATOM STATES We have now described the four quantum numbers that define the stationary states o f the hydrogen atom, and we have shown how each of them can be interpreted physi cally. Although we did not prove it, it is nevertheless true that they emerge from Schrodinger’s wave equation, with an important assist from Dirac’s equation in the case of the electron spin. In solving the Schrodinger wave equa tion, the quantum numbers and the other information come tumbling out in a natural way. Our next task is to see whether we can arrange the hydrogen atom states in some orderly fashion. Consider first the principal quantum number n. All states with the same value of n have the same energy, and we say that the assembly of such states forms a shell. Equation 24 tells us that the number of different values of / that are possible for a given value of n is just equal to n. Thus for « = 3 we can have three values of / ( = 0 , 1 , and 2 ). The shells can be further subdivided. Within a given shell, all states with the same value of / have the same angular momentum and are said to form a subshell. For example, the shell defined hy n = 3 contains three sub shells, each with the same energy but a different angular momentum. Within the subshells, the electrons may have different states of motion because of the different ways the angular momentum vector can be oriented. For a given /, Eq. 26 tells us that there are 21 + 1 values of m/. In our example, then, the subshell with 1 = 2 contains 5 ( = 2 X 2 + 1 ) states, and those with / = 1 and 0 contain 3 states and 1 state, respectively. This adds up to a total of 9 (= 5 + 3 + 1 ) states in the shell with n = 3. The effect o f the spin quantum number is simply to double the number of states. Each combination of «, /, and mi that we have identified can now be associated with either = + i or thus producing two states where one existed before. To continue with our example, there are not 9 but 18 states in the shell with n = 3. Table 4 summarizes this classification of hydrogen atom states into shells and subshells.
TABLE 4
Do the numbers o f states in the shells (that is, 2, 8 , and 18) in the bottom row of Table 4 seem familiar? They are the lengths of the horizontal rows (periods) in the periodic table of the elements! As Appendix E shows, period 1 contains 2 elements, periods 2 and 3 contain 8 elements each, and periods 4 and 5 contain 18. In Chapter 52 we shall see in detail just how the order of the elements in the periodic table arises from wave-mechanical principles.
Sample Problem 7 A certain shell has a principal quantum number n of 4. {a) How many subshells does this shell contain? (b) What is the number of states in each of these subshells? (c) What is the number of states in the shell? Solution (a) If « = 4, we know from Eq. 24 that the allowed values of / are 0, 1,2, and 3. This is a total of four values, in agreement with the fact that the number of allowed values of / for a given n is just equal to n. Each value of n defines a shell and each value of /defines a subshell within that shell. Thus there are four subshells in the « = 4 shell. (b) The number of states in a subshell is given by 2(2/ + 1), the factor of 2 coming from the two allowed values of the spin magnetic quantum number. For the numbers of states in the various subshells in the « = 4 shell we .then have
/
2(2/ + 1)
0 1 2
2 6 10
3
14
(c) The number of states in the « = 4 shell is found by adding up the numbers in the subshells that it contains. From the table above we have 2 + 6 + 10 + 14 or 32. Note that 32 is the num ber of elements in horizontal row 6 of the periodic table of the elements. See Appendix E. Can you prove that, in general, the number of states in a shell defined by principal quantum number n is given by 2n^l That works out in this case because 2 X 4^ = 32, as we found by ex plicit counting.
STATES OF THE HYDROGEN ATOM« 2
n
1
1
0
0
mi
0
0
3
1
0
1
o,±i
0
0 ,± 1
0 , ± 1,± 2
1
2
m,
±i
±i
±i
±i
±i
±i
Number of states in the subshells
2
2
6
2
6
10
Number of states in the shells
2
' Complete to n = 3 only.
8
18
Section 51-7
51-7 THE GROUND STATE OF HYDROGEN___________________ In this section, without going into the mathematical de tails, we present the results of using the Schrodinger equa tion to study the ground state of hydrogen. The procedure involves solving the Schrddinger equation in spherical polar coordinates when the electron and the proton inter act through the Coulomb force, given by the potential energy
U{r) = -
1 4neo r '
(38)
If we insert this potential energy function into the Schrodinger equation and carry out the needed mathe matical manipulations, we are able to derive an expres sion for the energies o f the allowed stationary states o f the atom and also for the wave functions that describe those states. The expression for the energies o f the stationary states turns out to be exactly Eq. 18, the expression derived from Bohr theory. We focus our attention here on the ground state o f the hydrogen atom, that is, on the state of lowest energy. The wave function for the ground state also emerges from the Schrodinger equation and turns out to depend only on the single variable r. It is given by Wir) = -
1
-r/ao
(39)
in which Uq is the Bohr radius. This wave function has spherical symmetry, by which we mean that it depends only on the magnitude of the vector r (which defines the point at which the wave func tion is evaluated), but not on its direction. This is perhaps
The Ground State o f Hydrogen
1085
not surprising. The potential energy function (Eq. 38) is also spherically symmetric, so that the atom has no builtin preferred direction. Like a billiard ball, the atom in its ground state looks the same in all directions. The square o f the wave function, which we have called the probability density, has the property that y/\r) dV gives the probability of finding the electron in a volume element t/K located at a position defined by the position vector r. From Eq. 39 we have -Ir/OQ TTflo
(40)
Figure 16^ is a “dot plot” representation of Eq. 40, the density of the dots suggesting the probabilistic nature of the electron’s location. A circle of one Bohr radius has been drawn to show the scale. Another useful quantity for representing the electron’s position is the radial probability density Pr(r). This is defined so that P,{r) dr gives the probability that, regard less o f direction, the electron will be found to lie between two spherical shells whose radii are r and r + dr. The volume between those shells is {47tr^){dr) so that we can write
P,{r) dr = \!/\r) dV = y/\r){4nr^){drX which, combined with Eq. 40, leads to
Pri^) = ii/\r){ 4 nr'^) =
r'^e~^r/ao (4 1 ) ^0 Figure 16b shows a plot o f Eq. 41. We note (see Sample Problem 8) that the maximum value of P^(r) occurs at r = aQ.ln wave mechanics we do not say that the electron in the ground state of the hydrogen atom moves in an orbit of one Bohr radius. We say instead that the electron is more likely to be found in a thin shell at this distance from the central nucleus than in a shell of equal thickness at any other distance, either larger or smaller. The so-
Figure 16 (a) A “dot plot” representation of the probability density for the ground state of the hydrogen atom, given by Eq. 40. A circle has been drawn at the radius r = Gq. (b) The radial probability density, given by Eq. 41. The filled triangle marks the maximum probability at r = Uq. The line marked “90%’’ shows the radius of a sphere containing 90% of the probability density.
1086
Chapter 51
The Structure o f Atomic Hydrogen
called 90% radius is also indicated in the figure. It defines a sphere such that the probability that the electron will be found inside is 90%. The probability that it will be found outside is, of course, 10%. We see that the answer to the question, “How big is the hydrogen atom?” is not so simple. You can say that it is one Bohr radius, but (see Problem 56) 68% of the times that you measure it, you will find that the electron is farther away than this. A more reasonable answer is to give the radius of the 90%probability sphere, which turns out to be 2.7 Bohr radii. The inherent quantum fuzziness of the atom simply does not permit us to answer the ques tion any more precisely. We have said nothing so far about the role of the spin of the electron in the ground state of the hydrogen atom. As Table 4 reminds us, the shell corresponding to n = 1 in the hydrogen atom contains two states, corresponding to the two allowed values (= ± i ) of the spin quantum num ber m,. These two states, however, have exactly the same energy and, for an isolated atom, there is no way to tell them apart experimentally. The atom in its ground state can be either in the state with = -I-i or in the state with w, = —}. The energy of the atom and the probability density curve of Fig. \(>b are the same for each of these states. If you really want to distinguish between these two spin orientation states you can do so by putting the atom in an external magnetic field. This not only provides a natural reference axis with respect to which the electron’s spin angular momentum vector (and its spin magnetic dipole moment vector) can orient itself, but it also separates the two states in energy. This, in fact, is exactly what was done in the Stem-Gerlach experiment.
Solution obtain
Differentiating
dP = —A 4 2 f)e -^i‘-‘ + 4 r f % dr al ^0 V ^0/ ^0
3 --------
51-8
(
1
E 3 2 TT
/\
0
0
TH E EXCITED STA TES OF H YDR O G EN_____________________
The « = 2, / = 0 Subshell
The wave function for this state is Wioo^r) = -
1
:(2 -
A'}Inal
rlao)e-'i ^^,
(42)
in which the subscripts 200 represent the quantum num ber sequence « = 2, / = 0, and aw/ = 0. This state, just like the ground state, has spherical symmetry in that it is a function of r only and involves no angles as variables. The probability density y/\r) and the radial probability density P^{r) are given by
i
\ n = 2 ,l = 0
/
5
10
15
r/OQ
(a)
^0 /
The state next highest in energy above the ground state is called the first excited state. Its energy (= —3.40 eV) is found by putting m= 2 in the energy equation (Eq. 18). As Table 4 reminds us, the « = 2 shell contains two sub shells, corresponding to / = 0 and to / = 1. We deal with each in turn.
/— X /
V
At the maximum of the curve we must have dPJdr = 0 and, as inspection of this equation shows, this does indeed occur at r = do. Note that we also have dPJdr = 0 at r = 0 and as r —►oo. These conditions are quite consistent with Fig. \6b and corre spond to minima rather than maxima.
4
1
Eq. 41 with respect to r, we
^
Sample Problem 8 Verify that the maximum of the radial probability density curve of Fig. 166 falls at r = Uo, where Oq is the Bohr radius.
^
P,{r) in
( 6)
Figure 17 (a) A “dot plot” representation of the probability density of atomic hydro gen for the excited state with « = 2 and / = 0. A circle has been drawn at the radius r = . {b) The radial probability density.
Section 51-8
\2 p -rfa o
(43)
and f,W -r V X 4 ir * ) -r/ao
(44)
Figure 17a is a “dot plot” of Eq. 43, the probability den sity y/\r). Figure 17Z> is a plot of Eq. 44, the radial proba bility density. Note that the latter curve has two maxima and goes to zero at r = 2ao, as simple inspection of Eq. 44 makes clear. The remarks about spin at the end of Section 51 -7 apply with equal force here. This state also has complete spheri cal symmetry. Its angular momentum is i , in units of h, due (as before) entirely to the spin of the electron.
The n = 2 , 1 = I Subshell The states that comprise this subsheil do have orbital an gular momentum, its z-axis projections being given by rriih, where m/ can take on the values of 0 or ± 1. The wave functions for these three states are not spherically symmetric, being functions not only of r but also of the polar angle 0, defined in Fig. 1ia. Figure 1ia shows “dot plots” of the three probability densities, The sub scripts represent the quantum number sequence n, /, and m/. All three plots have rotational symmetry about the z axis, the plots for m/ = — 1 and for m/ = + 1 being identi cal. You are entitled to be a little puzzled about the lack of spherical symmetry shown in Fig. 18^. After all, the po tential energy function that we inserted into the Schrodinger equation depended only on r. Does the elec
The Excited States o f Hydrogen
1087
tron in a state with n = 2 J = 1, and m/ = 0 really like to cluster about the z axis, avoiding the equatorial plane? How is the direction of this axis chosen? The answer to this puzzle comes when we realize that the three states in question have the same energy and, in the absence of an external magnetic field, there is no way to isolate them experimentally. If we assume that— on average— the atom spends one-third of its time in each o f the three states shown in Fig. 18^, we can calculate a weighted average probability density for the subshell as a whole. The result is
Vl\^r) = i
-I- i if/2ioir,d) + 1
% nal
Ip-rlQo r^e
(45)
The subscript on the probability density gives the values of n (= 2) and /( = ! ) . Note that the angular variable 6 has disappeared from the final result! The probability density for the subshell as a whole depends only on r and has the spherical symmetry that we expect it to have. This means that if you superimpose the three cylindrically sym metrical dot plots of Fig. 18fl, the resulting dot plot (imag ined in three dimensions) will be spherically symmetric. We now find the radial probability density for this sub shell, proceeding as we did in Section 51-7 for the ground state; namely, PAr) = ^|/\^{r){Anr^)
1
(46)
24aJ
Figure 1ib shows a plot of this radial probability density. Note that the maximum of the distribution occurs at r = 4^0, which (see Eq. 19) is just the radius of the second Bohr orbit.
(a)
Figure 18 (a) “Dot plot” representations of the probability density of atomic hydrogen for the excited state with n = 2 and / = 1. To obtain the full three-dimensional picture, imagine each plot rotated about the z axis, (b) The radial probability density. The filled triangle shows the location of the maximum at r = Aqq,
( 6)
1088
Chapter 51
The Structure o f Atomic Hydrogen
51-9 DETAILS OF ATOMIC STRUCTURE (Optional) So far in this chapter we have outlined basic aspects of the quan tum theory applied to the structure of atomic hydrogen, which allows us to understand such details of its properties as the Balmer series (and other series of emitted radiations). Here we mention briefly some additional details of atomic structure that we can similarly understand.
2J+1
4
n = 3, / = 1
2
E c 06 00
If)
Fine Structure When we study the spectral lines under high resolution, we find that what appears to be a single line is often a pair of very closely spaced lines (a doublet). This is called the fine structure of the spectrum. The effect is usually very small. In the case of the transition between the first excited state and the ground state in hydrogen (E = 10.2 eV), the energy difference between the two components due to the fine structure is 4.5 X 10“ ^ eV. The fine structure splitting increases rapidly with atomic number, how ever; in sodium, it is responsible for the splitting of the yellow D-lines, which differ in wavelength by about 0.6 nm out of 590 nm, or about 1 part in 10^. The fine structure splitting is usually analyzed in terms of the total angular momentum of the electron, obtained from the sum of the orbital and spin contributions. In the excited / = 1 state, the possible values of the total angular momentum quantum number are, according to the rules for adding angular momen tum in quantum mechanics, 7 = /±5 = 1± i = |or^.
Loosely speaking, these two possibilities correspond respec tively to the L and S vectors being parallel or antiparallel. These two different orientations have slightly different energies, which gives an energy splitting in the atom between the states corre sponding to these combinations. (You can think of the parallel combination as corresponding to two tiny bar magnets aligned side by side and parallel to each other, with their like poles repelling one another; in the antiparallel configuration, the mag nets are aligned in the opposite direction, with neighboring N and S poles attracting each other. The latter arrangement lowers the energy of the atom, that is, makes it more tightly bound.) The total angular momentum has properties similar to the orbital and spin angular momenta. Specifically, there are 2j + 1 different orientations of the J vector, corresponding to the dif ferent possible values of its z component = m jh , where mj ranges from -hj to —j in integer steps. That is, for j = have mj = + L or — while for 7 = i we have mj = or -i. Figure 19 shows a representation of the fine structure splitting in sodium. Note that, not considering the fine structure splitting, the / = 1excited state includes six substates (three corresponding tom/ = + 1, 0 , and —1, each of which can havem, = + i o r —^). Considering the fine structure, there are still six substates, four associated with 7 = | and two with 7 = i . The ground state, with / = 0 , can have only 7 = i .
E c eg O) If) O) 00 If)
n = 3, 1 = 0 -
Figure 19 The fine structure splitting of the energy levels in sodium that emit the light of the familiar doublet. The draw ing is not to scale; the actual splitting of the upper levels is about 1/1000 of the energy difference between the upper and lower levels. To the right is shown the number of different ori entations of the J vector (that is, the number of different values of mj for each level).
ment, because the equipment available to him did not have sufficient resolution to observe this small effect. About 30 years later in 1896, the Dutch physicist Pieter 2^eman repeated the experiment with more sensitive apparatus and observed that the spectral lines were measurably broadened in a strong magnetic field. With higher fields and better resolution, it is possible to see the lines dividing into components whose splitting increases in proportion to the field. For this work, Zeeman shared the 1902 Nobel prize in physics. Figure 20 shows an example of the Zeeman effect. In an intense magnetic field, the splitting of the lines in the Zeeman effect may be about 1 part in 10^ of the energy or wavelength of the lines. To understand the origin of the Zeeman effect, consider the energy-level diagram for sodium shown in Fig. 21, which shows only the 7 = | excited state and the 7 = ^ ground state. When the magnetic field is turned on, the four possible orientations of the J vector of the excited state (corresponding to the four different mj values) give different energies, which can be calculated from the magnetic moment of the state. Similarly, the 7 = i ground state splits into two substates. With the field off, there is only one possible transition between the excited state and the ground state; with the field on, there might be transitions from any substate of the excited state to any substate of the ground state, giving eight possible transitions. Two of these transitions are forbidden to occur by the rules of quantum mechanics, leaving six individual components. The number of components and their separations observed in the Zeeman effect can be calculated using wave mechanics. As Fig. 20 illustrates, different lines in a given spectrum may show different patterns of splitting. It is a triumph of wave mechanics that these details of the Zeeman effect, along with the relative intensities of the components and even their polarizations, can be calculated and agree precisely with experiment.
The Zeeman Effect Michael Faraday had the intuitive idea that the light from a source would change if you put the source in a strong magnetic field. Faraday was not successful in attempting to do this experi-
Reduced Mass Our derivation of the energy levels of hydrogen assumed that the electron revolves about a stationary nucleus. Actually, the elec-
Section 51-9
Details o f Atomic Structure (Optional)
1089
Figure 20 The Zeeman effect in rhodium. The bottom spectrum shows the splitting of the spectral lines when the magnet is turned on.
tron and proton each orbit about the center of mass of the sys tem. One convenient way of taking this into account is to assign the electron an effectively smaller mass called the reduced mass (see Section 15-10), defined according to m=
m .M M A -m ^
1+ m JM ’
where M is the mass of the nucleus. That is, in all expressions involving the electron mass, we should replace it with its reduced mass. In the case of hydrogen, Eq. 47 gives
(47) m=■
m.
1 1+ 1836.15
Magnet OFF
Magnet ON
and the corresponding value of the ground-state energy changes from —13.6057 eV (corresponding to an infinitely massive nu cleus) to —13.5983 eV. All spectral lines scale correspondingly. For example, the first line in the Lyman series would have an energy of 10.2043 eV for an infinitely massive nucleus, while in hydrogen the observed energy is 10.1972 eV. These differences, amounting to about 1 part in 10^, are easily observable with spectroscopes. About one hydrogen atom in 6000 is deuterium, or “heavy hydrogen,” whose nucleus contains one proton and one neu tron, making it about twice as massive as ordinary hydrogen. Most of the chemical and physical properties of heavy hydrogen are identical to those of ordinary hydrogen, except for those properties that specifically depend on mass. As we have seen, the energy of the spectral lines depends slightly on the mass of the nucleus. In an atom of heavy hydrogen, the reduced mass of the electron is m= •
m.
1 1+ 3670.48
I I I I I Wavelength
Wavelength
(a)
( 6)
Figure 21 An energy-level diagram illustrating the Zeeman splitting of one member of the fine-structure doublet in so dium. When the magnetic field is applied, the single spectral line of (a) splits into the six closely spaced components shown in (^).
m. 1.000545 ’
m. l.(X)0272
and the corresponding ground-state energy is —13.6020 eV. The first line of the Lyman series would have an energy of 10.2015 eV. If we examined the spectral lines from a sample of hydrogen, we would find that each line consisted of a doublet, separated by an interval o f0.027% of the energy (or wavelength) and with relative intensities of about 6000 to 1. By increasing the concentration of heavy hydrogen (by distillation, for example) the intensity ratio can be varied. Using this procedure, deute rium was discovered in 1932 by H. C. Urey, who was awarded the 1934 Nobel prize in chemistry for his discovery. ■
1090
Chapter 51
The Structure o f Atomic Hydrogen
QUESTIONS 1. Discuss the analogy between the Kepler-Newton relation ship in the development of Newton’s law of gravitation and the Balmer-Bohr relationship in developing the Bohr theory of atomic structure. 2. Why was the Balmer series, rather than the Lyman or Paschen series, the first to be detected and analyzed in the hy drogen spectrum? 3. Any series of atomic hydrogen yet to be observed will proba bly be found to be in what region of the spectrum? 4. In Bohr’s theory for the hydrogen atom orbits, what is the implication of the fact that the potential energy is negative and is greater in magnitude than the kinetic energy? 5. Can a hydrogen atom absorb a photon whose energy ex ceeds its binding energy (13.6 eV)? 6 . On emitting a photon, an isolated hydrogen atom recoils to conserve momentum. Explain the fact that the energy of the emitted photon is slightly less than the energy difference between the energy levels involved in the emission process. 7. Why are some lines in the hydrogen spectrum brighter than others? 8 . Radioastronomers observe lines in the hydrogen spectrum that originate in hydrogen atoms that are in states with n = 350 or so. Why can’t hydrogen atoms in states with such high quantum numbers be produced and studied in the laboratory? 9. Only a relatively small number of Balmer lines can be ob served from laboratory discharge tubes, whereas a large number are observed in stellar spectra. Explain this in terms of the small density, high temperature, and large volume of gases in stellar atmospheres. 10. According to classical mechanics, an electron moving in an orbit should be able to do so with any angular momentum whatever. According to Bohr’s theory of the hydrogen atom, however, the angular momentum is quantized according to L = nh/ln. Reconcile these two statements, using the corre spondence principle. 11. Why does the concept of Bohr orbits violate the uncertainty principle? 12. Consider a hydrogen-like atom in which a positron (a posi tively charged electron) circulates about a (negatively charged) antiproton. In what way, if any, would the emis sion spectrum of this “antimatter atom’’ differ from the spectrum of a normal hydrogen atom? 13. If Bohr’s theory and Schrodinger’s wave mechanics predict the same result for the energies of the hydrogen atom states, then why do we need wave mechanics, with its greater com plexity? 14. Compare Bohr’s theory and wave mechanics. In what re spects do they agree? In what respects do they differ? 15. How would you show in the laboratory that an atom has angular momentum? That it has a magnetic dipole mo ment? 16. Why don’t we observe space quantization for a spinning top? 17. The angular momentum of the electron in the hydrogen
18.
19.
20.
21.
22. 23.
24.
25.
26.
atom is quantized. Why isn’t the linear momentum also quantized? {Hint: Consider the implications of the uncer tainty principle.) Angular momentum is a vector and you might expect that it would take three quantum numbers to describe it, corre sponding to the three space components of a vector. Instead, in an atom, only two quantum numbers characterize the angular momentum. Explain why. Justify the statement that, in the Einstein-de Haas effect, the angular momentum of the iron bar as a whole must be conserved when the bar is suddenly magnetized. In the Einstein-de Haas experiment (see Sample Problem 5), can you justify the fact that the predicted period of rota tion of the cylinder depends only on the cylinder radius and not, for example, on its height? What assumptions were made in deriving the expression for the period of rotation? Convince yourself that the directions of the arrows in Fig. 10^ representing the current in the solenoid, the magnetic field, the atomic angular momenta, and the direction of rotation of the cylinder are consistent with each other. Does the Einstein - de Haas effect provide any evidence that angular momentum is quantized? A beam of circularly polarized light, viewed as a beam of photons whose spins are aligned, can exert a torque on an absorbing screen. Develop the analogy to the Einstein-de Haas experiment. A beam of neutral silver atoms is used in a Stem-Gerlach experiment. What is the origin of both the force and the torque that act on the atom? How is the atom affected by each? What determines the number of subbeams into which a beam of neutral atoms is split in a Stem-Gerlach experi ment? If in a Stem-Gerlach experiment an ion beam is resolved into five component beams, then what angular momentum quantum number does each ion have?
27. In a Stem-Gerlach apparatus, is it possible to have a mag netic field configuration in which the magnetic field itself is zero along the beam path but the field gradient is not? If your answer is yes, can you design an electromagnet that will produce such a field configuration? 28. The silver atoms in the Stem-Gerlach experiment of Sam ple Problem 6 are uncharged. Suppose that a silver atom in the apparatus were suddenly to lose an electron, becoming a silver ion. What would be the nature and the relative magni tude of the forces acting on it (a) before and (b) after this event? 29. How do we arrive at the conclusion that the spin magnetic quantum number can have only the values ± i? What kinds of experiments support this conclusion? 30. Why is the magnetic moment of the spinning electron di rected opposite to its spin angular momentum? 31. Discuss how good an analogy the rotating Earth revolving about the Sun is to a spinning electron moving about a proton in the hydrogen atom.
Problems 32. An atom in a state with zero angular momentum has spheri cal symmetry as far as its interaction with other atoms is concerned. It is sometimes called a “billiard-ball atom.” Explain. 33. “If the angular momentum of electrons in atoms were not quantized, the periodic table of the elements would not be what it is.” Discuss this statement. 34. How would the properties of helium differ if the electron had no spin, that is, if the only operative quantum numbers were «, /, and /W/? 35. We assert that the number of quantum numbers needed for a complete description of the motion of the electron in the hydrogen atom is equal to the number of degrees of freedom that the electron possesses. What is this number? How can you justify it? 36. Define and distinguish among the terms wave function, probability density, and radial probability density. 37. What are the dimensions and the SI units of a wave func tion, a probability density, and a radial probability density? Are the dimensions what you expect? 38. In the hydrogen atom state with / = 1, the spin and the orbital angular momentum vectors can be aligned either
39.
40.
41.
42.
43.
1091
parallel or antiparallel. Which arrangement has the greater energy and why? How can you account for the fact that in the state of the hydrogen atom with n = 2 and / = 0 , the probability density is a maximum at r = 0 but the radial probability density is zero there? See Fig. 17. Figure 1 shows the three probability densities for the hy drogen atom states with n = 2 and / = 1. What determines the direction in space that we choose for the z axis? Consider the three probability density “dot plots” of Fig. 18fl, each of which is a figure of revolution about the z axis. Do you see any connection between these figures and the semiclassical vector model of the atom (Fig. 9) for the case of / = 1? Use Heisenberg’s uncertainty principle to show that the probability densities in an / = 2 state have cylindrical sym metry about the z axis. Explain how the interaction between the spin and the orbital motions of the valence electron in sodium leads to the split ting of the spectral lines of sodium, producing the familiar sodium doublet. See Fig. 19.
PROBLEMS Section 51-1 The Bohr Theory 1. (a) By direct substitution of numerical values of the funda mental constants, verify that the energy of the ground state of the hydrogen atom is —13.6 eV; see Eq. 18. (^) Similarly, from Eq. 17 show that the value of the Rydberg constant R is 0.01097 nm“ *. (c) Also verify the numerical value of Oqby direct computation of its expression given in Eq. 19. 2. Answer the questions of Sample Problem 2, but for the Lyman series. 3. Using the Balmer-Rydberg formula, Eq. 3, calculate the five longest wavelengths of the Balmer series. 4. What are the (a) wavelength, (b) momentum, and (c) energy of the photon that is emitted when a hydrogen atom under goes a transition from the state n = 3 to n = 1? 5. Show, on an energy-level diagram for hydrogen, the quan tum numbers corresponding to a transition in which the wavelength of the emitted photon is 121.6 nm. 6 . (a) If the angular momentum of the Earth due to its motion around the Sun were quantized according to Bohr’s relation L = nh/2n, what would the quantum number be? (b) Could such quantization be detected if it existed? 7. Calculate the binding energy of the hydrogen atom in the first excited state. 8 . Find the value of the quantum number for a hydrogen atom that has an orbital radius of 847 pm. 9. Light with a wavelength of 1281.8 nm is emitted by a hydro gen atom, (a) What transition of the hydrogen atom is re sponsible for this radiation? (b) To what series does this radiation belong? (Hint: See Fig. 2.)
10. A hydrogen atom is excited from a state with n = \ to one with n = 4. (a) Calculate the energy that must be absorbed by the atom, (b) Calculate and display on an energy-level diagram the different photon energies that may be emitted if the atom returns to the « = 1 state. 11. The lifetime of an electron in the state n = 2 in hydrogen is about 10 ns. What is the uncertainty in the energy of the n = 2 state? Compare this with the energy of this state. 12 A diatomic gas molecule consists of two atoms of mass m separated by a fixed distance d rotating about an axis as indicated in Fig. 22. Assuming that its angular momentum is quantized as in the Bohr atom, determine (a) the possible angular velocities and (b) the possible rotational energies, (c) Calculate, according to this model, the ground-state en ergy, in eV, of an O 2 molecule for which d = 121 pm and m = 16.0 u. 13 If an electron is revolving in an orbit at frequency Vq, classi cal electromagnetism predicts that it will radiate energy not
Figure 22
Problem 12.
1092
Chapter 51
The Structure o f Atomic Hydrogen
only at this frequency but also at 2vq, 3vq, 4 vq, and so on. Show that this is also predicted by Bohr’s theory of the hydrogen atom in the limiting case of large quantum num bers. 14. In Table 2 show that the quantity in the last column is given by
24.
100( v - V q) ^ _ l^ V n for large quantum numbers. 15. A neutron, with kinetic energy of 6.0 eV, collides with a resting hydrogen atom in its ground state. Show that this collision must be elastic (that is, energy must be conserved). (Hint: Show that the atom cannot be raised to a higher excitation state as a result of the collision.) 16. (a) Calculate, according to the Bohr model, the speed of the electron in the ground state of the hydrogen atom, (b) Cal culate the corresponding de Broglie wavelength, (c) Com paring the answers to (a) and (Z?), find a relation between the de Broglie wavelength Aand the radius Oqof the ground-state Bohr orbit. 17. According to the correspondence principle, as we expect classical results in the Bohr atom. Hence the de Bro glie wavelength associated with the electron (a quantum result) should get smaller compared with the radius of the Bohr orbit as n increases. Indeed, we expect that A/r 0 as /2 —►00. Show that this is the case. 18. A hydrogen atom in a state having a binding energy (the energy required to remove an electron) of 0.85 eV makes a transition to a state with an excitation energy (the difference in energy between the state and the ground state) of 10.2 eV. (a) Find the energy of the emitted photon, (b) Show this transition on an energy-level diagram for hydrogen, labeling with the appropriate quantum numbers. 19. From the energy-level diagram for hydrogen, explain the observation that the frequency of the second Lyman-series line is the sum of the frequencies of the first Lyman-series line and the first Balmer-series line. This is an example of the empirically discovered Ritz combination principle. Use the diagram to find some other valid combinations. 20. Calculate the recoil speed of a hydrogen atom, assumed initially at rest, if the electron makes a transition from the « = 4 state directly to the ground state. (Hint : Apply conser vation of linear momentum.) 21. (a) How much energy is required to remove the electron from a He'*’ ion in its ground state? (b) From a Li^'*' ion in a state with « = 3? (Hint: See Eq. 18.) 22. In stars the Pickering series is found in the He'*' spectrum. It is emitted when the electron in He'*' jumps from higher levels to the level with « = 4. (a) Show that the wavelengths of the lines in this series are given by
A=
1 Rn^-\6'
in which n = 5 ,6 ,1 , . . . .(b) Calculate the wavelength of the first line in this series and of the series limit, (c) In what region(s) of the spectrum does this series occur? 23. Show that in Bohr’s semiclassical one-electron model of the atom (a) the orbital speeds are quantized as i;„ =
25. 26.
27.
28.
(Ze^/2eoh)(\/n) and (b) the orbital angular momenta are quantized 2&L„ = (h!2n)n. In the ground state of the hydrogen atom, according to Bohr’s theory, what are (a) the quantum number, (b) the orbit radius, (c) the angular momentum, (d) the linear mo mentum, (e) the angular velocity, ( / ) the linear speed, (^) the force on the electron, (h) the acceleration of the electron, (i) the kinetic energy, (j) the potential energy, and (k) the total energy? How do the quantities (b) to (k) in Problem 24 vary with the quantum number n? Suppose that we wish to test the possibility that electrons in atoms move in orbits by “viewing” them with photons with sufficiently short wavelength, say 10.0 pm. (a) What would be the energy of such photons? (b) How much energy would such a photon transfer to a free electron in a head-on Comp ton collision? (c) What does this tell you about the possibility of confirming orbital motion by “viewing” an atomic elec tron at two or more points along its path? Assume that the speed of the electron is 0 . 10c. Bohr proposed that, as an alternative to the correspondence principle, the quantization expression for the angular mo mentum (Eq. 20) could be taken as a basic postulate. Start ing from this point, and using only classical results, derive Bohr’s expression for the quantized energies of the station ary states of the hydrogen atom (Eq. 18). (a) Calculate the wavelength intervals over which the Lyman, the Balmer, and the Paschen series extend. (The interval extends from the longest wavelength to the series limit.) (b) Find the corresponding frequency intervals.
Section 51~3 Angular Momentum 29. Verify that = 9.274 X 10‘ 2^ J/T = 5.788 X 10-^ eV/T. as reported in Eq. 30. 30. If an electron in a hydrogen atom is in a state with / = 5. what is the smallest possible angle between L and L^? 31. For a hydrogen atom in a state with / = 3, calculate the allowed values of (a) L^, (b) p^, and (c) 0. Find also the magnitudes of (d) L and (e) p. Where appropriate, express answers in units of h and p^. 32. (a) Show that the magnetic moments of the electrons in the various Bohr orbits are given, according to the Bohr theory, by l^ = nps in which p^ is the Bohr magneton and n = \, 2, 3, . . . . (b) How does this expression compare with the actual values? 33. (a) Show, by reanalyzing Problem 12 for a diatomic mole cule with the angular momentum quantized by Eqs. 23 and 24, that the energy levels can be written as Ei =
h^KlA- 1)
/ = 1 , 2 , 3,
(b) Calculate the energies of the lowest three levels of the O 2 molecule, for which the two atoms are 121 pm apart. The mass of the oxygen atom is 16.0 u. Compare your result with Problem 12. 34. Show that Eq. 28 is a plausible version of the uncertainty
Problems principle Ap*Ax = h /ln . {Hint: Multiply by r/r; associate p with mv, and L with mvr.) Section 51-4 The Stern-Gerlach Experiment 35. O f the three scalar components of L, one, is quantized, according to Eq. 25. In view of the restrictions imposed by Eqs. 23 and 24, taken together, show that the most that can be said about the other two components of L is V l J + Z j = V/(/+
.
Note that these two components are not separately quan tized. Show also that /Ih ^ V 7 I+ L 2 < v /(/+ i)ft . Correlate these results with Fig. 9. 36. Suppose a hydrogen atom (in its ground state) moves 82 cm in a direction perpendicular to a magnetic field that has a gradient, in the vertical direction, of 16 mT/m. (a) What is the force on the atom due to the magnetic moment of the electron, which we take to be 1 Bohr magneton? {b) Find its vertical displacement if its speed is 970 m/s. 37. Calculate the acceleration of the silver atom as it passes through the deflecting magnet in the Stem -Gerlach experi ment of Sample Problem 6. 38. Assume that in the Stem - Gerlach experiment described for neutral silver atoms the magnetic field B has a magnitude of 520 mT. (a) What is the energy difference between the orien. tations of the silver atoms in the two subbeams? (^) What is ^ the frequency of the radiation that would induce a transition between these two states? (c) What is its wavelength, and to what part of the electromagnetic spectmm does it belong? The magnetic moment of a neutral silver atom is 1 Bohr magneton. Section 51-6 Counting the Hydrogen Atom States 39. Write down the quantum numbers for all the hydrogen atom states belonging to the subshell for which /2 = 4 and 1= 3. 40. A hydrogen atom state is known to have the quantum num ber / = 3. What are the possible n, mi, and m^ quantum numbers? 41. A hydrogen atom state has a maximum m/ value of + 4. What can you say about the rest of its quantum numbers? 42. How many hydrogen atom states are there with /2 = 5? How are they distributed among the subshells? 43. What are the quantum numbers n, /, m/, m, for the two electrons of the helium atom in its ground state? 44. Calculate the two possible angles between the electron spin angular momentum vector and the magnetic field in Sam ple Problem 6 . Bear in mind that the orbital angular mo mentum of the valence electron is zero. 45. Label as true or false these statements involving the quan tum numbers n, /, W/. {a) One of these subshells cannot exist: w = 2, / = 1; « = 4, / = 3 ; « = 3, / = 2 ; « = 1 , / = 1 . (b) The number of values of m/ that are allowed depends only on / and not on n. (c) The n = 4 shell contains four subshells, {d) The smallest value of n that can go with a given / is / 4- 1. (^) All states with / = 0 also have m/ = 0, regardless of the value of n. ( / ) Every shell contains n subshells.
1093
46. What is the wavelength of a photon that will induce a transi tion of an electron spin from parallel to antiparallel orienta tion in a magnetic field of magnitude 190 mT? Assume that 1= 0 . 47. The proton as well as the electron has spin i . In the hydrogen atom in its ground state, with n = 1 and / = 0 , there are two energy levels, depending on whether the electron and the proton spins are in the same direction or in opposite direc tions. The state with the spins in the opposite direction has the higher energy. If an atom is in this state and one of the spins “flips over,” the small energy difference is released as a photon of wavelength 21 cm. This spontaneous spin-flip process is very slow, the mean life for the process being about 10^ y. However, radio astronomers observe this 21cm radiation from interstellar space, where the density of hydrogen is so small that an atom can flip before being disturbed by collisions with other atoms. Calculate the ef fective magnetic field (due to the magnetic dipole moment of the proton) experienced by the electron in the emission of this 21-cm radiation. 48. Show that the number of states in any shell is given by 2n^.
Section 51-7 The Ground State o f Hydrogen 49. In the ground state of the hydrogen atom, evaluate the square of the wave function, n /\r \ and the radial probability density P^(r) for the positions (a) r = 0 and {b) r = Gq. Ex plain what these quantities mean. 50. In Fig. \6b, verify the plotted values of P^{r) at (a) r = 0, {b)r = a o ,2ind (c)r = 2 go. 51. Find the ratio of the probabilities of finding the electron in the hydrogen atom in a thin shell at the Bohr radius to that of finding it in a shell of the same thickness at twice that dis tance. 52. A spherical region of radius 0.05^0 is located a distance Gq from the nucleus of a hydrogen atom in its ground state. Calculate the probability that the electron will be found inside this sphere. (Assume that y/ is constant inside the sphere.) 53. For a hydrogen atom in its ground state, calculate the proba bility of finding the electron between two spheres of radii r = l.OOfloand r= l.Olflo* 54. In atoms there is a finite, though very small, probability that, at some instant, an orbital electron will actually be found inside the nucleus. In fact, some unstable nuclei use this occasional appearance of the electron to decay by electron capture. Assuming that the proton itself is a sphere of radius 1.1 X 10“ ‘^ m and that the hydrogen atom electron wave function holds all the way to the proton’s center, use the ground-state wave function to calculate the probability that the hydrogen atom electron is inside its nucleus. {Hint: When X
1094
Chapter 5 1
The Structure o f Atomic Hydrogen
P = 1 - e ~ ^ ( \ + 2 x + 2x2), in which x = r/aQ. (b) Evaluate the probability that, in the ground state, the electron lies within a sphere of radius Qq. 57. Use the result of Problem 56 to calculate the probability that the electron in a hydrogen atom, in the ground state, will be found between the spheres r = Uq and r = I uq. 58. For an electron in the ground state of the hydrogen atom, calculate the radius of a sphere for which the probability that the electron will be found inside the sphere equals the proba bility that the electron will be found outside the sphere. {Hint: See Problem 56.) Section 51-8 The Excited States o f Hydrogen 59. For the state « = 2, / = 0, (a) locate the two maxima for the radial probability density curve of Fig. 1lb, and (b) calculate the values of the radial probability density at the two max ima; compare with Fig. \lb . 60. Using Eq. 46, show that, for the hydrogen atom state with n = 2 and / = 1,
/:
P fr ) d r = 1.
What is the physical interpretation of this result? 61. For a hydrogen atom in a state with n = 2 and 1 = 0, calcu late the probability of finding the electron between two spheres of radii r = 5.00flo and r = 5.01flo« 62. For a hydrogen atom in a state with n = 2 and / = 0, what is the probability of finding the electron somewhere within the smaller of the two maxima of its radial probability density function? See Fig. \lb . Section 51-9 Details o f Atomic Structure 63. Potassium ( Z = 19), like sodium ( Z = 11), is an alkali metal, its single valence electron moving around a filled
18-electron argon-like core. As in sodium, there is a potas sium doublet, its wavelengths being 764.5 nm and 769.9 nm. The quantum numbers of the levels that give rise to these lines are just the same as for sodium (see Fig. 19) except th a t« = 4. Calculate {a) the energy splitting between the two upper states and (b) the energy difference between the uppermost state and the ground state. 64. The wavelengths of the lines of the sodium doublet (see Fig. 19) are 588.995 nm and 589.592 nm. (a) What is the differ ence in energy between the two upper levels in that figure? (b) This energy difference comes about because the elec tron’s spin magnetic dipole moment can be oriented either parallel or antiparallel to the internal magnetic field asso ciated with the electron’s orbital motion. Use the result you have just calculated to find the strength of this internal mag netic field. The electron’s spin magnetic dipole moment has a magnitude of 1 Bohr magneton. 65. Apply Bohr’s model to a muonic atom, which consists of a nucleus of charge Ze with a negative muon (an elementary particle with a charge q = —eand a mass m = 201 , where Wg is the electron mass) circulating about it. Calculate {a) the m uon-nucleus separation in the first Bohr orbit, (b) the ionization energy, and (c) the wavelength of the most energetic photon that can be emitted. Assume that the muon is circulating about a hydrogen nucleus (Z = 1). See “The Muonium Atom,” by Vernon W. Hughes, Scientific American, April 1966, p. 93. 66. Apply Bohr’s model to the positronium atom. This consists of a positive and a negative electron revolving around their center of mass, which lies halfway between them, (a) What relationship exists between this spectrum and the hydrogen spectrum? {b) What is the radius of the ground-state orbit? {Hint: Calculate the reduced mass of the atom.) See “Exotic Atoms,” by E. H. S. Burhop, Contemporary Physics, July 1970, p. 335.
CHAPTER 52 ATOMIC PHYSICS ♦
In the preceding three chapters, we have developed the foundations o f wave mechanics and used its principles to understand the structure o f the hydrogen atom. In this chapter, we broaden the development by considering the structure o f atoms beyond hydrogen. We begin by considering the emission o f x rays by atoms, which historically provided the first definitive means to measure the number o f electrons in an atom. We then consider the rules for determining how to construct atoms with more than one electron, and we consider how those rules and the resulting structure determine the arrangement o f elements in the fam iliar periodic table. We use information from atomic structure to analyze the operation o f the helium -neon laser, and we conclude with a brief look at how we can extend our knowledge o f atomic structure and wave functions to learn about the structure o f molecules.
52-1 THE X-RAY SPECTRUM_______ So far we have dealt w ith the behavior o f single electrons in atom s, either the lone electron o f hydrogen o r the single valence electron o f sodium . W e now shift o u r attention to the behavior o f electrons deep w ithin the atom . W e m ove from a region o f relatively low binding energy (5 eV for the w ork required to rem ove the valence electron from sodium , for exam ple) to a region o f higher energy (70 ke V for the w ork required to rem ove an innerm ost electron from tungsten, for exam ple). T he radiations we deal with, though o f course still p art o f the electrom agnetic spec tru m , differ drastically in wavelength, for exam ple, from 6 X 10"^ m for the sodium doublet lines to 2 X 10“ “ m for one o f the tungsten characteristic radiations, a ratio o f about 30,000. W e are now speaking o f x rays. T he usefulness o f these penetrating radiations in m edi cal an d dental diagnostics and in therapy is well know n, as are their m any industrial applications, such as exam ining welded jo in ts in pipe lines. In Section 4 7-4 we described how X rays can be used to deduce the atom ic structures o f crystalline m aterials. T he structures o f such com plex sub stances as insulin an d D N A have been w orked o u t by these m ethods. In astronom y, x-ray satellites have shown us an entirely new view o f o u r universe through images o f the X rays em itted by stars and galaxies.
We saw in Section 47-4 th at x rays are produced w hen energetic electrons strike a solid target and are brought to rest in it. Figure 1 shows the wavelength spectrum o f the x rays produced w hen 35-keV electrons strike a m olybde num target.
The Continuous X-Ray Spectrum W e first exam ine the continuous spectrum o f Fig. 1, ignoring— for the tim e b eing— the two p ro m in en t peaks th at rise from it. C onsider an electron o f kinetic energy K th at scatters from the nucleus o f one o f the m olybdenum atom s in the target, as in Fig. 2. In such a collision, m o m entum is transferred to the atom , and the electron loses kinetic energy. (Because the atom is so massive, the m o m entum im parted to it by the electron results in a negligi ble kinetic energy.) The energy lost by the electron ap pears as the energy hv o f an x-ray p h oton th at radiates away from the site o f the encounter. T his process is called brem sstrahlung (G erm an, “ braking radiation” ), and it accounts for the continuous x-ray spectrum . Suppose electrons are accelerated through a potential difference V and fall on a thick target. D ue to brem sstrah lung processes in the target, the electrons can lose any am o u n t o f energy from 0 to their m axim um energy o f eV. T he brem sstrahlung photons have a corresponding con tinuous spectrum from 0 to
1095
1096
Chapter 52 Atomic Physics
The Characteristic X-Ray Spectrum
30
40
50
60
70
80
90
Wavelength (pm)
Figure 1 The wavelength spectrum of x rays produced when 35-keV electrons strike a molybdenum target (1 pm = 10"'^ m).
A prominent feature of the continuous spectrum of Fig. 1 is the sharply defined cutoff wavelength below which the continuous spectrum does not exist. This mini mum wavelength corresponds to a decelerating event in which one o f the incident electrons (with initial kinetic energy eV) loses all this energy in a single encounter, radiating it away as a single photon. Thus
ev=hv
We now turn our attention to the two peaks o f Fig. I, labeled K„ and K^. These peaks are characteristic o f the target material and, together with other peaks that appear at longer wavelengths, form the characteristic x-ray spec trum of the element in question. Here is how these x-ray photons arise. (1) An energetic incoming electron strikes an atom in the target and knocks out one of its deep-lying electrons. If the electron is in the shell with n = 1 (called, for historical reasons, the K shell) there remains a vacancy, or a “hole” as we shall call it, in this shell. (2) One of the outer electrons moves in to fill this hole and, in the process, the atom emits a char acteristic x-ray photon. If the electron falls from the shell with n = 2 (called the L shell), we have the line of Fig. 1; if it falls from the next outermost shell (called the A/shell) we have the line, and so on. O f course, such a transition will leave a hole in either the L or the M shell, but this will be filled by an electron from still further out in the atom; in the process, yet another characteristic x-ray spectrum line is emitted. Figure 3 shows an x-ray atomic energy-level diagram for molybdenum, the element to which Fig. 1 refers. The base line {E = 0) represents the energy o f a neutral molyb denum atom in its ground state. The level marked K (E = 20 keV) represents the energy of a molybdenum atom with a hole in its K shell. Similarly, the level marked L (£■ = 2.7 keV) represents the energy of an atom with a hole in its L shell, and so on. Note that in representing the
^ m in
or A ^min = — •
( 1)
Equation 1 shows that if A 0, then A^in ^ 0, which is the prediction of classical theory. The existence of a mini mum wavelength is a quantum phenomenon. Note that as you change the target material, perhaps from molybdenum to copper, the general shape and in tensity of the continuous spectrum may change but the cutoff wavelength will not change. This wavelength de pends only on the kinetic energy of the electrons that bombard the target and not at all on the target material.
K
Incident electron
Figure 2 An electron passing near the nucleus of a target atom is accelerated and radiates a photon, losing part of its ki netic energy in the process.
Figure 3 An atomic energy-level diagram for molybdenum, showing the transitions that give rise to the characteristic X rays of that element. (All levels, except the K level, contain a number of close-lying components, which are not shown in the figure.)
Section 52-2 X Rays and the Numberling o f the Elements
energy levels for the hydrogen atom (see Fig. 4 of Chapter 51), we chose a different base line. Rather than the atom in its ground state, there we selected the atom with its electron removed to infinity as our E = Q configuration. Actually, the atomic configuration for which we choose to put £■ = 0 does not matter. Only differences in energy are physically significant, and these are the same no matter what our choice of an £* = 0 base line is. The transitions and Kp in Fig. 3 show the origin of the two peaks in Fig. 1. The line, for example, origi nates when an electron from the L shell of molybdenum — moving upward on the energy-level diagram— fills the hole in the K shell. This is the same as saying that a hole— moving downward on the diagram— moves from the K shell to the L shell. It is easier to keep track of a single hole than o f the 41 electrons in ionized molybdenum that are potentially available to fill it. We have drawn the arrows in Fig. 3 from the point of view of hole transitions.
atomic structure to the ordering of the elements in the periodic table. In his investigation of the atomic number concept, Moseley generated characteristic x rays by using as many elements as he could find— he found 38— as targets for electron bombardment in a special evacuated x-ray tube of his own design. He measured the wavelengths of a number of the lines of the characteristic x-ray spectrum by the crystal diffraction method described in Section 47-4. He then sought, and readily found, regularities in the spectra as he moved from element to element in the peri odic table. In particular, he noted that if, for a given spec trum line such as he plotted the square root o f its frequency (= >/v = V ^ ) against the position of the asso ciated element in the periodic table, a straight line re sulted. Figure 4 shows a portion of his data. We shall see below why it is logical to plot the data in this way and why a straight line is to be expected. Moseley’s conclusion from the full body of his data was:
Sample Problem 1 Calculate the wavelength A^in for the con tinuous spectrum of x rays emitted when 35-keV electrons fall on a molybdenum target, as in Fig. 1.
We have here a proof that there is in the atom a fundamen tal quantity, which increases by regular steps as we pass from one element to the next. This quantity can only be the charge on the central atomic nucleus.
Solution /
1097
From Eq. 1, we have _ /zc _ ( 4 . 1 4 X 10-'5eV*sK3.00X 10® m/s) eV 35.0 X 10^ eV = 3.54 X 10“ “ m = 35.4 pm.
This is in agreement with the experimental result shown by the vertical arrow in Fig. 1. Note that Eq. 1contains no reference to the target material. For a given accelerating potential all targets, no matter what they are made of, exhibit the same cutoff wave length.
52-2 X RAYS AND THE NUMBERING OF THE ELEM ENTS____________________ In this section, we consider what x rays can teach us about the structure of the atoms that emit or absorb them. We focus on the work of the British physicist H. G. J. Mose ley,* who, by x-ray studies, developed the concept of atomic number and gave physical meaning in terms of
* Henry G .J. Moseley (1887 -1915) joined Ernest Rutherford’s laboratory at the University of Manchester in 1910. Through a brilliant series of experiments, Moseley showed that characteris tic x-ray frequencies increased regularly with the atomic number of the element, and he was able to locate gaps in the sequences corresponding to elements not yet discovered in his time. Mose ley’s promising research career was cut short when he died at age 27 at the battle of Gallipoli in World War I.
Moseley’s achievement can be appreciated all the more when we realize the status of understanding of atomic structure at that time (1913). The nuclear model of the atom had been proposed by Rutherford only 2 years ear lier. Little was known about the magnitude of the nuclear charge or the arrangement of the atomic electrons; Bohr published his first paper on atomic structure only in that same year. An element’s place in the periodic table was at that time assigned by atomic mass, although there were several cases in which it was necessary to invert this order to fit the demands of the chemical evidence. The table had several empty squares, and a surprisingly large number of claims for the discovery of new elements had been ad vanced; the rare earth elements, because of the problems caused by their similar chemical properties, had not yet been properly sorted out. Due to Moseley’s work, the characteristic x-ray spec trum became the universally accepted signature o f an ele ment. Through such studies it became possible to string the elements in a line, so to speak, and to assign consecu tive numbers to them, all without the slightest need to know anything about their chemical properties. It is not hard to see why the characteristic x-ray spec trum shows such impressive regularities from element to element and the optical spectrum does not. The key to the identity of an element is the charge on its nucleus. This determines the number of its atomic electrons and thus its chemical and physical properties. Gold, for example, is what it is because its atoms have a nuclear charge of + 79^. If it had just one more unit of charge, it would not be gold but mercury; if it had one fewer, it would be platinum. The K electrons, which play such a large role in
1098
Chapter 52 Atomic Physics
2.5
Figure 4 A Moseley plot of the line of the characteristic x-ray spectra of 2 1 elements. The frequency is determined from the measured wavelength.
2.0
^ 1.5 X
oO
*>
1.0
0.5
10
20
30
40
50
Number of the element in the periodic table
the production of the characteristic x-ray spectrum, lie very close to the nucleus and are sensitive probes of its charge. The optical spectrum, on the other hand, is asso ciated with transitions of the outermost or valence elec trons, which are heavily screened from the nucleus by the remaining Z — 1 electrons of the atom; they are not sensi tive probes o f the nuclear charge.
using Eq. 18 of Chapter 51 for the energy levels. For the
Ka transition of Fig. 3, we can replace Z h y Z —b and substitute 1 for m and 2 for n. Doing so yields
Taking the square root of each side leads to yTv
Bohr Theory and the Moseley Plot Bohr’s theory works well for hydrogen but fails for atoms with more than one electron (in part because it does not include the repulsive interaction between the electrons). Nevertheless, it provides an excellent first approximation in accounting for the Moseley plot of Fig. 4. Consider an electron in the L shell of an atom that is about to move into a hole in the K shell, emitting a x-ray photon in the process. Using Gauss’ law (see Eq. 17 o f Chapter 29), we find that the electric field at the loca tion o f the L electron is determined by the charge enclosed in an imaginary sphere of radius equal to the radial coordi nate of the L electron. This sphere encloses a charge + Ze from the nucleus and a charge —e from the single re maining K electron. We say that the K electron “screens” the charge o f the nucleus. In part because o f this screening and in part because o f readjustments that take place in the electron cloud as a whole, the effective atomic number for the transition turns out to be Z — (>, where = 1. Bohr’s formula for the frequency of the radiation corresponding to a transition in a hydrogen-like atom between any two atomic levels differing in energy by A £ is
_ (
Yi l / 2
(Z-b),
which we can write in the form
•Jv = aZ —ab.
=
8 eS*> \ m ‘
nV ’
' '
(4)
where a is the indicated constant and b== I. Equation 4 represents a straight line, in full agreement with the experimental data of Fig. 4. If this plot is ex tended to higher atomic numbers, however, it turns out to be not quite straight but somewhat concave upward. Nev ertheless, the quantitative agreement with Bohr theory is surprisingly good, as Sample Problem 2 shows.
Sample Problem 2 Calculate the value of the quantity a in Eq. 4 and compare it with the measured slope of the straight line in Fig. 4. Solution
Comparing Eqs. 3 and 4 allows us to write
>/3 (9.11 X 1Q-J‘ kg)‘^(1.60 X IQ-'* Q " 4>/2 (8.85 X 10-'^ F/mX6.63 X 10"^ J-s)^^^
v = —
(3)
= 4.95X 10’ H z '” .
Section 52-3 Careful measurement of Fig. 4, using the triangle hgj, yields hg
(40-11)
which is in agreement with the value predicted by Bohr theory within the uncertainty of the graphical measurement. Note also that the intercept in Fig. 4 is in fact close to 1, as expected from our screening argument. The agreement with Bohr theory is not nearly as good for other lines in the x-ray spectrum, corresponding to the transi tions of electrons farther from the nucleus; here we must rely on calculations based on wave mechanics.
Sample Problem 3 A cobalt target is bombarded with elec trons, and the wavelengths of its characteristic spectrum are measured. A second, fainter, characteristic spectrum is also found, because of an impurity in the target. The wavelengths of the Ka lines are 178.9 pm (cobalt) and 143.5 pm (impurity). What is the impurity? Solution Let us apply Eq. 4 both to cobalt and to the impurity. Putting c/X for v (and assuming b = 1), we obtain aZ co^^
and
\ — = a Z x — ci.
Dividing yields A c.
Z x-1 ^Co 1
Building Atoms
1099
earth, or lanthanide, series o f elem ents, all cram m ed into one square o f the table. In short, wave m echanics, supple m ented by certain guiding principles th at we discuss in this section, accounts for every feature o f this table and thus, essentially, for all o f chem istry. Let us im agine th a t— in T inker Toy fashion— we are going to construct a typical atom for each o f the m ore than 100 elem ents th at m ake up the periodic table. O ur starting m aterials will be a supply o f nuclei, each charac terized by a charge + Ze, w ith Z ranging by integers from 1 to over 100. We also need an am ple supply o f electrons. O ur plan is to add Z electrons to each nucleus in such a way as to produce a neutral atom in its ground state. Success follows only if we observe these three principles o f atom building:
1. The quantum number principle. T he electron in a hydrogen atom m a y — to m ention one possibility— be in a state described by the q u an tu m num bers n = 2 , 1= 1, m/ = -h 1, and = —i. It tu rn s out th at a particular elec tron in any atom m ay also be fully identified by this sam e set o f q u an tu m num bers. T h at is not to say th at the elec trons in these different cases will m ove in the sam e way, because they will not. P ut an o th er way, although these electrons m ay share the sam e set o f q u an tu m num bers, the potentials in which they m o v e— and thus their wave fun ctio n s— will be quite different. Specifically, the q u an tu m n u m b er principle asserts:
Substituting gives us 178.9 p m _ Z x - 1 143.5 pm 27-1
V
Solving for the unknown, we find Zx = 30.0; a glance at the periodic table identifies the impurity as zinc.
52-3 BUILDING ATOMS In the preceding section we saw how, by m easuring the w avelengths o f the characteristic x-ray spectrum o f an elem ent, we could assign an atomic number Z to each elem ent an d thus string them in a line according to a logical principle. H ere we go a step further. W e try to see how far the principles o f wave m echanics can take us in breaking up this line in to a series o f segm ents, corresponding to the horizontal periods o f the periodic table o f the elem ents. T he attem p t m eets w ith essentially total success. Every detail o f the table (see A ppendix E) can be accounted for, including ( 1) the n um bers o f elem ents in the seven hori zontal periods into w hich the table is divided, ( 2 ) the sim ilarity o f the chem ical properties o f the elem ents in the various vertical c o lu m n s— the alkali m etals an d the inert gases, for ex am p le— an d (3) the existence o f the rare
The hydrogen atom quantum numbers can be used to describe electron states and to assign electrons to shells and subshells, in any atom, no matter how many electrons it contains. Furthermore, the restric tions among the quantum numbers discussed in Sec tion 51-6 remain in force. 2. The Pauli exclusion principle. T his powerful principle was p u t forward by the A ustrian-born physicist W olfgang Pauli in 1925. Speaking generally, it tells us th at no two electrons can be in the sam e state o f m otion at the sam e tim e. M ore specifically, it asserts that:
In a multielectron atom there can never be more than one electron in any given quantum state. That is, no two electrons in an atom can have the same set of quantum numbers. If this principle did not hold, all the electrons in an atom w ould pile up in its K shell, and chem istry as we know it w ould not exist. Y ou w ould not be here to read this sentence, and we w ould n o t have been here to have w ritten it. P au li’s exclusion principle is no trivial asser tion! 3. The minimum energy principle. As we fill subshells with electrons in the course o f atom building, the question arises: In w hat order shall we fill them ? T he answ er is:
1100
Chapter 52 Atomic Physics
When one subshell is filled, put the next electron in which ever vacant subshell will lead to an atom lowest in energy. To do otherwise would be to depart from our stated aim o f building atoms in their ground states. The lowest-energy subshell can be identified with the help o f the following rule, which we first state and then try to make reasonable:
For a given principal quantum number n in a multi electron atom, the order of increasing energy of the subshells is the order of increasing 1. Table 1 helps to clarify this rule. Consider first a hydro gen atom whose single electron is in a state with « = 4. There are four allowed values of /, namely, 0, 1,2, and 3. For electrons in true one-electron atoms— such as hydrogen— the energy does not depend on / at all but only on n, being given by Eq. 18 of Chapter 51,
E= -
m ^e^ Se^h^ n^ ’
« = 1, 2, 3, . . .
(5 )
Recall that this relation is predicted not only by Bohr theory but also by wave mechanics. Putting Z = 1 and n = 4 in this relation yields, for hydrogen, £ = —0.85 eV, as Table 1 shows. Consider now a lead nucleus (Z = 82) around which only a single electron circulates, again in a state with n = 4. Equation 5 also applies to this (admittedly rather un likely) one-electron atom. For Z = 82 and n = 4, the table shows that we have E = —5720 eV, once more indepen dent o f /. The electron moves in the field of a nucleus with a charge o f + %2e\ furthermore, it is drawn in very close to this nucleus, the equivalent Bohr orbit radius (see Eq. 19 o f Chapter 51) being 82 times smaller than for hydrogen. Finally, let us construct a normal, neutral lead atom by “sprinkling in” the missing 81 electrons. The outermost or valence electrons in lead have n = 6, so that an electron with n = 4 would lie somewhere in the middle o f the smeared-out electron cloud surrounding the lead nucleus. Equation 5 no longer holds for this multielectron atom, but we can find the energies of the four « = 4 subshells
TABLE 1 Orbital Quantum Number 1
0 1 2 3
experimentally from x-ray studies. Their approximate values are shown in the last column o f Table 1. We see at once that they lie higher in energy (that is, the binding energies are smaller) than for our hypothetical oneelectron lead “atom” and that they vary with / just as the minimum energy rule predicts. That the electrons in lead become more loosely bound when the entire electron cloud is present follows because some substantial fraction o f this cloud screens the nucleus electrically. A typical« = 4 electron no longer “sees” the full positive nuclear charge, but rather sees this charge reduced by the negative charge of that part of the electron cloud that lies between the nucleus and the effective radius of the electron in question. As for the variation o f energy with /, let us ask ourselves what an / = 0 orbit would have to look like under the mechanical constraints of the Bohr picture. Truly to have no angular momentum, the electron would have to oscil late back and forth on a straight line segment passing directly through the nucleus. This does not happen, of course. The equivalent wave-mechanical statement is that an electron with / = 0 must spend a larger fraction of its time near the nucleus than do electrons with higher values of /. Such electrons would then, on the average, “see” a higher effective nuclear charge and would be more tightly bound; they would lie lower in energy, just as the minimum energy principle and Table 1 predict. It is inter esting to compare Figs. 17 and 18 o f Chapter 51, which show the « = 2, / = 0 and n = 2 , l = I states of hydrogen. In the / = 0 state there is indeed a marked tendency for the electron to cluster near the nucleus— note the close-in secondary maximum— just as our qualitative argument suggests that it would.
52-4 THE PERIODIC TABLE________ Figure 5 shows how the periodic table is put together, using the three rules for atom building that we have de scribed in the previous section. Energy increases upward
ENERGY LEVELS FOR ELECTRONS WITH « = 4, IN THREE DIFFERENT ATOMS Energy (eV) Hydrogerf Z= 1
''Lead”^ Z=82
LeacE Z=82
-0 .8 5 -0 .8 5 -0 .8 5 -0 .8 5
-5 7 2 0 -5 7 2 0 -5 7 2 0 -5 7 2 0
-8 9 0 -7 1 0 -4 2 0 -1 4 0
“ A neutral hydrogen atom; see Eq. S. * A hypothetical one-electron atom with Z = 82; see Eq. 5. An actual neutral lead atom (Z = 82); data from experiment.
Section 52-4
The Periodic Table
6d Lr Rf Ha
Periods
103104 105106107 108109
uT
5/- Ac Th Pa U Np AmC Pu mBk Cf Es FmMd No
1101
89 90 91 92 93 94 95 96 97 98 99 100101102
J
^ Fr I Ra 87 88
5d
u/
Lu Hf Ta W Re Os lr Pt Au Hg 71 72 73 74 75 76 77 78 79 80
uT*
4f La Ce Pr Nd PmSm Eu Gd Tb Dy Ho Er Tm Yb
Tl Pb Bi Po At w 81 82 83 84 85 m
57 58 59 60 61 62 63 64 65 66 67 68 69 70 1=3 (fourteen states)
4d
Y Zr Nb Mo Tc Ru Rh Pd Ag. Cd 39 40 41 42 43 44 45. 46 47 48
3d
Sc Ti V Cr Mn Fe Co Ni Cu Zn 21 22 23 24 25 26 27 28 29 30
c 1=2 (ten states)
L Inert gases ISJ (end of periods)
J2s 1=
1
Li Be 3 4
Is H w IIs l J
Alkali metals (beginning of periods)
JL JL /=0
(two states)
Figure 5 Starting with hydrogen at the bottom, the curved line shows the sequence of the seven horizontal periods of the periodic table. Each period starts with an alkali metal and ends with an inert gas.
in this figure. States with the same value of / have been displaced to the left for clarity and grouped into columns according to their / value. Before we look more closely at this table, we introduce a new notation for the angular momentum quantum num ber /. For historical reasons* the values of / have been
* The letters s, p,
given letter equivalents, according to this scheme: /
0 1
2 3
4
5...
Symbol
s p
d f
g h . . .
In this notation, a state with n = 1 and / = 0 is called a “ 1s” state. Similarly, a state with n = 4 and / = 3 is called a “4 / ” state, and so on. These states are also known as
subshells. The dependence of energy on / is a dominant feature o f Fig. 5. Look, for example, at the sequence of states 4i, 4p,
1102
Chapter 52
Atomic Physics
Ad, and Af. They lie in the figure in the order of increasing energy, just as the minimum energy rule requires. In fact, the Afstates lie so high that they are above the 5^ and the 5p states, which have a higher value of n. The term shell is used to designate a group of states, lying close together in energy, that has a particular stabil ity when those states are fully occupied. When dealing with the hydrogen atom (for which the energy depends only on the principal quantum number n), we identified the shells by giving the value of that quantum number. We now see that, in many-electron atoms, the principal quantum number alone is no longer a good indicator. As Fig. 5 shows, the shell we have labeled “ 6 ,” corresponding to the sixth horizontal period of the periodic table, does indeed contain all the 6s and 6p states, corresponding to n = 6 . However, it also contains all the Af and 5d states. Moreover, the 6d states do not lie in this shell at all but in the shell above. By starting with hydrogen in Fig. 5 and following the curved line, we can see how the seven horizontal periods o f the periodic table are built up, each starting with an alkali metal and ending with an inert gas. Consider again the long sixth period, which starts with the alkali metal cesium (Z = 55) and ends with the inert gas radon (Z = 8 6 ). The order in which the subshells are filled, as the curved line indicates, is 6 5 , Af 5d, and 6p. The sixth period contains a run of 15 elements (Z = 5 7 to Z = 71), listed separately at the bottom of the periodic table in Appendix E. These elements are called the rare earths or lanthanides (after the element lanthanum that begins the series). Their chemical properties are so similar that they are all grouped into a single square of the table. This similarity arises because, while the Af state is being filled deep within the electron cloud, an outer screen of one or two 6s electrons remains in place. It is these outer most electrons that determine the chemical properties of the atom. A similar series (the actinides) occurs in the seventh period. The maximum number of electrons permitted in any subshell is 2(2/ + 1). This follows from the Pauli princi ple; for any value o f /, there are 21 + 1 different w , values, and for each of those there are 2 values. There are thus 2(2/ + 1) different possible labels for electrons in any sub shell, and by the Pauli principle each electron in an atom must have a different label. If you count the number of elements in each of the labeled subshells o f Fig. 5, you will find there to be 2(2 / + 1); that is, 2 for 5 subshells (/ = 0), 6 for p subshells ( / = 1), 10fori/subshells(/= 2), and 14for / subshells (1—3).
Electron Configurations We can describe the lowest energy state of an atom by giving its electron configuration, that is, by specifying the number o f electrons in each occupied state. For example, for lithium (Z = 3) we have, from Fig. 5, ls^2s', where
the superscript indicates the number of electrons in that state. Consider these three configurations: F (Z = 9): Ne(Z=10): Na(Z=ll):
\s^2s^2p^ ls^2s^2p^ ls^2s^2p^3s'
Neon has a filled 2p state, a particularly stable configura tion. It takes considerable energy to break this configura tion, and therefore neon does not readily give up elec trons. There is a particularly large gap between neon and the next element (sodium), which indicates that neon is also reluctant to accept another electron. Neon is corre spondingly an inert gas\ under most circumstances, it does not form compounds with other elements. The ele ments in the column above Ne in Fig. 5 (or below Ne in Appendix E) are also inert gases. (This property depends on the filling of an entire shell, not just a particular sub shell. The elements Zn, Cd, and Hg, for example, all have filled d states, but all form compounds readily.) Fluorine, on the other hand, lacks one electron from a filled 2p state. Since the filled state is a stable configura tion, fluorine readily accepts an electron from another atom to form compounds. Elements in the column above fluorine in Fig. 5 (or below F in Appendix E) behave similarly; they are collectively known as halogens. For another example, sodium has a single electron in the 3s state. This electron is not particularly tightly bound, and sodium can give up that electron in forming chemical compounds with other atoms (as in NaCl, for example). The elements with a single s electron, called the alkali elements, have similar properties.
Ionization Energy The energy needed to remove the outermost electron from an atom is called its ionization energy. Figure 6 shows the ionization energies of the elements. Note the regular behavior that is consistent with the electron config urations. For each shell, the ionization energy rises gradu ally and reaches a maximum at an inert gas, and there is a sharp drop for the alkali element that follows. The ele ment Na, for example, has an ionization energy of 5.14 eV. To remove a second electron from Na, however, takes nearly an order o f magnitude more energy (47.3 eV); with one electron removed, an Na ion has an electron configuration similar to inert Ne, consisting of a filled shell, in which the electrons are more tightly bound. There are occasionally small irregularities in the order in which the states are filled. For example, from Fig. 5 we would expect copper (Z = 29) to have the outer configura tion As‘^3d^. However, it is energetically favorable for one o f the 45 electrons to complete the filling of the 3i/state; as a result, the outer configuration of Cu is 45'3i/'°. The single 45 electron is responsible for the large electrical conductivity o f copper. A similar situation occurs for silver (Z = 47) and gold (Z = 79).
Section 52-4
The Periodic Table
1103
Figure 6 The ionization energies of the elements plotted against their atomic number. Subshell labels are indicated.
Excited States and Optical Transitions So far we have discussed only the minimum-energy or ground-state configuration of atoms. When we add en ergy to the atom, such as when we place it in an electric discharge tube or illuminate it with radiation, we can cause the electrons to move to higher states. If we supply sufficient energy, it is possible to remove an electron com pletely, thereby ionizing the atom. If an inner electron is removed, the ensuing filling of levels gives the characteris tic X rays, as we have discussed in Section 52-1. The energy differences are typically o f the order o f eV between the minimum-energy state of the outer electron and the next higher-lying states to which it can be excited. When the electron drops back to its lowest energy, the atom emits radiation o f energy in the eV range, that is.
visible light. For this reason, such changes in the state of the electron are called optical transitions. Using the wave functions corresponding to the various states, it is possible to calculate the relative probability for different transi tions to occur. When we do so, we find that transitions that change / by one unit are strongly favored over transi tions that change / by any other amount. Such a conclu sion is called a selection rule. For all electromagnetic tran sitions in atoms (optical, x-ray, and so forth), the selection rule is A/ = ± l . (6) Selection rules are often not absolute. In atoms it is possi ble to observe transitions corresponding to other changes in /; they are just far less likely to occur. Figure 7 shows some excited states in sodium and some
Figure 7 The excited states of sodium. Some emitted radia tions are indicated. Note the operation of the A/ = ± 1 selec tion rule.
0>
1104
Chapter 52
Atomic Physics
transitions th at m ay occur (not all o f w hich are in the optical region). M ost o f the states are actually close-lying doublets, which give two spectral lines o f nearly equal energies, such as the fam iliar sodium doublet, which, as you can see from Fig. 7, corresponds to the electron m ak ing a transition from the first excited state (3/?) back to the ground state.
52-5 LASERS AND LASER LIGHT In the late 1940s and again in the early 1960s q u an tu m physics m ade two eno rm o u s contrib u tio n s to technology, the transistor and the laser. T he first stim ulated the growth o f electronics, w hich deals with the interaction (at the q u a n tu m level) betw een electrons and bulk m atter. T he laser has led to a new field— som etim es called photonics— which deals with the interaction (again at the q u a n tu m level) betw een photons and bulk m atter. T o see the im portance o f lasers, let us look at som e of the characteristics o f laser light (see Fig. 8 ). W e shall com pare it as we go along with the light em itted by such sources as a tungsten filam ent lam p (continuous spec tru m ) o r a neon gas discharge tube (line spectrum ). We shall see th at referring to laser light as “ the light fantastic” goes far beyond whimsy. 1. L aser light is highly m onochrom atic. T ungsten light, spread over a contin u o u s spectrum , gives us no basis for com parison. T he light from selected lines in a gas dis charge tube, however, can have wavelengths in the visible region th at are precise to about 1 part in 10^. T he sharp-
Figure 9 The NOVA laser room at the Lawrence Livermore National Laboratory. These lasers, with a power of about lO'"* W, are used in controlled thermonuclear fusion research (see Section 55-10).
ness o f definition o f laser light can easily be a thousand tim es greater, or 1 part in 10’. 2. Laser light is highly coherent. W avetrains for laser light m ay be several hundred kilom eters long. Interfer ence fringes can be set up by com bining two beam s that have followed separate paths whose lengths differ by as m uch as this am ount. T he corresponding coherence length for light from a tungsten filam ent lam p or a gas discharge tube is typically considerably less th an 1 m. 3. L aser light is highly directional. A laser beam departs from strict parallelism only because o f diffraction effects, determ ined (see Section 46-4) by the wavelength and the diam eter o f the exit aperture. Light from other sources can be m ade into an approxim ately parallel beam by a lens or a m irror, b u t the beam divergence is m uch greater than for laser light. For exam ple, focused light from a tungsten filam ent source form s a beam , the angular diver gence o f which is determ ined by the spatial extent o f the filam ent. 4. Laser light can be sharply focused. This property is related to the parallelism o f the laser beam . As for star light, the size o f the focused spot for a laser beam is lim ited only by diffraction effects and not by the size o f the source. Flux densities for focused laser light o f 10*^ W /cm ^ are readily achieved. An oxyacetylene flame, by contrast, has a flux density o f only 10^ W /cm^.
Figure 8
Laser beams light up the sky.
The sm allest lasers, used for telephone com m unication over optical fibers, have as their active m edium a semi-
Section 52-6
Einstein and the Laser
1105
Figure 10 A Michelson interferometer using lasers at the National Institute of Standards and Technology, used to mea sure the jc, z coordinates of a point in space with extreme precision.
conducting gallium arsenide crystal ab o u t the size o f a pinhead. T he largest lasers, used for laser fusion research (see Fig. 9), fill a large building. They can generate pulses o f laser light o f 10“ s du ratio n , which have a pow er level o f 10*"* W during the pulse. T his is about 100 tim es the total pow er-generating capacity o f all the electric power stations on Earth. O ther laser uses include spot-w elding detached retinas, drilling tiny holes in diam o n d s for draw ing fine wires, cutting cloth (50 layers at a tim e, with no frayed edges) in the garm ent industry, precision surveying, precise length m easurem ents by interferom etry, precise fluid-flow veloc ity m easurem ents using the D oppler effect, an d the gener ation o f hologram s (see Section 47-5). Figure 10 shows an o th er exam ple o f laser technology, nam ely, a facility at the N ational Institute o f Standards an d Technology used to m easure the x, y, and z coordi nates o f a point, by laser interference techniques, with a precision o f ± 2 X 10"® m ( = ± 2 0 n m ) . It is used for m easuring the dim ensions o f special three-dim ensional gauges, which, in tu rn , are used in industry to check the dim ensional accuracy o f com plicated m achined parts. A nu m b er o f laser beam s, visible by scattered light, appear in the figure.
not appear until 1960, the groundw ork for its invention was put in place by Einstein’s work. T he im portance o f stim ulated em ission is indicated by the nam e “ laser,” which is an acronym for light am plification by the stim u lated em ission o f radiation. W hat was Einstein w orking on w hen the concept o f stim ulated em ission occurred to him ? N othing other than the cavity radiation problem , which in the hands o f Planck and others established the new science o f q u an tu m m echanics. In 1917 Einstein succeeded in deriving the Planck radiation law in term s o f beautifully sim ple as sum ptions and in a way th at m ade quite clear the role o f energy quantization and the photon concept.* It is interesting th at Einstein was also thinking deeply about this sam e fundam ental cavity radiation problem when, in 1905, he first proposed the concept o f the photon and realized th at the photoelectric effect could be ex plained with its use. W e learn from both o f these exam ples th at practical devices o f m ajor im portance can flow from a concern over problem s th at seem to have no relevance to technology. W hen you next see a photoelectric elevator door opener or listen to a com pact-disc stereo system, think o f Einstein. Now let us take a look at three processes th at involve the interaction between m atter and radiation. Tw o o f them , absorption and spontaneous em ission, have long been fam iliar; the third is stim ulated em ission.
52-6 EINSTEIN AND THE LASER In 1917 Einstein introduced into physics a new concept, th at o f stim u la ted em ission, w hich we shall define and discuss below. Even though the first operating laser did
* See Robert Resnick and David Halliday, Basic Concepts in Relativity and Early Quantum Theory, 2nd edition (Wiley, 1985), Supplementary Topic E.
1106
Chapter 52
Atomic Physics
1. Absorption. Figure 11a suggests an atomic system in the lower o f two possible states, of energies E^ and E^. A continuous spectrum o f radiation is present. Let a photon from this radiation field approach the two-level atom and interact with it, and let the associated frequency v o f the photon be such that
hv = E2 —E f
(7 )
The result is that the photon vanishes and the atomic system moves to its upper energy state. We call this pro cess absorption. 2. Spontaneous emission. In Fig. 116 the atomic system is in its upper state and there is no radiation nearby. After a mean time r, this (isolated) atomic system moves of its own accord to the state of lower energy, emitting a photon o f energy hv{= E2 —Ei)in the process. We call this pro cess spontaneous emission, in that no outside influence triggered the emission. Normally the mean life t for spontaneous emission by excited atoms is of the order of 10“* s. However, there are some states for which t is much longer, perhaps 10“^ s. We call such states metastable\ they play an essential role in laser operation. (They have such long lifetimes because they can emit radiation only through processes that vio late the selection rule of Eq. 6.) The light from a glowing lamp filament is generated by spontaneous emission. Photons produced in this way are totally independent of each other. In particular, they have different directions and phases. Put another way, the light they produce has a low degree of coherence. 3. Stimulated emission. In Fig. 11 c the atomic system is again in its upper state, but this time radiation o f fre quency given by Eq. 7 is present. As in absorption, a pho ton o f energy hv interacts with the system. The result is
Before
that the system is driven down to its lower state, and there are now two photons where only one existed before. We call this process stimulated emission. The emitted photon in Fig. 11 c is in every way identical with the “triggering” or “stimulating” photon. It has the same energy, direction, phase, and state of polarization. Furthermore, each of these two photons can cause an other stimulated emission event, giving a total o f four photons, which can cause additional stimulated emis sions, and so on. We can see how a chain reaction of similar processes could be triggered by one such event. This is the “amplification” of the laser acronym. The photons have identical energies, directions, phases, and states of polarization. This is how laser light acquires its characteristics. Figure 11 refers to the interaction with radiation o f a single atom. In the usual case, however, we find ourselves dealing with a large number of atoms. For the two-level system of Fig. 11, how many of these atoms will be in level El and how many in level £ 2 ? In any system at thermal equilibrium, the number occupying a state at energy E is determined by the exponential factor in the Maxwell-Boltzmann distribution (see Eqs. 27 and 32 of Chapter 24). The ratio o f the number of atoms in the upper level to the number in the lower level is n(E 2)/n(E i) =
Figure 12a illustrates this situation. The quantity kTis the mean energy of agitation o f an atom at temperature T, and we see that the higher the temperature the more atoms— on long-term average— will be “bumped up” by thermal agitation to the level £ 2 - Because £ 2 > £ 1 , the ratio n (£ 2)/« (£ i) will always be less than unity, which means that there will always be fewer atoms in the higher
After
Process
-E2
-E2
ia) 'V Wh vW - ^
None
Absorption
-El
-El
-£2
(h)
■^’2
. hv "W W V -^
Spontaneous emission
None
-El
-E l
-£2
I \ U’) ' v w w - ^
Stimulated emission
-^1 J
Radiation
Figure 11
.
'\A A A A A /-^ |
-■------ E l J
L
Matter
Matter
hv
L
Radiation
The interaction of matter and radiation for the processes of (a) absorption,
(b) spontaneous emission, and (c) stimulated emission.
( 8)
Section 52-7 How a Laser Works ••
El
ia)
. E2
■E2. ••
.£1
ih)
Figure 12 (a) The normal thermal equilibrium distribution of atomic systems occupying one of two possible states, (b) An inverted population distribution, which can be obtained using special techniques.
energy level than in the lower. This is what we would expect if the level populations are determined only by the action of thermal agitation. If we expose a system like that of Fig. 1 2a to radiation, the dominant process— by sheer weight of numbers— will be absorption. However, if the level populations were inverted, as in Fig. 126, the dominant process in the pres ence of radiation would be stimulated emission, and with it the generation of laser light. A population inversion like that o f Fig. 126 is not a situation that is obtained by ther mal processes; we must use clever tricks to bring it about.
52-7 HOW A LASER WORKS_______ Figure 13 shows schematically how a population inver sion can be achieved so that laser action— or “lasing” as it is called— can occur. Atoms from the ground state are “pumped” up to an excited state £ 3 , for example by the absorption of light energy from an intense, continuousspectrum source that surrounds the lasing material. From £ 3 the atoms decay rapidly to a state of energy £ 2 • For lasing to occur this state must be metastable; that is, it must have a relatively long mean life against decay by spontaneous emission. If conditions are right, state £ 2 can then become more heavily populated than state £ , , thus providing the needed population inversion. A stray pho ton o f the right energy can then trigger an avalanche of stimulated emission events, resulting in the production of laser light. A number of lasers using crystalline solids
Pumping (optical)
Laser light
Figure 13 The basic three-level scheme for laser operation. Metastable state £2 has a greater population than the ground state £ ,.
1107
(such as ruby) as a lasing material operate in this threelevel mode. Figure 14 shows the elements of a type of laser that is often found in student laboratories. The glass discharge tube is filled with an 80%-20% mixture of the inert gases helium and neon, the helium being the “pumping” me dium and the neon the “lasing” medium. Figure 15 is a simplified version of the level structures for these two atoms. Note that four levels, labeled £ q, £ 1 , £ 2 , and £ 3 , are involved in this lasing scheme, rather than three levels as in Fig. 13. Pumping is accomplished by setting up an electrically induced gas discharge in the helium -neon mixture. Elec trons and ions in this discharge occasionally collide with helium atoms, raising them to level £ 3 in Fig. 15. This level is metastable, spontaneous emission to the ground state (level £ q) being very infrequent. Level £ 3 in helium (= 20.61 eV) is, by chance, very close to level £ 2 in neon (= 20.66 eV), so that, during collisions between helium and neon atoms, the excitation energy of the helium can readily be transferred to the neon. In this way level £ 2 in Fig. 15 can become more highly populated than level £ , in that figure. This population inversion is maintained because ( 1 ) the metastability of level £ 3 ensures a ready supply of neon atoms in level £ 2 and (2 ) level £ , decays rapidly (through intermediate stages not shown) to the neon ground state, £ 0 . Stimulated emission from level £ 2 to level £ 1 predominates, and red laser light of wavelength 632.8 nm is generated. Most stimulated emission photons initially produced in the discharge tube of Fig. 14 will not happen to be parallel to the tube axis and will be quickly stopped at the walls. Stimulated emission photons that parallel to the axis, however, can move back and forth through the dis charge tube many times by successive reflections from mirrors and M2 . These photons can in turn cause other stimulated emissions to occur. A chain reaction thus builds up rapidly in this direction, and the inherent parallelism of the laser light results. Rather than thinking in terms of the photons bouncing back and forth between the mirrors, it is perhaps more useful to think of the entire arrangement of Fig. 14 as an optical resonant cavity that, like an organ pipe for sound waves, can be tuned to be sharply resonant at one (or more) wavelengths. The mirrors and M 2 are concave, with their focal points nearly coinciding at the center o f the tube. Mirror M 1 is coated with a dielectric film whose thickness is care fully adjusted to make the mirror as close as possible to totally reflective at the wavelength of the laser light; see Section 45-4. Mirror M2 , on the other hand, is coated so as to be slightly “leaky,” so that a small fraction of the laser light can escape at each reflection to form the useful beam. The windows W in Fig. 14, which close the ends of the discharge tube, are slanted so that their normals make an
1108
C hapters!
Atomic Physics
angle 0p, the Brewster angle, with the tube axis, where tan 0p = n.
E , - E , = hv = f j
(9)
_ (6.63 X 10-^ J-SX3.00 X 10* m/s) (550 X 10-’ mX1.60X 10-'’ J/eV)
n being the index of refraction of the glass at the wave length of the laser light. In Section 48-3 we showed that such windows transmit light without loss by reflection, provided only that the light is polarized with its plane of polarization in the plane of Fig. 14. If the windows were square to the tube ends, beam loss by reflection (about 4% from each surface of each window) would make laser operation impossible.
= 2.26 eV. The mean energy of thermal agitation is equal to k T ^ (8.62 X 10-5 eV/KX300 K) = 0.0259 eV. From Eq. 8 we have then, for the desired ratio, =
Sample Problem 4 A three-level laser of the type shown in Fig. 13 emits laser light at a wavelength o f550 nm, near the center of the visible band, (a) If the optical pumping mechanism is shut off, what will be the ratio of the population of the upper level (energy F j) to that of the lower level (energy E i)? Assume that T = 300 K. {b) At what temperature for the conditions of (a) would the ratio of populations be i? Solution (a) From the Bohr frequency condition, the energy difference between the two levels is given by
^-(2.26 eV)/(0.0259 eV) = ^-87.3 = J 3 X 10“ 5®.
This is an incredibly small number. It is not unreasonable, how ever. An atom whose mean thermal agitation energy is only 0.0259 eV will not often impart an energy of 2.26 eV (87 times as great) to another atom in a collision. (b) Setting the ratio in Eq. 8 equal to taking the natural logarithm of each side, and solving for T yields ^ _ E ,-E ,_ 2.26 eV A:(ln 2) (8.62 X 10” 5 eV/KXO.693) = 37,800 K.
Figure 15 The atomic levels involved in the operation of a H e-N e gas laser.
Section 52-8 This is much hotter than the surface of the Sun. It is clear that, if we are to invert the populations of these two levels, a special mechanism is needed. Without population inversion, lasing is not possible.
Sample Problem 5 A pulsed ruby laser has as its active element a synthetic ruby crystal in the form of a cylinder 6 cm long and 1 cm in diameter. Ruby consists of AI2O 3 in which— in this case— one aluminum ion in every 3500 has been replaced by a chromium ion, C t^^. It is in fact the optical absorption proper ties of this small chromium “impurity” that account for the characteristic color of ruby. These same ions also account for the lasing ability of ruby, which occurs— by the three-level mecha nism of Fig. 13— at a wavelength of 694.4 nm. Suppose that all the Cr^"^ ions are in a metastable state corre sponding to state £*2 of Fig- i 3 and that none are in the ground state £ j . How much energy is available for release in a single pulse of laser light if all these ions revert to the ground state in a single stimulated emission chain reaction episode? Our answer will be an upper limit only because the conditions postulated cannot be realized in practice. The density p of AI2O 3 is 3700 kg/m^, and its molar mass M is 0.102 kg/mol. Solution
The number of AF*^ ions is _ 2 N ^ m _ 2N^pV M M
where m is the mass of the ruby cylinder and the factor 2 ac counts for there being two aluminum ions in each “molecule” of AI2O 3. The volume F is
Molecular Structure
1109
52-8 MOLECULAR STRUCTURE Understanding the structure of atoms is the first step in the process that eventually leads to the understanding of the structure of macroscopic objects. The next step is to understand how atoms join together to form molecules. The force responsible for binding atoms together in molecules is the same electrostatic force that binds elec trons in atoms. However, atoms are ordinarily electrically neutral and would thus exert no electrostatic force on one another. To have molecular bonds between atoms, there must therefore occur some readjustment of the electronic structure of the atoms. Consider, for example, a molecule o f hydrogen, H 2 (Fig. 16a). The separation between the two protons in a molecule o f H 2 is measured to be 0.074 nm, which is comparable to the radius of the lowest electronic orbit in atomic hydrogen, 0.0529 nm. The single electron in atomic hydrogen would like to acquire a partner to fill the b shell, and therefore the electron from one H atom can be regarded as pairing with the other electron in the same shell. (Of course, the electrons are identical, and in accord ance with the quantum rules for indistinguishable parti cles, we can no longer speak o f the electrons from the two original atoms as having separate identities. Instead, a pair o f electrons, represented by a two-electron wave func-
F = (7r/4X 1.0 X 10-2 m)2(6.0 X 10” 2 m) = 4.7X 10-"^ m^ Thus _ (2X6.0 X 102VmolX3.7 X 10^ kg/m^X4.7 X 10"^ m^) 0.102 kg/mol = 2.1 X 1023.
(a )
Separation (jistance (nm)
The number of Cr^^ ions is then
The energy of the stimulated emission photon is , E=
he (4.1 X 10-” eV-sX3.0X 10*m/s) = T ----------------694X 10-’ m----------------
„
and the total available energy per laser pulse is U = N c r E = ( 6. 0X 10*’X1.8 eVX1.6X 10-'’ J/e V )= 17 J. Such large pulse energies have indeed been achieved, but only by much more elaborate laser arrangements than that described here. In this example we have postulated an ideal circumstance, namely, a total population inversion, in which the ground state remains virtually unpopulated. The actual population inversion in a working ruby laser will be very much less than total. For this and other reasons the pulse energy in practice will be very much less than the upper limit calculated above.
Figure 16 (a) The overlap of the s electrons in H is responsi ble for the formation of the H 2 molecule, (b) The total energy of the two electrons in the bound state of the H 2 molecule, as a function of the atomic separation distance. When the sepa ration is large, the energy is —27.2 eV (twice the energy of the single electron in atomic hydrogen, —13.6 eV). The mini mum energy of the bound molecule is —31.7 eV when the separation is 0.074 nm.
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Chapter 52 Atomic Physics
tion, m oves in the com bined electrostatic field o f the two protons.) W e can regard this binding as originating from a shar ing o f electrons. T he electrons “ belong to ” neither proton. In effect, each p roton attracts the pair o f electrons, and this attraction is sufficient to overcom e the C oulom b re pulsion o f the protons. (A sim ilar effect, involving a very different type o f shared particle, is responsible for the binding o f tw o protons in the nucleus o f an atom .) T his type o f m olecular bonding, based on shared elec trons, is called covalent bonding. M olecules containing tw o atom s o f the sam e elem ent are co m m o n exam ples o f those having covalent bonds. A m easure o f the strength o f the bond is the dissociation energy, the energy necessary to break the m olecule into tw o neutral atom s. In H 2, the dissociation energy is 4.5 eV, as indicated in Fig. 16ft. It is also possible to have covalent bonding in atom s in which the o u ter electrons are in the p shell, such as in the case o f N 2 o r O 2. In the case o f N 2, the sharing o f the three 2p electrons from each ato m gives a total o f six 2p electrons, a configuration th at w ould (in a single ato m ) correspond to a filled shell; the N 2 m olecule is therefore very stable (the energy needed to dissociate the m olecule is 9.8 eV). In O 2, on the oth er hand, there are eight 2/? electrons, w hich is a less stable configuration; the dissociation en ergy o f O 2 is only 5.1 eV. In practical term s, this difference m akes O 2 m ore reactive th an N 2. M olecules o f O 2 can be broken by relatively m odest chem ical reactions, as, for exam ple, the oxidation o f m etals exposed to air. T he F 2 m olecule (ten 2p electrons) is even less stable than O 2; its dissociation energy is only 1.6 eV, less th an the energy o f p hotons o f visible light, and as a result, F 2 can be disso ciated by exposure to light. C ovalent m olecular bonds can also be form ed between dissim ilar atom s, even in cases in w hich the shared elec tro n s originate from different atom ic shells. Bonds be tw een s an d /7 electrons are com m on, such as in H 2O {s electrons from H an d p electrons from O, as illustrated in Fig. 17a) an d in hydrocarbon co m pounds (s electrons from H an d p electrons from C). W e can regard the p electrons as having wave functions w ith lobes o f high probability along the coordinate axes (see Fig. 18 o f C hap ter 51, for exam ple). A n 5 electron can be attached at each o f these lobes. In H 2O, for instance, an s electron from each H ato m attaches to two o f the different p electrons. W e w ould therefore expect the angle betw een the bonds in H 2O to be 90 ®; the m easured angle is 104 % indicating th at there is som e C oulom b repulsion o f the H atom s th at spreads the b ond angle. A m m o n ia (N H 3) is an o th er exam ple o f this type o f structure, illustrated in Fig. 17ft. In carbon, the 25 an d 2p electrons are m ixed, giving C an effective valence o f 4. These four electrons can form a variety o f covalent bonds w ith oth er atom s, w hich is re sponsible for the diversity o f organic com pounds, from sim ple m olecules such as m ethane (C H 4) to the com plex m olecules th a t form the basis o f living things.
Figure 17 {a) The overlap of 5 electrons from H and p elec trons from O in a molecule of HjO. (ft) The overlap of s elec trons from H and p electrons from N in a molecule of NH 3.
(a)
Figure 18 {a) Ionic bonding in NaCl. Note that there is no appreciable overlap of the electron distributions, (ft) Binding energy in NaCl. The zero of energy corresponds to Na and Cl atoms separated by a large distance. The dashed line repre sents the energy of Na"^ and Cl“ ions separated by a large dis tance.
Questions
At the other extreme from covalent bonds are those in which the electrons are not shared but belong to one atom or another. In a molecule {not a solid crystal) of NaCl, the Cl lacks one electron from a complete p shell, while the Na has a single valence electron in the s shell. As neutral atoms o f Na and Cl are brought close together, it becomes energetically favorable for the valence electron from Na to be transferred to Cl, thereby filling its p shell. As a result, we have ions Na"^ and Cl“, which then exert electro static forces on one another and bind together in a mole cule of NaCl (Fig. \Sa), The atoms are prevented from approaching too close to one another, because the Pauli principle does not allow the filled p shells to overlap. The stable equilibrium separation is 0.236 nm, and the bind ing energy (the energy needed to split the molecule into its neutral atoms) is 4.26 eV, as shown in Fig. 1 Molecules
1111
of this type, based on the bonding o f ions, are called ionic molecules. Molecular bonding has analogs in the bonding o f atoms in solids. There are ionic solids (such as NaCl), which we can regard as being made of assemblies of positive and negative ions. There are also covalent solids, such as dia mond, whose structure depends on the overlap of electron wave functions. Other types include molecular solids (such as ice), in which the molecules retain their elec tronic structure and are bound by much weaker forces based on electric dipole interactions, and metallic solids, in which each atom contributes one or more electrons to a “sea” of electrons that are shared throughout the entire solid. In the next chapter, we consider some solids whose properties can be understood on the basis of this structure.
QUESTIONS 1. What is the origin of the cutoff wavelength of Fig. 1? Why is it an important clue to the photon nature of x rays? 2. In Fig. 2, why is the emitted photon shown moving off in the direction that it is? Could it be shown moving off in any other direction? Explain. 3. What are the characteristic x rays of an element? How can they be used to determine the atomic number of an element? 4. Compare Figs. 1 and 3. How can you be sure that the two prominent peaks in Fig. 1 do indeed correspond numeri cally with the two transitions similarly labeled in Fig. 3? 5. Can atomic hydrogen be caused to emit x rays? If so, de scribe how. If not, why not? 6 . How does the x-ray energy-level diagram of Fig. 3 differ from the energy-level diagram for hydrogen, displayed in Fig. 4 of Chapter 51? In what respects are the two diagrams similar? 7. When extended to higher atomic numbers, the Moseley plot of Fig. 4 is not a straight line but is concave upward. Does this affect the ability to assign atomic numbers to the ele ments? 8 . Why is it that Bohr theory, which does not work very well even for helium (Z = 2), gives such a good account of the characteristic x-ray spectra of the elements, or at least of that portion that originates deep within the atom? 9. Why does the characteristic x-ray spectrum vary in a system atic way from element to element but the spectrum in the visible range does not? 10. Why do you expect the wavelengths of radiations generated by transitions deep within the atom to be shorter than those generated by transitions occurring in the outer fringes of the atom? 11. Given the characteristic x-ray spectrum of a certain ele ment, containing a number of lines, how would you go about identifying and labeling them? 12. On what quantum numbers does the energy of an electron in {a) a hydrogen atom and (b) a vanadium atom depend?
13. The periodic table of the elements was based originally on atomic mass, rather than on atomic number, the latter con cept having not yet been developed. Why were such early tables as successful as they proved to be? In other words, why is the atomic mass of an element (roughly) proportional to its atomic number? 14. How does the structure of the periodic table support the need for a fourth quantum number, corresponding to elec tron spin? 15. If there were only three quantum numbers (that is, if the electron had no spin), how would the chemical properties of helium be different? 16. Explain why the effective radius of a helium atom is less than that of a hydrogen atom. 17. Why does it take more energy to remove an electron from neon (Z = 10) than from sodium (Z = 11)? 18. What can Fig. 5 tell you about why the inert gases are so chemically stable? 19. Does it make any sense to assign quantum numbers to a vacancy in an otherwise filled subshell? 20. Why do the lanthanide series of elements (see Appendix E) have such similar chemical properties? How can we justify putting them all into a single square of the periodic table? Why is it that, in spite of their similar chemical properties, they can be so easily sorted out by measuring their charac teristic x-ray spectra? 21. In your own words, state the minimum energy principle for atom building and give a physical argument in support of it. 22. Figure 5 shows that the 25 state is lower in energy than the 2p state. Can you explain why this should be so, basing your argument on the radial probability densities of the two states (see Figs. 17 and 18 of Chapter 51)? 23. If you start with a bare nucleus and fill in the electrons to form an atom in its ground state, the energies of the unfilled levels change as you proceed. Why do they change? Do they increase or decrease in energy as electrons are added?
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Chapter 52 Atomic Physics
24. Why is focused laser light inherently better than focused light from a tiny incandescent lamp filament for delicate surgical jobs such as spot-welding detached retinas? 25. Laser light forms an almost parallel beam. Does the inten sity of such light fall off as the inverse square of the distance from the source? 26. In what ways are laser light and star light similar? In what ways are they different? 27. Arthur Schawlow, one of the laser pioneers, invented a type writer eraser, based on focusing laser light on the unwanted character. Can you imagine what its principle of operation is? 28. In what ways do spontaneous emission and stimulated emission differ? 29. We have spontaneous emission and stimulated emission. From symmetry, why don’t we also have spontaneous and stimulated absorption? Discuss in terms of Fig. 11. 30. Why is a population inversion necessary between two atomic levels for laser action to occur? 31. What is a metastable state? What role do such states play in the operation of a laser?
32. Comment on this statement: “Other things being equal, a four-level laser scheme such as that of Fig. 15 is preferable to a three-level scheme such as that of Fig. 13 because, in the three-level scheme, one-half of the population of atoms in level £■, must be moved to state £2 before a population inversion can even begin to occur.’’ 33. Comment on this statement: “In the laser of Fig. 14, only light whose plane of polarization lies in the plane of that figure is transmitted through the right-hand window. There fore, half of the energy potentially available is lost.” (Hint: Is this second statement really true? Consider what happens to photons whose effective plane of polarization is at right angles to the plane of Fig. 14. Do such photons participate fully in the stimulated emission amplification process?) 34. A beam of light emerges from an aperture in a “black box” and moves across your laboratory bench. How could you test this beam to find out the extent to which it is coherent over its cross section? How could you tell (without opening the box) whether or not the concealed light source is a laser? 35. Why is it difficult to build an x-ray laser?
PROBLEMS Section 52-1 The X-Ray Spectrum 1. Show that the short-wavelength cutoff in the continuous x-ray spectrum is given by
8.
Amin = 12 40 pm/F, where V is the applied potential difference in kilovolts. 2. Determine Planck’s constant from the fact that the mini mum x-ray wavelength produced by 40.0-keV electrons is 31.1 pm. 3. What is the minimum potential difference across an x-ray tube that will produce x rays with a wavelength of 0.126 nm? 4. In Fig. 1, the X rays shown are produced when 35.0-keV electrons fall on a molybdenum target. If the accelerating potential is maintained at 35.0 kV but a silver target (Z = 47) is substituted for the molybdenum target, what values of (^) Amin» (b) ^ud (c) result? The K, L, and A/atomic x-ray levels for silver (compare with Fig. 3) are 25.51, 3.56, and 0.53 keV. 5. Electrons bombard a molybdenum target, producing both continuous and characteristic x rays as in Fig. 1. In that figure the energy of the incident electrons is 35.0 keV. If the accelerating potential applied to the x-ray tube is increased to 50.0 kV, what values of (a) A^in, (h) , and (c) Aj^^result? 6 . The wavelength of the from iron is 19.3 pm. (a) Find the energy difference between the two states of the iron atom (see Fig. 3) that give rise to this transition, (b) Find the corresponding energy difference for the hydrogen atom. Why is the difference so much greater for iron than for hydrogen? (Hint: In the hydrogen atom the K shell corre sponds to n = I and the L shell to n = 2 .) 7. From Fig. 1, calculate approximately the energy difference
9.
10.
11.
E ^ — E m for the x-ray atomic energy levels of molybdenum. Compare with the result that may be found from Fig. 3. Find the minimum potential difference that must be applied to an x-ray tube to produce x rays with a wavelength equal to the Compton wavelength of the electron. (See Problem 54 of Chapter 49.) X rays are produced in an x-ray tube by a target potential of 50.0 kV. If an electron makes three collisions in the target before coming to rest and loses one-half of its remaining kinetic energy on each of the first two collisions, determine the wavelengths of the resulting photons. Neglect the recoil of the heavy target atoms. A tungsten target (Z = 74) is bombarded by electrons in an x-ray tube, (a) What is the minimum value of the accelerat ing potential that will permit the production of the charac teristic Kp and lines of tungsten? (b) For this same acceler ating potential, what is the value of A^jn? (c) Calculate Aj^^ and A^^. The K, L, and M atomic x-ray levels for tungsten (see Fig. 3) are 69.5, 11.3, and 2.3 keV, respectively. A molybdenum target (Z = 42) is bombarded with 35.0keV electrons and the x-ray spectrum of Fig. 1 results. Here = 63 p m and A^^^ = 71 pm. (a) What are the corre sponding photon energies? (b) It is desired to filter these radiations through a material that will absorb the Kp line much more strongly than it will absorb the line. What substance(s) would you use? The K ionization energies for molybdenum and for four neighboring elements are as fol lows: Z Element £^(keV )
41 42 40 43 Zr Nb Mo Tc 18.00 18.99 20.00 21.04
44 Ru 22.12
Problems
12.
13.
14.
15.
(Hint: A substance will selectively absorb one of two x radia tions more strongly if the photons of one have enough en ergy to eject a K electron from the atoms of the substance but the photons of the other do not.) The binding energies of /f-shell and L-shell electrons in cop per are 8.979 keV and 0.951 keV, respectively. If a x ray from copper is incident on a sodium chloride crystal and gives a first-order Bragg reflection at 15.9® when reflected from the alternating planes of the sodium atoms, what is the spacing between these planes? A 20.0-keV electron is brought to rest by undergoing two successive bremsstrahlung events, thus transferring its ki netic energy into the energy of two photons. The wavelength of the second photon is 130 pm greater than the wavelength of the first photon to be emitted, (a) Find the energy of the electron after its first deceleration, (b) Calculate the wave lengths and energies of the two photons. In an x-ray tube an electron moving initially at a speed of 2.73 X 10® m/s slows down in passing near a nucleus. A single photon of energy 43.8 keV is emitted. Find the final speed of the electron. (Relativity must be taken into ac count; ignore the energy imparted to the nucleus.) Show that a moving electron cannot spontaneously emit an x-ray photon in free space. A third body (atom or nucleus) must be present. Why is it needed? (Hint: Examine conser vation of total energy and of momentum.)
Section 52~2 X Rays and the Numbering o f the Elements 16. Using the Bohr theory, calculate the ratio of the wavelengths of the Ka line for niobium (Nb) to that of gallium (Ga). Take needed data from the periodic table. 17. Here are the wavelengths of a few elements: Ti V Cr Mn Fe
27.5 pm 25.0 22.9
21.0 19.3
Co Ni Cu Zn Ga
17.9 pm 16.6 15.4 14.3 13.4
Make a Moseley plot (see Fig. 4) and verify that its slope agrees with the value calculated in Sample Problem 2. Section 52-4 The Periodic Table 18. If a uranium nucleus (Z = 92) had only a single electron, what would be the radius of its ground-state orbit, according to Bohr’s theory? 19. Two electrons in lithium (Z = 3) have as their quantum numbers n, AW/, m ,, the values 1, 0 , 0, ± i . (a) What quan tum numbers can the third electron have if the atom is to be in its ground state? (b) If the atom is to be in its first excited state? 20. By inspection of Fig. 5, what do you think might be the atomic number of the next higher inert gas above radon ( Z = 86)? 21. If the electron had no spin, and if the Pauli exclusion princi ple still held, how would the periodic table be affected? In particular, which of the present elements would be inert gases?
1113
22. Suppose there are two electrons in the same system, both of which have n = 2 and / = 1. (a) If the exclusion principle did not apply, how many combinations of states are conceiva bly possible? (b) How many states does the exclusion princi ple forbid? Which ones are they? 23. In the alkali metals there is one electron outside a closed shell, (a) Using the Bohr theory, calculate the effective charge number of the nucleus as seen by the valence electron in sodium (ionization energy = 5.14 eV) and potassium (ionization energy = 4.34 eV). (b) For each element, what fraction is this of the actual nuclear charge Z? Needed quan tum numbers can be found on Fig. 5.
Section 52-7 How a Laser Works 24. A ruby laser emits light at wavelength 694.4 nm. If a laser pulse is emitted for 12.0 ps and the energy release per pulse is 150 mJ, (a) what is the length of the pulse, and (b) how many photons are in each pulse? 25. Lasers have become very small as well as very large. The active volume of a laser constructed of the semiconductor GaAlAs has a volume of only 200 (//m)^ (smaller than a grain of sand) and yet it can continuously deliver 5.0 mW of power at 0.80-//m wavelength. Calculate the production rate of photons. 26. A He - Ne laser emits light at a wavelength o f632.8 nm and has an output power of 2.3 mW. How many photons are emitted each minute by this laser when operating? 27. It is entirely possible that techniques for modulating the frequency or amplitude of a laser beam will be developed so that such a beam can serve as a carrier for television signals, much as microwave beams do now. Assume also that laser systems will be available whose wavelengths can be precisely “tuned” to anywhere in the visible range, that is, in the range 400 nm < A < 700 nm. If a television channel occupies a bandwidth of 10 MHz, how many channels could be ac commodated with this laser technology? Comment on the intrinsic superiority of visible light to microwaves as carriers of information. 28. A hypothetical atom has energy levels evenly spaced by 1.2 eV in energy. For a temperature o f2000 K, calculate the ratio of the number of atoms in the 13th excited state to the number in the 11th excited state. 29. A particular (hypothetical) atom has only two atomic levels, separated in energy by 3.2 eV. In the atmosphere of a star there are 6.1 X 10*^ of these atoms per cm^ in the excited (upper) state and 2.5 X 10’^ per cm^ in the ground (lower) state. Calculate the temperature of the star’s atmosphere. 30. A population inversion for two levels is often described by assigning a negative Kelvin temperature to the system. Show that such a negative temperature would indeed corre spond to an inversion. What negative temperature would describe the system of Sample Problem 4 if the population of the upper level exceeds that of the lower by 10.0%? 31. An atom has two energy levels with a transition wavelength of 582 nm. At 300 K, 4.0 X 10^ atoms are in the lower state, (a) How many occupy the upper state, under condi tions of thermal equilibrium? (b) Suppose, instead, that 7.0 X 10^® atoms are pumped into the upper state, with
1114
Chapter 52 Atomic Physics
4.0 X 10^® in the lower state. How much energy could be released in a single laser pulse? 32. The mirrors in the laser of Fig. 14 form a cavity in which standing waves of laser light are set up. In the vicinity of 533 nm, how far apart in wavelength are the adjacent al lowed operating modes? The mirrors are 8.3 cm apart. 33. A high-powered laser beam (A = 600 nm) with a beam diam eter of 11.8 cm is aimed at the Moon, 3.82 X 10^ km dis tant. The spreading of the beam is caused only by diffraction effects. The angular location of the edge of the central dif fraction disk (see Eq. 11 in Chapter 46) is given by sin 6 =
1.22 A
where d is the diameter of the beam aperture. Find the diameter of the central diffraction disk at the Moon’s sur face. 34. The beam from an argon laser (A = 515 nm) has a diameter d of 3.00 mm and a power output of 5.21 W. The beam is focused onto a diffuse surface by a lens of focal length / = 3.50 cm. A diffraction pattern such as that of Fig. 13 in Chapter 46 is formed, (a) Show that the radius of the central disk is given by 1.22/A ^ ------The central disk can be shown to contain 84% of the inci dent power. Calculate (b) the radius R of the central disk,
and the average power flux density (c) in the incident beam and (d) in the central disk. 35. The use of lasers for defense against ballistic missiles is being studied. A laser beam of intensity 120 MW/m^ would proba bly bum into and destroy a hardened (nonspinning) missile in about 1 s. (a) If the laser has a power output of 5.30 MW, a wavelength of 2.95 pm, and a beam diameter of 3.72 m (a very powerful laser indeed), would it destroy a missile at a distance of 3000 km? (b) If the wavelength could be changed, what minimum value would work? (c) If the wave length of the laser could not be changed, what would be the destmctive range of the laser in (a)l Use the equation for the central disk given in Problem 34 and take the focal length to be the distance to the target. 36. The active medium in a particular mby laser (A = 694 nm) is a synthetic mby crystal 6.00 cm long and 1.0 cm in diame ter. The crystal is silvered at one end and— to permit the formation of an external beam— only partially silvered at the other, (a) Treat the crystal as an optical resonant cavity in analogy to a closed organ pipe and calculate the number of standing-wave nodes there are along the crystal axis. (b) By what amount Av would the beam frequency have to shift to increase this number by one? Show that Av is just the inverse of the travel time of light for one round trip back and forth along the crystal axis, (c) What is the corresponding fractional frequency shift Av/v? The appropriate index of refraction is 1.75.
CHAPTER 53 ELECTRICAL CONDUCTION IN SOLIDS We have seen in the previous two chapters how well quantum theory works when we apply it to individual atoms. In this chapter we show that this powerful theory works equally well when we apply it to collections o f atoms in theform o f solids. Every solid has an enormous range o f properties that we can choose to examine: Is it soft or hard? Can it be hammered into a thin sheet or drawn into a fine wire? Is it transparent? What kind o f waves travel through it and at what speeds? Does it conduct heat? What are its magnetic properties? What is its crystal structure? And so on. In each case, we should like to use quantum theory to understand the measured properties. In this chapter, we focus on one particular property o f solids: conduction o f electricity. We discuss the classification o f solids into conductors, insulators, semiconductors, and superconductors, and we show how quantum theory provides the framework for understanding why some materials behave one way and some another.
53-1 CONDUCTION ELECTRONS IN A METAL An isolated copper atom has 29 electrons. In solid copper, 28 o f these are held close to their lattice sites by electro magnetic forces and are not free to move throughout the volume o f the solid. The remaining electron is free to so move and, if we apply an em f between the ends o f a copper wire, it is these conduction electrons (one per atom) that constitute the current that is set up in the wire. In Section 32-5 we looked at this problem from the point of view o f classical physics, comparing the conduc tion electrons in a metal cube to the atoms of a gas con fined to a cubical box. Using this (classical) free electron gas model, we derived an expression for the resistivity of the metal. It is (see Eq. 20 of Chapter 32)
P= -
m
( 1)
in which m is the mass and e the charge of the electron, n is the number o f conduction electrons per unit volume, and T is the average time between collisions o f the electrons with the lattice. We showed in Section 32-5 that r is essentially con stant, independent of whether or not an electric field has
been set up inside the cube by an externally applied em f Thus the resistivity p is independent of the applied electric field, which is another way of saying that metals obey Ohm’s law. Although this derivation of the form of Ohm’s law is a fine achievement for classical physics, it is no simple mat ter to go much further. Also, there is one problem— the heat capacities o f metals— about which this classical theory does have something to say, but unfortunately its predictions do not agree with experiment. Looking beyond this level of concern, it is hard to imagine how we would explain something as complicated as a transistor on the basis of the classical free electron gas model. We had better see what wave mechanics has to offer. The first step in solving any wave mechanical problem is to specify the potential energy o f the particle— which we take to be a single conduction electron— as a function of its position. As Fig. 6 of Chapter 51 reminds us, we need this information to substitute into the Schrodinger equa tion. We start with the simplest reasonable assumption, namely, that the potential energy is zero for all points within the cubical metal sample and that it approaches an infinitely great value for all points outside. We are still dealing with a free electron gas, but it is now one that is governed by quantum— rather than classical— rules. This potential energy reminds us of the problem o f an
1115
1116
Chapter 53
Electrical Conduction in Solids
electron trapped in an infinite well that we solved in Sec tion 50-7. Note, however, two differences: the present problem is three-dimensional, and it involves a well of macroscopic, rather than atomic, dimensions. We represent a single conduction electron trapped in its metal cube by a (standing) matter wave in which r is a position vector, and we impose the condition that the probability density ip\r) be zero both at the surface of the cube and at all outside points. This is our way of recogniz ing that the electron is truly trapped inside the metal cube. Figure 18 o f Chapter 50 reminds us that we proceeded in just the same way in the one-dimensional case. If we impose these boundary conditions on the wave function, the Schrodinger equation tells us that the total energy E of the electron will be quantized, just as it was for an electron trapped in a one-dimensional well. There is a big difference, however. Because our metal cube is so very large on the scale of atomic dimensions, the number of standing matter waves that we can fit into the volume of the cube and still satisfy the boundary requirements is enormous, and the allowed electron energies are ex tremely close together. Sample Problem 1 shows that, for a cube 1 cm on edge, there are about 10^®quantized states that lie between £ = 5 eV and £ = 5.01 eV! Compare this with the limited array of well-spaced levels shown, for example, for the hydrogen atom in Fig. 4 of Chapter 51. We cannot possibly deal with this vast number of states one at a time; we must use statistical methods. Instead of asking, “What is the energy of this state?” we must ask, “How many states have energies that lie in the range £ to
E ^dE T We have met situations like this before. For example, in describing the speeds of the molecules of an ideal gas in Section 24-3, we saw that the only way to proceed was to pose the question: “How many molecules have speeds that lie in the range vtov-\- dvT For the conduction electrons, the number of states (per unit volume of the solid) whose energies lie in the range £ to E + dE can be written as n{E)dE, where n{E) is a function called the density of states. For our (quantum) free electron gas it can be shown to be* « ( £ ) = ----- ^5-----
( 2)
At this stage we are simply counting the states that are available to a single conduction electron. Note that there is nothing in Eq. 2 that depends on the material of which our sample is made. Fitting patterns of standing waves into a cubical box is a purely geometrical problem. In the following section we shall see how to go about filling those states.
* See Quantum Physics o f Atoms, Molecules, Solids, Nuclei, and Particles, by Robert Eisberg and Robert Resnick (Wiley, 1985) 2nded., Section 11-11.
You may well ask, “If the energies of the allowed states are so close together, why don’t we just forget about the quantization and assume a continuous distribution in en ergy?” The answer, as we shall see in the next section, rests on the fact that the Pauli exclusion principle applies to electrons wherever we find them, whether as orbital elec trons in atoms or as conduction electrons in metals. Even though there are many states in our problem, there are also many conduction electrons available to occupy them, and the Pauli principle allows us to put only one electron in each of these states. Thus, even though we cannot easily detect directly the quantized nature of the energies of the conduction electrons, the fact o f quantiza tion remains an absolutely central feature and has impor tant consequences.
Sample Problem 1 A cube of copper is 1 cm on edge. How many states are available for its conduction electrons in the energy interval between E = 5.00 eV and 5.01 eV? Assume that the conduction electrons behave like a (quantum) free electron gas. Solution These energy limits are so close together that we can safely say that the answer, on a per unit volume basis, is m(£)A £, where £ = 5 eV and A £ = 0.01 eV. From Eq. 2 we have 8
n(E) = -
y flT tm ^ ^ ^
.£• 1/2
h^
(8V27r)(9.11 X 10-^' kg)^/^ [(5 eVK1.6X 10- ‘’ J/eV )]'/2 (6.63 X 10-^J*s)^ = 9.48 X 10^ m-3 J - ' = 1.52 X lO^* m -^eV "'. Note that we must express the energy £ in joules before substi tuting into Eq. 2, even though we wish our final result to be given in terms of electron volts. The actual number N of states that lie in the range from £ = 5.00 eV to £ = 5.01 eV in our cube is, if a is the length of the cube edge, N=n(E)AEa^ = (1.52 X 10^8 m-^eV-'XO.Ol eVXl X \0~^ m)^ = 1.52X 1020. That is, there are 1.52 X lO^o individual energy states between 5.00 eV and 5.01 eV. The average energy interval A£ 3^3between adjacent levels in this interval follows readily from A£ A£adj =
0.01 eV «=7X 10-23 eV. 1.52X1020
We conclude that, even in this narrow energy band, there are very many states and they lie exceedingly close together in en ergy. Our conclusions are completely independent of the material of the sample. Nor is it important that the sample is cubical; any other shape enclosing the same volume would give the same final result. What we have assumed to be true is that the conduc tion electrons behave like a (quantum) free electron gas. That is, we have assumed that their potential energy is constant (which
Section 53-2 Filling the Allowed States we have taken to be zero) for all points within the sample. For actual metals this assumption is never strictly true. Nevertheless, our central conclusion holds: many states are available to a con duction electron in a metal and they lie very close together in energy.
or, carrying out the integral, 8 fln m '^'^
n= -
Solving for £p gives
Sm\n I
53-2 FILLING THE ALLOWED STATES________________________ Now that we have seen how many states there are, we are ready to start Ailing them with electrons. We went through this process in Section 52-3 in connection with building up the periodic table o f the elements. There we saw the central importance o f Pauli’s exclusion principle, which tells us that we can allocate only one electron to a given state. This powerful principle is just as important for our present problem. Figure \a shows the density of states given by Eq. 2. This function gives the number of possible states in any energy interval. However, not all those states are occu pied. We fill the available states in a metal just as we filled the available states in an atom: we add electrons, one per quantum state, starting at the lowest energy and ending when we have added all the necessary electrons to the metal. Let us first consider conditions at the absolute zero of temperature. This represents the lowest energy state of our sample, and we achieve it by placing the conduction electrons into the unfilled states that lie lowest in energy. This process is suggested by Fig. 1^, which shows the probability function p(E). This function gives the proba bility o f the state at the energy £ to be occupied. At T = 0, all states below a certain energy are filled {p = 1) and all states above that energy are vacant (p = 0). The highest occupied state under these conditions is called the Fermi level, and its energy, marked £p in Fig. \b, is called the Fermi energy. The Fermi energy for copper, for example, is 7.06 eV. If we multiply the density n(E) of available states by the probability p{E) that those states are occupied, the result is the density of occupied states, nJiE), or
n,(E) = n(E)p(E).
■
(5)
A glance at Fig. Ic should be enough to shatter the popular misconception that all motion ceases at the abso lute zero o f temperature. We see that, entirely because of Pauli’s exclusion principle, the electrons are stacked up in energy from zero to the Fermi energy. The average energy for the conditions of Fig. 1c turns out to be about 4.2 eV. By comparison, the average translational kinetic energy of a molecule of an ideal gas at room temperature is only 0.025 eV. The conduction electrons in a metal have plenty of energy at the absolute zero!
(a)
Energy (eV) ( 6)
(3)
This quantity is plotted in Fig. \c. The shaded area in Fig. Ic represents the total number o f occupied states (per unit volume). Finding this area and equating it to the density n of conduction electrons in the metal gives a means to find the Fermi energy. Inte grating between the limits of £ = 0 and £ = £p to find the area, we obtain
(E)dE,
1117
(4)
Figure 1 (a) The density of states n(E) plotted as a function of the energy E. (b) The probability function p(E) at T = 0. (c) The density of occupied states nJ^E), equal to the product of n(E) and p{E). All states below £p are occupied and all states above Ep are vacant.
1118
Chapter 53
Electrical Conduction in Solids
It seems clear that the molecules of a gas at ordinary temperatures and the conduction electrons in a metal behave in quite different ways. Formally, we say that the gas molecules obey the (classical) Maxwell-Boltzmann statistics and that the conduction electrons obey the (quantum) Fermi-Dirac statistics. The word “statistics” here refers to the formal rules for counting particles. In Maxwell-Boltzmann statistics, for example, we assume that we can tell identical particles apart, but in FermiDirac statistics we assume that we cannot. Again, in Maxwell-Boltzmann statistics, Pauli’s exclusion princi ple plays no role, but in Fermi-Dirac statistics its role, as we have seen, is vital. See Section 24-6 for a discussion of these statistical distributions. What happens to the electron distribution of Fig. 1 as we raise the temperature? Only a small change occurs in the distribution, but that small change has important con-
Energy (eV) (6)
sequences. Figure 2 shows what the distributions o f Fig. 1 would look like at T = 1000 K, a temperature at which a metal sample would glow brightly in a dark room. The striking feature of Fig. 2 is how little it differs from Fig. 1, the distributions at absolute zero. At T = 0, the probability function p ( E ) was strictly unity below fp and strictly zero above E^. As Fig. 2b shows, at T = 1000 K there is a small probability to have a few vacant states below Ey and a few occupied states above The density n J,E ) of occupied states, again given by Eq. 3 as the product of n{E) and p { E \ is shown in Fig. 2c. Because a few states above Ep are occupied, the average energy is a little larger than it was at the absolute zero but it is not much larger. This is again in striking contrast with the behavior of an ideal gas, for which the average kinetic energy of the molecules is proportional to the tempera ture. By comparing conditions at T = 0 and at T = 1000 K, we see that all the “action” takes place for conduction electrons whose energies are close to the Fermi energy. The motion of most of the electrons remains unchanged as the temperature is raised, their large store o f energy being effectively locked up. Let us see why this is so. Figure Ic displays the magni tude of /cT, a measure of the energy available from ther mal agitation; its value at 1000 K is only 0.086 eV. No electron can hope to have its energy changed by more than a few times this relatively small amount by thermal agitation alone. Because of the exclusion principle, only electrons whose energies are near the Fermi energy have vacant states close enough to them for such thermal tran sitions to occur. An electron with an energy of, say, 2 eV can neither gain nor lose energy because all states close enough to it in energy are already filled; it simply has nowhere to go. In analogy with waves on the ocean, ther mal agitation of the electrons normally causes only ripples on the surface of the “Fermi sea”; the vast depths of that sea lie undisturbed. The probability function p { E) plotted in Figs. \ b and 2b is called the F erm i - Dirac probability function and can be shown to be 1
P{E) = ^{E-E^VkT ^ J
Figure 2 Same as Fig. 1, but for T = 1000 K. Note how little the plots differ from those of Fig. 1. (These plots are some what idealized in that they assume the electrons move in a re gion of uniform potential. Measured density of states plots in real metals do not have this simple shape.)
( 6)
in which E’p is the Fermi energy, now defined (see Fig. 2b) as the energy corresponding \o p = \. Note that Eq. 6 yields the rectangular plot in Fig. \b for r = 0. As r ^ 0, the exponent {E — E ^ ) / k T in Eq. 6 ap proaches —00 if £■ < £p and H-oo if £* > Ep. In the first case we have p { E) = 1 and in the second case p { E) = 0, just as required. Equation 6 also shows us that the important quantity is not the energy E but rather E —Ep, the energy interval between E and the Fermi energy. We see further that, because of the exponential nature of the term in the de nominator of Eq. 6, p ( E ) is very sensitive to small changes
Section 53-3
in E —Ep. This confirms our assertion that electrons whose energies are close to the Fermi energy are the only ones that play an active role. As we shall see, the first question in dealing with the electrons in a solid— be it a conductor, a semiconductor, or an insulator— is likely to be: “On an energy scale, where is the Fermi level?”
Electrical Conduction in Metals
1119
energy of the available states? (c) For this same energy, what is «o(£), the distribution in energy of the occupied states? Solution
(a) Substitution into Eq. 6 yields ^ ^^EfkT^ j ^
in which A E = E — Ep. A little algebra leads to A E / k T = —2.20 so that Sample Problem 2 Calculate the Fermi energy for copper, given that the number of conduction electrons per unit volume (see Sample Problem 2, Chapter 32) is 8.49 X 10^® m“ ^.
A E = - 2 . 2 0 k T = -(2.20X8.62 X 10‘ 5 eV/KXlOOO K) = -0 .1 9 e V . For copper, assuming that Ep = 7.06 eV, we have
Solution
From Eq. 5 we obtain
E = Ep + A E = 7.06 eV - 0.19 eV = 6.87 eV.
\2/3
(b) Carrying out a calculation just like that of Sample Problem 1 for £ = 6.87 eV yields n(E) = 1.78 X 10^® m"^ e V .
2 m \n ) (6.63 X 10- ^ J -s )^ r (3X8.49 X 10^« (8X9.11 X 10 kg) L ^ = 1.13 X 10-'»J = 7.06 eV.
(c) From Eq. 3 we have, again for E = 6.87 eV, J
n,(E) = n(E)p(E) = (1.78X 102® m-^eV-'X0.90) = 1.60X 102®m -5 e V -‘.
Sample Problem 3 What is the probability of occupancy for a state whose energy is (a) 0.1 eV above the Fermi energy, (b) 0.1 eV below the Fermi energy, and (c) equal to the Fermi energy? Assume a temperature of 800 K. Solution
(a) The (dimensionless) exponent in Eq. 6 is E -E p_ 0.1 eV = 1.45. kT (8.62 X 10-5 eV/KX800 K)
Inserting this exponent into Eq. 6 yields 1
Thus the occupancy probability for this state is 19%. {b) For an energy 0.1 eV below the Fermi energy, the expo nent in Eq. 6 has the same numerical value as above but is negative. Thus, from Eq. 6, 1 1
= 0.81.
The occupancy probability for this state is 81%.
53-3 ELECTRICAL CONDUCTION IN METALS____________________ Figure 3 represents the Fermi distribution of velocities in a metal. The Fermi speed Vp is the speed of an electron whose kinetic energy equals Ep, the Fermi energy. With no applied electric field, electrons have speeds ranging from 0 to approximately Vp, corresponding to energies ranging from 0 to approximately Ep, The distribution in Fig. 3 represents a typical v^/oc/7y component, rather than the speed. This illustrates that there are equal numbers of electrons moving in opposite directions, so that the net current is zero in the absence of an electric field. When an electric field is applied, the electrons are accel erated by the field and acquire a small increase in velocity in a direction opposite to the field. (Because electrons are
(c) For E — Ep the exponent in Eq. 6 is zero and that equation becomes /I T .-
1
1
1+ 1
p{v)
= 0.50.
Note that this result does not depend on the temperature. Note also that none of these three results depends on the actual value of the Fermi energy, only on the energy interval between the Fermi energy and the energy of the state in question.
Sample Problem 4 (a) For copper at 1000 K, find the energy at which the probability p(E) that a conduction electron state will be occupied is 90%. (Assume that the conduction electrons in copper behave like a free electron gas, with a Fermi energy of 7.06 eV.) (b) For this energy, what is n(E), the distribution in
(f
“ — —
/
k __________
-Velocity
Figure 3 The Fermi distribution of velocities. With no elec tric field (solid line) states up to the Fermi speed Vp are filled. When an electric field E is applied in the direction shown, the distribution shifts to the right (dashed line) as the electrons are accelerated by the field.
1120
Chapter 53
Electrical Conduction in Solids
negatively charged, the force on an electron is F = — eE, which is in a direction opposite to E.) T he entire velocity distribution in the presence o f a field is shifted slightly to the right in Fig. 3. However, m ost o f the electrons still add pairwise to zero velocity an d do not contrib ute to the conduction. T he electrons th at contribute to the conduction are those in a small group o f velocities near % . T he electric field causes states having velocities ju st below % in the direction o f E to becom e unoccupied, while states having velocities ju st above Vp in a direction opposite to E be com e occupied. Y ou can see from Fig. 3 why the drift velocity (the average velocity o f all the electrons) is m uch sm aller th an % , because in the averaging process m any positive an d negative velocities will cancel one another. T he drift speed is d eterm ined prim arily by the sm all n u m ber o f electrons m oving from states below speed % to states above speed Vp u n d er the action o f the electric field. T he resistivity o f the m etal to the flow o f these electrons is determ ined by collisions m ade by the electrons w ith the ion cores o f the lattice. In Sam ple Problem 5 we show that, for copper at room tem perature, the Ferm i speed, which is the average speed o f the conduction electrons between collisions, is 1.6 X 10^ m /s, a significant fraction o f the speed o f light. T he m ean tim e betw een collisions is 2.5 X lO"*'^ s an d the m ean free p ath is 41 nm , which is about 150 nearest-neighbor distances in the copper lattice. Y ou m ay be surprised that, at room tem perature, a conduction electron can m ove so far through a copper lattice w ithout hitting an ion core. At lower tem peratures — where the resistivity is low er— it can m ove even m uch further. It is, in fact, a perhaps unexpected prediction o f wave m echanics th at a perfectly periodic lattice at the absolute zero o f tem perature would be totally transparent to conduction electrons. T here w ould never be any colli sions! T here are, however, no perfectly periodic lattices. Va can t lattice sites and im purity atom s are always present, no m atter how hard we try to elim inate them . F u rth er m ore, at tem peratures above the absolute zero the lattice is vibrating, and these m otions also spoil the periodicity o f the lattice. At room tem p eratu re the “ collisions” o f which we have spoken are largely interactions betw een the con duction electrons and the vibrations o f the lattice.
Sample Problem 5 Take the Fermi energy of copper to be 7.06 eV. (a) What is the speed of a conduction electron with this kinetic energy? {b) The resistivity of copper at room temperature is 1.7 X 10~* Q-m. What is the average time t between colli sions? (c) What mean free path A may be calculated from the results of (a) and (b)l Solution (a) Throughout this section we have assumed that the conduction electrons are moving in a region in which their po tential energy is zero. Thus their total energy E is all kinetic and
we can write,
E = Ep, Ef =
in which Vf is the Fermi speed. Solving for Vf yields = \ w /
= r (2X7-06 eVXl -6 X lQ-'» J/eV )1 '^ [ 9 .1 1 X 1 0 -” kg
J
= 1.6 X 10‘ m/s. You should not confuse this speed with the drift speed of the conduction electrons, which is of the order of 10“^ m/s and is thus smaller by about a factor of 1O'®. As we explained in Section 32-5, the drift speed is the average speed at which electrons actually drift through a conductor when an electric field is ap plied; the Fermi speed is their average speed between collisions. (b) Solving Eq. 1 for t yields T=
m ne^p 9.11 X 10-^' kg (8.49 X 1Q2* m-3)(1.60X lO"*’ C)2(1.7 X 10-®^2•m)
= 2.5 X 10-'^ s. (c) To find the mean free path, we have A = ypT = (1.6 X 10^ m/sK2.5 X 10"'^ s) = 4.1 X 10"* m = 41 nm. In the copper lattice the centers of neighboring ion cores are 0.26 nm apart. Thus a typical conduction electron can move a substantial distance through a copper lattice at room tempera ture without making a collision.
53-4
B A N D S A N D G APS_____________
Figure 4a suggests the potential energy variation th at we have been using to describe a conduction electron in a m etal. T he potential energy is zero inside the m etal, and it rises to infinity at the surface. However, there are prob lem s w ith this m odel. F or exam ple, it tells us that, because o f the infinite potential barrier, an electron could never escape from inside the sam ple through its surface. W e know th at this isn’t true, because electrons can be “ boiled o u t” o f a m etal by raising its tem perature, as in the heated filam ent o f a vacuum tube (therm ionic em ission). They can also be “ kicked o u t” if we shine light o f high enough frequency on the m etal surface (photoelectric effect). Figure 4b shows th at we can take care o f this difficulty easily enough by m aking the potential energy at the sur face finite. W e m ade the sam e realistic adjustm ent (see Fig. 20 o f C hapter 50) for the electron trapped in an ato m sized, one-dim ensional well. T he q u an tity (f> in Fig. 4b is the w orkfunction o f the m etal, defined as the least a m o u n t o f energy th at m ust be supplied to an electron to rem ove it from the sam ple. Figure 4b has been draw n to bring o u r energy scale into agreem ent w ith th a t used for the hydrogen atom . T h at is.
Section 53-5
Fermi energy
- U
= 0
(a) Surface
Fermi energy
- U
= 0
(c)
Figure 4 (a) The potential energy variation assumed for a metal in the free electron gas model, (b) A more realistic vari ation, showing a finite change in potential energy at the sur face of the sample, (c) A still more realistic variation, taking the lattice of ion cores into account. This curve is a one dimensional cut along a line of ion cores (shown as dots at the bottom of the figure). The shaded regions are the energy bands permitted for the electrons. Electrons are not permitted to have energies corresponding to the gaps.
we have chosen the £ = 0 configuration to represent an electron at rest far outside the sample. It is possible to make this change because the potential energy always contains an arbitrary additive constant and we are more concerned, in any case, with changes in the total energy E than with E itself. On our new energy scale, the total energies o f electrons trapped in the sample are negative, just as they are for the hydrogen atom. By far the largest difficulty remaining in Fig. 4Z? is that it assumes that the potential energy of a conduction elec tron is constant throughout the volume of the sample. This ignores the fact that the conduction electrons move about among an array of positively charged ion cores. It is rather remarkable, in fact, that we have been able to learn as much as we have about the resistivity of metals without
Conductors, Insulators, and Semiconductors
1121
taking into account the potential energy variations caused by the ion cores of the lattice. We have not, however, been able to answer such questions as: “Why is copper a con ductor and diamond a nonconductor?” If we take the lattice periodicity into account, we shall be able to answer this question and to go far beyond. Figure 4c shows a potential energy curve that takes the ion cores into account. Substituting this potential energy (or some approximation to it) into the Schrodinger equa tion brings out an interesting new phenomenon. As Fig. 4c shows, the allowed states are now grouped into bands, with energy gaps between them in which no states exist.
Note that electrons just below the Fermi level are free to move throughout the lattice, but electrons with lower en ergies, the core electrons, are not. Let us see if we can understand these bands and gaps in physical terms. The distance between nearest neighbors in a copper lattice is 0.26 nm. Consider, however, two copper atoms separated by a much greater distance, say, 50 nm, so that we may describe them as “isolated”; see Fig. 5a. In each atom the 29 electrons are assigned to the levels shown in Fig. Sb. Now let us bring the two atoms closer together so that an outer electron in either atom can be influenced, how ever slightly, by forces exerted on it by the other atom. In the language of wave mechanics we say that their wave functions begin to overlap. We state without proof that the two overlapping wave functions can be combined in two independent ways, describing two states having (slightly) different energies, as shown in the second col umn of Fig. 6. Because the overlap is greater for the outer electrons, the energy splitting will be greater for them than for the inner electrons. By extension, if we bring N copper atoms together to form a copper lattice, each level of the isolated atom be comes N levels of the solid. Thus the 15 level of the atom becomes the l5 band of the solid and so on. Figure 6 suggests the process. From this point of view the forbidden gaps are not so hard to understand, being familiar from the level struc ture of the isolated atom. Indeed, we can say that Niels Bohr, even before wave mechanics, “invented” energy gaps when he said, in effect: “I assume that atoms can exist without radiating in a discrete set of stationary states of definite energy, states o f interm ediate energy being fo r bidden.''
53-5 CONDUCTORS, INSULATORS, AND SEMICONDUCTORS___________ Figure l a represents the band structure of a conductor, such as copper. Its central feature is that the most ener getic band that contains any electrons at all is only par-
1122
Chapter 53
Electrical Conduction in Solids
Figure 5 (a) Two neutral copper atoms of diameter d, separated by a distance r, with r » d. (b) The atoms are independent systems, and in its ground state each has the same quan tum number assignments for its electrons, as shown. The energy scale is symbolic only.
-E = 0
N
Number of atoms
Figure 6 As atoms are brought together to form a lattice, the levels of the isolated atoms split, eventually forming bands of closely lying levels. For the case shown, the upper bands overlap in energy.
Section 53-5
Conduction
bm6 t
Conductors, Insulators, and Semiconductors
1123
an insulator and a sem iconductor is som ew hat arbitrary. T here is no hesitation, however, in calling diam ond (£'g = 5.5 eV) an insulator and silicon {E^ = 1.1 eV ) a sem iconductor.
Valence band
Sample Problem 6 What is the probability at room tempera ture that a state at the bottom of the conduction band is occupied in diamond and in silicon? Take the Fermi energy to be at the middle of the gap between the conduction and valence bands. Conductor (a)
Insulator ( 6)
Semiconductor (c)
Figure 7 An idealized representation of the energy bands for (a) a conductor, (b) an insulator, and (c) a semiconductor. Filled bands are shown in colored shading, and empty bands in gray shading. The black triangle marks the Fermi level for the conductor.
Solution For a state at the bottom of the conduction band, the energy difference £ —Ep is 0.5Eg, if the Fermi energy is at the middle of the gap. At room temperature (300 K), k T = 0.026 eV. We therefore have E — E^':> kT, and so we can ap proximate the Ferm i-Dirac probability function (Eq. 6) as
For diamond, tially filled. T here are vacant states above the Ferm i level so that, if you apply an electric field E, every electron in this b and is able to increase its m o m en tu m in the —E direction, and there will be a current. T he lower energy bands are com pletely filled and can n o t contrib ute to the conduction process, all velocities adding pairwise to zero. Figure l b represents an insulator. Its central feature is th at the m ost energetic b an d th at contains any electrons at all is com pletely filled, an d the forbidden energy gap lying im m ediately above it, m arked in the figure, is substan tial. By “ substantial” we m ean th a t :> k T , so th at the probability th a t an electron will be lifted, by therm al agita tion, into the em pty b and th at lies above the gap is negligi ble. If you set up an electric field w ithin an insulator, there is no way for any o f the electrons to respond to it so th at there will be no current. C arbon in its d iam o n d form is an excellent insulator, its energy gap being 5.5 eV, m ore th an 200 tim es the value o f /cT at room tem perature. Figure I c represents a sem iconductor. It differs from an insulator in th at its energy gap is sm all enough so th at therm al excitation o f electrons across it can occur to a useful extent at room tem perature. T his puts som e elec trons into the (nearly em pty) b an d labeled conduction band in the figure and leaves an equal n u m b er o f vacant states, o r holes, in the (nearly filled) valence band. In a band th at is nearly full, it tu rn s o u t to be m ore convenient to analyze its co n trib u tio n to the electrical conduction in term s o f the m otion o f holes, w hich behave like positively charged particles. Silicon is o u r prototype sem iconductor. It has the sam e crystal structure as d iam o n d b u t its gap w idth (= 1.1 eV) is considerably sm aller. At the absolute zero o f tem pera ture, where therm al agitation is absent, all sem iconduc tors are insulators. At any higher tem p eratu re the proba bility th at an electron will be raised across the gap is very sensitive to the gap w idth. T h u s the distinction between
1
p{E) = -
. «
p - (E - E r ) lk T =
p -E j2 k T
= 5.5 eV and so
P ( E )
=
^ - ( 5 .5 e V ) /2 ( 0 .0 2 6 e V )
= J 2 X 10"^.
For silicon, E^ = 1.1 eV and so p(E) =
‘ ev)/2(0.026cv)
6 5 X 10-'0
In a cubic centimeter of material, containing roughly 10^^ atoms, there will be a negligible probability to find even one electron in the conduction band of diamond, while there may be roughly 10*^ electrons in the conduction band (and an equal number of holes in the valence band) available for electrical conduction in silicon. This calculation illustrates the extreme difference in conductivity that results from small variations in the gap energy, and it clearly shows the distinction between insulators and semiconductors. In a cubic centimeter of a con ductor, on the other hand, there might be 10^^electrons available for electrical conduction.
Semiconductors In the previous sam ple problem , we com pared a property o f a sem iconductor w ith th at o f an insulator. Table 1 com pares som e properties o f a typical sem iconductor (sili con) and a typical co n ductor (copper). Let us now discuss these properties in m ore detail.
TABLE 1
SOM E ELEC TR IC A L PR O PE R T IE S O F C O PPE R A N D SILICON*
Type of material Density of charge carriers^ n (m ^) Resistivity /? (D • m) Temperature coefficient of resistivity a (K *)
Copper
Silicon
Conductor
Semiconductor
9 X 10“ 2 X 10-* -1-4 X 10-’
All values refer to room temperature. ^ Includes, for silicon, both electrons and holes.
1 X 10“ 3X 10» - 70X 10-^
1124
Chapter 53
Electrical Conduction in Solids
1. T he density o f charge carriers, n. C opper has m any m ore charge carriers th an does silicon, by a factor o f about 10'^. F or copijer the carriers are the conduction electrons, about one per atom . Figure I c shows that, at th e absolute zero o f tem perature, silicon w ould have no charge carriers at all. At room tem perature, to w hich Table 1 refers, charge carriers arise only because, at therm al equilibrium , therm al agitation has caused a certain (very sm all) n u m ber o f electrons to be raised to the conduction band, leav ing an equal n u m b er o f vacant states (holes) in the valence band. T he holes in the valence b an d o f a sem iconductor also serve effectively as charge carriers because they perm it a certain freedom o f m ovem ent to the electrons in th at band. If an electric field is set up in a sem iconductor, the electrons in the valence band, being negatively charged, drift in the direction o f —E. T he holes drift in the direction o f the field and behave like particles carrying a charge -\-e, w hich is exactly how we shall regard them . C onduction by holes is an im p o rtan t characteristic o f sem iconductors.
2. The resistivity, p. At room tem p eratu re the resistivity o f silicon is considerably higher th an th a t o f copper, by a factor o f ab o u t 1 0 ". F o r both elem ents, the resistivity is determ ined by Eq. 1. As th at equation shows, the resistiv ity increases as n, the density o f charge carriers, decreases. T he vast difference in resistivity betw een copper an d sili con can be accounted for by the vast difference in n. (T he m ean collision tim e t will also be different for copper and for silicon, b u t the effect o f this on the resistivity is over w helm ed by the en o rm o u s difference in the density o f charge carriers.) F or com pleteness, we m ention th at the resistivity o f a good insulator (fused q u artz o r diam o n d , for exam ple) m ay be as high as 1 Q • m , a b o u t 10^* tim es higher than th at o f copper at room tem perature. Few physical proper ties have as wide a range o f m easurable values as the electrical resistivity. 3. T he tem perature coefficient o f resistivity, a. This q u an tity (see Eq. 16 o f C h ap ter 32) is the fractional change in the resistivity p p er u n it change o f tem perature, or
1 dp p dT T he resistivity o f copper an d o th er m etals increases w ith tem p eratu re {d p /d T > 0). T his happens because col lisions o ccur m ore frequently the higher the tem perature, th u s reducing t in Eq. 1. F or metals, the density o f charge carriers n in th at equation is independent o f tem perature. O n the o th er hand, the resistivity o f silicon (and other sem iconductors) decreases w ith tem p eratu re {d p Id T < 0 ). T his happens because the density o f charge carriers n in Eq. 1 increases rapidly with tem perature. T he decrease in T m entioned above for m etals also occurs for sem iconduc
tors, b u t its effect on the resistivity is overw helm ed by the very rapid increase o f the density o f charge carriers. In the laboratory, you can identify a sem iconductor by its large resistivity p and, especially, by its large— and negative— tem perature coefficient o f resistivity a , both quantities being com pared to values for a typical m etal.
53-6 DOPED SEMICONDUCTORS T he perform ance o f sem iconductors can be substantially changed by deliberately introducing a sm all n u m b er o f suitable replacem ent atom s as im purities into th e sem i co nductor lattice, a process called doping. W e describe the sem iconductor th at results as extrinsic, to distinguish it from the pure undoped or intrinsic m aterial. Essentially all sem iconducting devices today are based on extrinsic m aterial. Figure 8a is a tw o-dim ensional representation o f a lat tice o f pure silicon. Each silicon ato m has 4 valence elec trons an d form s a tw o-electron bon d w ith each o f its four nearest neighbors, the electrons involved in th e bonding m aking u p the valence band o f the sam ple. In Fig. Sb one o f the silicon atom s has been replaced by an atom o f phosphorus, w hich has 5 valence electrons. F o u r o f these electrons form bonds w ith the 4 neighboring silicon atom s, b u t the fifth electron is loosely b o u n d to the phosphorus ion core, as Fig. 8 Z>suggests. It is far easier for this electron to be therm ally excited in to the conduction band th an it is for one o f the silicon valence electrons to be so excited. T he phosphorus atom is called a donor a to m because it readily donates an electron to the conduction band. T he “ extra” electron in Fig. 86 can be said to lie in a localized d o n o r level, as Fig. 9a shows. T his level is separated from the bottom o f the conduction band by an energy gap E ^, w here usually E^. A dding d o n o r atom s can greatly increase the density o f electrons in the conduction band. Sem iconductors doped w ith d o n o r atom s are called ntype sem iconductors, the standing for “ negative” be cause the negative charge carriers (electrons) greatly o u t nu m b er the positive charge carriers (holes). In «-type sem iconductors, the electrons in the conduction band are called the m ajority carriers, while th e holes in the valence band are called the m inority carriers. Figure 8c shows a silicon lattice in w hich a silicon atom has been replaced by an alu m in u m atom , w hich has 3 valence electrons. In this case there is a “ m issing” elec tron, and it is easy for the alu m in u m ion core to “ steal” a valence electron from a nearby silicon atom , th u s creating a hole in the valence band. T he alu m in u m atom is called an acceptor ato m because it so readily accepts an electron from the valence band. T he electron so accepted m oves in to a localized acceptor level, as Fig. 9b shows. T his level is separated from the to p
Section 53-6
Doped Semiconductors
1125
Figure 8 {a) A two-dimensional representation of the silicon lattice. Each silicon ion (core charge = +4^) is bonded to each of its four nearest neighbors by a shared two-electron bond. The dots show those valence electrons, (b) A phosphorus atom (valence = 5) is substituted for a silicon atom, creating a donor site, (c) An aluminum atom (valence = 3) is substituted for a silicon atom, creating an acceptor site.
of the valence band by an energy gap , for which E^ Eg, Adding acceptor atoms can greatly increase the num ber o f holes in the valence band. Semiconductors doped with acceptor atoms are called p-type semiconductors, the “p” standing for “positive” because the positive charge carriers (holes) greatly out number the negative carriers (electrons). In p-type semi conductors the majority carriers are the holes in the va lence band and the minority carriers are the electrons in the conduction band. Table 2 summarizes the properties of a typical n-type and a typical p-type semiconductor. Note particularly that the donor and acceptor ion cores, although they are charged, are not charge carriers because, at normal tem peratures, they remain fixed in their lattice sites.
temperature, the thermal agitation is effective enough so that essentially every phosphorus atom donates its “extra” electron to the conduction band.) Solution The density np of the phosphorus atoms must be about (10*^ m“ ^)( 10^) or 10^^ m“ \ The density of silicon atoms in a pure silicon lattice may be found from _N ^d M
«Si =
in which is the Avogadro constant, d (=2330 kg/m^) is the density of silicon, and M (=28.1 g/mol) is the molar mass of silicon. Substituting yields
«Si
^ (6.02 X 10^^ mol-X2330 kg/m^) _ ^ ^ 0.0281 kg/mol
The ratio of these two number densities is the quantity we are looking for. Thus "si_5XJ0^^nr^_
Conduction band
i
i Donor levels
t
(a)
1
o
o
^
T
E
8
1
o Valence band
(6)
*
-5X10.
Sample Problem 7 The density of conduction electrons in pure silicon at room temperature is about 10'^ m“ ^. Assume that, by doping the lattice with phosphorus, you want to increase this number by a factor of 10^. What fraction of the silicon atoms must you replace by phosphorus atoms? (Assume that, at room
We see that if only one silicon atom infive million is replaced by a phosphorus atom, the number of electrons in the conduction band will be increased by a factor of 10^.
.....is:....;....:;.......................... Acceptor levels O
1
____L _ t
Figure 9 {a) An «-type semiconductor, showing donor levels that have contributed electrons (majority carriers) to the conduc tion band. The small number of holes (mi nority carriers) in the valence band is also shown, {b) A p-type semiconductor, showing acceptor levels that have contributed holes (majority carriers) to the valence band. The small number of electrons (minority carriers) in the conduction band is also shown.
1126
Chapter 53
TABLE 2
Electrical Conduction in Solids
TWO TYPICAL EXTRINSIC SEMICONDUCTORS
Matrix material Dopant Type of dopant Dopant valence Dopant energy gap Majority carriers Minority carriers Dopant ion core charge
Thus r = (12)(52.9 pm) = 630 pm.
n-Type
p-Type
Silicon Phosphorus Donor 5 (= 4 + 1 ) 0.045 eV Electrons Holes -\-e
Silicon Aluminum Acceptor 3 (= 4 -1 ) 0.057 eV Holes Electrons —e
How can such a tiny admixture of phosphorus atoms have such a big effect? The answer is that, for pure silicon at room temperature, there were not many conduction electrons there to start with! The density of conduction electrons was 10*^ m“^ before doping and 10^^ m“ ^ after doping. For copper, however, the conduction electron density (see Table 1) is about 10^’ m"^. Thus, even after doping, the conduction electron density of sili con remains much less than that of a typical conductor such as copper.
Sample Problem 8 Assume that the “extra” electron in a phos phorus donor atom moves in a Bohr orbit around the central phosphorus ion core, as in Fig. 8^. Calculate (a) the binding energy and (b) the orbit radius for this electron. Solution (a) The Bohr theory expression for the binding en ergy £b of the « = 1 state is (see Eq. 18 of Chapter 51) mZV*
(7)
Here we put Z = 1 because the orbiting electron “sees” a net central charge o f+ ^ . We derived the Bohr energy by considering a hydrogen-like atom, its orbiting electron moving in a vacuum. In this case, however, the electron moves through a silicon lattice. One effect of this is to reduce the electrostatic force by a factor of k^, the dielectric constant of silicon. To realize this force reduction quantitatively, we must replace €o in Coulomb’s law by K^eQ. Making the same replacement in Eq. 7 leads to
in which the factor in parentheses is just 13.6 eV, the binding energy of the hydrogen atom. For silicon we have k , = 12, that
This is comparable to the atomic spacing in the silicon lattice (540 pm). We should also replace the electron mass m in Eqs. 7 and 8 by an effective electron mass to take partially into account the periodic nature of the silicon lattice potential. Doing so reduces the estimated binding energy and increases the estimated orbit radius, both changes being in the direction of improving the agreement with experiment.
53-7 THE p n JU N CTIO N ____________ In the next few sections we describe some commonly used semiconducting devices, such as diode rectifiers, lightemitting diodes, and transistors. There is no end to the number of such devices that we could have chosen to describe. With today’s technology, in fact, it is possible to tailor-make complex semiconducting devices to meet spe cific needs. Essentially all semiconducting devices involve one or more pn junctions. Consider a hypothetical plane through a rod of a pure crystalline semiconducting material such as silicon. On one side of the plane, the rod is doped with donor atoms (thus creating «-type material), and on the other side it is doped with acceptor atoms (thus creating p-type material). This combination gives a pn junction.* Figure 10a represents a pn junction at the imagined moment of its creation. There is an abundance of elec trons in the «-type material and of holes in the p-type material. Electrons close to the junction plane will tend to diffuse across it, for much the same reason that gas molecules will diffuse through a permeable membrane into a vacuum beyond it. The diffusing electrons in the conduction band, which move from right to left in Fig. 10a, will readily combine with the holes in the valence band on the other side of the junction plane. Similarly, the holes in the ptype region diffuse across the junction plane from left to right and combine with electrons in the ^-region. For every such diffusion-recombination event, the portion of the bar on the right side of this plane acquires a
so
13.6eV = 0.094 eV. ( 12)^ This result is in rough order-of-magnitude agreement with the value of 0.045 eV listed in Table 2. (b) The orbit radius follows from Eq. 19 of Chapter 51. Substi tuting as in part (a) leads to ' \ nm e^/
(8)
The factor in parentheses is just the Bohr radius (= 52.9 pm).
* In common practice, to make a pn junction one starts with, say, p-type material, made by adding acceptor atoms to the mol ten silicon from which the solid silicon crystal is drawn. Donor atoms are then diffused into the solid sample at high tempera ture in a special furnace, overcompensating the acceptor atoms to a certain (controllable) depth below the surface and creating the A2-type region. The junction that we analyze here is idealized in that we assume that the n-type and the p-type regions are separated by a well-defined plane; in practice, these regions blend into each other gradually.
Section 53-7
(a)
p
n
6
( )
V {x)
— En
(c) Conduction band ■m
..*. m m
O^O ^O O ^ > V oO oOo 'NX
Valence band (d )
‘^ rt d -l^d. (e )
Figure 10 (a) A pw junction at the imagined moment of its creation, (b) Diffusion of majority carriers across the junction plane causes a space charge of fixed donor and acceptor ions to appear, (c) The space charge establishes a potential differ ence Vq and a corresponding electric field E q across the junc tion plane, (d) The electron energy bands near the junction. The arrows show the diffusion of the majority carriers, {e) In equilibrium, the diffusion of majority carriers across the junc tion plane is just balanced by the drift of minority carriers in the opposite direction.
positive charge and the portion on the left side a negative charge. These chargest cause a potential difference Vq to build up across the ju n c tio n , as Fig. 1Oc shows. Related to the potential difference (by the equation E = —d V jd x ) is an internal electric field th a t appears across the ju n c
t The fixed charges, which are close to — and separated by— the junction plane, are those of the donor and the acceptor ion cores, which, we recall, are not mobile. Normally the charges of these ion cores are compensated by the (opposite) charges of the mo bile charge carriers. However, when charge carriers cross the junction plane, the ion cores are no longer fully compensated and are, so to speak, uncovered.
The pn Junction
1127
tion plane, pointing as shown in Fig. 10c. This field exerts a force on the electrons, opposing their m otion o f diffu sion. Put an o th er way, for an electron to succeed in diffus ing from right to left or a hole from left to right in Fig. 1Ob, it m ust be energetic enough to overcom e the potential barrier represented by Fig. 10c. T his is represented in Fig. \0d, which shows the electron energy bands. T o diffuse from the A2-type region to the /?-type region, an electron m ust “ clim b” the hill o f height c F q. A hole m ust also “ clim b” a hill o f this sam e height to diffuse from left to right. T he diffusion o f both electrons and holes gives a current whose direction, in the usual conventional sense, is from left to right in Fig. 10. W e call this cu rren t the diffusion current It is, o f course, not possible to have an isolated silicon rod resting on a shelf with a current flowing indefinitely along its length. Som ething m ust happen to stop, or to com pensate, this current. T o find out w hat it is we tu rn o u r attention to the m inority carriers. As Fig. 9 and T able 2 show, although the m ajority carri ers in «-type m aterial are electrons, there are nevertheless also a few holes, the m inority carriers. Likewise in /?-type m aterial, although the m ajority carriers are holes, there are also a few electrons. The m inority carriers are shown in Fig. \0d. A lthough the electric field in Fig. 10c acts to retard the m otions o f the m ajority carriers— being a barrier for th e m — it is a dow nhill trip for the m inority carriers, be they electrons or holes. W hen, by therm al agitation, an electron close to the ju n ctio n plane is raised from the valence band to the conduction band o f the ;?-type m ate rial, it drifts steadily from left to right across the ju n ctio n plane, swept along by the electric field E q, Similarly, if a hole is created in the A2-type m aterial, it too drifts across to the other side. The space-charge region shown in Fig. \0b is effectively swept free o f charge carriers by this process and, for th at reason, we call it the depletion zone. T he current represented by the m otions o f the m inority carri ers, called the drift current i^^ft»is in the opposite direction to the diffusion cu rren t and ju st com pensates it at equilib rium , as Fig. \0 e shows. Thus, at equilibrium , a pn ju n ctio n resting on a shelf develops a contact potential difference Vq betw een its ends. T he diffusion current i^^ff th at m oves through the ju n ctio n plane from the direction p i o n is ju st balanced by a drift cu rrent i^^n th at m oves in the opposite direction. An electric field E q acts across the depletion layer, whose w idth is Jo-
Sample Problem 9 A silicon-based pn junction has an equal concentration /Jq of donor and acceptor atoms. Its depletion zone, of width J, is symmetrical about the junction plane, as Fig. 1\a shows, (a) Derive an expression for Fmax^ the maximum intensity of the electric field in the depletion zone, (b) Derive an expression for Fq, the potential difference that exists across the
1128
Chapter 53
Electrical Conduction in Solids
\
eno
)
r (4X12X8.85 X 10-*2 F/mX0.60 V )]>/2 (1.60 X 10-*’ CX3 X 10^2 m - 2)
L
J
= 2.3 X 10-2 m = 230 nm. {d) Substituting in Eq. 9 leads to
2/Ce€o _ (3 X 1022 m-^Xl.60 X 10-*’ CX2.3 X 10-^ m) (2X12X8.85 X 10-'2F/m ) Figure 11 Sample Problem 9. {a) The depletion zone of a pn junction. The rectangle represents the cross section of a Gaus sian surface with end caps of area A. (b) The variation of the electric field in the depletion zone.
depletion zone; see Fig. 10c. (c) Assume that n^ = 3 X IO22 m-^ and that Vq is measured to be 0.6 V. Calculate the width of the depletion zone, (d) Using this value of d, calculate the value of
Solution (a) The electric field may be taken as zero in the A2-type and the p-type material outside the depletion zone. The field points from right to left within the depletion zone and, from symmetry, has its maximum value in the center of this zone; see Fig. 11^. Let us apply Gauss’ law to the closed “box” (Gaussian sur face) shown in Fig. 1\a. This law is
= 5.2 X 10^ V/m. What assumptions were made in this problem that might lead to different values of the calculated quantities under practical laboratory conditions?
The Diode Rectifier Although a pn junction can be used in many ways, it is basically a rectifier. That is, if you connect it across the terminals of a battery, the current (a few picoamperes) in the circuit will be very much smaller for one polarity of the battery connection than for the other. Figure 12 shows that, for a typical silicon-based pn junction diode, the current for the reverse-biased connection ( F < 0) is negli gible by comparison with the current for the forwardbiased connection ( F > 0).
€0 f KcE •dA = q, in which (= 12) is the dielectric constant of silicon and q [=nQeA(dl2)] is the free charge contained within the box. The integral is to be taken over the surface of the box. The only contribution to the integral comes from the face of the box that lies in the junction plane so that the integral has the value k^E^^A, Making these substitutions and solving for E ^ yields F
2k,€o ’
(9 )
the relationship we seek. (b) As Fig. 11 b shows, the electric field drops linearly from its central value of E^^ax to zero at each edge of the depletion zone. Its average value throughout the zone is thus \ E ^ . The poten tial difference Vqis equal to the work per unit charge required to carry a test charge qQfrom one face of the depletion zone to the other. Thus if F is the average force acting on the test charge, _ W _Fd_(iiE^qo)d_ Qo
Qo
Qo
Substituting for E ^ ^ from Eq. 9 above leads to _ nped^ " 4fc.Co ■ (c) find
Solving Eq. 10 for
( 10)
and substituting the given values, we
Figure 12 A current - voltage plot for a typical pn junction, showing that it conducts easily in the forward direction but is essentially nonconducting in the reverse direction. The dots refer to Problem 43.
Section 53- 7
The pn Junction
1129
Figure 13 A pn junction diode connected as a rectifier. The diode conducts easily in the for ward direction (positive sections of input wave) but not at all in the reverse direction (nega tive sections of input wave).
Figure 13 shows one o f m any possible applications o f a diode rectifier. A sine wave in p u t potential generates a half-wave o u tp u t potential, the diode rectifier acting as essentially a short circuit for one polarity o f the in put potential an d as essentially an open circuit for the other. An ideal diode rectifier, in fact, has only these two m odes o f operation. It is either on (zero resistance) o r off (infinite resistance). Figure 13 displays the conventional sym bol for a diode rectifier. T he arrow head corresponds to the p-type term i nal o f the device an d points in the direction o f “ easy” conventional cu rren t flow. T h a t is, the diode is on w hen the term inal w ith the arrow head is (sufficiently) positive w ith respect to the o th er term inal. Figure 14 shows details o f the tw o connections. In Fig.
1Aa (the reverse-biased arrangem ent) the battery e m f sim ply adds to the contact potential difference, th u s increas ing the height o f the barrier th at the m ajority carriers m ust surm ount. Fewer o f them can do so and, as a result, the diffusion current decreases m arkedly. T he drift current, however, senses no barrier an d thus is independent o f the m agnitude or direction o f the external potential. T he current balance th at existed at zero bias (see Fig. 10^) is thus upset and, as shown in Fig. 14^, a c u rre n t— b u t a very sm all o n e — appears in the circuit. A nother effect o f reverse bias is to widen the depletion zone, as a com parison o f Figs. 106 an d \Aa shows. This seems reasonable because the positive battery term inal, connected to the «-type end o f the ju n ctio n , tends to pull electrons o u t o f the depletion zone back into the «-type
Figure 14 (a) The reverse-biased con nection of a pn junction, showing the wide depletion zone, the energy bands, and the corresponding small back current /‘b. (6) The forward-biased connection, showing the narrow depletion zone, the energy bands, and the large forward current />. Note that the drift current is the same in each case.
1130
Chapter 53
Electrical Conduction in Solids
m aterial an d to repel holes back into the p-type m aterial. Because the depletion zone contains very few charge carri ers, it is a region o f high resistivity. T h u s its substantially increased w idth m eans a substantially increased resist ance, consistent w ith the sm all value o f the current in a reverse-biased diode. Figure \4 b shows the forw ard-biased connection, the positive term inal o f the battery being connected to the p-type end o f the ju n ctio n . H ere the applied e m f sub tracts from the contact potential, the diffusion current rises substantially, an d a relatively large net forward cu rren t results. T he depletion zone becom es narrower, its low resistance being consistent with the large forward current.
53-8 OPTICAL ELECTRONICS W e are all fam iliar w ith the brightly colored n um bers th at flash and glow at us from cash registers, gasoline pum ps, an d electronic equipm ent. In nearly all cases this light is em itted from an assem bly o f pn ju n ctio n s operating as light-em itting diodes (LEDs). Figure 15a shows the fam iliar seven-segm ent display from which the num bers are form ed. Figure 15b shows th at each elem ent o f this display is the end o f a flat plastic lens, at the oth er end o f which is a sm all LED, possibly ab o u t 1 mm^ in area. Figure 15c shows a typical circuit, in which the LED is forw ard biased. H ow can a d isju n ctio n em it light? W hen an electron at the bo tto m o f the conduction b and o f a sem iconductor falls into a hole at the to p o f the valence band, an energy is released, where is the gap w idth. W hat happens to this energy? T here are at least tw o possibilities. It m ight be transform ed into internal energy o f the vibrating lattice and, with high probability, th a t is exactly w hat happens in a silicon-based sem iconductor. In som e sem iconducting m aterials, however, the em it ted energy can also appear as electrom agnetic radiation.
A = - = - ^ V E Jh
he
( 11)
C om m ercial LEDs designed for the visible region are usu ally based on a sem iconducting m aterial th at is a g a lliu m -a rse n ic -p h o s p h o ru s com pound. By adjusting the ratio o f phosphorus to arsenic, the gap w id th — and thus the wavelength o f the em itted light— can be varied. If light is em itted w hen an electron falls from the con duction band to the valence band, then light o f th at sam e w avelength will be absorbed w hen an electron m oves in the other direction, th at is, from the valence ban d to the conduction band. T o avoid having all the em itted pho tons absorbed, it is necessary to have a great surplus o f both electrons and holes present in the m aterial, in m uch greater num bers th an w ould be generated by therm al agi tation in the intrinsic sem iconducting m aterial. These are precisely the conditions th at result w hen m ajority carriers — be they electrons or holes— are injected across the cen tral plane o f a pn ju n ctio n by the action o f an external potential difference. T hat is why a sim ple intrinsic sem i co nductor will not serve as an LED. Y ou need a pi? ju n c tion! T o provide lots o f m ajority carriers— an d thus lots o f p h o to n s— it should be heavily doped and strongly for ward biased.
Sample Problem 10 An LED is constructed from a j u n c t i o n based on a certain semiconducting material whose energy gap is 1.9 eV. What is the wavelength of its emitted light? Solution
From Eq. 11 we have /jc _ (6 .6 3 X 10-^J*sK3.00X 10« m/s) (1.9eVK1.60X lO -'^J/eV ) = 6.53 X 10“ ^ m = 653 nm.
Light of this wavelength is red.
Figure 15 {a) The familiar seven-segment number display, activated to show the num ber “7.” (b) One segment of such a display, (c) An LED connected to an external source of em f
3n
(a)
the w avelength being given by
(6)
Section 53-8
Optical Electronics
1131
The Diode Laser The dropping down of an electron from the conduction band to fill a hole in the valence band, with the emission of a photon, bears a strong resemblance to the dropping down o f electrons in transitions between atomic states we considered in Chapter 52. There is an important applica tion based on this similarity: by injecting electrons into the conduction band and holes into the valence band, it is possible to create a population inversion analogous to that considered in our discussion of lasers in Section 52-6. In this way it is possible to make a diode laser, in which the lasing medium is not a gas but a solid semiconductor. Diode lasers are commonly used in compact disk players and other optical data retrieval systems. Figure 16 shows a representation of the energy levels in a diode laser. The lasing material is sandwiched between layers o f /7-type and «-type material, which have slightly larger gap energies. Electrons are injected by an external circuit into the «-type material; some of these excess elec trons drift into the lasing layer, where they are prevented from drifting into the p-type material by a potential barrier. Similarly, holes are injected into the p-type mate rial, drift into the lasing layer, and are trapped there. The excess of electrons (and holes) in the active region gives the lasing action. The physical construction of the device is illustrated schematically in Fig. 17, and Fig. 18 shows a photograph o f a diode laser. The lasing material is a narrow (0.2 pm) layer o f a material such as GaAs (gallium arsenide), and the p-type and w-type material on each side may be layers of GaAlAs (gallium aluminum arsenide) a few microme ters in thickness. The ends of the material are cleaved to create mirror-like surfaces that reflect a portion of the light wave to enable stimulated emission in the active region. The device illustrated in Fig. 18 emits at 840 nm (in the infrared region). Diode lasers at this wavelength are commonly used in communication to send signals
Figure 17 The physical construction of a diode laser. The lasing action occurs in the narrow GaAs layer.
along optical fibers. Other materials can be used in similar fashion to give visible radiation. Among the advantages of diode lasers are their small size and low power input (in the range of 10 milliwatts, compared with the standard HeNe laser that may require several watts of electrical power). Like other semiconduc tor devices, the diode laser can be powered by batteries. Efficiencies of the order of 20% are possible (that is, 20% of the electrical power supplied to the device appears in the laser beam), compared with 0.1% in the HeNe laser. The light signal can easily be modulated by controlling the injection current, and thus we have an optical device that
region
Figure 16 The energy bands in a diode laser. The active re gion has a smaller energy gap than the «-type and p-type ma terials on either side. When electrons in the conduction band of the active region drop down to fill holes in the valence band, light is emitted.
Figure 18 A diode laser, compared in size with a grain of table salt on the right.
1132
Chapter 53
Emitter Base Collector
(a)
Electrical Conduction in Solids Emitter Base Collector
( 6)
Figure 19 (a) An junction transistor. At the bottom are shown the energy bands and majority carriers in the three re gions. (b) The em itter-base junction is forward biased, and the base-collector junction is reverse biased. Electrons that move from the emitter to the base either recombine with holes or (far more likely) continue to the collector.
can respond at the rapid switching tim es (< 100 ps) char acteristic o f electronic circuits.*
53-9 THE TRANSISTOR T he ju n c tio n diodes we have considered so far are twoterm inal devices. H ere we consider a device with three (or m ore) term inals, called a transistor.^ A transistor often operates in the following m ode: a cu rren t established be tw een tw o o f the term inals is regulated by a cu rrent or voltage at the third term inal. O ne co m m o n variety o f transistor is the junction tran sistor, w hich consists o f three layers o f doped sem iconduc tors, such as npn o r pnp. Figure 19a shows a typical config uration for a npn ju n c tio n transistor. T he three sections are called the em itter, base, an d collector. T he conduction an d valence bands are show n, an d only the m ajority carri ers are indicated. T he e m itte r-b a s e an d b a se -co llecto r ju n ctio n s behave m uch like ordinary junctions. In norm al operation, as illustrated in Fig. 196, the e m itte r-b a s e ju n c tio n is forw ard biased and the b a s e collector ju n c tio n is reverse biased, which gives the energy bands show n in the figure. Electrons flow from the heavily doped w-type em itter into the base. Because the base is very narrow , m ost o f these electrons reach the collector, b u t a few recom bine
* See “Applications of Lasers,” by Elsa Garmire, in Fundamen tals o f Physics by David Halliday and Robert Resnick (Wiley, 1988), essay 19. t The transistor was invented in 1947 at what is now the AT&T Bell Laboratories by John Bardeen, Walter Brattain, and Wil liam Shockley, who shared the 1956 Nobel prize in physics for their discovery.
with holes in the p-type region. T o replenish the holes in the base, electrons from the valence band in the base m ust leave the transistor through the external circuit as the small base current /’b. A small change in the base current /‘b can result in a large change in the collector cu rren t z^. In this configuration, the transistor serves as a cu rren t am pli fier, and the current gain i^/iy, can have typical values in excess o f 100. A second type o f transistor is called a field-effect transis tor (FET). Figure 20 illustrates the basic geom etry. Elec trons flow through the zz-type region from the source to the drain w hen there is an external potential difference Kls betw een the drain and the source. T he p-type regions are heavily doped, and the depletion layers form ed at the tw o in ju n c tio n s determ ine the w idth o f the n-type chan nel. A n external voltage applied to the p-type region (the gate) changes the w idth o f the depletion region and consequently changes the w idth o f the n-type channel. This in tu rn changes the current through the device, be cause the ability o f cu rren t to flow along th e n-channel depends on the w idth o f the channel. A sm all change in the gate voltage changes the w idth o f the channel and causes a large change in the cu rren t through the nchannel, so th at the device can operate as an am plifier. If the gate voltage is m ade large enough, the n-channel w idth can becom e zero, an d the FE T stops conducting. H ere the transistor is acting like a switch: it is either con ducting (on) or n o t conducting (off). T he cu rren t can be switched on or off very rapidly by the signal applied to the gate; switching tim es sm aller th an 1 ns ( 10“ * s) are com m on. A com m on type o f FET widely used in digital circuits is the m etal-oxide-sem iconductor FE T (M O SFE T ), which is fabricated by depositing an d etching successive layers in a p-type substrate. A cross section o f a M O SFET is shown in Fig. 2 1. T he n-region an d the n-channel are m ade by etching a m ask o n to the p-type substrate an d diffusing d o n o r atom s a know n distance into the substrate. An oxide layer (Si02 ) is then deposited, a n d a m etal layer is then deposited to form the contacts for the n-region and the gate.
p Drain
Source
JCate
Figure 20 The basic structure of a field-effect transistor. Electrons travel along the narrow zz-channel from the source to the drain. The width of the channel can be controlled by varying the voltage at the gate.
Section 53-10 Figure 21
Gate
Drain
Superconductors
1133
The structure of a MOSFET.
Source
n-Channel Substrate
^ Metal I |p-Type semiconductor [^Insulator (Si02) [ |n-Type semiconductor
53-10 SUPERCONDUCTORS_________ The resistivity o f a typical metallic conductor decreases as the temperature decreases. However, the resistivity does not fall to zero, even as T approaches 0 K. As we have seen in Chapter 32, the resistivity of a conductor originates with collisions made by the conducting electrons as they move through the lattice. Impurities and lattice defects increase the chances for electrons to have collisions, and collisions o f electrons with atoms displaced from their lattice sites by vibrational motion contribute to the resis tivity. In certain materials called superconductors (see Section 32-8), the resistance falls gradually with decreasing tem perature, as expected; however, at a certain critical temper ature Tc the resistivity drops suddenly to zero (Fig. 22). Below Tc, electrons move unimpeded through the mate rial. Table 3 shows a selection o f some superconductors and their critical temperatures. Superconductivity has been observed in 27 elements and in numerous compounds, but it has not been ob served for the best metallic conductors (Cu, Ag, Au). We conclude that a superconductor is not merely a good con ductor getting better, and we are led to suspect that the mechanism that causes superconductivity may differ from the mechanism that causes the conductivity o f ordi-
Resistivity
nary metals. As we shall see, superconductivity results from a strong coupling between conduction electrons and the lattice. Normal conduction in the best conductors occurs when there is a weak coupling between the valence electron and the lattice. Consider an electron moving through a lattice. As it moves, it pulls the positive ion cores toward it and changes the charge density in its vicinity. It leaves a some what higher positive charge density in its wake than would otherwise be there. This positive charge attracts other electrons. The electrons interact with one another through the intermediary of the lattice, somewhat like two boats on a lake interacting through their wakes. The net result is a slight attraction of the electrons for each other. The BCS (Bardeen-Cooper-Schrieffer) theory* of su perconductivity shows that the electron system has the lowest possible energy if the electrons are bound together in pairs, called Cooper pairs. When no current exists in a superconductor, the two electrons of a Cooper pair have momenta of equal magnitude but exactly opposite direc tions, so that the total momentum and the electric current both vanish. When a current is generated, both electrons in a pair acquire the same increase in momentum, result ing in a motion of the center of mass of the pair. All Cooper pairs acquire the same momentum.
* This theory of superconductivity was developed in 1957 by John Bardeen, Leon N. Cooper, and J. Robert Schrieffer, who were awarded the 1972 Nobel prize in physics for their work. Bardeen also shared the 1956 Nobel prize for his research on semiconductors and the discovery of the transistor.
TABLE 3
Figure 22 Comparison of the dependence of resistivity on temperature for a normal conductor and a superconductor. The resistivity of a normal conductor falls gradually with de creasing temperature. In superconducting materials, the resis tivity drops suddenly to zero at the critical temperature .
PROPERTIES OF SOME SUPERCONDUCTORS
Material
t;(K )
Pairing Energy (meV)
Cd A1 Sn Hg Pb Nb N b 3Sn YBa2Cu 307 Tl 2Ba2Ca 2Cu 3O 10
0.56 1.19 3.75 4.16 7.22 9.46 18.1 90 125
0.21 0.34 1.15 1.65 2.73 3.05
1134
Chapter 53
Electrical Conduction in Solids
Superconductivity is a cooperative p h enom enon. If som e C ooper pairs have been form ed, the reduction in energy th at occurs for the next pair is greater th an if no pairs were previously form ed. O nce the tem perature drops below and som e pairs are form ed, a small addi tional reduction in tem perature causes m any m ore pairs to form . T he change from the norm al to the supercon ducting state is q uite precipitous. T he cooperative m o tions o f the C ooper pairs also force every pair to have the sam e m o m en tu m . T he C ooper pairs have a binding energy A, called the pairing energy, w hich is typically in the range o f lO” "^ to 10"^ eV, as shown in T able 3. N ote th at critical tem pera tures o f 1 - 1 0 K (typical for m ost o f the superconductors show n in T able 3) correspond to energies kT^ in the sam e range o f 10“ ^ to 10“ ^ eV. T he critical tem perature o f a su perconductor is directly related to the pairing energy. Above T c the pairs are broken an d the m aterial has nor m al electrical resistance. T he binding energy o f a C ooper p air introduces a pair ing gap 2A in to the density o f states n ( E) near the Ferm i energy. (Figure \ a shows an exam ple o f the density o f states for a norm al conductor.) It is energetically favor able for electrons near the F erm i energy in a supercon d u cto r to b ind together in C ooper pairs. As a result, the density o f states decreases to zero w ithin an interval o f ± A o f Ef:, with a corresponding increase in a2(£') ju st above an d below E^, Figure 23 shows the resulting density o f states and the pairing gap 2A. Above the density o f states o f a superconductor m ight be as show n in Fig. \a. T he gap begins to open as the superconductor is cooled below T^\ the gap energy increases as the tem perature decreases, reaching its m ax im u m as T approaches 0 K. T he occupation probability o f electron states in a su perconductor can be found from the prod u ct o f the den sity o f states, shown in Fig. 23, an d a F e rm i-D ira c distri b ution function, as was show n in Fig. 2b. T his leads to a high occupation probability o f the superconducting states ju st below the gap. Above the gap, a sm all density o f norm al (unpaired) states occurs for T > 0 . Beginning in 1986, a new class o f superconductors was discovered* with unusually high values o f . T he last two
* See “Superconductors Beyond 1-2-3,” by Robert J. Cava, Sci entific American, August 1990, p. 42.
Figure 23 The density of states in a superconductor below its transition temperature. There is an energy gap of 2A, within which the density of states is zero. The scale of this drawing has been exaggerated; typically the Fermi energy E^ is a few electron-volts, while the pairing gap is 10“^ to 10“ ^ eV.
entries in Table 3 are exam ples o f these com pounds, which are ceram ic m aterials th at (unlike the m ore fam il iar types o f ceram ics) are conductors at room tem pera ture. Since the highest tem perature at which supercon ductivity had previously been observed was ab o u t 20 K, these new m aterials represent a substantial leap in technol ogy. In particular, they allow superconductivity to be at tained at tem peratures characteristic o f cooling with liq uid nitrogen (77 K) rather th an at those characteristic of the m ore expensive and less convenient liquid helium (4 K). This ju m p o f a factor o f 6 in holds o u t the hope that, with an o th er ju m p o f a factor o f less th an 3, it m ight be possible to achieve superconductivity at room tem pera ture. These high-tem perature superconductors are oxides of copper in com bination w ith various other elem ents. The theory o f operation o f these m aterials is not yet under stood; it is not clear w hether there is a BCS-type m echa nism involved. It seems apparent th at the superconducti vity resides with the copper oxides; although elem ental copper is not superconducting, the copper oxide com bina tions are. T he crystal structure o f these com pounds places the copper and oxygen in planes anchored between the other elem ents, and it is likely th at these planes provide the pathw ay for the electrons th at carry the superconduct ing current.
QUESTIONS 1. Do you think that any of the properties of solids listed in the opening to this chapter are related to each other? If so, which? 2. The conduction electrons in a metallic sphere occupy states of quantized energy. Does the average energy interval be
tween adjacent states depend on {a) the material of which the sphere is made, (b) the radius of the sphere, (c) the energy of the state, or (d) the temperature of the sphere? 3. What role does the Pauli exclusion principle play in account ing for the electrical conductivity of a metal?
Questions 4. In what ways do the classical model and the quantum me chanical model for the electrical conductivity of a metal differ? 5. If we compare the conduction electrons of a metal with the atoms of an ideal gas, we are surprised to note (see Fig. Ic) that so much kinetic energy is locked into the conduction electron system at absolute zero. Would it be better to com pare the conduction electrons, not with the atoms of a gas, but with the inner electrons of a heavy atom? After all, a lot of kinetic energy is also locked up in this case, and we don’t seem to find that surprising. Discuss. 6 . What features of Fig. 2 make it specific for copper, for which it was drawn? What features are independent of the identity of the metal? 7. Why do the curves in Figs. 1c and 2c differ so little from each other? 8 . Distinguish carefully among the density of states function n {E \ the density of occupied states function n J,E \ and the Fermi - Dirac probability function p { E \ all of which appear in Eq. 3. 9. Does the Fermi energy for a given metal depend on the volume of the sample? If, for example, you compare a sam ple whose volume is 1 cm^ with one whose volume is twice that, the latter sample has just twice as many available con duction electrons; it might seem that you would have to go to higher energies to fill its available levels. Do you? 10. In Section 25-4 we showed that the (molar) heat capacity of an ideal monatomic gas is \R . If the conduction electrons in a metal behaved like such a gas, we would expect them to make a contribution of about this amount to the measured specific heat of a metal. However, this measured specific heat can be accounted for quite well in terms of energy absorbed by the vibrations of the ion cores that form the metallic lattice. The electrons do not seem to absorb much energy as the temperature of the specimen is increased. How does Fig. 2 provide an explanation of this prequantum-days puzzle? 11. Give a physical argument to account qualitatively for the existence of allowed and forbidden energy bands in solids. 12. Is the existence of a forbidden energy gap in an insulator any harder to accept than the existence of forbidden energies for an electron in, say, the hydrogen atom? 13. On the band theory picture, what are the essential require ments for a solid to be (a) a metal, (b) an insulator, or (c) a semiconductor? 14. What can band theory tell us about solids that the classical model (see Section 32-5) cannot? 15. Distinguish between the drift speed and the Fermi speed of the conduction electrons in a metal. 16. Why is it that, in a solid, the allowed bands become wider as one proceeds from the inner to the outer atomic electrons? 17. Do pure (undoped) semiconductors obey Ohm’s law? 18. At room temperature a given applied electric field will gener ate a drift speed for the conduction electrons of silicon that is about 40 times as great as that for the conduction electrons of copper. Why isn’t silicon a better conductor of electricity than copper? 19. Consider these two statements: (a) At low enough tempera tures silicon ceases to be a semiconductor and becomes a
20.
21.
22.
23.
24.
1135
rather good insulator, (b) At high enough temperatures sili con ceases to become a semiconductor and becomes a rather good conductor. Discuss the extent to which each statement is either true or not true. Does the electrical conductivity of an intrinsic (undoped) semiconductor depend on the temperature? On the energy gap Eg between the full and empty bands? How do you account for the fact that the resistivity of metals increases with temperature but that of semiconductors de creases? The energy gaps for the semiconductors silicon and germa nium are 1.1 eV and 0.67 eV, respectively. Which sub stance do you expect would have the higher density of charge carriers at room temperature? At the absolute zero of temperature? Discuss this sentence: “The distinction between a metal and a semiconductor is sharp and clear-cut, but that between a semiconductor and an insulator is not.” The Hall effect is much greater in semiconductors than in metals. Why? What practical use can be made of this result?
25. Does a slab of /i-type material carry a net negative charge? 26. Suppose that a semiconductor contains equal numbers of donor and acceptor impurities. Do they cancel each other in their electrical effects? If so, what is the mechanism? If not, why not? 27. Why does an «-type semiconductor have so many more electrons than holes? Why does a p-type semiconductor have so many more holes than electrons? Explain in your own words. 28. What elements other than phosphorus are good candidates to use as donor impurities in silicon? What elements other than aluminum are good candidates to use as acceptor im purities? Consult the periodic table. 29. Does one distinguish between majority and minority carri ers for an intrinsic semiconductor such as silicon or germa nium? If not, why not? If so, what criterion do you use? 30. In preparing «-type or p-type semiconductors by doping, why is it extremely important to avoid contamination of the sample with even very small concentrations of unwanted impurities? 31. Would you expect doping to change the resistivity of silicon by very much? 32. When a current flows through a p-type material, positive holes move toward the negative terminal of the battery and combine with electrons in the ohmic electrode connected to the boundary of the crystal. Why doesn’t the crystal become negatively charged? 33. Why is silicon often preferred to germanium for making semiconductor devices? 34. Germanium and silicon are similar semiconducting materi als whose principal distinction is that the gap width Eg is 0.67 eV for the former and 1.1 eV for the latter. If you wished to construct a p« junction (see Fig. 10) in which the back current is to be kept as small as possible, which mate rial would you choose and why? 35. In a p« junction (see Fig. 10) we have seen that electrons and holes may diffuse, in opposite directions, through the junc tion region. What is the eventual fate of each such particle as it diffuses into the material on the opposite side of the junc
1136
36.
37.
38.
39.
40.
Chapter 53
Electrical Conduction in Solids
tion? Why is it that the electrons and positive holes do not all recombine, thus removing the possibility of conduction? Consider two possible techniques for fabricating a junc tion (see Fig. 10). {a) Prepare separately an n-type and a p-type sample and join them together, making sure that their abutting surfaces are planar and highly polished. (b) Prepare a single «-type sample and diffuse an excess acceptor impurity into it from one face, at high temperature. Which method is preferable and why? The pn junction shown in Fig. 1\a has equal dopant con centrations on each side of its junction plane. Suppose, how ever, that the donor concentration were significantly greater than the acceptor concentration. Would the depletion zone still be symmetrically located about the junction plane? If not, would the central plane of the zone move toward the «-type or toward the p-type face of the junction? Give your reason. Why can’t you measure the contact potential difference gen erated at a pn junction by simply connecting a voltmeter across it? In Fig. lOZ?, why does the depletion zone build up close to the junction plane? Why does it not spread out throughout the volume of the sample? What does it mean to say that a /?«junction is biased in the forward direction?
41. (a) Discuss the motions of the majority carriers (both elec trons and holes) in a forward-biased junction, (b) Discuss the motions of the minority carriers in this same junction. 42. Explain in your own words how the thickness of the deple tion zone of a p /2junction can be decreased by (a) increasing the forward-bias voltage and (b) increasing the dopant con centration. 43. If you increase the temperature of a reverse-biased junc tion, what happens to the current (see Fig. 14a)? Is the effect larger for silicon or for germanium? (The intrinsic energy gap Eg for silicon is larger than that for germanium.) 44. Does the diode rectifier whose characteristics are shown in Fig. 12 obey Ohm’s law? What is your criterion for deciding? 45. We have seen that a simple intrinsic (undoped) semiconduc tor cannot be used as a light-emitting diode. Why not? Would a heavily doped n-type or p-type semiconductor work? 46. Explain in your own words how the MOSFET device of Fig. 21 works. 47. Do you think that there is a correlation between the critical temperature of a superconductor (Table 3) and its electrical conductivity (inverse of resistivity) at room temperature?
PROBLEMS Section 53-1 Conduction Electrons in a Metal 1. (a) Show that Eq. 2 can be written as n(E) = CE'^\
2.
3.
4.
5.
where C = 6.81 X 10^^ m“ ^ •eV“ ^/^. (b) Use this relation to verify a calculation of Sample Problem 1, namely, that for £ = 5 .0 0 e V , n{E)= 1.52 X lO^* m -^-eV ” *. Calculate the density n(E) of conduction electron states in a metal for £" = 8.00 eV and show that your result is consist ent with the curve of Fig. \c. Gold is a monovalent metal with a molar mass of 197 g/mol and a density of 19.3 g/cm^ (see Appendix D). Calculate the density of charge carriers. At what pressure would an ideal gas have a density of mole cules equal to that of the density of the conduction electrons in copper (= 8.49 X 10^* m“ ^)? Assume that T = 297 K. The density and molar mass of sodium are 971 kg/m^ and 23.0 g/mol, respectively; the radius of the ion Na"^ is 98 pm. (a) What fraction of the volume of metallic sodium is avail able to its conduction electrons? (b) Carry out the same calculation for copper. Its density, molar mass, and ionic radius are, respectively, 8960 kg/m^, 63.5 g/mol, and 96 pm. (c) For which of these two metals do you think the conduction electrons behave more like a free electron gas?
Section 53-2 Filling the Allowed States
6 . Calculate the probability that a state 0.0730 eV above the Fermi energy is occupied at (a) T = 0 K and (b) T = 320 K.
7. The Fermi energy of silver is 5.5 eV. {a) At T = 0®C, what are the probabilities that states at the following energies are occupied: 4.4 eV, 5.4 eV, 5.5 eV, 5.6 eV, 6.4 eV? {b) At what temperature will the probability that a state at 5.6 eV is occupied be 0.16? 8 . Prove that the occupancy probabilities for two states whose energies are equally spaced above and below the Fermi en ergy add up to one. 9. The density of gold is 19.3 g/cm^. Each atom contributes one conduction electron. Calculate the Fermi energy of gold. See Appendix D for the molar mass of gold. 10. Figure 2c shows the density of occupied states nJiE) of the conduction electrons in copper at 1000 K. Calculate nJ^E) for copper for the energies E = 4.00, 6.75, 7.00, 7.25, and 9.00 eV. The Fermi energy of copper is 7.06 eV. 11. In Section 50-7 we considered the situation of an electron trapped in an infinitely deep well. Suppose that 100 elec trons are placed in a well of width 120 pm, two to a level with opposite spins. Calculate the Fermi energy of the sys tem. (Note: The Fermi energy is the energy of the highest occupied level at the absolute zero of temperature.) 12. The conduction electrons in a metal behave like an ideal gas if the temperature is high enough. In particular, the tempera ture must be such that k T :> the Fermi energy. What temperatures are required for copper (Ep = 7.06 eV) to sat isfy this requirement? Compare your answer with the boi ling point of copper; see Appendix D. Study Fig. Ic in this connection and note that we have k T c E^: for the condi tions of that figure. This is just the reverse of the require ment cited above.
Problems 13. Show that Eq. 5 can be written as
where the constant A has the value 3.65 X 10"
1137
conduction electrons excited to energies greater than the Fermi energy, •eV.
14. The Fermi energy of copper is 7.06 eV. (a) For copper at 1050 K, find the energy at which the occupancy probability is 0.910. For this energy, evaluate {b) the density of states and (c) the density of occupied states. 15. Show that the density of states function given by Eq. 2 can be written in the form n(E) = Explain how it can be that n(E) is independent of material when the Fermi energy Ep (=7.06 eV for copper, 9.44 eV for zinc, etc.) appears explicitly in this expression. 16. Show that if £ » f p , the distribution in energy of the occu pied states rio(E) can be written as n^{E) * in which C is a constant. Compare this result with that calculated for the Maxwell-Boltzmann distribution in Sec tion 24-4. What do you conclude? 17. Show that the probability Ph that a hole exists at a state of energy E is given by 1
Ph = ^-(£-£F)//cr _|_ j (Hinv, The existence of a hole means that the state is unoc cupied; convince yourself that this implies that = 1 —p.) 18. The Fermi energy of aluminum is 11.66 eV; its density is 2.70 g/cm^ and its molar mass is 27.0 g/mol (see Appendix D). From these data, determine the number of free electrons per atom. 19. White dwarf stars represent a late stage in the evolution of stars like the Sun. They become dense enough and hot enough that we can analyze their structure as a solid in which all Z electrons per atom are free. For a white dwarf with a mass equal to that of the Sun and a radius equal to that of the Earth, calculate the Fermi energy of the electrons. Assume the atomic structure to be represented by iron atoms, and T = 0 K. 20. A neutron star can be analyzed by techniques similar to those used for ordinary metals. In this case the neutrons (rather than electrons) obey the probability function, Eq. 6. Consider a neutron star of 2.00 solar masses with a radius of 10.0 km. Calculate the Fermi energy of the neutrons. 21. Estimate the number of conduction electrons in a metal that have energies greater than the Fermi energy as follows. Strictly, N is given by N - / ■ n(E)p{E)dE. By studying Fig. 2c, convince yourself that, to a good degree of approximation, this expression can be written as CEr +A N n{E ^m dE . J e^ By substituting the density of states function, evaluated at the Fermi energy, show that this yields for the fraction / of
^
n
__ 3A:772 £p *
Why not evaluate the first integral above directly without resorting to an approximation? 22. Use the result of Problem 21 to calculate the fraction of excited electrons in copper at temperatures of {a) absolute zero, {b) 300 K, and (c) 1000 K. 23. At what temperature will the fraction of excited electrons in lithium equal 0.0130? The Fermi energy of lithium is 4.71 eV. See Problem 21. 24. Silver melts at 962 °C. At the melting point, what fraction of the conduction electrons are in states with energies greater than the Fermi energy of 5.5 eV? See Problem 21. 25. Show that, at the absolute zero of temperature, the average energy E of the conduction electrons in a metal is equal to ^£p, where £p is the Fermi energy. (Hint: Note that, by definition of average, £ = (!/«) fEnJ^E)dE.) 26. (a) Using the result of Problem 25, estimate how much energy would be released by the conduction electrons in a penny (assumed all copper; mass = 3.1 g) if we could sud denly turn off the Pauli exclusion principle, (b) For how long would this amount of energy light a 100-W lamp? Note that there is no known way to turn off the Pauli principle! Section 53-3 Electrical Conduction in Metals 27. Silver is a monovalent metal. Calculate (a) the number of conduction electrons per cubic meter, (b) the Fermi energy, (c) the Fermi speed, and (d) the de Broglie wavelength corre sponding to this speed. Extract needed data from Appen dix D. 28. Zinc is a bivalent metal. Calculate (a) the number of con duction electrons per cubic meter, (b) the Fermi energy, (c) the Fermi speed, and (d) the de Broglie wavelength corre sponding to this speed. See Appendix D for needed data on zinc. 29. For silver, calculate (a) the mean free path of conduction electrons and (b) the ratio of the mean free path to the distance between neighboring ion cores. Silver has a Fermi energy of 5.51 eV and a resistivity of 1.62 X 10"* Q • m. See Problem 27. Section 53-5 Conductors, Insulators, and Semiconductors 30. Repeat the calculation of Sample Problem 6 for a tempera ture of (a) 1000 K and (b) 4.0 K. 31. The Ferm i-Dirac distribution function can be applied to semiconductors as well as to metals. In semiconductors, E is the energy above the top of the valence band. The Fermi level for an intrinsic semiconductor is nearly midway be tween the top of the valence band and the bottom of the conduction band. For germanium these bands are separated by a gap of 0.67 eV. Calculate the probability that (a) a state at the bottom of the conduction band is occupied and (b) a state at the top of the valence band is unoccupied at 290 K. 32. The band gap in pure germanium is 0.67 eV. Assume that the Fermi level is at the middle of the gap. (a) Calculate the probability that a state at the bottom of the conduction band
Chapter 53
1138
Electrical Conduction in Solids
is CKCupied at 16°C. (^) At what temperature will the occu pation probability of this state be 3.0 times the probability at Ib^’C? In a simplified model of an intrinsic semiconductor (no 33. doping), the actual distribution in energy of states is re placed by one in which there are states in the valence band, all these states having the same energy and states in the conduction band, all these states having the same energy E^. The number of electrons in the conduction band equals the number of holes in the valence band. (a) Show that this last condition implies that
^(E-Er)/kT^
J
^-(E,-Ef)lkT ^
J
’
{Hint: See Problem 17.) (b) If the Fermi level is in the gap between the two bands and is far from both bands compared to kT, then the exponentials dominate in the denominators. Under these conditions, show that £p = and therefore that, if center of the gap.
+ ^v) +
{k T \n {N J N X
the Fermi level is close to the
Section 53-6 Doped Semiconductors 34. Identify the following as p-type or «-type semiconductors: (a) Sb in Si; (b) In in Ge; (c) A1 in Ge; {d) As in Si. 35. Pure silicon at 300 K has an electron density in the conduc tion band of 1.5 X 10'^ m“ ^and an equal density of holes in the valence band. Suppose that one of every 1.0 X 10^ sili con atoms is replaced by a phosphorus atom, (a) What charge carrier density will the phosphorus add? Assume that all the donor electrons are in the conduction band. (See Appendix D for needed data on silicon.) (b) Find the ratio of the charge carrier density in the doped silicon to that for the pure silicon. 36. What mass of phosphorus would be needed to dope a 1.0-g sample of silicon to the extent described in Sample Problem 7?
»
12.00
Band 2
1
37. A silicon crystal is doped with phosphorus to a concentra tion of 10^^ phosphorus atoms per cubic meter. On average, how far apart are these atoms? See Sample Problem 7. 38. A sample of very pure germanium has one impurity atom to 1.3 X 10’ atoms of germanium. Calculate the distance be tween impurity atoms. 39. In Fig. 24 two energy bands of a hypothetical solid are repre sented. The bands are filled to level E^, which may be in either band 1 or band 2. There may be an impurity level at Ei. Indicate whether the solid is a conductor, insulator, in trinsic semiconductor, or extrinsic semiconductor. The im purity type may be donor, acceptor, or none, and extrinsic semiconductors may be either p-type or n-type. Complete the table.
Type E^ (eV) 3.00 3.00 3.00 1.49 4.40 3.00
E, (eV)
Ey, (eV)
____
9.00 4.10 4.10 9.00 4.10 4.10
4.06 — — —
3.04
Solid
Impurity
Extrinsic Semiconductor
40. Doping changes the Fermi energy of a semiconductor. Con sider silicon, with a gap of 1.1 eV between the valence and conduction bands. At 290 K the Fermi level of the pure material is nearly at the midpoint of the gap. Suppose that it is doped with donor atoms, each of which has a state 0.15 e V below the bottom of the conduction band, and suppose further that doping raises the Fermi level to 0.084 eV below the bottom of that band, (a) For both the pure and doped silicon, calculate the probability that a state at the bottom of the conduction band is occupied, {b) Also calculate the prob ability that a donor state in the doped material is occupied. See Fig. 25. 41. A silicon sample is doped with atoms having a donor state 0.11 eV below the bottom of the conduction band, {a) If each of these states is occupied with probability 4.8 X 10“ ^ at temperature 290 K, where is the Fermi level relative to the top of the valence band? {b) What then is the probability that a state at the bottom of the conduction band is occu pied? The energy gap in silicon is 1.1 eV.
> Gap
Conduction band -Fermi level - Donor level ,3 .0 0
1.1 eV
Band 1 Valence band
1 Figure 24
Problem 39.
Figure 25
Problem 40.
Problems Section 53-7 Thepn Junction 42. When a photon enters the depletion region of a junction, electron-hole pairs can be created as electrons absorb part of the photon’s energy and are excited from the valence band to the conduction band. These junctions are thus often used as detectors for photons, especially for x rays and nu clear gamma rays. When a 662-keV gamma-ray photon is totally absorbed by a semiconductor with an energy gap of 1.1 eV, on the average how many electron-hole pairs are created? 43. Calculate and compare the resistances of the diode rectifier for the two points shown on the characteristic curve of Fig. 12. The current for the left-hand dot (too small to show in the figure) is 50 pA. 44. For an ideal /?«-junction diode, with a sharp boundary be tween the two semiconducting materials, the current / is related to the potential difference V across the diode by I=
1),
where /q, which depends on the materials but not on the current or potential difference, is called the reverse satura tion current. V is positive if the junction is forward biased and negative if it is reverse biased, (a) Verify that this expres sion predicts the behavior expected of a diode by sketching i as a function of K over the range—0.12 V < F < H-0.12 V. Take T = 290 K and /q = 5.0 nA. (b) For the same tempera ture, calculate the ratio of the current for a 0.50-V forward bias to the current for a 0.50-V reverse bias. 45. A drop of lead (work function = 3.4 eV) is in close contact with a sheet of copper (work function = 4.5 eV). Find the contact potential difference that appears across the leadcopper interface. How might you measure it? Draw an en
1139
ergy diagram, showing (in the style of Fig. Ab) the relative Fermi levels both before and after the two metals are joined together. Can such a junction serve as a diode rectifier? 46. (a) A capacitance is associated with a junction. Explain why. {b) Derive an expression for the capacitance of the pn junction of Sample Problem 9. Section 53-8 Optical Electronics 47. (a) Calculate the maximum wavelength that will produce photoconduction in diamond, which has a band gap of 5.5 eV. (b) In what part of the electromagnetic spectrum does this wavelength lie? 48. In a particular crystal, the highest occupied band of states is full. The crystal is transparent to light of wavelengths longer than 295 nm but opaque at shorter wavelengths. Calculate the width, in electron-volts, of the gap between the highest occupied band and the next (empty) band. 49. The KCl crystal has a band gap of 7.6 eV above the topmost occupied band, which is full. Is this crystal opaque or trans parent to radiation of wavelength 140 nm? 50. (a) Fill in the seven-segment display shown in Fig. 15a to show how all 10 numbers may be generated, (b) If the num bers are displayed randomly, in what fraction of the displays will each of the seven segments be used? 51. Section 53-8 discussed the mode of operation of a lightemitting diode, in which light is emitted when charge carri ers are injected across the central plane of a junction by an external potential. The reverse device, a photodiode, is also a possibility. That is, you can shine light on a pw junc tion and a current will develop across the junction plane. Discuss how such a device might operate. Would it be best to operate it in a forward- or a reverse-biased mode?
CHAPTER 54 NUCLEAR PHYSICS
Deep within the atom lies its nucleus, occupying only 10~^^ o f the volume o f the atom but providing most o f its mass as well as the force that holds it together. The next goal in our study o f physics is to understand the structure o f the nucleus and the substructure o f its components. Our task is made easier by the many similarities between the study o f atoms and the study o f nuclei. Both systems are governed by the laws o f quantum mechanics. Like atoms, nuclei have excited states that can decay to the ground state through the emission o f photons (gamma rays). In certain circumstances, as we shall see, nuclei can exhibit shell effects that are very similar to those o f atoms. We shall also see that there are differences between the study o f atoms and the study o f nuclei that keep us from achieving as complete an understanding o f nuclei as we have o f atoms. In this chapter we study the structure o f nuclei and their constituents. We consider some experimental techniques for studying their properties, and we conclude with a description o f the theoretical basis for understanding the structure o f nuclei.
54-1 DISCOVERING THE NUCLEUS______________________ In the first years of the 20th century not much was known about the structure o f atoms beyond the fact that they contained electrons. This particle had been discovered (by J. J. Thomson) only in 1897, and its mass was un known in those early days. Thus it was not possible even to say just how many electrons a given atom contained. Atoms are electrically neutral so they must also contain some positive charge, but at that time nobody knew what form this compensating positive charge took. How the electrons moved within the atom and how the mass of the atom was divided between the electrons and the positive charge were also open questions. In 1911, Ernest Rutherford, interpreting some experi ments carried out in his laboratory, was led to propose that the positive charge of the atom was densely concen trated at the center of the atom and that, furthermore, it was responsible for most of the mass of the atom. He had discovered the atomic nucleus! Until this step had been taken, all attempts to under
stand the motions of the electrons within the atom were doomed to failure. Only 2 years after Rutherford’s pro posal, Niels Bohr used the concept of the nuclear atom to develop the semiclassical theory of atomic structure that we described in Chapter 51. This early work by Ruther ford and Bohr marks the beginning o f our understanding of the structure of atoms. How did Rutherford come to make this proposal? It was not an idle conjecture but was based firmly on the results of an experiment suggested by him and carried out by his collaborators, Hans Geiger (of Geiger counter fame) and Ernest Marsden, a 20-year-old student who had not yet earned his bachelor’s degree. Rutherford’s idea was to probe the forces acting within an atom by firing energetic alpha (a) particles through a thin target foil and measuring the extent to which they were deflected as they passed through the foil. Alpha par ticles, which are about 7300 times more massive than electrons, carry a charge of -\-2e and are emitted spon taneously (with energies of a few MeV) by many radioac tive materials. We now know that these useful projectiles are the nuclei of the atoms of ordinary helium. Figure 1 shows the experimental arrangement of Geiger and Mars-
1141
1142
Chapter 54
Nuclear Physics
Alpha source
Figure 1 The experimental arrangement used in Ruther ford’s laboratory to study the scattering of a particles by thin metal foils. The detector can be rotated to various scattering angles 0.
den. T he experim ent consists in counting the n u m b er o f a particles deflected through various scattering angles 6, (See Section 29-7.) Figure 2 shows their results. N ote especially th at the vertical scale is logarithm ic. W e see th at m ost o f the a particles are scattered through rather sm all angles, b u t— an d this was the big su rprise— a very sm all fraction o f them is scattered through very large angles, approaching 180°. In R u th erfo rd ’s words: “ It was quite the m ost in credible event th at ever h appened to m e in m y life. It was alm ost as incredible as if you had fired a 15-inch shell at a piece o f tissue paper and it cam e back an d hit you.’’ W hy was R utherford so surprised? At the tim e o f these
10'^ . 10®
10^ 1Q4 103 1Q2
10
20®
40®
60®
80®
100®
120°
140°
160°
Scattering angle, $
Figure 2 The dots show the a-particle scattering results from the experiments of Geiger and Marsden, and the solid curve is computed according to Rutherford’s theory of the nucleus. Note that the vertical axis is marked in powers of 10.
Figure 3 The angle through which an a particle is scattered depends on how close its extended incident path lies to the nucleus of an atom. Large deflections result only from very close encounters.
experim ents, m any physicists believed in a m odel o f the atom th a t had been proposed by J. J. T hom son. In T h o m son’s m odel, the positive charge o f the atom was thought to be spread o u t through th e entire volum e o f the atom . T he electrons were thought to be distributed th roughout this volum e, som ew hat like seeds in a w aterm elon, an d to vibrate about their equilibrium positions w ithin this sphere o f charge. T he m axim um deflecting force acting on the a particle as it passes through such a positive sphere o f charge proves to be far too small to deflect the a particle by even as m uch as one degree. T he electrons in the ato m w ould also have very little effect on th e massive, energetic a particle. They would, in fact, be them selves strongly deflected, m uch as a sw arm o f gnats w ould be brushed aside by a stone throw n through them . T here is sim ply no m echanism in T h o m son’s atom m odel to account for the backw ard deflection o f an a particle. R utherford saw th at to produce such a laige deflection there m ust be a large force, w hich could be provided if the positive charge were concentrated tightly at th e center o f the atom , instead o f being spread throughout its volum e. O n this m odel the incom ing a particle can get very close to the center o f the positive charge w ithout penetrating it, resulting in a large deflecting force; see Sam ple Problem 1. Figure 3 shows the paths taken by typical a particles as they pass through the atom s o f th e target foil. As we see, m ost are deflected only slightly o r n o t a t all, b u t a few (those whose extended incom ing paths pass, by chance, close to a nucleus) are deflected through large angles. From an analysis o f the data, R utherford concluded that the dim ensions o f th e nucleus m ust be sm aller th an the diam eter o f an ato m by a factor o f ab o u t 10^. T he atom is m ostly em pty space! It is n o t often th a t the piercing in-
Section 54-2
sight of a gifted scientist, supported by a few simple calcu lations,* leads to results of such importance.
Sample Problem 1 A 5.30-MeV a particle happens, by chance, to be headed directly toward the nucleus of an atom of gold (Z = 79). How close does it get before it comes momentarily to rest and reverses its course? Neglect the recoil of the (relatively massive) gold nucleus. Solution Initially the total mechanical energy of the two inter acting particles is just equal to (=5.30 MeV), the initial ki netic energy of the a particle. At the moment the a particle comes to rest, the total energy is the electrostatic potential energy of the system of two particles. Because energy must be con served, these two quantities must be equal, or K “
*
d '
in which q (= 2e) is the charge of the a particle, Q (= 19e) is the charge of the gold nucleus, and d is the distance between the centers of the two particles. Substituting for the charges and solving for d yield d=
qQ
- (8.99 X 10’ N • mVC^)
(2X79X1.60 X 10-'’ C)2 MeVXl.60 X lO” ' ’ J/MeV)
= 4.29X 10-'‘'m = 42.9fin. This is a small distance by atomic standards but not by nuclear standards. As we shall see in the following section, it is considera bly larger than the sum of the radii of the gold nucleus and the a particle. Thus the a particle reverses its course without ever “touching” the gold nucleus. If the positive charge associated with the gold atom had been spread uniformly throughout the volume of the atom, the maxi mum retarding force acting on the a particle would have oc curred at the moment the a particle began to touch the surface of the atom. This force (see Problem 2) would have been far too weak to have had much effect on the motion of the a particle, which would have gone barreling right through such a “spongy” atom.
54-2 SOM E NUCLEAR PROPERTIES___________________ The nucleus, tiny as it may be, has a structure that is every bit as complex as that of the atom. Nuclei are made up of protons and neutrons. These particles (unlike the elec tron) are not true elementary particles, being made up of
* For an analysis of this scattering experiment, see Kenneth S. Krane, Modern Physics (Wiley, 1983), Chapter 6.
Some Nuclear Properties
1143
other particles, called quarks. However, nuclear physics — the subject of this chapter— is concerned primarily with studies of the nucleus that do not involve the internal structure of the protons and neutrons themselves. The fundamental nature of these two particles is a topic in the field of elementary particle physics, which we consider in Chapter 56.
Nuclear Systematics Nuclei are made up of protons and neutrons. The number of protons in the nucleus is called the atom ic num ber emd is represented by Z. The number of neutrons is called the neutron number, and we represent it by N. Aside from the difference in their electric charges {q = -\-e for the proton, ^ = 0 for the neutron), the proton and the neutron are very similar particles: they have nearly equal masses and experience identical nuclear forces inside nuclei. For this reason, we classify the proton and neutron together as nucleons. The total number of nucleons (= Z + A^) is called the m ass number, and we represent it by By specifying Z and A (and therefore N) we uniquely identify a particular nuclear species or nuclide. We use A, the total number of nucleons, as an identifying super script in labeling nuclides. In ®*Br, for example, there are 81 nucleons. The symbol “Br” tells us that we are dealing with bromine, for which Z = 35. The remaining 46 nu cleons are neutrons, so that, for this nuclide, Z = 35, N = 46, and = 81. Two nuclides with the same Z but differ ent N and A, such as ®‘Br and *^Br, are called isotopes. Figure 4 shows a chart of the known nuclides as a plot of Z against N. The dark shading represents stable nuclides; the lighter shading represents known radioactive nu clides, or radionuclides. Table 1 shows some properties of a few selected nuclides. Note that there is a reasonably well-defined zone of stability in Fig. 4. Unstable radionuclides lie on either side of the stability zone.
The Nuclear Force The force that controls the electronic structure and prop erties of the atom is the familiar Coulomb force. To bind the nucleus together, however, there must be a strong attractive force of a totally new kind acting between the neutrons and the protons. This force must be strong enough to overcome the repulsive Coulomb force be tween the (positively charged) protons and to bind both neutrons and protons into the tiny nuclear volume. Ex periments suggest that this strong force, as it is simply called, has the same character between any pair of nuclear constituents, be they neutrons or protons. The “strong force” has a short range, roughly equal to 10“ *^ m. This means that the attractive force between pairs of nucleons drops rapidly to zero for nucleon separa tions greater than a certain critical value. This in turn
1144
Chapter 54
Nuclear Physics
Consider a nucleus with 238 nucleons. If it were to lie on the Z = TVline, it would have Z = TV= 119. However, such a nucleus, if it could be assembled, would fly apart at once because of Coulomb repulsion. Relative stability is found only if we replace 27 of the protons by neutrons, thus greatly diluting the Coulomb repulsion effect. We then would have the nuclide which has Z = 92 and TV= 146, a neutron excess of 54. Even in Coulomb effects are evident in that (1) this nuclide is radioactive and emits a particles, and (2) it can easily break up (fission) into two fragments. Both of these processes reduce the Coulomb energy more than they do the energy in the strong-force bonds. Figure 4 A plot of the known nuclides. The dark shading in dicates stable nuclides and the light shading shows radioactive nuclides. Note that light stable nuclides have essentially equal numbers of protons and neutrons, while > Z for heavy nuclei.
means that, except in the smallest nuclei, a given nucleon cannot interact through the strong force with all the other nucleons in the nucleus but only with a few of its nearest neighbors. By contrast, the Coulomb force is not a shortrange force. A given proton in a nucleus exerts a Coulomb repulsion on all the other protons, no matter how large their separation; see Problem 12. Figure 4 shows that the lightest stable nuclides tend to lie on or close to the line Z = N. The heavier stable nu clides lie well below this line and thus typically have many more neutrons than protons. The tendency to an excess of neutrons at large mass numbers is a Coulomb repulsion effect. Because a given nucleon interacts with only a small number o f its neighbors through the strong force, the amount of energy tied up in strong-force bonds between nucleons increases just in proportion to A. The energy tied up in Coulomb-force bonds between protons in creases more rapidly than this because each proton inter acts with all other protons in the nucleus. Thus the Cou lomb energy becomes increasingly important at high mass numbers.
TABLE 1
Nuclear Radii We have used the Bohr radius Aq(= 5.29 X 10“ ** m) as a convenient unit for measuring the dimensions o f atoms. Nuclei are smaller by a factor of about 10"*, and a conve nient unit for measuring distances o f this scale is the ferntometer (= 10“ ‘^ m). This unit is often called the fermi and shares the same abbreviation. Thus 1 fermi = 1 femtometer = 1 fm = 10“ *^ m. We can learn about the size and structure o f nuclei by doing scattering experiments, much as suggested by Fig. 1, using an incident beam of high-energy electrons. The energy of the incident electrons must be high enough (> 200 MeV) so that their de Broglie wavelength will be small enough for them to act as structure-sensitive nu clear probes. In effect, these experiments measure the dif fraction pattern of the scattered particles and so deduce the shape of the scattering object (the nucleus). From a variety of scattering experiments, the nuclear density has been deduced to be of the form shown in Fig. 5. We see that the nucleus does not have a sharply defined surface. It does, however, have a characteristic mean radius R. The density p{r) has a constant value in the nuclear interior and drops to zero through the fuzzy sur face zone. From these experiments it has been found that
SOME PROPERTIES OF SELECTED NUCLIDES
Nuclide
z
N
A
Stability^
Atomic Mass (u)
’Li “*N
3 7 15 37 50 64 79 94
4 7 16 51 70 93 118 145
1 14 31
92.5% 99.6% 100% 18 m 32.4% 15.7% 100% 24,100 y
7.016003 14.003074 30.973762 87.911326 119.902199 156.923956 196.966543 239.052158
3 .p
*‘Rb •“ Sn ■” Gd •” Au « ’Pu
88 120 157 197 239
^ For stable nuclides the isotopic abundance is given; for radionuclides, the half-life.
Radius (fm) 2.30 2.89 3.77 5.34 5.92 6A7 6.98 7.45
Binding Energy per Nucleon (MeV) 5.61 7.48 8.48
Spin (h/2n) i
1
8.68
i 2
8.50
0
8.20
i
7.92 7.56
i
Magnetic Moment (Ps) + 3.26 +0.403 + 1.13 +0.508
0 -0 .3 4 0 +0.146 +0.203
Section 54-2 Som e Nuclear Properties
Figure 5 The variation with radial distance of the density of a nucleus of '” Au.
/? = (1.2fm)(63)'/^ = 4.3fm . By comparison, the mean radius of a copper ion in a lattice o f solid copper is 1.8 Bohr radii, about 2 X 10“ times larger.
Nuclear Masses and Binding Energies Atomic masses can be measured with great precision using modem mass spectrometer and nuclear reaction techniques. We recall that such masses are measured in unified atomic mass units (abbreviation u), chosen so that the atomic mass {not the nuclear mass) of '^C is exactly 12 u. The relation of this unit to the SI mass standard is 1 u = 1.6605 X 10-2’ kg Note that the mass number (symbol A ) identifying a nu clide is so named because this number is equal to the atomic mass o f the nuclide, rounded to the nearest in teger. Thus the mass number of the nuclide '^’Cs is 137; this nuclide contains 55 protons and 82 neutrons, a total o f 137 particles; its atomic mass is 136.907073 u, which rounds off numerically to 137. In nuclear physics, as contrasted with atomic physics, the energy changes per event are commonly so great that Einstein’s well-known mass-energy relation £ = Amc’ is an indispensable work-a-day tool. We shall often need to use the energy equivalent of 1 atomic mass unit, and we find it from ,
,
(2)
w<,c2 -I- £ b =
(m<, -I- We)c2 -I- £ b =
(1)
in which A is the mass number and R q is a constant with a value o f about 1.2 fm. For “ Cu, for example,
^
As an example, consider the deuteron, the nucleus o f the heavy hydrogen atom. A deuteron consists o f a proton and a neutron bound together by the strong force. The energy that we must add to the deuteron to tear it apart into its two constituent nucleons is called its binding en ergy. In effect, the binding energy is the total internal energy of the nucleus, due in part to the strong force between the nucleons, the Coulomb force between the nucleons, and the kinetic energies of the nucleons relative to the center of mass of the entire nucleus. From the conservation of energy we can write, for this pulling-apart process. If we add WeC’, the energy equivalent o f one electron mass, to each side of this equation, we have
R increases with A approximately as R = R o A '^ \
1145
(1.6605 X 10-2’ kg)(2.9979X 10* m/s)2 -------------- 1.6022 X 1 0 - J/MeV-----------
+ {m„ A- m^)c^.
or w (2H)c2-I-£ b = w „£'^ + w ('H )c2.
(3)
Here w f’H) and w ('H ) are the masses of the neutral heavy hydrogen atom and the neutral ordinary hydrogen atom, respectively. They are atomic masses, not nuclear masses. Solving Eq. 3 for yields
^B = I"J„ + m ('H )-w (2H )]c2 = Awc2,
(4)
in which Aw is the mass difference. In making calcula tions of this kind we always use atomic, rather than nu clear, masses, as this is what is normally tabulated. As in this example, the electron masses conveniently cancel.* For the deuteron calculation the needed masses are w„ = 1.008665 u,
w ('H ) = 1.007825 u,
w(2H) = 2.014102 u. Substituting into Eq. 4 and replacing c ’ by its equivalent, 931.5 MeV/u, we find the binding energy to be £ b = (1.008665 u -I- 1.007825 u -2 .0 1 4 1 0 2 u)(931.5 MeV/u) = (0.002388 u)(931.5 MeV/u) = 2.224 MeV. Compare this with the binding energy of the hydrogen atom in its ground state, which is 13.6 eV, about five orders of magnitude smaller. If we divide the binding energy of a nucleus by its mass number, we get the average binding energy per nucleon, a property we have listed in Table 1. Figure 6 shows a plot of this quantity as a function of mass number. The fact that this binding energy curve “droops” at both high and low mass numbers has practical consequences of the greatest importance, t
= 931.5 MeV. This means that we can write c ’ as 931.5 MeV/u and can thus easily find the energy equivalent (in MeV) o f any mass or mass difference (in u), or conversely.
• See, however, Problem 51, for an exception. A The Curve o f Binding Energy has even been adopted as the title of a book (by John McPhee) about the possibilities of nuclear terrorism!
1146
Chapter 54
Nuclear Physics
is replaced by the proton mass m^. That
electron mass is, // n =
t
^
4nm p
= 3.15X 10-*eV/T.
Because the magnetic moment of the free electron is (very closely) one Bohr magneton, it might be supposed that the magnetic moment of the free proton would be (very closely) one nuclear magneton. It is not very close, however, the measured value being -1-2.7929 To un derstand the magnetic moments of the proton and neu tron, it is necessary to consider their internal structure. The magnetic moments of heavier nuclei can in turn be analyzed in terms of the magnetic moments of the constit uent protons and neutrons.
Sample Problem 2 What is the approximate density of the nuclear matter from which all nuclei are made? Figure 6 The binding energy per nucleon over the range of mass numbers. Some of the nuclides of Table 1 are identified, along with a few others. The region of greatest stability corre sponds to mass numbers from about 50 to 80.
The drooping of the binding energy curve at high mass numbers tells us that nucleons are more tightly bound when they are assembled into two middle-mass nuclei rather than into a single high-mass nucleus. In other words, energy can be released in the nuclearfission of a single massive nucleus into two smaller fragments. The drooping of the binding energy curve at low mass numbers, on the other hand, tells us that energy will be released if two nuclei of small mass number combine to form a single middle-mass nucleus. This process, the re verse of fission, is called nuclear fusion. It occurs inside our Sun and other stars and is the mechanism by which the Sun generates the energy it radiates to us.
Nuclear Spin and Magnetism Nuclei, like atoms, have an intrinsic angular momentum whose maximum component along any chosen z axis is given by Jh . Here 7 is a quantum number, which may be integral or half-integral, called the nuclear spin; some values for selected nuclides are shown in Table 1. Again as for atoms, a nuclear angular momentum has a nuclear magnetic moment associated with it. Recall that, in atomic magnetism, the Bohr magneton defined as
eh
= 5.79X lO-'^eV/T, Pb = 4nm ^
Solution We know that this density is large, because virtually all the mass of the atom resides in its tiny nucleus. The volume of the nucleus, approximated as a uniform sphere of radius R, is given by Eq. 1 as
The density p^ of nuclear matter, expressed in nucleons per unit volume, is then Pn
T/ V
(4nH)RlA
The mass of a nucleon is 1.7 X 10“ ^^ kg. The nuclear matter is then
density/?„ of
p^ = (0.14 nucleons/fm^Xl-7 X 10“ ^^ kg/nucleon) X (1 fm/10“ ‘^ m)^ = 2.4X 10''kg/m ^ or 2.4 X 10'^ times the density of water! Unlike the orbital elec trons, the nuclides have a density nearly independent of the number of their nucleons. To some extent nucleons are packed in like marbles in a bag. Sample Problem 3 Imagine that a typical middle-mass nu cleus such as '^°Sn is picked apart into its constituent protons and neutrons. Find (a) the total energy required and (b) the energy per nucleon. The atomic mass of *^°Sn is 119.902199 u; see Table 1. Solution (a) '^®Sn contains 50 protons and 120 — 50 = 70 neutrons. The combined atomic mass of these free particles is A /= ZWp -h Nm„ = 50 X 1.007825 u -h 70 X 1.008665 u = 120.997800 u.
is a unit of convenience. In nuclear physics the corre sponding unit of convenience is the nuclear magneton p^, defined similarly to the Bohr magneton except that the
This exceeds the atomic mass of '^®Sn by Am = 120.997800 u - 119.902199 u = 1.095601 u.
Section 54-3 Converting this to a rest energy yields the total binding energy, £b=
= (1.0956 u)(931.5 MeV/u) = 1020.6 MeV.
Radioactive Decay
1147
We are often more interested in the activity or decay rate R (= —dNldt) of the sample than we are in N, Differ entiating Eq. 6 yields
(b) The binding energy E per nucleon is
R = R qC-
^ 1020.6 MeV , £ = —7 = ----------------= 8.50 MeV/nucleon. A 120 This agrees with the value that may be read from the curve of Fig. 6.
54-3 RADIOACTIVE DECAY_________
(7)
in which R q (=AA o) is the decay rate at / = 0. Note also that R = XNsiX any time t. We assumed initially that the ratio o f R to N is constant, so we are not surprised to confirm that they both decrease with time according to the same exponential law. A quantity of interest is the time /j/2 , called the half-life, after which both A^and R are reduced to one-half o f their initial values. Putting R = \ R q in Eq. 7 gives iR o = R o e -^ '-,
As Fig. 4 shows, most of the nuclides that have been identified are radioactive. That is, they spontaneously emit a particle, transforming themselves in the process into a different nuclide. In this chapter we discuss the two most common situations, the emission of an a particle (alpha decay) and the emission of an electron (beta decay). No matter what the nature of the decay, its main feature is that it is statistical. Consider, for example, a 1-mg sam ple o f uranium metal. It contains 2.5 X 10** atoms of the very long-lived alpha emitter The nuclei of these atoms have existed without decaying since they were cre ated (before the formation of our solar system) in the explosion o f a supernova. During any given second about 12 of the nuclei in our sample will decay, emitting an a particle in the process. We have absolutely no way of predicting, however, whether any given nucleus in the sample will be among those that do so. Every single nucleus has exactly the same probability as any other to decay during any 1-s observation period, namely, 12/(2.5 X 10**), or one chance in 2 X 10*^. In general, if a sample contains radioactive nuclei, we can express the statistical character of the decay process by saying that the ratio of the decay rate R {= —dN/dt) to the number o f nuclei in the sample is equal to a constant, or
-dN Idt ■= A, N
(5)
which leads readily to U/ 2 ~
( 8)
a relationship between the half-life and the disintegration constant. The following two sample problems show how Acan be measured for decay processes with relatively short halflives and also with relatively long half-lives.
Sample Problem 4 In short-lived decays, it is possible to meas ure directly the decrease in the decay rate R with time. The following table gives some data for a sample of '^*1, a radionu clide often used medically as a tracer to measure the iodine uptake rate of the thyroid gland. Find (a) the disintegration constant A and (b) the half-life /,/2 from these data. Time (min)
R (counts/s)
Time (min)
R (counts/s)
4 36 68 100
392.2 161.4 65.5 26.8
132 164 196 218
10.9 4.56 1.86 1.00
Solution (a) If we take the natural logarithm of each side of Eq. 7, we find that
In R = In R q —A/.
in which A, the disintegration constant, has a different characteristic value for each radioactive nuclide. We can rewrite Eq. 5 as
(6.06 - 0) (220 min —0) ’ or
which integrates readily to
A = 0.0275 m in-'
( 6)
Here Nqis the number of radioactive nuclei in the sample at r = 0. We see that the decrease of with time follows a simple exponential law.
(9)
Thus if we plot the natural logarithm of R against t, we should obtain a straight line whose slope is —A. Figure 7 shows such a plot. Equating the slope of the line to —A yields
dN
N = N oe~ ^,
In 2 A
(b) Equation 8 yields for /,/ 2: In 2 ^/2 “ ■
0.693 0.0275 min
= 25.2 min.
1148
Chapter 54
Nuclear Physics From Eq. 5 we have - d N / d t _ R _ 1600 s - ' = 1.69X lO -'^sN 9.49 X 10” and the half-life, from Eq. 8, is _ In 2 _ / 0.693 \ / 1y \ A \1 .6 9 X 10"'’ s - ' / \3 .1 6 X lO’ s / = 1.30 X 10’ y.
50
100
150
200
\
This is of the order of magnitude of the age of the universe. No wonder we cannot measure the half-life of this nuclide by wait ing around for its decay rate to decrease! (Interestingly, the potas sium in our own bodies has its normal share of the isotope. We are all slightly radioactive.)
Time (min)
Figure 7 Sample Problem 4. A logarithmic plot of the decay data is fitted by a straight line, showing the exponential nature of the decay. The disintegration constant Acan be found from the slope of the line.
54-4 A LPH A DECAY__________________ The radionuclide a typical alpha emitter, decays spontaneously according to the scheme
Sample Problem 5 A 1.00-g sample of pure KCl from the chemistry stock room is found to be radioactive and to decay at an absolute rate R of 1600 counts/s. The decays are traced to the element potassium and in particular to the isotope ^K , which constitutes 1.18% of normal potassium. What is the half-life for this decay? Solution In the case of long-lived decays it is not possible to wait long enough to observe a measurable decrease in the decay rate R with time. We must find A by measuring both N and —dNIdt in Eq. 5. The molar mass of KCl is 74.9 g/mol, so the number of potassium atoms in the sample is
. (6.Q2XKy-mol-XI.00i). , „ 74.9 g/mol The number of
atoms is 1.18% of A^k , or
= (0.0118X8.04 X 10^') = 9.49 X 10'’.
238U ^ 2MJ^^ + 4He,
( 10)
with a half-life of 4.47 X 10’ y. In Sample Problem 6 we show that, in every such decay, an energy of 4.27 MeV is emitted, appearing as kinetic energy shared between the a particle (’ He) and the recoiling residual nucleus (^^Th). We now ask ourselves: “If energy is released in every such decay event, why did the nuclei not decay shortly after they were created?” The creation process is believed to have occurred in the violent explosions o f ancestral stars (supernovas), predating the formation of our solar system. Why did these nuclei wait so very long before getting rid o f their excess energy by emitting an a particle? To answer this question, we must study the de tailed mechanism of alpha decay. We choose a model in which the a particle is imagined
Figure 8 A potential energy function representing the emis sion of a particles by The shaded area represents the po tential barrier that inhibits the decay process. The horizontal lines represent the decay energies of (4.27 MeV) and (6.81 MeV).
Section 54~5
to exist preformed inside the nucleus before it escapes. Figure 8 shows the approximate potential energy function U(r) for the a particle and the residual ^^Th nucleus as a function o f their separation. It is a combination of a po tential well associated with the (attractive) strong nuclear force that acts in the nuclear interior (r < /?,) and a Cou lomb potential associated with the (repulsive) electro static force that acts between the two particles after the decay has occurred (r > /?,). The horizontal line marked = 4.27 MeV shows the disintegration energy for the process, as calculated in Sam ple Problem 6. Note that this line intersects the potential energy curve at two points, and /?2 - We now see why the a particle is not immediately emitted from the nucleus! That nucleus is surrounded by an impressive potential barrier, shown by the shaded area in Fig. 8. Visualize this barrier as a spherical shell whose inner radius is R , and whose outer radius is >its volume being forbidden to the a particle under the laws of classical physics. If the a particle found itself in that region, its potential energy U would exceed its total energy £*, which would mean, classically, that its kinetic energy K {= E — U) would be negative, an impossible situation. Indeed, we now change our question and ask: “How can the nucleus emit an a particle?” The a particle seems permanently trapped inside the nucleus by the barrier. The answer is that, as we learned in Section 50-8, in wave mechanics there is always a chance (described by Eq. 19 of Chapter 50) that a particle can tunnel through a barrier that is classically insurmountable. In fact, the ex planation of alpha decay by wave mechanical barrier tun neling was one of the very first applications of the new quantum physics. For the long-lived decay of the barrier is actually not very “leaky.” We can show that the a particle, pre sumed to be rattling back and forth within the nucleus, must present itself at the inner surface of the barrier about 10^®times before it succeeds in tunneling through. This is about 10^® times per second for about 10’ years! We, of course, are waiting on the outside, taking note of only those a particles that do manage to escape. We can test this barrier tunneling explanation of alpha decay by looking at other alpha emitters, for which the barrier would be different. For an extreme contrast, con sider the alpha decay of another uranium nuclide, which has a disintegration energy of 6.81 MeV, as shown in Fig. 8. The barrier in this case is both thinner (compare the lengths of the dashed lines in Fig. 8) and lower (compare the heights of the barrier above the dashed lines); if our barrier tunneling notions are correct, we would expect alpha decay to occur more readily for than for Indeed it does. As Table 2 shows, the half-life o f is only 550 s! We recall from Section 50-8 that the transmission coefficient of a barrier— because of the exponential nature of Eq. 19 of Chapter 50— is very
TABLE 2
Beta Decay
1149
THE ALPHA DECAY OF 23»U AND 22«u
Nuclide 23«U 22«U
Half-life
Qa 4,21 MeV 6.81 MeV
4.5 X 10’ y 550 s
sensitive to small changes in the dimensions of the barrier. We see that an increase in by a factor o f only 1.6 produces a decrease in half-life (that is, in the effectiveness of barrier tunneling) by a factor of 3 X 10*"^.
Sample Problem 6 {a) Find the energy released during the alpha decay of Show that this nuclide cannot spontane ously emit a proton. The needed atomic masses are 238U
238.050785 u
23^Th 234.043593 u
^He 4.002603 u ‘H
1.007825 u
237pa 237.051143 u. Solution (a) In the alpha decay process of Eq. 10 the total atomic mass of the decay products (= 238.046196 u) is less than the atomic mass of by Am = 0.004589 u, whose energy equivalent is = Am c^ = (0.004589 uK931.5 MeV/u) = 4.27 MeV. This disintegration energy is available to share as kinetic energy between the a particle and the recoiling ^^^Th atom. (b) If ^gj-e to emit a proton, the decay process would be — 237pa-h«H .
In this case the mass of the decay products exceeds the mass of 23*U by Aw = 0.008183 u, the energy equivalent being —7.622 MeV. The minus sign means that we must add energy to split 238U into 237pa plus a proton. Thus 23«u is stable against spontaneous proton emission.
54-5
BETA DECAY
A nucleus that decays spontaneously by emitting an elec tron (either positive or negative) is said to undergo beta decay.* Here are two examples: +
+ y
(t,/2= 14.3 d)
( 11)
and ‘'*Cu ^ ‘^Ni-I-e+-I-V
(t,/2= 12.7h).
(12)
The symbols v and v represent the neutrino and its anti particle, the antineutrino, neutral particles that are emit-
* Beta decay also includes electron capture, in which a nucleus decays by absorbing one of its orbital electrons. We do not con sider that process here.
1150
Chapter 54
Nuclear Physics
ted from the nucleus along w ith the electron or positron (positive electron) d uring the decay process. N eutrinos interact only very weakly w ith m atter a n d — for th at reaso n — are so extrem ely difficult to detect that, for m any years, their presence w ent unnoticed. W e consider the fundam ental n atu re and im portance o f these elusive particles in C hapter 56. It m ay seem surprising th at nuclei can em it electrons (and neutrinos) in view o f the fact th at we have said that nuclei are m ade up o f n eutrons an d pro to n s only. H ow ever, we saw earlier th a t atom s em it photons, and we certainly do n o t say th a t atom s “ c o n tain ” photons. We say th at the p hotons are created during the em ission pro cess. So it is w ith the electrons an d the neutrinos em itted from nuclei during beta decay. They are both created during the em ission process, a n eu tro n transform ing itself into a pro to n w ithin the nucleus (or conversely) accord ing to n - ^ p + e“ + v
{p~ decay)
(13)
decay).
(14)
or p —►n + e'^ -h v
These are the basic beta-decay processes. In any decay process, the am o u n t o f energy released is uniquely determ ined by the difference in rest energy be tw een the initial nucleus and the final nucleus plus decay products (see, for exam ple. Sam ple Problem 6). In a par ticular alpha-decay process, such as th at o f every em itted a particle carries the sam e kinetic energy. In beta decay, however, the kinetic energy o f the em itted elec tro n s is not uniquely determ ined. Instead, the em itted electrons have a con tin u o u s sjjectrum o f energies, from zero up to a m axim um as Fig. 9 illustrates for the beta decay o f ^ C u (Eq. 12). F or m any years, before the n eu trin o was identified, curves such as th at o f Fig. 9 were a challenging puzzle. They suggested th at som e energy was “ m issing” in the decay process and led m any reputable physicists, includ ing Niels Bohr, to speculate th a t perhaps the law o f con servation o f energy m ight be valid only statistically in such decays.
The answ er to this puzzle lies in the em ission o f the neutrino or antineutrino, which carries a share o f the decay energy. If we were to m easure the energies o f both particles (electron and an tin eu trin o or positron and neu trino) in a particular decay process an d add them up, we would com e out every tim e with the sam e fixed value, equal to the disintegration energy. Energy is indeed con served in each individual decay process. T he existence o f an undetected particle as a solution to the missing energy problem was proposed by Pauli in 1931, and the n eu trino was m ade a part o f a form al theory o f beta decay by Ferm i in 1934. Nevertheless, it took an o th er 20 years before neutrinos were detected in the laboratory. T he difficulty in their m easurem ent results from their exceedingly weak interactions with m a tte r— their m ean free path through solid m atter is o f the order o f several thousand light years. Today n eutrino physics is an im p o rtan t subfield o f nuclear and particle physics, and its practitioners study not only neutrinos from radioactive sources b u t also those em itted in great quantities by the Sun an d those th at were created during the form ation o f the universe (which have a present density o f ab out 100 per cm^). Figure 10 shows evidence o f a burst o f neutrinos detected on Earth resulting from the 1987 supernova in the nearby Large M agellanic Cloud (see Fig. 17 o f C hapter 8). Because the detector was located in Japan and the supernova occurred in the southern sky, the neutrinos had to travel com pletely through the Earth to reach the detector. O u r study o f alpha and beta decay perm its us to look at the nuclidic chart o f Fig. 4 in a new way. Let us construct a three-dim ensional surface by plotting the m ass o f each nuclide in a direction at right angles to the N Z plane o f th at figure. T he surface so form ed gives a graphic represen tation o f nuclear stability. As Fig. 11 shows (for the light nuclides), it describes a “ valley o f the nuclides,” the stabil ity zone o f Fig. 4 running along its bottom . N uclides on
40
30
.20
10
•• •
•
0 Kinetic energy (MeV)
-6 0
i -3 0
0
30
60
Time (s)
Figure 9 The kinetic energy distribution of the positrons emitted in the beta decay of ^C u. The maximum kinetic en ergy is 0.653 MeV.
Figure 10 Evidence for a burst of neutrinos from the super nova SN 1987A.
Section 54-6
Measuring Ionizing Radiation
1151
The disintegration energy for the ^^P decay is then G = Am = (31.973907 u - 31.972071 uX931.5 MeV/u) = 1.71 MeV. This is just equal to the measured value of the maximum energy of the emitted electrons. Thus although 1.71 MeV is released every time a ^^P nucleus decays, in essentially every case the electron carries away less energy than this. The neutrino gets the rest, carrying it away from the laboratory undetected. (A negligible share, of the order of eV, also goes to the nucleus in order to conserve momentum in the decay.)
54-6 MEASURING IONIZING _______ RADIATION*___________________
%
o0 Figure 11 A portion of the valley of the nuclides, showing only the lightest nuclides. The quantity plotted on the vertical axis is the mass excess, defined as {m — A)c^, where m is the atomic mass in u.
the headwall o f the valley (a region not displayed in Fig. 11) decay into it largely by chains of alpha decay and by spontaneous fission. Nuclides on the proton-rich side of the valley decay into it by emitting positive electrons and those on the neutron-rich side do so by emitting negative electrons.
Sample Problem 7 Calculate the disintegration energy Q in the beta decay of as described by Eq. 11. The needed atomic masses are 31.973907 u for and 31.972071 u for Solution Because of the presence of the emitted electron, we must be especially careful to distinguish between nuclear and atomic masses. Let m ' represent the nuclear masses of ^^P and and let m represent their atomic masses. We take the disinte gration energy 0 to be Am where
Am = m'C^P) - [mV^S)
H- w j ,
m^ being the mass of the electron and the neutrino being as sumed to be massless. If we add and subtract 1Sm^ on the righthand side, we have
Am = [m'C^P) +
1 5 We] -
The quantities in brackets are the atomic masses. Thus we have
Am = mC^P) — mC^S). If we subtract the atomic masses in this way, the mass of the emitted electron is automatically taken into account.* * This is not the case for positron decay or for electron capture; see Problems 51 and 52. Note also that in this sample problem we neglect the (small) difference in the binding energies of the atomic electrons before and after the beta decay.
When radiations such as x rays, gamma rays, beta parti cles, or alpha particles encounter an atom, they can cause the atom to eject electrons and to become ionized. Be cause ionization can damage individual cells o f living tis sue, the effects of ionizing radiations have become a mat ter of general public interest. Such radiations arise in nature from the cosmic rays and also from radioactive elements in the Earth’s crust. Artificially produced radia tions also contribute, including diagnostic and therapeu tic Xrays and radiations from radionuclides used in medi cine and in industry. The disposal of radioactive waste and the evaluation of the probabilities of nuclear acci dents continue to be dealt with at the level o f national policy. It is not our task here to explore the various sources of ionizing radiations but simply to describe the units in which the properties and effects of these radiations are expressed. There are four such units, and they are often used loosely or incorrectly in popular reporting. 1. The curie (abbreviation Ci). This is a measure of the activity or rate of decay of a radioactive source. It was originally defined as the activity of 1 g of radium in equi librium with its by-products, but it is now defined sim ply as 1 curie = 3.7 X 10*® disintegrations per second. This definition says nothing about the nature o f the decays. Note also that this unit is not appropriate to de scribe the ionizing effects of x rays from, say, a medical x-ray machine. The radiations must be emitted from a radionuclide. An example of the proper use of the curie is the state ment: “One milligram of ^^’Pu has an activity o f 62 //Ci.” The fact that ^^’Pu is an alpha emitter does not enter.
* See “ Radiation Exposure in Our Daily Lives,” by Stewart C. Bushong, The Physics Teacher, March 1977, p. 135.
1152
Chapter 54
Nuclear Physics
2. The roentgen (abbreviation R). T his is a m easure o f exposure, th a t is, o f the ability o f a beam o f x rays or gam m a rays to produce ions in a particular substance. Specifically, one roentgen is defined as th at exposure that w ould produce 1.6 X 10*^ ion pairs per gram o f air, the air being dry an d at standard tem perature an d pressure. We m ight say, for exam ple: “ In 0.1 s, this dental x-ray beam provides an exposure o f 30 m R .” T his says nothing about w hether ions are actually produced o r w hether or not there is a p atient in the chair. 3. The rad. T his is an acronym for radiation absorbed dose and is a m easure, as its nam e suggests, o f the dose actually delivered to a specific object, in term s o f the energy transferred to it. An object, which m ight be a per son (whole body) o r a specific p art o f the body (the hands, say) is said to have received an absorbed dose o f 1 rad w hen 10"^ J/g have been delivered to it by ionizing radia tions. A typical statem ent to show the usage is: “ A wholebody gam m a-ray dose o f 300 rad will cause death in 50% o f the population exposed to it.” By way o f com fort we note th at the present average exposure to radiation from both natural and artificial sources is abo u t 0.2 rad (= 200 m rad) per year. 4. The rem. T his is an acronym for roentgen equivalent in m an an d is a m easure o f dose equivalent. It takes ac co u n t o f the fact that, although different types o f radiation (gam m a rays and neutrons, say) m ay deliver the sam e energy per u n it m ass to the body, they do not have the sam e biological effect. T he dose equivalent (in rem s) is found by m ultiplying the absorbed dose (in rads) by a quality fa cto r QF, w hich m ay be found tabulated in various reference sources. F or x rays an d electrons, Q F = 1. F or slow neutrons, Q F = 5, an d so on. Personnel m oni toring devices such as film badges are designed to register the dose equivalent in rems. A n exam ple o f correct usage o f the rem is: “ T he recom m endation o f the N ational C ouncil on R adiation Protec tion is th at no individual w ho is (nonoccupationally) ex posed to radiations should receive a dose equivalent greater th an 500 m rem (= 0.5 rem ) in any one year.” This includes radiations o f all kinds, using the appropriate quality factors.
Sample Problem 8 A dose of 300 rad is lethal to 50% of the population that receives it. If the equivalent amount of energy were absorbed directly as heat, what temperature increase would result? Assume that c, the specific heat capacity of the human body, is the same as that of water, namely, 4180 J/kg* K. Solution An absorbed dose of 300 rad corresponds to an ab sorbed energy per unit mass of (300 rad)
/ l0 -^ J/k g \ _ 3J/kg. V 1 rad /
The temperature increase that would result from such an influx of heat is found from 3J/kg = 7.2 X 10-^ K. 4180 J/kg-K We see from this tiny temperature increase that the damage done by ionizing radiation has very little to do with thermal heating. The harmful effects arise because the ionizing radiation succeeds in breaking molecular bonds and thus interfering with the normal functioning of the tissue in which it has been ab sorbed.
54-7 NATURAL RADIOACTIVITY All the elem ents beyond hydrogen and helium were m ade in nuclear reactions in the interiors o f stars or in explosive supernovas. Both radioactive and stable nuclides are cre ated in these processes. T he solar system is com posed o f nuclides th at were form ed about 4.5 X 10’ years ago. (H ow this is determ ined is discussed later in this section.) M ost o f the radioactive nuclides th at were form ed at th at tim e have half-lives th at are far shorter than a billion years, and so they have long since decayed to stable n u clides through alpha or beta em ission. A few o f the origi nal radioactive nuclides, however, have half-lives th at are not short in com parison with the age o f the solar system. T he decay o f these nuclides can still be observed, and these decays form part o f the background o f natural radio activity in our environm ent. Som e o f these radioactive species are part o f decay chains th at start with heavy nuclides, such as ^^^Th (t^f2 = 1.4 X 10*° y) o r (/,/2 = 4.5 X 10’ y). These nuclides decay through a sequence o f alpha and beta decays, even tually reaching stable end products (respectively, ^°®Pb and ^°^Pb). T he interm ediate nuclei in these decay chains have m uch shorter half-lives; the rate at which the original nuclide disappears and is replaced with the stable end product is determ ined by the longest-lived m em ber o f the chain. These decay processes have presum ably been going on since the solar system was form ed, and so (as we dis cuss later) the relative am o u n ts o f the initial nuclide and stable decay products present in a m aterial can give a m easure o f the age o f the m aterial. These decays are also thought to contribute to the internal heating o f the planets. N orm ally, the products o f these decays rem ain in place in the rocks or m inerals containing the parent nuclide. However, one o f the interm ediate substances produced in these decay chains, radon, is a gas. N atural decays th at occur near the surface o f the Earth (and in building m ateri als, such as concrete) release radioactive radon gas into the atm osphere. T he hazards o f breathing this radon gas are currently the subject o f active research. R adon gas can
Section 54-8
TABLE 3
SOME NATURAL RADIOACTIVE ISOTOPES
Isotope 40V
M/2
(y)
1.28 X 10’ 4.8 X 10'® 9 X 10” 4.4 X 10” 1.3 X 10” 3.6 X 10'® 5X 10'®
4b "^Cd "Mn ■“ U ■’‘Lu '*’Re
also be released from the fracture of rocks beneath the surface; therefore the detection o f radon gas has been used as a way o f predicting earthquakes. In addition to the heavy elements, other long-lived ra dioactive nuclides are present in natural substances. Some o f these are listed in Table 3. Other radioactive nuclides are continually produced by natural processes, generally in the Earth’s atmosphere by reactions o f molecules o f the air with cosmic rays (highenergy protons from space). Notable among these is “*0 (t ,/2 = 5730 y), which has important applications in radio active dating o f organic materials.
Radioactive Dating Suppose we have an initial radionuclide / that decays to a final product F with a known half-life . At a particular time / = 0, we start with Vqinitial nuclei and none o f the final product nuclei. At a later time t, we find N, of the original nuclei remain, while Np(=No —N,) of the prod uct nuclei have appeared. The initial nuclei decay accord ing to
by these methods all seem to have common ages of around 4.5 X 10’ y, which we take to be the age of the solar system. The radioactive isotope is present in the atmo sphere; about 1 carbon atom in 10*^ is radioactive Each gram of carbon has an activity of about 12 decays per minute due to Living organisms can absorb this activity by aspiration of CO2 or by eating plants that have done so. When the organism dies, it stops absorbing and the present at its death begins to decay. By mea suring the decay rate of we can determine the age of the sample. For example, if we examine a sample and it shows 6 decays per minute per gram of carbon, we know that the original activity has been reduced by half, and the sample must be one half-life (5730 y) old. This method of radiocarbon dating (which was devel oped in 1947 by Willard Libby, who was awarded the 1960 Nobel prize in chemistry for this work) is useful for samples of organic matter that are less than about 10 half-lives in age. In 10 half-lives, the activity o f a sample drops by a factor of 2“ or about 10“ ^, and the decay rate becomes too small to be determined with precision. The practical upper limit on the age of samples that can be dated by this method is about 50,000 y. In recent years, a new technique has been developed in which an accelera tor is used as a mass spectrometer to determine the ratio to high precision. In this way the usefulness of radiocarbon dating has been extended to samples as old as 100,000 y.
Sample Problem 9 In a sample of rock, the ratio of ^®^Pb to nuclei is found to be 0.65. What is the age of the rock?
and thus /=
N,
In 2
In ^®
or, substituting Ni + Nf for Nq, '
1153
Solution From Eq. 15, using 4.5 X 10^ y for the half-life of we have
N , = Noe-
\_ A
Nuclear Reactions
V ;)‘
That is, a measurement of the present ratio of product and original nuclei can determine the age of the sample. This calculation has been based on the assumption that none of the product nuclei were present at / = 0. This assumption may not always be valid, but there are tech niques for radioactive dating that can correct for the pres ence of these original product nuclei. This method can be used to determine the time since the formation of the solar system; examples include the ratios of to »^Rb to «^Sr, and to ^Ar. Terrestrial rocks. Moon rocks, and meteorites analyzed
4.5 X 10’ y ln (l +0.65) = 3.3 X 10’ y. 0.693
This rock is somewhat younger than the maximum age of 4.5 X 10’ y that we determine for rocks in the solar system. This may suggest that the rock did not solidify until 3.3 X 10’ y ago. The ^^Pb that was formed prior to that time was “boiled ofT’ from the molten rock. Only after the rock solidified could the ^^Pb begin to accumulate.
54-8
NUCLEAR REACTIONS
We can represent a nuclear reaction by
X+a-* Y+b
(16)
or, in more compact notation,
X(a,b)Y.
( 17)
1154
Chapter 54
Nuclear Physics
Usually, particle a is the projectile nucleus and particle X is the target nucleus, which is often at rest in the labora tory. If the projectile is a charged particle, it m ay be raised to its desired energy in a V an de G raaff accelerator (see Section 30-11) or a cyclotron (see Section 34-3). T he pro jectile m ay also be a neu tro n from a nuclear reactor. It is custom ary to designate p roduct particle Y as the heavier residual nucleus an d b as the lighter em erging nucleus. T he reaction energy Q is defined as 0 =
- (m y +
(18)
82
197pb 198pb 199pb 200pb 201pb 202pb 203pb 8min 2.4 h 1.5 h 21.5 h 9.42 h 5250 y 52.0 h
81
196JI 197TI 198t | 1.84 h 2.83 h 5.3 h
80
199Hg 200Hg 20lHg 198Hg 195Hg 196Hg 9.5 h 0.15% 64.1 h 10.0% 16.9% 23.1% 13.2%
N 79
(19)
in which K represents the kinetic energy. E quations 18 and 19 are valid only when Y and b are in th eir ground states. As we discuss later in this section, if either nuclide is produced in an excited state, the reaction energy is reduced by the excitation energy. A typical reaction is ‘’ F (p ,a )'^ 0 , for which 0 = 8.13 M eV. E quations 18 an d 19 tell us th at, in this reaction, the system loses rest energy and gains kinetic energy, in am o u n t 8.13 M eV per event. R e actions, like this one, for which 0 > 0 are called exother mic. R eactions for w hich 0 < 0 are called endotherm ic, such reactions will n o t “ go” unless a certain m inim um kinetic energy (the threshold energy) is carried into the system by the projectile. If a and b are identical particles, which requires th at X and Y also be identical, we describe the reaction as scat tering. If the kinetic energy o f the system is the sam e both before and after the event (which m eans th a t Q = 0 and all nuclides rem ain in their ground states), we have elastic scattering. If these energies are different {Q # 0), we have inelastic scattering, in w hich case Y or b m ay be left in an excited state. W e can easily keep track o f nuclear reactions by plot ting them on a nuclidic chart like th at o f Fig. 4. Figure 12 shows an enlarged portion o f such a chart, centered arbi trarily on the nuclide *’^Au. Stable nuclides are shaded, and their isotopic abundances are shown. T he unshaded squares represent radionuclides, their half-lives being shown. Figure 13 suggests a tran sp aren t overlay th at we can place over a nuclidic chart such as th at o f Fig. 12. If the shaded central square o f Fig. 13 overlays a p articular tar get on the chart o f Fig. 12, the residual nuclides resulting from the various reactions printed on the overlay are identified. T hus if we chose *’^Au as a target, a (p ,a ) reaction will produce (stable) and either an (n,y) o r a (d,p) reac tion will produce the radionuclide *’®Au, whose half-life is 2.70 d.
196Au 197Au 198Au 199Au 200au 183 d 6.18 d 100% 2.70 d 3.14 d 48.4 min
195Au
78
193pt 50 y
77
192|r 193j, I94|r I95|r 74.2 d 62.7% 19.2 h 2.5 h
76
19l0s
U sing energy conservation, we can write Eq. 18 as Q = {Ky + K , ) - { K ^ ^ K , \
39.5 h
199TI 200TI 201JI 202t| 7.4 h 26.1 h 73.6 h 12.2 d
194pt 195pt 196pt 197pt 198pt 199pt 32.9% 33.8% 25.3 % 18.3 h 7.2% 30.8 min I96|r 197,r l98|r 52 s 5.8 min 8s
193os 19^0s 1950s 1920S 15.4 d 41.0% 30.5 h 6.0 y 6.5 min 35 min 115
118
117
116
120
119
121
Neutron number, N
Figure 12 (Fig. 4).
An expanded portion of the chart of the nuclides
Nuclei, like atom s, have stationary states o f definite energy, an d reaction studies can be used to identify them . C onsider, for exam ple, the reaction 2’Al(d,p)2«Al,
Q = 5.49 MeV,
in which a th in alu m in u m target foil is bom barded with 2.10-M eV deuterons. In th e laboratory the em erging pro to n s are seen to com e off w ith a n u m b er o f well-defined discrete energies an d are accom panied by gam m a rays. Figure 14 shows the energy distribution o f the em erging protons. In every reaction event we know th a t an energy equal to the kinetic energy o f the incident deuteron (= 2 .1 0 M eV)
a ,n
p,n
P ,Q !
7 , Of
P ,7 d, n
d,7 Q f ,d
7,n
n ,7
P,d
d,p
7 id
n ,d
d,of
7- P
O f,7
Q f ,p
n ,p
n ,a
Figure 13 Placing this as an overlay on Fig. 12, with the shaded central square over a particular target nuclide, shows the residual nuclides that result from the indicated reactions.
Section 54-8
Nuclear Reactions
1155
studying the energies and other properties of their station ary states.
Sample Problem 10
In the reaction
protons (' H) with kinetic energy 5.70 MeV are incident on ’H at rest, (a) What is the Q value for this reaction? (b) Find the kinetic energies of the deuterons emitted along the direction of the inci dent proton. Solution
(a) From Eq. 18 we have Q = [m('H) -I- m(’H) - ml^H) - w(^H)]c" = (1.007825 u -I- 3.016049 u - 2.014102 u - 2.014102 uX931.5 MeV/u) = -4 .0 3 MeV.
This reaction is endothermic; the final products have the greater mass and correspondingly the smaller kinetic energy by Eq. 19. {b) Using Eq. 19, with AT= 0 for the initial *H, we have K ,+ K ^ = Q + K„ = -4 .0 3 MeV -I- 5.70 MeV = 1.67 MeV. Figure 14 The energy distribution of protons resulting from the reaction ^’Al(d,p)^*Al. The incident deuteron has an en ergy of 2.10 MeV. The protons are detected as they emerge from the target at right angles to the incident beam.
(20)
Here the subscripts 1 and 2 refer to the two product nuclei. Conservation of momentum along the direction of the incident protons gives Pi + P 2 = Pp= V2w('H)/i:p = V2(938 MeV/c^X5.70 MeV) ( 21)
= 103.4 MeV/c.
plus the reaction energy Q (= 5.49 MeV) is available to be shared between the two reaction products, that is, be tween the residual nucleus “ A1 and the emerging proton p. How is this total energy (2.10 MeV -I- 5.49 MeV = 7.59 MeV) to be shared between these two particles? It all depends on whether the residual nucleus ^*A1 is produced in its ground state or in one o f its excited sta tionary states. In the former case, the emerging proton will have the maximum possible energy, corresponding to the peak on the extreme right o f the proton spectrum in Fig. 14. If, however, the residual nucleus is formed in an excited state, that nucleus will retain more o f the available energy and there will be less eneigy left for the emerging proton. The residual nucleus will not remain in its excited state very long but will get rid of its excess energy, such as by emitting a gamma ray. Every proton peak in the spectrum o f Fig. 14 corre sponds to a stationary state o f the residual nucleus “ Al. Figure 15 shows the energy levels that may be deduced by analyzing this spectrum. You can see the correspondence between the peaks o f Fig. 14 and the energy levels of Fig. 15. We have seen that our understanding of the way atoms are put together rests on the measured energies of the hydrogen atom states as its firm foundation. In the same way, we can learn how nuclei are put together by
2
-
0> 5
1-
28aI Figure 15 Energy levels of ^®A1, deduced from data such as those of Fig. 14.
1156
Chapter 54
Nuclear Physics
Equations 20 and 21 can be solved as two equations in two unknowns (either and P2 or and K ^. The results are K, = 0.24 MeV and
= 1.43 MeV.
Note that we have used nonrelativistic dynamics in solving this problem. Is this a good approximation?
54-9 NUCLEAR MODELS (Optional) The structure of atoms is now well understood. The Coulomb force is exerted by the massive center (the nucleus) on the elec trons, and (given enough computer time) we can use the meth ods of quantum mechanics to calculate properties of the atom. Things are not quite so well understood in the case of nuclei. The force law is complicated and cannot be written down explic itly in full detail. Nor is there a natural force center to simplify the calculations. To understand nuclear structure, we face a many-body problem of great complexity. In the absence of a comprehensive theory of nuclear structure, we try instead to construct nuclear models. Physicists use models as simplified ways of looking at a complex system to give physi cal insight into its properties. The usefulness of a model is tested by its ability to make predictions that can be verified experimen tally in the laboratory. Two models of the nucleus have proved useful. One model describes situations in which we can consider all the protons and neutrons to behave cooperatively, while the other model ne glects all but one proton or neutron in determining the proper ties of the nucleus. These two models represent quite opposing views of nuclear structure, but they can be combined to create a single unified model of the nucleus.
The Collective Model In the collective model, we ignore the motions of individual nucleons and treat the nucleus as a single entity. This model, originally called the “liquid drop model,” was developed by Niels Bohr to explain nuclear fission. We imagine the nucleus as a body analogous to a liquid drop, in which the nucleons interact with each other like molecules in the liquid. The equilibrium shape of the liquid drop is determined by the interactions of its molecules, and similarly the equilibrium shape of a nucleus is determined by the interactions of its nu cleons. Many nuclei have spherical equilibrium shapes, while others may be ellipsoidal. Like a liquid drop, a nucleus can absorb energy by the entire nucleus rotating about an axis or vibrating about its equilibrium shape. Through radioactive decay or nuclear reaction experi ments, it is possible to study the spectra of these excited states. Figure 16 shows examples of the two kinds of situations. The rotational energy can be written in terms of the angular momentum L (=/cu) as L}/!!, Writing the quantized angular momentum according to Eq. 23 of Chapter 51 as L = yfJ(jT~\)h, where J is the rotational angular momentum quan tum number of the entire nucleus, we obtain Ej = ^ J { J + \ ) .
2 ■ 1 ■
0
J
n
(a )
ib)
-
Figure 16 (a) Rotational excited states, labeled with the an gular momentum quantum number J. (h) Vibrational states, labeled with the vibrational quantum number n.
The vibrational states have energies given by E„ = nho) = nhv,
w = 1, 2, 3, . . . ,
(23)
where v is the vibrational frequency. This is the same expression that was used by Planck to describe the quantized vibrations of the atomic oscillators in the theory of cavity radiation (see Eq. 7 and Fig. 6 of Chapter 49). Note that the vibrational states in Fig. \6b are equally spaced, as given by Eq. 23. Evidence for collective structure can also be found in nuclear reactions. In a certain class of reactions A" -h a —►T +/?, an in termediate state is formed when X and a coalesce into a single entity C*, called a compound nucleus, which then breaks apart into Y b. The energy carried by projectile a into target X is quickly shared more or less equally in the random motion of the nucleons of the compound nucleus. (In the context of the liquid drop model, think of two drops coming together to form a larger drop whose molecules have a higher mean kinetic energy, corre sponding to a higher temperature for the combined drop.) The compound nucleus may exist for as long as 10“ s, a very long time by the standards of nuclear reactions, which may typi cally last for only 10“ ^^ s. Eventually, a nucleon or a group of nucleons will, by a statistical fluctuation, acquire enough energy to break free of the compound nucleus. We observe this outgoing particle b and the residual nucleus Y. A central feature of this hypothesis is that, once the energy of the projectile is shared among the nucleons, the compound nu cleus “forgets” how it was formed and decays purely according to statistical considerations. Figure 17 represents this process, in which a compound nucleus ^°Ne* is formed in any of three ways
Formation
Compoun(j nucleus
Decay
( 22)
Note that the spacing between the states grows as the angular momentum increases.
Figure 17 A few of the many possible formation and decay modes of the compound nucleus ^®Ne*.
Section 54-9
Nuclear Models (Optional)
1157
and decays in any of three different ways. Experimentally, we observe that the relative probability of the different decay modes has the same value for any of the combinations of projectile and target. This confirms the compound nucleus interpretation and provides another example of the collective behavior of nucleons in the nucleus.
The Independent Particle Model In the liquid drop model, the nucleons move around at random and bump into each other frequently. The independent particle model, however, considers that each nucleon moves in a welldefined orbit within the nucleus and hardly makes any collisions at all! The nucleus— unlike the atom — has no fixed center of charge, and we assume in this model that each nucleon moves in a potential that is determined by the smeared-out motions of all the other nucleons. A nucleon in a nucleus, like an electron in an atom, has a set of quantum numbers that defines its state of motion. Also nu cleons, again like electrons, obey the Pauli exclusion principle. That is, no two nucleons may occupy the same state at the same time. In considering nucleon states, the neutrons and the pro tons are treated separately, each having its own array of available quantized states. The fact that nucleons obey the Pauli principle helps us to understand the relative stability of the nucleon states. If two nucleons within the nucleus are to collide, the energy of each of them after the collision must correspond to the energy of an unoccupied stationary state. If these states (or even just one of them) are filled, the collision simply cannot occur. In time, any given nucleon will find it possible to collide, but meanwhile it will have made enough revolutions in its orbit to give meaning to the notion of a stationary nucleon state with a quantized energy. In the atomic realm, the essence of the periodic table of the elements is that it is periodic. That is, certain properties of the elements repeat themselves in a regular fashion as one proceeds through the table. These repetitions are associated with the fact that the atomic electrons arrange themselves in shells and sub shells that have a special stability when they are fully occupied. We can take the atomic numbers of the inert gases, 2, 10, 18,36,54,86, . . . , as magic electron numbers that mark the completion of such shells. Nuclei also show shell effects, associated with certain magic nucleon numbers:
Neutron number
Figure 18 The variation in nuclear radius as a function of neutron number. The variation is expressed relative to the “standard” variation expected from the “collective” structure o f R = RoA^f'^. The sudden jumps indicate shell structure.
Evidence for atomic shell structure can be found, for example, from measurements of the ionization energies or mean radii of atoms. Figure 6 of Chapter 52 shows the variation in the ioniza tion energy of atoms as a function of the number of electrons. If we plot the atomic radii as a function of electron number, we find a gradual decrease as one shell is filled and then a sudden jump as we begin filling the next shell, because the radius de pends primarily on the principal quantum number n. These sudden jumps in the radius and in the ionization energy occur when the number of electrons is equal to one of the magic elec tron numbers. In the case of nuclei, we can gather similar evidence for nu clear shell structure. Sample Problem 12 gives an example of the change in “ionization energy” (the energy needed to remove a single proton or neutron from the nucleus) at closed shells. Fig ure 18 shows the variation in the nuclear radius as a function of neutron number. Just as in the atomic case, the radius gradually decreases within a shell and then increases suddenly as we begin filling the next shell. The sudden jumps occur when either the proton number or neutron number is equal to one of the magic nucleon numbers. Similar evidence for shell structure can be found in other nuclear properties, including alpha-decay halflives, magnetic dipole moments, cross sections for capture of neutrons and scattering of electrons, and energies of excited states. ■
2 ,8 ,2 0 ,2 8 ,5 0 ,8 2 , 126,. . . . Any nuclide whose proton number Z or neutron number N has one of these values turns out to have a special stability that may be made apparent in a variety of ways. Examples of “magic” nuclides are (Z = 8), ^C a (Z = 20, A^= 20), ’^Mo (A = 50), and 2o«Pb (Z = 82, A^= 126). Both ^C a and ^°®Pb are said to be “doubly magic” because they con tain filled shells of both protons and neutrons. The magic number 2 shows up in the exceptional stability of the a particle (^He), which, with Z = = 2, is doubly magic. For example, the binding energy per nucleon for this nuclide stands well above that of its neighbors on the binding energy curve of Fig. 6. The a particle is so tightly bound, in fact, that it is impossible to add another particle to it; there is no stable nuclide with A = 5.
Sample Problem 11
Consider the neutron-capture reaction
‘^Ag + n —> ' '°Ag* —►‘‘°Ag + y. Figure 19 shows its cross section as a function of the energy of the incident neutron. Analyze this figure in terms of the compound nucleus concept and the uncertainty principle. Solution The cross-section curve of Fig. 19 is sharply peaked, reaching a maximum cross section of 12,500 bams.f This “reso
t The cross section for a reaction is a measure of the probability for the reaction to occur. A common unit for expressing cross section is the barn, which is equivalent to 10“ ^®m^.
1158
Chapter 54
Nuclear Physics Sample Problem 12 The nuclide *^°Sn (Z = 50) has a filled proton shell, 50 being one of the magic nucleon numbers. The nuclide '^‘Sb (Z = 51) has an “extra” proton outside this shell. According to the shell concept, this extra proton should be easier to remove than a proton from the filled shell. Verify this by calculating the required energy in each case. Use the following mass data:
z
N
Atomic Mass (u)
50+1 50 50-1
70 70 70
120.903821 119.902199 118.905819
Nuclide •2'Sb ■»Sn " ’In
4.5 4.7 4.9 5.1 5.3 5.5 5.7 5.9 Neutron energy (eV)
Figure 19 Sample Problem 11. The cross section for the re action '°’Ag(n,y)“ °Ag as a function of the energy of the inci dent neutron. The width of the peak at half its maximum is about 0.20 eV.
nance peak” suggests that we are dealing with a single excited level in the compound nucleus ' ‘°Ag*. When the available en ergy just matches the energy of this level above th e ' *°Ag ground state, we have “resonance,” and the reaction really “goes.” However, the resonance peak is not infinitely sharp. From the figure we can measure that it has an approximate width at half maximum (that is, at 6250 bams) of 0.20 eV. We account for this by saying th a t' *°Ag in its excited state is not sharply defined in energy; it is “fuzzy,” with an energy uncertainty A£* of 0.20 eV. We can use the uncertainty principle, written in the form A E -A t ~ h jln
(24)
to tell us something about any state of an atomic or nuclear system. We have seen that A £ is a measure of the uncertainty of our knowledge of the energy of the state. The quantity At is interpreted as the time available to measure the energy of the state; it is in fact the mean life of the state before it decays. For the excited state ' ‘°Ag* we have, from Eq. 24, A t-
h /ln _ 6.58 X 10->"eV»s = 3.3X lO-'^s. AE 0.20 eV
This is the order of magnitude of the lifetime that is characteris tic of a compound nucleus.
The proton atomic mass is 1.007825 u. Solution cess
Removing the “extra” proton corresponds to the pro
The required energy E follows from £ = [w(‘20Sn) + m ('H) - w('2>Sb)]c2 = (119.902199 u + 1.007825 u - 120.903821 u) X (931.5 MeV/u) = 5.8 MeV. Removing the proton from the filled shell corresponds to •20Sn — “ ^In + p. The required energy follows from E = [m(‘'^In) + m(‘H) - mC^oSn)]^^ = (118.905819 u + 1.007825 u - 119.902199 u) X (931.5 MeV/u) = 10.7 MeV. This is considerably greater than the energy required to remove an “extra” proton (= 5.8 MeV), just as the shell model predicts. In much the same way the energy needed to remove an electron from a filled electron shell (= 22 e V for the filled shell of neon) is much greater than that needed to remove an “extra” electron from outside such a filled shell (= 5 eV for the “extra” electron from sodium).
QUESTIONS 1. When a thin foil is bombarded with a particles, a few of them are scattered back toward the source. Rutherford con cluded from this that the positive charge of the atom — and also most of its mass— must be concentrated in a very small “nucleus” within the atom. What was his line of reasoning? 2. In what ways do the so-called strong force and the electro static or Coulomb force differ? 3. Why does the relative importance of the Coulomb force compared to the strong nuclear force increase at large mass numbers?
4. In your body, are there more neutrons than protons? More protons than electrons? Discuss. 5. Why do nuclei tend to have more neutrons than protons at high mass numbers? 6. Why do we use atomic rather than nuclear masses in analyz ing most nuclear decay and reaction processes? 7. How might the equality 1 u = 1.6605 X 10“ ^^ kg be arrived at in the laboratory? 8. The atoms of a given element may differ in mass, have different physical characteristics, and yet not vary chemi cally. Why is this?
Problems 9. The deviation of isotopic masses from integer values is due to many factors. Name some. Which is most responsible? 10. How is the mass of the neutron determined? 11. The most stable nuclides have a mass number A near 60 (see Fig. 6). Why don’t all nuclides have mass numbers near 60? 12. If we neglect the very lightest nuclides, the binding energy per nucleon in Fig. 6 is roughly constant at 7 to 8 MeV/nucleon. Do you expect the mean electronic binding energy per electron in atoms also to be roughly constant throughout the periodic table? 13. Why is the binding energy per nucleon (Fig. 6) low at low mass numbers? At high mass numbers? 14. In the binding energy curve of Fig. 6, what is special or notable about the nuclides ^He, ^^Ni, and ^^’Pu? 15. The magnetic moment of the neutron is —1.9130//n . What is a nuclear magneton and how does it differ from a Bohr magneton? What does the minus sign mean? How can the neutron, which carries no net charge, have a magnetic moment in the first place? 16. A particular nucleus was created in a massive stellar explosion, perhaps 10‘° y ago. It suddenly decays by a emis sion while we are observing it. After all those years, why did it decide to decay at this particular moment? 17. Can you justify this statement: “In measuring half-lives by the method of Sample Problem 4, it is not necessary to measure the absolute decay rate R; any quantity propor tional to it will suffice. However, in the method of Sample Problem 5 an absolute rate is needed.”
25.
26.
27. 28.
29. 30.
31.
32.
18. Does the temperature affect the rate of decay of radioactive nuclides? If so, how? 19. You are running longevity tests on light bulbs. Do you ex pect their “decay” to be exponential? What is the essential difference between the decay of light bulbs and of radionu clides? 20. Generally clocks exhibit complete regularity of some peri odic process. Considering that radioactive decay is com pletely random, how can it nevertheless be used for the measurement of time? 21. Can you give a justification, even a partial one, for the barrier tunneling phenomenon in terms of basic ideas about the wave nature of matter? 22. Explain why, in alpha decay, short half-lives correspond to large disintegration energies, and conversely. 23. A radioactive nucleus can emit a positron, e"^. This corre sponds to a proton in the nucleus being converted to a neu tron. The mass of a neutron, however, is greater than that of a proton. How then can positron emission occur? 24. In beta decay the emitted electrons form a continuous spec
33.
34.
35. 36. 37.
1159
trum, but in alpha decay they form a discrete spectrum. What difficulties did this cause in the explanation of beta decay, and how were these difficulties finally overcome? How do neutrinos differ from photons? Each has zero charge and (presumably) zero rest mass and travels at the speed of light. The decay of radioactive elements produces helium, which eventually passes into the Earth’s atmosphere. The amount of helium actually present in the atmosphere, however, is very much less than the amount released in this way. Ex plain. The half-life of is 4.5 X 10’ y, about the age of the solar system. How can such a long half-life be measured? In radioactive dating with how do you get around the fact that you don’t know how much was present in the rocks to begin with? (Hint: What is the ultimate decay prod uct of 23«U?) Make a list of the various sources of ionizing radiation en countered in our environment, whether natural or artificial. Which of these conservation laws apply to all nuclear reac tions? Conservation of (a) charge, (b) mass, (c) total energy, (d) rest energy, (e) kinetic energy, ( / ) linear momentum, (g) angular momentum, and (h) total number of nucleons. Small temperature changes have a large effect on the rate of chemical reactions but generally have a negligible effect on the rate of nuclear reactions. Explain. In the development of our understanding of the atom, did we use atomic models as we now use nuclear models? Is Bohr’s theory such an atomic model? Are models now used in atomic physics? What is the difference between a model and a theory? What are the basic assumptions of the liquid drop and the independent particle models of nuclear structure? How do they differ? Are there similarities between them? Does the collective model of the nucleus give us a picture of the following phenomena: (a) acceptance by the nucleus of a colliding particle; (b) loss of a particle by spontaneous emis sion; (c) fission; (d) dependence of stability on energy con tent? What is so special (“magic”) about the magic nucleon num bers? Why aren’t the magic nucleon numbers and the magic elec tron numbers the same? What accounts for each? The average number of stable (or very long-lived) isotopes of the inert gases is 3.7. The average number of stable nuclides for the four magic neutron numbers, however, is 5.8, consid erably greater. If the inert gases are so stable, why were not more stable isotopes of them created when the elements were formed?
PROBLEMS Section 54~1 Discovering the Nucleus 1. Calculate the distance of closest approach for a head-on collision between a 5.30-MeV a particle and the nucleus of a copper atom.
2. (a) Calculate the electric force on an a particle at the surface of a gold atom, presuming that the positive charge is spread uniformly throughout the volume of the atom. Ignore the atomic electrons. A gold atom has a radius of 0.16 nm; treat
1160
Chapter 54
Nuclear Physics
the a particle as a point particle, (b) Through what distance would this force, presumed constant, have to act to bring a 5.30-MeV a particle to rest? Express your answer in terms of the diameter of a gold atom. 3. Assume that a gold nucleus has a radius of 6.98 fm (see Table 1), and an a particle has a radius of 1.8 fm. What energy must an incident a particle have to just touch the gold nucleus? 4. When an a particle collides elastically with a nucleus, the nucleus recoils. A 5.00-MeV a particle has a head-on elastic collision with a gold nucleus, initially at rest. What is the kinetic energy (a) of the recoiling nucleus and (b) of the rebounding a particle? The mass of the a particle may be taken to be 4.00 u and that of the gold nucleus 197 u. Section 54-2 Some Nuclear Properties 5. Locate the nuclides displayed in Table 1 on the nuclidic chart of Fig. 4. Which of these nuclides are within the stabil ity zone? 6. The radius of a nucleus is measured, by electron-scattering methods, to be 3.6 fm. What is the likely mass number of the nucleus? 7. Arrange the 25 nuclides given below in squares as a section of the nuclidic chart similar to Fig. 4. Draw in and label (a) all isobaric (constant A) lines and (b) all lines of con stant neutron excess, defined as N — Z. Consider nuclides ii8-i22Te, i'7-'2isb, “ ^-'20Sn, and ‘'"-“ «Cd. 8. A neutron star is a stellar object whose density is about that of nuclear matter, as calculated in Sample Problem 2. Sup pose that the Sun were to collapse into such a star without losing any of its present mass. What would be its expected radius? 9. Verify that the binding energy per nucleon given in Table 1 for ^^^Pu is indeed 7.56 MeV/nucleon. The needed atomic masses are 239.052158 u (^^’Pu), 1.007825 u ('H ), and 1.008665 u (neutron). 10. Calculate the average binding energy per nucleon of ^^Ni, which has an atomic mass o f61.928346 u. This nucleus has the greatest binding energy per nucleon of all the known stable nuclei. 11. The atomic masses of 'H , '^C, and are 1.007825 u, 12.000000 u (by definition), and 238.050785 u, respec tively. {a) What would these masses be if the mass unit were defined so that the mass of *H was (exactly) 1.000000 u? (b) Use your result to suggest why this perhaps obvious choice was not made. 12. (a) Convince yourself that the energy tied up in nuclear, or strong-force, bonds is proportional to .4, the mass number of the nucleus in question, {b) Convince yourself that the en ergy tied up in Coulomb-force bonds between the protons is proportional to Z (Z — 1). (c) Show that, as we move to larger and larger nuclei (see Fig. 4), the importance of (b) increases more rapidly than does that of (a). 13. In the periodic table, the entry for magnesium is:
There are three isotopes: ^^Mg, atomic mass = 23.985042 u. ^^Mg, atomic mass = 24.985837 u. ^^Mg, atomic mass = 25.982594 u. The abundance of ^'‘Mg is 78.99% by mass. Calculate the abundances of the other two isotopes. 14. Because a nucleon is confined to a nucleus, we can take its uncertainty in position to be approximately the nuclear radius R. What does the uncertainty principle yield for the kinetic energy of a nucleon in a nucleus with, say, A = 100? (Hint: Take the uncertainty in momentum Ap to be the actual momentum p.) 15. You are asked to pick apart an a particle (^He) by removing, in sequence, a proton, a neutron, and a proton. Calculate (a) the work required for each step, (b) the total binding energy of the a particle, and (c) the binding energy per nu cleon. Needed atomic masses are ^He 4.002603 u 3.016049 u n
2.014102 u 'H
1.007825 u
1.008665 u.
16. To simplify calculations, atomic masses are sometimes ta bulated, not as the actual atomic mass m but as (m —A)c^, where A is the mass number expressed in mass units. This quantity, usually reported in MeV, is called the mass excess, symbol A. Using data from Sample Problem 3, find the mass excesses for (a) ‘H, (b) the neutron, and (c) *^°Sn. 17. (a) Show that the total binding energy of a nuclide can be written as £ b = ZA„ + A^A„-A, where Ah , A„, and A are the appropriate mass excesses; see Problem 16. (b) Using this method calculate the binding energy per nucleon for ’’^Au. Compare your result with the value listed in Table 1. The needed mass excesses are Ah = + 7.289 MeV, A„ = + 8.071 MeV, and A ,97 = -3 1 .1 7 MeV. Ah is the mass excess of ‘H. Note the economy of calculation that results when mass excesses are used in place of the actual masses. 18. A penny has a mass of 3.00 g. Calculate the nuclear energy that would be required to separate all the neutrons and pro tons in this coin. Ignore the binding energy of the electrons. For simplicity assume that the penny is made entirely of ^^Cu atoms (mass = 62.929599 u). The atomic masses of the proton and the neutron are 1.007825 u and 1.008665 u, respectively. 19. Nuclear radii may be measured by scattering high-energy electrons from nuclei, (a) What is the de Broglie wavelength for 480-MeV electrons? (b) Are they suitable probes for this purpose? Relativity must be taken into account. 20. Because the neutron has no charge, its mass must be found in some way other than by using a mass spectrometer. When a resting neutron and a proton meet, they combine and form a deuteron, emitting a gamma ray whose energy is 2.2233 MeV. The atomic masses of the proton and the deu teron are 1.007825 u and 2.014102 u, respectively. Find the
Problems mass of the neutron from these data, to as many significant figures as the data warrant. 21. The spin and the magnetic moment (maximum z compo nent) of ^Li in its ground state (see Table 1) are ^ and + 3.26 nuclear magnetons, respectively. A free ^Li nucleus is placed in a magnetic field of 2.16 T. (a) Into how many levels will the ground state split because of space quantization? (b) What is the energy difference between adjacent pairs of levels? (c) What is the wavelength that corresponds to a transition between such a pair of levels? (t/) In what region of the electromagnetic spectrum does this wavelength lie? 22. (a) Show that the electrostatic potential energy of a uniform sphere of charge Q and radius R is given by U=
3(2' lOneoR
32.
33.
34.
35.
{Hint: Assemble the sphere from thin spherical shells brought in from infinity.) {b) Find the electrostatic potential energy for the nuclide ^^’ Pu, assumed spherical; see Table 1. (c) Compare its electrostatic potential energy per particle with its binding energy per nucleon of 7.56 MeV. (d) What do you conclude? Section 54~3 Radioactive Decay 23. The half-life of a radioactive isotope is 140 d. How many days would it take for the activity of a sample of this isotope to fall to one-fourth of its initial decay rate? 24. The half-life of a particular radioactive isotope is 6.5 h. If there are initially 48 X 10'^ atoms of this isotope in a partic ular sample, how many atoms of this isotope remain after 26 h? 25. A radioactive isotope of mercury, ’’^Hg, decays into gold, *’^Au, with a decay constant of 0.0108 h" *. (a) Calculate its half-life, (b) What fraction of the original amount will re main after three half-lives? (c) After 10 days? 26. From data presented in the first few paragraphs of Section 54-3, deduce (a) the disintegration constant A and (b) the half-life of 238U. 27. ^^Ga, atomic mass = 66.93 u, has a half-life o f78.25 h. Con sider an initially pure 3.42-g sample of this isotope, {a) Find its activity (decay rate), (b) Find its activity 48.0 h later. 28. Show that the law of radioactive decay (Eq. 6) can be written in the form
29. ^^^Ra decays by alpha decay with a half-life of 11.43 d. How many helium atoms are created in 28 d from an initially pure sample of ^^^Ra containing 4.70 X 10^' atoms? 30. The radionuclide ^C u has a half-life of 12.7 h. How much of an initially pure 5.50-g sample of ^C u will decay during the 2-h period beginning 14.0 h later? 31. The radionuclide (half-life = 14.28 d) is often used as a tracer to follow the course of biochemical reactions involv ing phosphorus, (a) If the counting rate in a particular exper imental setup is 3050 counts/s, after what time will it fall to 170 counts/s? (b) A solution containing is fed to the root system of an experimental tomato plant and the ^^P activity in a leaf is measured 3.48 d later. By what factor must this
36.
37.
38.
1161
reading be multiplied to correct for the decay that has oc curred since the experiment began? A 1.00-g sample of samarium emits a particles at a rate of 120 particles/s. "*^Sm, whose natural abundance in bulk samarium is 15.0%, is the responsible isotope. Calculate the half-life of this isotope. ^^’Pu, atomic mass = 239 u, decays by alpha decay with a half-life of 24,100 y. How many grams of helium are pro duced by an initially pure 12.0-g sample of ^^’Pu after 20,000 y? (Recall that an a particle is a helium nucleus, with an atomic mass of 4.00 u.) A source contains two phosphorus radionuclides, ^^P (/,/2 = 14.3 d) and ^^P (t„2 = 25.3 d). Initially 10.0% of the decays come from ^^P. How long must one wait until 90.0% do so? After a brief neutron irradiation of silver, two activities are present: '°®Ag (/,/2 = 2.42 min) with an initial decay rate of 3.1 X lOVs, and "°Ag (/,/2 = 24.6 s) with an initial decay rate of 4.1 X 1OVs. Make a plot similar to Fig. 7 showing the total combined decay rate of the two isotopes as a function of time from / = 0 until / = 10 min. In Fig. 7, the extraction of the half-life for simple decays was illustrated. Given only the plot of total decay rate, can you suggest a way to analyze it in order to find the half-lives of both isotopes? As of this writing there is speculation that the free proton may not actually be a stable particle but may be radioactive, with a half-life of about 1X10^^ y. If this turns out to be true, about how long would you have to wait to be reason ably sure that one proton in your body has decayed? Assume that you are made of water and have a mass of 70 kg. A certain radionuclide is being manufactured in a cyclotron, at a constant rate P. It is also decaying, with a disintegration constant A. Let the production process continue for a time that is long compared to the half-life of the radionuclide. Convince yourself that the number of radioactive nuclei present at such times will be constant and will be given by N = P/A. Convince yourself further that this result holds no matter how many of the radioactive nuclei were present initially. The nuclide is said to be in secular equilibrium with its source; in this state its decay rate is just equal to its production rate. The radionuclide ^^Mn has a half-life of 2.58 h and is pro duced in a cyclotron by bombarding a manganese target with deuterons. The target contains only the stable manga nese isotope ^^Mn and the reaction that produces ^^Mn is 55Mn + d — 5^Mn + p.
After being bombarded for a time » 2.58 h, the activity of the target, due to ^^Mn, is 8.88 X 10'° s“ '; see Problem 37. (a) At what constant rate Pare ^^Mn nuclei being produced in the cyclotron during the bombardment? {b) At what rate are they decaying (also during the bombardment)? (c) How many ^^Mn nuclei are present at the end of the bombard ment? {d) What is their total mass? The atomic mass of 5^Mn is 55.94 u. 39. A radium source contains 1.00 mg of ^^^Ra, which decays with a half-life of 1600 y to produce ^^^Rn, an inert gas. This radon gas in turn decays by alpha decay with a half-life of 3.82 d. (a) Calculate the decay rate of ^^^Ra in the source.
1162
Chapter 54
Nuclear Physics
(b) At what rate is the radon decaying when it has come to secular equilibrium with the radium source? See Problem 37. (c) How much radon is in secular equilibrium with the radium source? Section 54-4 Alpha Decay 40. Generally, heavier nuclides tend to be more unstable to alpha decay. For example, the most stable isotope of ura nium, has an alpha decay half-life of 4.5 X 10’ y. The most stable isotope of plutonium is ^^Pu with a 8.2 X 10^ y half-life, and for curium we have ^^*Cm and 3.4 X 10^ y. When half of an original sample of has decayed, what fractions of the original isotopes of (a) plutonium and {b) curium are left? 41. Consider a nucleus to be made up of an a particle (^He) and a residual nucleus (^^Th). Plot the electrostatic poten tial energy U(r\ where r is the distance between these parti cles. Cover the range 10 fm < r < 100 fm and compare your plot with that of Fig. 8. 42. A nucleus emits an a particle of energy 4.196 MeV. Calculate the disintegration energy Q for this process, taking the recoil energy of the residual ^^Th nucleus into account. The atomic mass of an a particle is 4.0026 u and that of the 2^Th is 234.04 u. Compare your result with that of Sample Problem 6a. 43. Heavy radionuclides emit an a particle rather than other combinations of nucleons because the a particle is such a stable, tightly bound structure. To confirm this, calculate the disintegration energies for these hypothetical decay pro cesses and discuss the meaning of your findings: 235U ^ 232J h + 3He,
Q 3;
235U — 23'Th + ‘'He,
e .;
+ ’He,
65 -
235U ^
The needed atomic masses are
2“ Th 232.038051 u
»He
231Th
“He 4.002603 u
231.036298 u
230Th 230.033128 u
’He
3.016029 u
5.01222 u .
235U 235.043924 u 44. Consider that a nucleus emits (a) an a particle or (b) a sequence of neutron, proton, neutron, proton. Calculate the energy released in each case, (c) Convince yourself both by reasoned argument and also by direct calculation that the difference between these two numbers is just the total binding energy of the a particle. Find that binding energy. Needed atomic masses are 23«U
238.050785 u
^He 4.002603 u
23?u
237.048725 u
‘H
1.007825 u
236pa
236.048890 u
n
1.008665 u.
235.045430 u 234Th 234.043593 u 45. Under certain circumstances, a nucleus can decay by emit ting a particle heavier than an a particle. Such decays are very rare. Consider the decays
223Ra — 209p5 + 14Q and 223Ra _
2 i9 R n +
4H e.
(a) Calculate the 0-values for these decays and determine that both are energetically possible, (b) The Coulomb barrier height for a particles in this decay is 30 MeV. What is the barrier height for ‘^C decay? Atomic masses are 223Ra
223.018501 u
'"C
^o’ Pb
208.981065 u
^He
2*’Rn
219.009479 u
14.003242 u 4.002603 u.
Section 54-5 Beta Decay 46. A certain stable nuclide, after absorbing a neutron, emits a negative electron and then splits spontaneously into two a particles. Identify the nuclide. 47. *^^Cs is present in the fallout from above-ground detona tions of nuclear bombs. Because it beta decays with a slow 30.2-y half-life into ‘^^Ba, releasing considerable energy in the process, it is an environmental concern. The atomic masses of the Cs and Ba are 136.907073 u and 136.905812 u, respectively. Calculate the total energy re leased in one decay. 48. A free neutron decays according to Eq. 13. Calculate the maximum energy of the beta spectrum. Needed atomic masses are: n
1.008665 u;
'H
1.007825 u.
49. An electron is emitted from a middle-mass nuclide (A = 150, say) with a kinetic energy of 1.00 MeV. (a) Find its de Broglie wavelength, (b) Calculate the radius of the emitting nucleus, (c) Can such an electron be confined as a standing wave in a “box” of such dimensions? (d) Can you use these numbers to disprove the argument (long since abandoned) that electrons actually exist in nuclei? 50. The radionuclide ^^P decays to as described by Eq. 11. In a particular decay event, a 1.71 -MeV electron is emitted, the maximum possible value. Find the kinetic energy of the recoiling atom in this event. The atomic mass of is 31.97 u. (Hint: For the electron it is necessary to use the relativistic expressions for the kinetic energy and the linear momentum. Newtonian mechanics may safely be used for the relatively slow-moving atom.) 51. The radionuclide "C decays according to •‘C — *'B + e-^ +
V,
= 20.3 min.
The maximum energy of the positron spectrum is 960.8 ke V. (a) Show that the disintegration energy Q for this process is given by
Q = ('Wc - 'Wb“ 2me)c2, where and me are the atomic mass of “ C and ‘' B, respec tively and is the electron (positron) mass, (b) Given that me = 11.011433 u, mg = 11.009305 u, and me = 0.0005486 u, calculate Q and compare it with the maxi mum energy of the positron spectrum, given above. (Hint: Let m'c and m^ be the nuclear masses and proceed as in Sample Problem 7 for beta decay. Note that positron decay
Problems is an exception to the general rule that, if atomic masses are used in nuclear decay processes, the mass of the emitted electron is automatically taken care of.) 52. Some radionuclides decay by capturing one of their own atomic electrons, a /f-electron, say. An example is V + e-
^’Ti +
V,
61.
/,/2 = 331 d.
Show that the disintegration energy Q for this process is given by 0 = (m v -W T i)c ^ -£ K , where and are the atomic masses of and "*’Ti, respectively, and is the binding energy of the vanadium /^-electron. (Hint: Put my and as the corresponding nuclear masses and proceed as in Sample Problem 7; see the footnote in that sample problem.) 53. Find the disintegration energy Q for the decay of by ^-electron capture, as described in Problem 52. The needed data are = 48.948517 u, = 48.947871 u, and = 5.47 keV.
62. 63.
Section 54-6 Measuring Ionizing Radiation 54. A Geiger counter records 8722 counts in 1 min. Calculate the activity of the source in Ci, assuming that the counter records all decays. 55. A typical chest x-ray radiation dose is 25 mrem, delivered by Xrays with a quality factor of 0.85. Assuming that the mass of the exposed tissue is one-half the patient’s mass of 88 kg, calculate the energy absorbed in joules. 56. A 75-kg person receives a whole-body radiation dose of 24 mrad, delivered by a particles for which the quality fac tor is 12. Calculate (a) the absorbed energy in joules and (b) the equivalent dose in rem. 57. An activity of 3.94 //Ci is needed in a radioactive sample to be used in a medical procedure. One week before treatment, a nuclide sample with a half-life of 1.82 X 10^ s is prepared. What should be the activity of the sample at the time of preparation in order that it have the required activity at the time of treatment? 58. The plutonium isotope ^^’Pu, atomic mass 239.05 u, is pro duced as a by-product in nuclear reactors and hence is accu mulating in reactor fuel elements. It is radioactive, decaying by alpha decay with a half-life of 2.411 X 10^ y. But pluto nium is also one of the most toxic chemicals known; as little as 2.00 mg is lethal to a human, (a) How many nuclei con stitute a chemically lethal dose? (b) What is the decay rate of this amount? (c) Its activity in curies? 59. Cancer cells are more vulnerable to x and gamma radiation than are healthy cells. Though linear accelerators are now replacing it, in the past the standard source for radiation therapy has been radioactive ^C o, which beta decays into an excited nuclear state of ^N i, which immediately drops into the ground state, emitting two gamma-ray photons, each of approximate energy 1.2 MeV. The controlling betadecay half-life is 5.27 y. How many radioactive ^C o nuclei are present in a 6000-Ci source used in a hospital? The atomic mass of ^C o is 59.93 u. 60. An airline pilot spends an average of 20 h per week flying at 12,000 m, at which altitude the dose equivalent rate due to
64.
65.
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cosmic and solar radiation is 12 //Sv/h (1 Sv = 1 sievert = 100 rem; the sievert is the SI unit of dose equivalent). Calcu late the annual equivalent dose in mrem. After long effort, in 1902, Marie and Pierre Curie succeeded in separating from uranium ore the first substantial quantity of radium, 1 decigram (dg) of pure RaCl2. The radium was the radioactive isotope ^^^Ra, which decays with a half-life of 1600 y. (a) How many radium nuclei had they isolated? (b) What was the decay rate of their sample, in Bq? (1 Bq = 1 becquerel = 1 decay/s.) (c) In curies? The molar mass of Cl is 35.453 g/mol; the atomic mass of the radium isotope is 226.03 u. Calculate the mass of 4.60 //Ci of ^K , which has a half-life of 1.28 X 10’ y and an atomic mass of 40.0 u. One of the dangers of radioactive fallout from a nuclear bomb is ’^Sr, which beta decays with a 29-y half-life. Be cause it has chemical properties much like calcium, the strontium, if eaten by a cow, becomes concentrated in its milk and ends up in the bones of whoever drinks the milk. The energetic decay electrons damage the bone marrow and thus impair the production of red blood cells. A 1-megaton bomb produces approximately 400 g of ’^Sr. If the fallout spreads uniformly over a 2000-km^ area, what area would have radioactivity equal to the allowed bone burden for one person of 0.002 mCi? The atomic mass o f ’^Sr is 89.9 u. The nuclide *’®Au, half-life = 2.693 d, is used in cancer ther apy. Calculate the mass of this isotope required to produce an activity of 250 Ci. An 87-kg worker at a breeder reactor plant accidentally in gests 2.5 mg of ^^’Pu dust. ^^’Pu has a half-life of 24,100 y, decaying by alpha decay. The energy of the emitted a parti cles is 5.2 MeV, with a quality factor of 13. Assume that the plutonium resides in the worker’s body for 12 h, and that 95% of the emitted a particles are stopped within the body. Calculate (a) the number of plutonium atoms ingested, (b) the number that decay during the 12 h, (c) the energy ab sorbed by the body, (d) the resulting physical dose in rad, and (e) the equivalent biological dose in rem.
Section 54-7 Natural Radioactivity 66. A rock is found to contain 4.20 mg of and 2.00 mg of Assume that the rock contained no lead at formation, all the lead now present arising from the decay of the ura nium. Find the age of the rock. The half-life of is 4.47 X 10’ y. 67. Two radioactive materials that are unstable to alpha decay, and ^^^Th, and one that is unstable to beta decay, ^K , are sufficiently abundant in granite to contribute signifi cantly to the heating of the Earth through the decay energy produced. The alpha-unstable isotopes give rise to decay chains that stop at stable lead isotopes. has a single beta decay. Decay information follows: Half-life Parent Decay Stable Q / Endpoint (MeV) (ppm) Nuclide Mode (y) 238U 51.7 4 4.47 X 10’ 206pb a 232Th 42.7 1.41 X 10'® 20»P5 a 13 40K 1.28 X 10’ 1.32 4 -“ Ca
1164
Chapter 54
Nuclear Physics
Q is the total energy released in the decay of one parent nucleus to the final stable endpoint a n d /is the abundance of the isotope in kilograms per kilogram of granite; ppm means parts per million, (a) Show that these materials give rise to a total heat production of 987 pW for each kilogram ofgranite.(Z?) Assuming that there is 2.7 X 10^^ kg of granite in a 20-km thick, spherical shell around the Earth, estimate the power this will produce over the whole Earth. Compare this with the total solar power intercepted by the Earth, 1.7 X 10'" W. 68. A particular rock is thought to be 260 million years old. If it contains 3.71 mg of how much should it con tain? 69. A rock, recovered from far underground, is found to contain 860 pg of 150 /ig of ^®^Pb, and 1.60 mg of'^^Ca. How much will it very likely contain? Needed half-lives are listed in Problem 67. Section 54-8 Nuclear Reactions 70. Fill in the missing nuclide in each of the following reactions: {a) "^Sn(?,p)""Sn; (b) ^Ca(a,n)?; and (c) ?(p,n)"Be. 71. Calculate Q for the reaction ^’Co(p,n)^’Ni. Needed atomic masses are 5’Co
58.933198 u
'H
1.007825 u
5^Ni
58.934349 u
n
1.008665 u.
72. Making mental use of the overlay of Fig. 13 applied to Fig. 12, write down the reactions by which the radionuclide *’"Pt Ui/2 18.3 h) can be prepared, at least in principle. Except in special circumstances, only stable nuclides can serve as practical targets for nuclear reactions. 73. The radionuclide ^C o (/,/2 = 5.27 y) is much used in cancer therapy. Tabulate possible reactions that might be used in preparing it. Limit the projectiles to neutrons, protons, and deuterons. Limit the targets to stable nuclides. The stable nuclides suitably close to ^C o are ^^Cu, 6o.6i.62jsjj^ ^’Co, and ^"'^*Fe. (Commercially, ^C o is made by bombarding ele mental cobalt, which consists only of the isotope ^’Co, with neutrons in a reactor.) 74. A beam of deuterons falls on a copper target. Copper has two stable isotopes, ^^Cu (69.2%) and ^^Cu (30.8%). Tabulate the residual nuclides that can be produced by the reactions (d,n), (d,p), (d,a), and (d,y). By inspection of Fig. 4, indicate which residual nuclides are stable and which are radioactive. 75. Prepare an overlay like that of Fig. 13 in which that figure is extended to include reactions involving the light nuclides (tritium) and ^He, considered both as projectiles and as emerging particles. 76. A platinum target is bombarded with cyclotron-accelerated deuterons for several hours and then the element iridium (Z = 77) is separated chemically from it. What radioiso topes of iridium are present and by what reactions are they formed? (Note: *^Pt and '^^Pt, not shown in Fig. 12, are stable platinum isotopes, but their isotopic abundances are so small that we may ignore their presence.) 77. Consider the reaction X(a,b)Y, in which X is taken to be at rest in the laboratory reference frame. The initial kinetic energy in this frame is
(a) Show that the initial velocity of the center of mass of the system in the laboratory frame is \m x ^ m j Is this quantity changed by the reaction? (b) Show that the initial kinetic energy, viewed now in a reference frame at tached to the center of mass of the two particles, is given by ^cn. = ^U b (
)•
Is this quantity changed by the reaction? (c) In the reaction ^Zr(d,p)’'Z r the kinetic energy of the deuteron, measured in the laboratory frame, is 15.9 MeV. Find (= v^), V, and Kcm • Ignore the small relativistic effects. 78. In an endothermic reaction (Q < 0), the interacting parti cles a and X must have a kinetic energy, measured in the center-of-mass reference frame, of at least |Q| if the reaction is to “go.” Show, using the result of Problem 77, that the threshold energy for particle a, measured in the laboratory reference frame, is K ^ = \Q\-
rrix
Is it reasonable that should be greater than \Qp. 79. Prepare an overlay like that of Fig. 13 in which two nucleons or light nuclei may appear as emerging particles. The reac tion ^^Cu(a,pn)^^Zn is an example. Consider the combina tions nn, np, and pd as possibilities. Section 54-9 Nuclear Models 80. A typical kinetic energy for a nucleon in a middle-mass nucleus may be taken as 5 MeV. To what effective nuclear temperature does this correspond, using the assumptions of the liquid drop model of nuclear structure? (Hint: See Eq. 30 in Chapter 24.) 81. An intermediate nucleus in a particular nuclear reaction decays within 1.2 X 10“ ^^ s of its formation, (a) What is the uncertainty A £" in our knowledge of this intermediate state? (b) Can this state be called a compound nucleus? See Sample Problem 11. 82. From the following list of nuclides, identify (a) those with filled nucleon shells, (b) those with one nucleon outside a filled shell, and (c) those with one vacancy in an otherwise filled shell. Nuclides: *^C, '«0, ^K , ^’Ti, ^'Zr, ’^Mo, ‘2'Sb, '^^Nd, '^Sm , ^o^t i , and ^o^Pb. 83. As Table 1 shows, the nuclide '’"Au has a nuclear spin of j. (a) If we regard this nucleus as a spinning rigid sphere with a radius given in Table 1, what rotational frequency results? (b) What rotational kinetic energy? Note that this picture is overly mechanistic. 84. Consider the three formation modes shown for the com pound nucleus ^°Ne* in Fig. 17. What energies must (a) the a particle, (b) the proton, and (c) the gamma-ray photon have to provide 25.00 MeV of excitation energy to the com pound nucleus? Needed atomic masses are 2°Ne
19.992435 u
a
4.002603 u
'^F
18.998403 u
'H
1.007825 u.
“O
15.994915 u
Problems 85. Consider the three decay modes shown for the compound nucleus in Fig. 17. If the compound nucleus is ini tially at rest and has an excitation energy of 25.0 MeV, what kinetic energies, measured in the laboratory, will (a) the deuteron, (b) the neutron, and (c) the ^He nuclide have when the nucleus decays? Needed atomic masses are 20Ne
19.992435 u
d
2.014102 u
>’Ne
19.001879 u
n
1.008665 u
18.000937 u
^He
3.016029 u.
16.999131 u 86. The nuclide is “doubly magic” in that both its proton number Z (= 82) and its neutron number N (= 126) repre sent filled nucleon shells. An additional proton would yield and an additional neutron ^®^Pb. These “extra” nu cleons should be easier to remove than a proton or a neutron from the filled shells of ^®®Pb. (a) Calculate the energy re quired to move the “extra” proton from ^^’Bi and compare it with the energy required to remove a proton from the filled proton shell of ^®®Pb. (b) Calculate the energy required to remove the “extra” neutron from ^^’ Pb and compare it with the energy required to remove a neutron from the filled neutron shell of ^°®Pb. Do your results agree with expecta-
1165
tion? Use these atomic mass data: Nuclide 20»Bi jospb 207-pj 20»pb 207pb
z
N
Atomic Mass (u)
82+1 82 8 2 -1 82 82
126 126 126 126+ 1 1 2 6 -1
208.980374 207.976627 206.977404 208.981065 206.975872
The atomic masses of the proton and the neutron are 1.007825 u and 1.008665 u, respectively. 87. The nucleus ’‘Zr (Z = 40, N = 5\) has a single neutron outside a filled 50-neutron core. Because 50 is a magic num ber, this neutron should perhaps be especially loosely bound, {a) Calculate its binding energy, (b) Calculate the binding energy of the next neutron, which must be extracted from the filled core, (c) Find the binding energy per particle for the entire nucleus. Compare these three numbers and discuss. Needed atomic masses are 9‘Zr 90.905644 u
n
1.008665 u
^Zr
89.904703 u
‘H
1.007825 u.
«’Zr
88.908890 u
CHAPTER 55 ENERGY FROM THE NUCLEUS In a system o f interacting particles, we can extract useful energy when the system moves to a lower energy state (that is, a more tightly bound state). In an atomic system, we can extract this energy through chemical reactions, such as burning. In a nuclear system, we can extract energy in a variety o f ways. For example, the energy released in radioactive decay has been used to provide electrical power to cardiac pacemakers and to space probes. In this chapter, we consider the two primary methods that are used to extract energy from the nucleus and convert it to useful purposes. In nuclear fission, a heavy nucleus is split into two fragments. In nuclear fusion, two light nuclei are combined into a heavier nucleus. Figure 6 o f Chapter 54 showed that either o f these processes can result in more tightly bound nuclei and therefore can release excess nuclear binding energy to be converted into other form s o f energy. Reactors based on nuclear fission today provide a significant share o f the world's electrical power. Research and engineering are actively underway to develop reactors based on nuclear fusion.
55-1 THE ATOM AND THE NUCLEUS______________________ W hen we get energy from coal by b urning it in a furnace, we are tinkering w ith atom s o f carbon an d oxygen, rearranging their o u ter electrons in m ore stable com bina tions. W hen we get energy from u ran iu m by consum ing it in a nuclear reactor, we are tinkering w ith its nucleus, rearranging its nucleons in m ore stable com binations. Electrons are held in atom s by the C oulom b force, and it takes a few electron volts to rem ove one o f the outer electrons. O n the o th er hand, nucleons are held in nuclei by the strong nuclear force, an d it takes a few m illion electron volts to pull one o f them out. T his factor is also reflected in o u r ability to extract ab o u t a m illion tim es m ore energy from a kilogram o f u ran iu m th an from a kilogram o f coal. In both the atom ic an d nuclear cases, the appearance o f energy is accom panied by a decrease in the rest energy o f the fuel. T he only difference betw een consum ing ura niu m and b urning coal is that, in the form er case, a m uch larger fraction o f the available rest energy (again, by a factor o f several m illion) is converted to other form s o f energy.
We m ust be clear about w hether ou r concern is for the quantity o f energy or for the rate at which the energy is delivered, th at is, the power. In the nuclear case will the kilogram o f u ranium b u m slowly in a pow er reactor or explosively in a bom b? In the atom ic case, are we thinking about exploding a stick o f dynam ite or digesting a jelly doughnut? (Surprisingly, the energy release is greater in the second case th an in the first!) Table 1 shows how m uch energy can be extracted from 1 kg o f m atter by doing various things to it. Instead o f reporting the energy directly, we m easure it by showing how long the extracted energy could operate a 100-W light bulb. Row 6, the total m utual annihilation o f m atter and antim atter, is the ultim ate in extracting energy from m at ter. W hen you have used up all the available m ass you can do no m ore. (How ever, no one has yet figured out an econom ical way to produce and store 1 kg o f an tim atter to use for energy production.) K eep in m ind th at the com parisons o f Table 1 are on a per-unit-m ass basis. K ilogram for kilogram we get several m illion tim es m ore energy from u ranium th an we do from coal or from falling water. O n the other hand, there is a lot o f coal in the E arth’s crust and there is a lot o f w ater backed up behind the Bonneville D am on the C olum bia River.
1167
1168
Chapter 55
TA B L E 1
Energy from the Nucleus
E N E R G Y FR O M 1 kg O F M A TTER
Form of Matter
Process
W ater Coal Enriched UO 2 (3%)
A 50-m waterfall Burning Fission in a reactor Complete fission Complete fusion Complete annihilation
235 y
Hot deuterium gas M atter and antim atter
Time‘s 5s 8h 680y 3 X lO'*y 3 X 10"*y 3 X 10^y
° These numbers show how long the energy generated could power a 100-W light bulb.
55-2 NUCLEAR FISSION: THE _______ BASIC PROCESS_______________ In 1932 the English physicist Jam es Chadw ick discovered the neutron. A few years later Enrico Ferm i and his collab orators in R om e discovered that, if various elem ents are bom barded by these new projectiles, new radioactive ele m ents are produced. Ferm i had predicted that the neu tron, being uncharged, w ould be a useful nuclear projec tile; unlike the proton or the a particle, it experiences no repulsive C oulom b force when it approaches a nuclear surface. Because there is no C oulom b barrier for it, the slowest neu tro n can penetrate and interact with even the m ost massive, highly charged nucleus. T herm al neutrons, which are neutrons in equilibrium with m atter at room tem perature, are convenient and effective bom barding particles. At 300 K, the m ean kinetic energy o f such neu trons is K = \ k T = K8.62 X 10-5 eV /K )(300 K) = 0.04 eV. In 1939 the G erm an chem ists O tto H ahn and Fritz Strassm ann, following work initiated by Ferm i and his collaborators, bom barded u ran iu m with therm al neu trons. They found by chem ical analysis that after the b o m b ard m en t a n u m b er o f new radioactive elem ents were present, am ong them one whose chem ical properties were rem arkably sim ilar to those o f barium . R epeated tests finally convinced these chem ists th at this “ new ” ele m ent was not new at all; it really was barium . How could this m iddle-m ass elem ent (Z = 56) be produced by b o m barding uran iu m (Z = 92) with neutrons? T he riddle was solved w ithin a few weeks by the physi
cists Lise M eitner and her nephew O tto Frisch. They showed that a uranium nucleus, having absorbed a neu tron, could split, with the release o f energy, into two roughly equal parts, one o f which m ight well be barium . They nam ed this process n u c le a rfssio n .i Figure 1 shows the tracks left in the gas o f a cloud cham ber by the two energetic fission fragm ents th at result from a fission event occurring near the center o f the cham ber. The fission o f therm al neutrons, a process o f great practical im portance, can be represented by + r+ /? n .
( 1)
in which as the asterisk indicates, is a com pound nucleus. H ere A'and Y stand for fission fragm ents, m iddlemass nuclei that are usually highly radioactive. T he factor b, which has the average value 2.47 for fission events o f this type, is the num ber o f neutrons released in such events. Figure 2 shows the distribution by mass n um ber o f the fission fragm ents X and Y. We see th at in only about 0.01% o f the events will the fragm ents have equal mass. The m ost probable mass num bers, occurring in about 7% o f the events, are A = \40 and A = 95. We can also tell from the difference in the length o f the two fission frag m ent tracks in Fig. 1 that the two fragm ents in this particu lar fission event do not have the sam e mass. The nuclide which is the fissioning nucleus in Eq. 1, has 92 protons and 236 — 92 or 144 neutrons, a
t See “The Discovery of Fission,” by O tto Frisch and John Wheeler, Physics Today, November 1967, p. 43, for a fascinat ing account of the early days of discovery.
Figure 1 When a fast charged particle passes through a cloud chamber, it leaves a track of liquid droplets. The two back-to-back tracks represent fission fragments, produced by a fis sion event that took place in a thin vertical uranium foil in the center of the chamber.
Section 55-3
10
Theory o f Nuclear Fission
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MeV. In the interm ediate range {A = 120, say), it is about 8.5 MeV. T he difference in total binding energy between a single nucleus {A = 240) and two fragm ents (assum ed equal) into which it m ay be split is then Q = 2(8.5 M eV ) - - (7.6 M qW)A
200 M eV.
Sam ple Problem 1 shows a m ore careful calculation, which agrees very well with this rough estim ate.
Sample Problem 1 Calculate the disintegration energy Q for the fission event of Eq. 2, taking into account the decay of the fission fragments. Needed atomic masses are 235U 235.043924 u n
1.008665 u
‘^C e
139.905433 u 93.906315 u.
Solution If we replace the fission fragments in Eq. 2 by their stable end products, we see that the overall transformation is 235|J ^ i40Ce + 94^r + n. Figure 2 The distribution in mass of the fission fragments X and Y (see Eq. 1) from the fission of by thermal neu trons. Note that the vertical scale is logarithmic.
n e u tro n /p ro to n ratio o f ab o u t 1.6. T he prim ary frag m ents form ed im m ediately after fission will retain this sam e n e u tro n /p ro to n ratio. Study o f the stability curve o f Fig. 4 o f C hap ter 54, however, shows th a t stable nuclides in the m iddle-m ass region {15 < A < 150) have a neu tro n /p ro to n ratio o f only 1.2 to 1.4. T he prim ary frag m ents will th u s be excessively neutron-rich an d will “ boil o ff” a sm all n u m b er o f neutrons, 2.47 o f them on the average for the reaction o f Eq. 1. T he fragm ents X and Y th a t rem ain are still too neutron-rich an d approach the stability line by a chain o f successive beta decays. A specific exam ple o f the generalized fission process o f Eq. 1 is 235U + n
236U*
i40Xe 4- ’^Sr 4 2n.
( 2)
T he fission fragm ents *"^Xe an d ’"‘Sr decay until each reaches a stable end product, as follows: '^ X e T14Ts - > * ^ C65s s^ '^ B a
The single neutron comes about because the (initiating) neutron on the left side of Eq. 2 cancels one of two neutrons on the right side of that equation. The mass difference for this reaction is Am = 235.043924 u - (139.905433 u 4 93.906315 u 4 1.008665 u) = 0.223511 u, and the corresponding energy is G = Am
= (0.223511 u)(931.5 MeV/u) = 208.2 MeV,
in good agreement with our previous rough estimate of 2(X) MeV. About 80% of the disintegration energy is in the form of the kinetic energy of the two fragments, the remainder going to the neutron and the radioactive decay products. If the fission event takes place in a bulk solid, most of the disintegration energy appears as an increase in the internal en ergy of the solid, which shows a corresponding rise in tempera ture. Five percent or so of the disintegration energy, however, is associated with neutrinos that are emitted during the beta decay of the primary fission fragments. This energy is carried out of the system and does not contribute to the increase in its internal energy.
'« C e (stable)
^"Zr (stable) »'‘Sr 75 s . 94y 19 min ► T he decays are P~ events, the half-lives being indicated at each stage. As for all beta decays, the m ass num bers (140 an d 94) rem ain unchanged as the decays continue. T he disintegration energy Q for fission is very m uch larger th an for chem ical processes. W e can support this by a rough calculation. F rom the binding energy curve o f Fig. 6 o f C hapter 54, we see th a t for heavy nuclides {A = 240, say) the binding energy per nucleon is ab o u t 7.6
55-3 THEORY OF NUCLEAR FISSION_______________________ Soon after the discovery o f fission, Niels Bohr and Jo h n W heeler developed a theory, based on the analogy be tween a nucleus and a charged liquid drop, th at explained its m ain features. Figure 3 suggests how the fission process proceeds.
1170
Chapter 55
Energy from the Nucleus
Neutron
i nucleus absorbs a thermal neutron
(a)
\
i J |
It forms nucleus, with excess energy; it oscillates violently ( 6)
The motion may produce a neck
(c)
Coulomb forces stretch it out
id)
Fission occurs
(e)
/
The fragments separate; prompt neutrons boil off
(/I
Figure 3 The stages in a fission process, according to the liquid-drop fission model.
When a heavy nucleus such as absorbs a slow neu tron, as in Fig. 3a, that neutron falls into the potential well associated with the strong nuclear forces that act in the nuclear interior. Its potential energy is then transformed into internal excitation energy, as Fig. Zb suggests. The amount of excitation energy that a slow neutron carries into the nucleus that absorbs it is equal to the work required to pull the neutron back out of the nucleus, that is, to the binding energy o f the neutron. In much the same way, the amount of excitation energy delivered to a well when a stone is dropped into it is equal to the work required to pull the stone back out of the well, that is, to the “binding energy” of the stone. In Sample Problem 2 we show that the binding energy £„ of a neutron in is 6.5 MeV. Figure 3c shows that the nucleus, behaving like an energetically oscillating charged liquid drop, will sooner or later develop a short “neck” and will begin to separate into two charged “globs.” If conditions are right, the elec trostatic repulsion between these two globs will force them apart, breaking the neck. The two fragments, each still carrying some residual excitation energy, then fly apart. Fission has occurred. So far this model gives a good qualitative picture of the fission process. It remains to be seen, however, whether it can answer a hard question: “Why are some heavy nu clides and ^^’Pu, say) readily fissionable by slow neutrons but other, equally heavy, nuclides and ^^^Am, say) are not?” Bohr and Wheeler were able to answer this question. Figure 4 shows the potential energy curve for the fission process that they derived from their model. The horizon tal axis displays the distortion parameter r, which is a rough measure o f the extent to which the oscillating nu cleus departs from a spherical shape. Figure Zd suggests how this parameter is defined before fission occurs. When the fragments are far apart, this parameter is simply the distance between their centers. The energy interval between the initial state and the final state o f the fissioning nucleus— that is, the disinte gration energy Q — is displayed in Fig. 4. The central fea ture o f that figure, however, is that the potential energy
curve passes through a maximum at a certain value o f r. There is a potential barrier o f height £b that must be surmounted (or tunneled) before fission can occur. This reminds us of alpha decay (see Fig. 8 of Chapter 54), which also is a process that is inhibited by a potential barrier. We see then that fission will occur only if the absorbed neutron provides an excitation energy E„ great enough to overcome the barrier or to have a reasonable probability of tunneling through it. Table 2 shows a test of fissionability by thermal neu trons applied to four heavy nuclides, chosen from dozens of candidates that might have been considered. For each nuclide both the barrier height £b and the excitation en ergy £„ are given. £b was calculated from the theory of Bohr and Wheeler, and £„ was computed (as in Sample Problem 2) from the known masses. For and ^^’Pu we see that E„> E^. This means that fission by absorbing a thermal neutron is predicted to occur for these nuclides. This is confirmed by noting, in the table, the large measured cross sections (that is, the reaction probabilities) for the process. For the other two nuclides (^“ U and ^“^Am), we have £„ < £b, so that there is not enough energy to surmount the barrier or to tunnel through it effectively. The excited nucleus (Fig. Zb) prefers to get rid o f its excitation energy by emitting a gamma ray instead o f by breaking into two large fragments. Table 2 shows, as we expect, that the cross sections for thermal neutron fission in these cases
Figure 4 The potential energy at various stages in the fission process, showing the disintegration energy Q and the barrier height £b-
Section 55-4
TABLE 2
Nuclear Reactors: The Basic Principles
1171
TEST OF THE nSSIONABILITY OF FOUR NUCLIDES
Target Nuclide 235U 23*U 239pu
Nuclide Being Fissioned 23«U 239U 240pu
Fission Cross Section^
(MeV)
E. (MeV)
(MeV)
(bams)
5.2 5.7 4.8 5.8
6.5 4.8 6.4 5.5
+ 1.3 - 0 .9 + 1.6 - 0 .3
584 2.7 X 10-‘ 742 <0.08
" The cross section is a measure of the probability for a nuclear reaction to occur. The cross section is measured in units of bams, where 1 bam = 10“^* m^.
are exceedingly small. These nuclides can be m ade to fission, however, if they absorb a substantially energetic (rather th an a therm al) neutron. F or for exam ple, the absorbed neutro n m ust have an energy o f at least 1.3 M e V for the fission process to “ go” with reasonable proba bility.
Sample Problem 2 Consider a nucleus in its ground state. How much energy is required to remove a neutron from it, leaving a nucleus behind? The needed atomic masses are 235.043924 u;
n
1.008665 u;
^36^
236.045563 u.
Solution The increase in mass of the system as the neutron is pulled out is Am = 1.008665 u + 235.043924 u - 236.045563 u = 0.007026 u.
(see Eq. 1) raises ju st this possibility ; the neutrons th at are produced can cause fission in nearby nuclei and in this way a chain o f fission events can propagate itself. Such a process is called a chain reaction. It can either be rapid and uncontrolled as in a nuclear bom b or controlled as in a nuclear reactor. Suppose th at we wish to design a nuclear reactor based, as m ost present reactors are, on the fission o f by slow neutrons. T he fuel in such reactors is alm ost always artifi cially “ enriched,” so th at m akes up a few percent o f the uran iu m nuclei rather th an the 0.7% th at occurs in natural uranium ; the rem aining 99.3% o f natural ura nium is which is not fissionable by therm al neu trons. A lthough on the average 2.47 neutrons are pro duced in fission for every therm al neutron consum ed, there are serious difficulties in m aking a chain reaction “go.” H ere are three o f the difficulties, together with their solutions:
This means that an energy equal to = Am c2 = (0.007026 uX931.5 MeV/u) = 6.545 MeV must be expended. This, by definition, is the binding energy of the neutron in the nucleus. When a nucleus absorbs a thermal neutron, 6.545 MeV is the amount of excitation energy that the thermal neutron brings into the nucleus. In effect, the nucleus is formed in an excited state 6.545 MeV above the ground state. The ex cited nucleus can get rid of this energy either by emitting gamma rays (which leaves a nucleus in its ground state) or by fission (see Eq. 1). It turns out that fission is about six times more likely than gamma-ray emission.
55-4 NUCLEAR REACTORS: THE BASIC PRINCIPLES____________ Energy releases per ato m in individual nuclear events such as alpha em ission are roughly a m illion tim es larger th an those o f chem ical events. T o m ake large-scale use o f nuclear energy, we m ust arrange for one nuclear event to trigger an o th er until the process spreads th ro u g hout bulk m atter like a flame through a burn in g log. T he fact that m ore neu tro n s are generated in fission th an are consum ed
1. The neutron leakage problem. A certain percentage o f the neutrons produced will sim ply leak out o f the reactor core and be lost to the chain reaction. If too m any do so, the reactor will not work. Leakage is a surface effect, its m agnitude being proportional to the square o f a typical reactor core dim ension (surface area = Anr'^ for a sphere). N eutron production, however, is a volum e effect, propor tional to the cube o f a typical dim ension (volum e = for a sphere). T he fraction o f neutrons lost by leakage can be m ade as small as we wish by m aking the reactor core large enough, thereby decreasing its surface-to-volum e ratio (= 3 /r for a sphere). 2. The neutron energy problem. Fission produces fa s t neutrons, with kinetic energies o f ab out 2 M eV, but fis sion is induced m ost effectively by slow neutrons. T he fast neutrons can be slowed dow n by m ixing the u ranium fuel with a substance th at has these properties: (a) it is effective in causing neutrons to lose kinetic energy by collisions and (b) it does not absorb neutrons excessively, thereby rem oving them from the fission chain. Such a substance is called a m oderator. M ost pow er reactors in this country are m oderated by water, in which the hydrogen nuclei (protons) are the effective m oderating elem ent. 3. The neutron capture problem. N eutrons m ay be cap tured by nuclei in ways th at do not result in fission. T he
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Chapter 55
Energy from the Nucleus
m ost co m m o n possibility is capture followed by the em is sion o f a gam m a ray. In particular, as the fast (M eV ) n eutrons generated in the fission processes are slowed dow n in the m oderator to therm al equilibrium (0.04 eV), they m ust pass through an energy interval ( 1 - 1 0 0 eV) in which they are particularly susceptible to nonfission cap tu re by T o m inim ize such resonance capture, as it is called, the u ran iu m fuel and the m od erato r (water, say) are not inti m ately m ixed but are “ clu m p ed ,” rem aining in close con tact w ith each other b u t occupying different regions o f the reactor volum e. T he hope is th a t a fast fission neutron, produced in a u ran iu m “ clu m p ” (which m ight be a fuel rod), will w ith high probability find itself in the m oderator as it passes through the “dangerous” resonance energy range. O nce it has reached therm al energies, it will very likely w ander back into a clum p o f fuel an d produce a fission event. T he task for reactor designers is to produce the m ost effective geom etrical arrangem ent o f fuel and m oderator. Figure 5 shows the neutro n balance in a typical power reactor operating with a steady output. Let us trace the behavior o f a sam ple o f 1000 therm al n eutrons in the reactor core. They produce 1330 n eutrons by fission in the fuel and 40 m ore by fast fission in the m aking a total o f 370 new neutrons, all o f them fast. Exactly this sam e n u m b er o f n eutrons is then lost to the chain by leakage from the core and by nonfission capture, leaving 1000 therm al n eutrons to con tin u e the chain. W hat has been gained in this cycle, o f course, is th at each o f the 370 n eutrons produced by fission has deposited Thermal neutron leakage
Thermal captures
Figure 5 A generation of 1000 thermal neutrons is followed through various stages in a reactor. At a steady operating level, the loss of neutrons due to captures (in the fuel, moder ator, and structural elements) and leakage through the surface is exactly balanced by the production of neutrons in the fis sion processes.
about 200 M eV o f energy in the reactor core, heating it up. An im p o rtan t reactor param eter is the m ultiplication factor k, the ratio o f the n u m b er o f neutrons present at the beginning o f a particular generation to the n u m ber present at the beginning o f the next generation. For the situation o f Fig. 5, the m ultiplication factor is exactly 1. For k = 1, the operation o f the reactor is said to be exactly critical, which is w hat we wish it to be for steady pow er production. Reactors are designed so th at they are inher ently supercritical ( k > 1); the m ultiplication factor is then adjusted to critical operation { k = 1) by inserting control rods into the reactor core. These rods, containing a m aterial such as cadm ium th at absorbs neutrons readily, can then be w ithdraw n as needed to com pensate for the tendency o f reactors to go subcritical as (n eu tro n absorbing) fission products build up in the core during continued operation. If you pulled out one o f the control rods, how fast would the reactor pow er level increase? T his response tim e is controlled by the fascinating circum stance th at a small fraction o f the neutrons generated by fission is n o t em itted prom ptly from the newly form ed fission fragm ents but is em itted from these fragm ents later, as they decay by beta em ission. O f the 370 “ new ” neutrons analyzed in Fig. 5, for exam ple, about 16 are delayed, being em itted from fragm ents following beta decays whose half-lives range from 0.2 to 55 s. These delayed neutrons are few in n u m ber b u t they serve the useful purpose o f slowing dow n the reactor response tim e to m atch h u m an reaction times. Figure 6 shows the broad outlines o f an electric pow er plant based on a pressurized-water reactor {P N /R \ a type in com m on use in the U nited States. In such a reactor, w ater is used both as the m oderator and as the heat transfer m edium . In the p rim ary loop, w ater at high tem perature and pressure (possibly 600 K and 150 atm ) cir culates through the reactor vessel and transfers heat from the reactor core to the steam generator, which provides high-pressure steam to operate the tu rb in e th at drives the generator. To com plete the secondary loop, low-pressure steam from the turbine is condensed to w ater an d forced back into the steam generator by a pum p. T o give som e idea o f scale, a typical reactor vessel for a 1000-M W (elec tric) plant m ay be 10 m high and weigh 450 tons. W ater flows through the prim ary loop at a rate o f about 300,000 gal/m in. An unavoidable feature o f reactor operation is the ac cum ulation o f radioactive wastes, including both fission products and heavy “ tran su ran ic” nuclides such as pluto nium and am ericium . O ne m easure o f their radioactivity is the rate at which they release energy in therm al form. Figure 7 shows the variation with tim e o f the therm al pow er generated by such wastes from one year’s operation o f a typical large nuclear plant. N ote th at both scales are logarithm ic. T he total activity o f the waste 10 years after its rem oval from the reactor is about 3 X 1 0 ^ Ci.
Section 55-4
Nuclear Reactors: The Basic Principles
1173
steam (high pressure) Electric power .— / -(__^Generator
Steam (low pressure)
:
Coolant in
i^Coolant out
Water (high pressure)
Primary loop
Water (low pressure)
Secondary loop
Figure 6 A simplified layout of a nuclear power plant based on a pressurized-water reactor.
Sample Problem 3 A large electric generating station is pow ered by a pressurized-water nuclear reactor. The thermal power in the reactor core is 3400 MW, and 1100 MW of electricity is generated. The fuel consists o f86,000 kg of uranium, in the form of 110 tons of uranium oxide, distributed among 57,000 fuel rods. The uranium is enriched to 3.0% (a) What is the plant efficiency? (b) At what rate R do fission events occur in the
reactor core? (c) At what rate is the fuel disappearing? As sume conditions at start-up. (d) At this rate of fuel consumption, how long would the fuel supply last? (e) At what rate is mass being lost in the reactor core? Solution (a) The efficiency e is the ratio between the power output (in the form of electric energy) to the power input (in the form of thermal energy), or electric output _ 11(X) MW ^ ' thermal input 3400 MW
10^
= 0.32 or 32%. 105
As for all power plants, whether based on fossil fuel or nuclear fuel, the efficiency is controlled by the second law of thermody namics. In this plant, 3400 MW - 1100 MW or 2300 MW of power must be discharged as thermal energy to the environment. (b) If P (= 3400 MW) is the thermal power in the core and Q (=200 MeV) is the average energy released per fission event, then, in steady-state operation, P
^
Q
3 . 4 X 1 0 M / S (2 0 0
MeV/fissionXl.6 X lO” *^ J/MeV)
= 1.06 X 10^° fissions/s.
Time (y)
Figure 7 Thermal power released by the radioactive wastes of one year’s operation of a typical large nuclear power plant, as a function of time after the fuel is removed. The curve rep resents the effect of many radionuclides with a range of halflives. Note that both scales are logarithmic.
(c) disappears by fission at the rate calculated in (b). It is also consumed by (nonfission) neutron capture at a rate about one-fourth as large. The total consumption rate is then (1.25)(1.06 X 10^° s“ ‘) or 1.33 X 10^° s“ ‘. We recast this as a mass rate as follows: ( 0.2 ,235 kg/mol \ ^ = (1.33 X 10“ $-') U .0 2 XX 10^^ atom s/m ol/ dt
= 5.19X 10-5 kg/s = 4.5 kg/d.
1174
Chapter 55
Energy from the Nucleus
(d) From the data given, we can calculate that, at start-up, about (0.03X86,000 kg) or 2600 kg of were present. Thus a somewhat simplistic answer would be 7- = ^ = 5 8 0 d . 4.5 kg/d In practice, the fuel rods are replaced (often in batches) before their content is entirely consumed. (e) From Einstein’s E = Am
relation, we can write
dM _dE /dt_ 3 .4 X 1 0 ’ W dt (3.00X 10«m/s)2 = 3.8X 10"«kg/s = 3.3g/d. The mass loss rate is about the mass of one penny every day! This mass loss rate (reduction in rest energy) is quite a different quan tity than the fuel consumption rate (loss of ^^^U) calculated in part (c).
55-5 A NATURAL REACTOR________ O n D ecem ber 2, 1942, w hen the reactor assem bled by Enrico F erm i and his associates first w ent critical, they had every right to expect th at they had p u t into operation the first fission reactor th a t had ever existed on this planet. A bout 30 years later it was discovered th at, if they did in fact th in k that, they were wrong. Som e tw o billion years ago, in a u ran iu m deposit now being m ined in G abon, W est Africa, a natural fission reac to r w ent into operation an d ran for perhaps several hund red th ousand years before shutting itself off. T he story o f this discovery is fascinating at the level o f the best detective thriller. M ore im p o rtan t, it provides a first-class exam ple o f the n ature o f the scientific evidence needed to back up w hat m ay seem at first to be an im prob able claim . It set a high standard for all w ho speculate ab o u t past events. W e consider here only tw o points.* 1. W as there enough fuel? T he fuel for a uranium -based fission reactor m ust be the easily fissionable isotope which constitutes only 0.72% o f natural uranium . This isotopic ratio has been m easured not only for terrestrial sam ples b u t also in M oon rocks and in m eteorites, in w hich the sam e value is always found. T he initial clue to the discovery in G ab o n was th at the u ran iu m from this deposit was deficient in som e sam ples having an ab u n dance as low as 0.44%. Investigation led to the specu lation th at this deficit in could be accounted for if, at som e tim e in the past, this isotope was partially consum ed by the operation o f a natural fission reactor.
* For the complete story, see “A Natural Fission Reactor,” by George A. Cowan, Scientific American, July 1976, p. 36.
T he serious problem rem ains that, with an isotopic abundance o f only 0.72%, a reactor can be assem bled (as Ferm i and his team learned) only with the greatest o f difficulty. T here seems no chance at all th at it could have happened naturally. However, things were different in the distant past. Both and are radioactive, with half-lives o f 0.704 X 10^ y and 4.47 X 10’ y, respectively. T hus the half-life o f the readily fissionable is about 6.4 tim es shorter than th at o f Because decays faster, there m ust have been m ore o f it, relative to in the past. Tw o billion years ago, in fact, this abundance was not 0.72%, as it is now, b u t 3.8%. This abundance happens to be ju st about the abundance to which natural u ranium is artificially enriched to serve as fuel in m odern pow er reactors. W ith this am o u n t o f readily fissionable fuel available in the distant past, the presence o f a natural reactor (provid ing certain other conditions are m et) is m uch less surpris ing. T he fuel was there. Tw o billion years ago, inciden tally, the highest order o f life form s th at had evolved were the blue-green algae. 2. W hat is the evidence?T ht m ere depletion o f in an ore deposit is not enough evidence on which to base a claim for the existence o f a natural fission reactor. M ore convincing pro o f is needed. If there were a reactor, there m ust also be fission prod ucts; see Fig. 2. O f the 30 or so elem ents whose stable isotopes are produced in this way, som e m ust still rem ain. Study o f their isotopic ratios could provide the convincing evidence we need. O f the several elem ents investigated, the case o f neody m ium is spectacularly convincing. Figure 8^ shows the isotopic abundances o f the seven stable neodym ium iso topes as they are norm ally found in nature. Figure Sb shows these abundances as they appear am ong the ulti m ate stable products o f the fission o f T he clear dif ferences are not surprising, considering their totally dif ferent origins. T he isotopes shown in Fig. 8^ were form ed in supernova explosions th at occurred before the form a tion o f o u r solar system. T he isotopes o f Fig. Sb were cooked up in a reactor by totally different processes. N ote particularly th at *^^Nd, the d o m in an t isotope in the n atu ral elem ent, is totally absent from the fission products. T he big question is: “ W hat do the neodym ium isotopes found in the uranium ore body in G abon look like?” W e m ust expect that, if a natural reactor operated there, iso topes from both sources (that is, natural isotopes as well as fission-produced isotopes) m ight be present. Figure 8c shows the results after this and other corrections have been m ade to the raw data. C om parison o f Figs, i b and 8c certainly suggests th at there was indeed a natural fission reactor at work!
Sample Problem 4 The isotopic ratio of to in natural uranium deposits today is 0.0072. What was this ratio 2.0 X
Free ebooks ==> www.Ebook777.com Section 55-6
142
143 144
145
146
148
150
142 143
Mass number, A
144 145 146
Thermonuclear Fusion: The Basic Process
148 150
142 143 144
Mass number, A
ib)
(o )
145
146 148
1175
150
Mass number, A (c)
Figure 8 The distribution by mass number of the isotopes of neodymium as they occur in (a) natural terrestrial deposits, (b) the spent fuel of a power reactor, and (c) the uranium mine in Gabon, West Africa. Note that (b) and (c) are virtually identical and quite different from (a).
10’ y ago? The half-lives of the two isotopes are 0.704 X 10’ y and 4.47 X 10’ y, respectively. Solution Consider two samples that, at a time t in the past, contained and N^(Q) atoms of and respectively. The numbers of atoms remaining at the present time are =
and
N^(t) =
respectively, in which Aj and Ag are the corresponding disinte gration constants. Dividing gives
N^{t)
^ ,(0 )
Expressed in terms of the isotopic ratio R = comes
this be
R{0) = R{t)e^^^-^^. The disintegration constants are related to the half-lives by Eq. 8 of Chapter 54, or As *
In 2 'i/ 2,5
0.693 = 0.984 X lO-’ y - ' 7.04 X 10«y
and , _ In 2 _ 0.693 = 0.155 X 10"’ y -'. Ag “ ■; ^1/2,8 4.47 X 10’ y Substituting in the expression for the isotopic ratio gives R{0) = = (0.0072V '« = 0.0378 or 3.78%. We see that, two billion years ago, the ratio of to in itafural uranium deposits was much higher than it is today. When the Earth was formed (4.5 billion years ago) this ratio was 30%.
55-6 THERMONUCLEAR FUSION: THE BASIC PROCESS_________ We pointed out in connection vv^ith the binding energy curve of Fig. 6 of Chapter 54 that energy can be released if light nuclei are combined to form nuclei of somewhat larger mass number, a process called nuclearfusion. How ever, this process is hindered by the mutual Coulomb repulsion that tends to prevent two such (positively) charged particles from coming within range of each other’s attractive nuclear forces and “fusing.” This re minds us of the potential barrier that inhibits nuclear fission (see Fig. 4) and also of the barrier that inhibits alpha decay (see Fig. 8 of Chapter 54). In the case of alpha decay, two charged particles— the a particle and the residual nucleus— are initially inside their mutual potential barrier. For alpha decay to occur, the a particle must leak through this barrier by the barriertunneling process and appear on the outside. In nuclear fusion the situation is just reversed. Here the two particles must penetrate their mutual barrier from the outside if a nuclear interaction is to occur. The interaction between two deuterons is o f particular importance in fusion. Sample Problem 5 gives a rough calculation of the potential barrier between two deuter ons, which works out to be about 200 keV. The corre sponding barrier for two interacting ^He nuclei (charge = + 2e) is about 1 MeV. For more highly charged particles the barrier, of course, is correspondingly higher. One way to arrange for light nuclei to penetrate their mutual Coulomb barrier is to use one light particle as a target and to accelerate the other by means o f a cyclotron or a similar device. To generate power in a useful way from the fusion process, however, we must have the inter action of matter in bulk, just as in the combustion of coal.
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Chapter 55
Energy from the Nucleus
T he cyclotron technique holds no prom ise in this direc tion. T he best hope for obtaining fusion in bulk m atter in a controlled fashion is to raise the tem perature o f the m aterial so th at the particles have sufficient energy to penetrate the barrier due to th eir therm al m otions alone. This process is called therm onuclear fim o n . T he m ean therm al kinetic energy o f a particle in equilibrium at a tem perature T is given, as we have seen in C hapter 23, by K = \k T ,
(3)
w h e re /c (= 8 .62 X 10"^eV /K ) is the M tz m a n n constant. At room tem perature ( T ~ 300 K), K = 0.04 eV, which is, o f course, far too sm all for o u r purpose. Even at the center o f the Sun, where 1.5 X 10^ K, the m ean therm al kinetic energy calculated from Eq. 3 is only 1.9 keV. This still seem s hopelessly sm all in view o f the m agnitude o f the C oulom b barrier o f 200 keV calcu lated in Sam ple Problem 5. Y et we know th at th erm o n u clear fusion not only occurs in the solar interior b u t is its central an d d o m in an t feature. T he puzzle is solved w ith the realization th at (1) the energy calculated from Eq. 3 is a m ean kinetic energy; particles w ith energies m uch greater th an this m ean value constitute the high-energy “ tails” o f the M axwellian speed distribution curves (see Fig. 10 o f C hapter 24). Also, (2) the barrier heights th a t we have quo ted represent only the p eaks o f the barriers. B arrier tunneling can occur to a significant extent at energies well below these peaks, as we saw in Section 54-4 in the case o f alpha decay. Figure 9 sum m arizes the situation by a q u antitative exam ple. T he curve m arked n{K) in this figure is a M ax well energy distribution curve draw n to correspond to the S u n ’s central tem perature, 1.5 X 10^ K. A lthough the sam e curve holds no m atter w hat particle is u n d er consid eration, we focus o u r atten tio n on protons, bearing in
m ind th at hydrogen form s ab o u t 35% o f the m ass o f the S un’s central core. _ For r = 1.5 X 10^ K, Eq. 3 yields A: = 1.9 keV, and this value is indicated by a vertical line in Fig. 9. N ote that there are m any particles whose energies exceed this m ean value. The curve m arked p{K) in Fig. 9 is the probability o f barrier penetration for two colliding protons. AX K = 6 keV, for exam ple, we have p = 2.4 X 10“ ^. T his is the probability th at two colliding protons, each w ith K = 6 keV, will succeed in penetrating th eir m utual C oulom b barrier and com ing w ithin range o f each o th er’s strong nuclear forces. P ut an other way, on the average, one o f every 42,000 such encounters will succeed. It tu rn s out th at the m ost probable energy for p ro to n proton fusion events to occur at the S un’s central tem pera ture is about 6 keV. If the energy is m uch higher, the barrier is m ore easily penetrated (that is, p is greater), but there are too few protons in the M axwellian “ tail” (n is smaller). If the energy is m uch lower, there are plenty o f protons b u t the barrier is now too form idable.
Sample Problem 5 The deuteron (^H ) has a charge + e, and its radius has been measured to be 2.1 fm. Two such particles are fired at each other with the same initial kinetic energy K. What must A be if the particles are brought to rest by their mutual Coulomb repulsion when the two deuterons are just “touch ing”? Solution Because the two deuterons are momentarily at rest when they “touch” each other, their kinetic energy has all been transformed into electrostatic potential energy associated with the Coulomb repulsion between them. If we treat them as point charges separated by a distance 2A, we have 2K =
1 Q\Qi _ 1 47T60 2A ’ 47160
which yields A=
167r€oA (1.6X 10-‘^C)2 / 1 keV \ (167t)(8.9X 10-'2CVJ*m)(2.1 X IQ-'^m ) \1 .6 X lO -'M /
« 200 keV. This quantity provides a reasonable measure of the height of the Coulomb barrier between two deuterons.
Figure 9 The curve marked n(K) gives the distribution in energy of protons in the core of the Sun, corresponding to a temperature of 1.5 X 10^ K. The vertical line indicates the mean kinetic energy per particle at that temperature. The curve marked p(K) gives the probability of barrier penetration in proton-proton collisions. The two curves are drawn to dif ferent arbitrary vertical scales.
55-7 THERMONUCLEAR FUSION IN STARS______________________ H ere we consider in m ore detail the th erm onuclear fusion processes th at take place in o u r Sun an d in o th er stars. In the S un’s deep interior, w here its m ass is concentrated and
Section 55-7 where m ost o f the energy p roduction takes place, the (central) tem p eratu re is 1.5 X 10^ K an d the central den sity is on the order o f 10^ kg/m^, ab o u t 13 tim es the den sity o f lead. T he central tem p eratu re is so high that, in spite o f the high central pressure (2 X 10“ atm ), the Sun rem ains gaseous throughout. T he present com position o f the S un’s core is about 35% hydrogen by mass, ab o u t 65% helium , an d abo u t 1% other elem ents. At these tem peratures the light elem ents are essentially totally ionized, so th at o u r picture is one o f an assem bly o f protons, electrons, an d a particles in random m otion. T he Sun radiates at the rate o f 3.9 X 10^^ W and has been doing so for as long as the solar system has existed, which is ab o u t 4.5 X 10^ y. It has been know n since the 1930s th at therm o n u clear fusion processes in the S un’s interior account for its prodigious energy o utput. Before analyzing this further, however, let us dispose o f tw o other possibilities th at had been p u t forward earlier. C onsider first chem ical reactions such as sim ple burning. If the Sun, whose m ass is 2.0 X 10^° kg, were m ade o f coal and oxy gen in ju st the right p roportions for burning, it w ould last only about 10^ y, which o f course is far too short (see Problem 47). T he Sun, as we shall see, does not b u m coal b u t hydrogen, and in a nuclear furnace, n o t an atom ic or chem ical one. A nother possibility is that, as the core o f the Sun cools and the pressure there drops, the Sun will shrink u n der the action o f its own strong gravitational forces. By transfer ring gravitational potential energy to internal energy (just as we do w hen we d ro p a stone o nto the E arth ’s surface), the tem perature o f the S u n ’s core will rise so th at radiation m ay continue. C alculation shows, however, th at the Sun could radiate from this cause for only abo u t 10® y, too short by a factor o f 30 (see Problem 51). T he S u n ’s energy is generated by the therm onuclear “ b u rn in g ” (that is, “ fusing” ) o f hydrogen to form helium . Figure 10 shows the p ro to n -p ro to n cycle by m eans o f which this is accom plished. N ote th at each reaction shown is a fusion reaction, in th at one o f the products (^H, ^He, or ^^He) has a higher m ass n u m b er th an any o f the reacting particles th at form it. T he reaction energy Q for each reaction shown in Fig. 10 is positive. T his character izes an exotherm ic reaction, w ith the net p roduction o f energy.
1h + 1h -^H +
e'"
+ »/ (Q= 0.42
e-^ + e" - 7 + 7
2H+^H- ^He +
MeV) e'^ + e
(Q = 1 0 2 MeV)
7
'—
(Q =
5.49
- 7
+
7
^—
3He + ^ H e - ' ‘ H e + * H + ^ H
’ (Q = 12.8 6 MeV)
(Q
n il
T he cycle is initiated by the collision o f tw o protons (^H + *H) to form a deuteron (^H), with the sim ulta neous creation o f a positron (e"^) and a neutrino (v). T he positron very quickly encounters a free electron (e") in the Sun and both particles annihilate, their rest energies ap pearing as two gam m a-ray photons (y), as we discussed in Section 8-7. In Fig. 10 we follow the consequences o f two such events, as indicated in the top row o f the figure. Such events are extrem ely rare. In fact, only once in about 10^^ p ro to n -p ro to n collisions is a deuteron form ed; in the vast m ajority o f cases the colliding protons sim ply scatter from each other. It is the slowness o f this process th at regulates the rate o f energy production and keeps the Sun from exploding. In spite o f this slowness, there are so very m any protons in the huge volum e o f the S un’s core th at deuterium is produced there in this way at the rate o f about 10*^ kg/s! O nce a deuteron has been produced, it quickly (within a few seconds) collides with an o th er proton and form s a ^He nucleus, as the second row o f Fig. 10 shows. Two such ^He nuclei m ay then eventually (w ithin about 10^ y) col lide, form ing an a particle C*He) and two protons, as the third row o f the figure shows. T here are other variations o f the p ro to n -p ro to n cycle, involving other light elem ents, but we concentrate on the principal sequence as repre sented in Fig. 10. T aking an overall view o f the p ro to n -p ro to n cycle, we see th at it am o u n ts to the com bination o f four protons and two electrons to form an a particle, two neutrinos, and six gam m a rays: 4 'H + 2 e - - ^ ^ H e + 2v + 6y.
(4)
Now, in a form al way, let us add two electrons to each side o f Eq. 4, yielding. 4 (‘H + e -) ^ (^He + 2e") + 2v + 6y.
(5)
The quantities in parentheses then represent atom s (not bare nuclei) o f hydrogen and o f helium . T he energy release in the reaction o f Eq. 5 is, using the atom ic masses o f hydrogen and helium , Q = A m c ^ = [4m (‘H ) - m(^HQ)]c^ = [4(1.007825 u) - 4.002603 u](931.5 M eV /u) = 26.7 MeV.
+ */ (Q = 0.42 MeV) (Q = 1.02 MeV)
+^
MeV)
Thermonuclear Fusion in Stars
= 5.49
MeV)
Figure 10 The proton - proton cycle that primarily accounts for energy production in the Sun.
1178
Chapter 55
Energy from the Nucleus
N eutrinos and gam m a-ray p hotons have no m ass and th u s do not en ter into the calculation o f the disintegration energy. T his sam e value o f Q follows (as it m ust) by ad d ing up the Q values for the separate steps o f the p ro to n pro to n cycle in Fig. 10. N ot q uite all this energy is available as internal energy inside the Sun. A bout 0.5 M eV is associated with the two neutrinos th a t are produced in each cycle. N eutrinos are so penetrating th at in essentially all cases they escape from the Sun, carrying this energy w ith them . Som e are inter cepted by the Earth, bringing us o u r only direct inform a tion about the S un’s interior. S ubtracting the neu trin o energy leaves 26.2 M eV per cycle available w ithin the Sun. As we show in Sam ple Problem 6, this corresponds to a “ heat o f co m b u stion” for the nuclear burning ofhydrogen into helium o f 6.3 X 10*^ J/k g o fh y d ro g en consum ed. By com parison, the heat o f com bustion o f coal is abo u t 3.3 X 10^ J/kg, som e 20 m il lion tim es lower, reflecting roughly the general ratio o f energies in nuclear an d chem ical processes. W e m ay ask how long the Sun can co n tin u e to shine at its present rate before all the hydrogen in its core has been converted into helium . H ydrogen b urning has been going on for abo u t 4.5 X 10’ y, and calculations show th a t there is enough available hydrogen left for ab o u t 5 X 10’ y m ore. At th at tim e m ajo r changes will begin to happen. T he S un’s core, w hich by then will be largely helium , will begin to collapse and to heat up while the o u ter envelope will expand greatly, perhaps so far as to encom pass the E arth ’s orbit. T he Sun will becom e w hat astronom ers call a red giant. If the core tem perature heats up to ab o u t 10* K, energy can be produced by burn in g helium to m ake carbon. H e lium does not b u m readily, the only possible reaction being ^He 4- ^He -h "H e
+ y
( 0 = + 7.3 M eV).
Such a three-body collision o f three a particles m ust occur w ithin 10“ *^ s if the reaction is to go. Nevertheless, if the density and tem perature o f the helium core are high enough, carbon will be m anufactured by the b urning o f helium in this way. As a star evolves an d becom es still hotter, other ele m ents can be form ed by o th er fusion reactions. However, elem ents beyond ^ = 56 can n o t be m anufactured by fur th er fusion processes. T he elem ents w ith A = 56 (^^Fe, ^^Co, ^^Ni) lie near the peak o f the binding energy curve o f Fig. 6 o f C hapter 54, an d fusion betw een nuclides beyond this point involves the co nsum ption, an d n o t the produc tion, o f energy. T he production o f the elem ents in fusion processes is discussed in C hapter 56.
Sample Problem 6 At what rate is hydrogen being consumed in the core of the Sun, assuming that all the radiated energy is generated by the proton-proton cycle of Fig. 10?
Solution We have seen that 26.2 MeV appears as internal en ergy in the Sun for every four protons consumed, a rate of 6.6 MeV/proton. We can express this as 6.6 MeV/proton 67 X 10“ ^^ kg/proton
6 X l O - 'M / M e V )
= 6.3X 10‘"J/kg, which tells us that the Sun radiates away 6.3 X 10'" J for every kilogram of protons consumed. The hydrogen consumption rate is then just the output power (=3.9 X 10^^ W) divided by this quantity, or dm _ 3.9 X 1026 W = 6.2X 10" kg/s. dt 6 .3X 10'" J/kg To keep this number in perspective, keep in mind that the Sun’s mass is 2.0 X 10^ kg.
55-8 CONTROLLED THERMONUCLEAR FUSION T herm onuclear reactions have been going on in the uni verse since its creation in the presum ed cosm ic “ big bang” o f som e 15 billion years ago. Such reactions have taken place on Earth, however, only since O ctober 1952, when the first fusion (or hydrogen) bom b was exploded. The high tem peratures needed to initiate the therm onuclear reaction in this case were provided by a fission bom b used as a trigger. A sustained and controllable th erm onuclear power source— a fusion reacto r— is proving m uch m ore diffi cult to achieve. T he goal, however, is being vigorously pursued because m any look to the fusion reactor as the ultim ate pow er source o f the future, at least as far as the generation o f electricity is concerned. T he p ro to n -p ro to n interaction displayed in Fig. 10 is not suitable for use in a terrestrial fusion reactor because the process displayed in the first row is hopelessly slow. T he reaction cross section is in fact so small th at it cannot be m easured in the laboratory. T he reaction succeeds under the conditions th at prevail in stellar interiors only because o f the enorm ous nu m b er o f protons available in the high-density stellar cores. T he m ost attractive reactions for terrestrial use appear to be the d e u te ro n -d e u te ro n (d-d) and the d e u te ro n triton (d-t) reactions: d-d:
2h -h 2H ^ ^He + n
( 0 = + 3 .2 7 M eV ),
(6)
d-d:
2 H - h 2 H ^ ^ H + 'H
( 0 = + 4 .0 3 M eV),
(7)
d-t:
2h -h ^H — "H e + n
( 0 = + 17.59 M eV). (8)
Here triton indicates ^H, the nucleus o f hydrogen with A = 5. N ote th at each o f these reactions is indeed a fusion reaction and has a positive 0 value. D euterium , whose isotopic abundance in norm al hydrogen is 0.015%, is
Section 55-9 available in unlim ited quantities as a co m p o n ent o f sea water. T ritiu m (atom ic ^H) is radioactive and is not nor m ally found in naturally occurring hydrogen. T here are three basic requirem ents for the successful operation o f a therm o n u clear reactor. 1. A high particle density n. T he n u m b er o f interacting particles (deuterons, say) per u n it volum e m ust be great enough to ensure a sufficiently high d e u te ro n -d e u te ro n collision rate. At the high tem peratures required, the deu terium gas w ould be com pletely ionized into a neutral plasm a consisting o f deuterons an d electrons. 2. A high plasm a tem perature T. T he plasm a m ust be hot. O therw ise the colliding deuterons will not be ener getic enough to penetrate the m utual C oulom b barrier th at tends to keep th em apart. In fusion research, tem pera tures are often reported by giving the corresponding value o f k T (not \k T ) . A plasm a tem p eratu re o f 33 keV, corre sponding to 2.8 X 10® K, has been achieved in the labora tory. This is m uch higher th an the S un’s central tem pera ture (1.3 keV, or 1.5 X 10^ K). 3. A long confinem ent tim e r. A m ajor problem is con taining the hot plasm a long enough to ensure th at its density an d tem p eratu re rem ain sufficiently high. It is clear th at no actual solid co n tain er can w ithstand the high tem peratures necessarily involved, so th at special tech niques, to be described later, m ust be em ployed. By use o f one such technique, confinem ent tim es greater th an 1 s have been achieved. It can be shown that, for the successful operation o f a therm o n u clear reactor, it is necessary to have nr > 10^° s • m “ ^
(9 )
a condition called L aw son s criterion. E quation 9 tells us, loosely speaking, th at we have a choice betw een confining a lot o f particles for a relatively short tim e or confining fewer particles for a som ew hat longer tim e. Beyond m eet ing this criterion, it is also necessary th at the plasm a tem perature be sufficiently high. T here are two techniques th at have been used to at tem p t to achieve the com bination o f tem p erature T and Law son’s p aram eter nx th at are necessary to produce fu sion reactions. M agnetic confinem ent uses m agnetic fields to confine the plasm a while its tem perature is increased. In inertial confinem ent, on the oth er hand, a small am o u n t o f fuel is com pressed an d heated so rapidly that fusion occurs before the fuel can expand an d cool. These techniques are discussed in the following tw o sections. D eriving L aw son's C riterion (O ptional) Let us see how Lawson’s criterion comes about. To raise a plasma to a suitably high temperature and to maintain it there against losses, energy must be added to the plasma at a rate per unit volume where the subscript stands for “heating.” The heating may be done by passing an electric current through the
Magnetic Confinement
1179
plasma, by firing a beam of energetic neutral particles into it, or in other ways. The denser the plasma, the greater the heating power required, in direct proportion, or ( 10)
where Q is a suitable constant. If thermonuclear fusion occurs in the plasma, there will be a certain rate of energy generation per unit volume Pf, where the subscript now stands for “fusion.” Pfis proportional to the con finement time T. It is also proportional to where n is the particle density. To see this, suppose that we double the particle density. Not only will a given particle make twice as many colli sions as it wanders through the plasma, but there will be twice as many wandering particles, giving an overall factor of four. Thus Pf = Cf/2^T.
( 11)
To have a net production power, we must have Pr>P^ or, from Eqs. 10 and 11, nr > Ch/Cf, which leads directly to Eq. 9 if the constants Q and Cf are suitably evaluated. The condition in which Pf = P^^ is called breakeven, m
55-9 MAGNETIC CONFINEM ENT Because a plasm a consists o f charged particles, its m otion can be controlled with m agnetic fields. F or exam ple, charged particles spiral ab out the direction o f a uniform m agnetic field. By suitably varying the field strength, it is possible to design a “ m agnetic m irro r” (see Fig. 14 o f C hapter 34) from which particles can be reflected. A n other design m akes use o f the toroidal geom etry, in which the particles spiral aro u n d the axis o f a toroid inside a “doughnut-shaped” vacuum cham ber. T he type o f fusion reactor based on this principle, which was first developed in Russia, is called a tokam ak, which com es from the Russian-language acronym for “ toroidal m agnetic cham ber.” Several large m achines o f this type have been built and tested. In a tokam ak, there are two com ponents to the m ag netic field, as illustrated in Fig. 11. T he toroidal field is the one we usually associate with a toroidal w inding o f wires; Fig. 11 shows one sm all section o f an external coil th at contributes to the toroidal field. Because the toroidal field decreases with increasing radius, it is necessary to add a second field com p o n en t to confine the particles. This poloidal co m ponent o f the field adds to the toroi dal co m ponent to give the total field a helical structure, as illustrated in Fig. 11. T he poloidal field is produced by a current /' in the plasm a itself, which is induced by a set o f windings not illustrated in the figure. This cu rren t also serves to heat the plasm a. A dditional m eans o f heating, such as by firing neutral beam s o f particles into the
1180
Chapter 55
Energy from the Nucleus 100
(1980).
10 Field Lines
(1978) •
Figure 11 The toroidal chamber that forms the basis of the tokamak. Note the plasma, the helical magnetic field B that confines it, and the induced current /' that heats it.
plasm a, are also necessary to achieve the desired plasm a tem perature. Figure 12 shows a w orker in the interior o f the toroidal vacuum cham ber o f the T o kam ak Fusion Test R eactor at the Princeton Plasm a Physics Laboratory. T he interior radius o f the vacuum ch am ber is ab o u t 2 m, and the m ajo r radius o f the toroid is 2.5 m. In designing m agnetic confinem ent devices such as the tokam ak, the goal is to increase both the Lawson confine m en t p aram eter nx an d the tem perature T o f the plasm a. At sufficiently high values o f these param eters, fusion reactions in the plasm a will produce enough energy to equal the energy th at m ust be supplied to heat the plasm a. T his condition is called “ breakeven.” At still higher values o f these param eters, the device will achieve “ igni-
• (1983)
10l6
(1983*
iQlS
1q20
Lawson number n r(s • m"^)
Figure 13 The approach to breakeven and ignition in con trolled fusion reactors, shown as a plot of Lawson number against temperature.
tio n ,” where self-sustaining fusion reactions will occur. Figure 13 illustrates the steady progress tow ard these goals th at has been m ade. D espite the approach to the break even condition, m any form idable engineering problem s rem ain to be solved, and the production o f electric power from fusion is likely m any decades away.
Sample Problem 7 The Tokamak Fusion Test Reactor at Princeton has achieved a confinement time o f400 ms. {a) What must be the density of particles in the plasma if Lawson’s crite rion is to be satisfied? (b) How does this number compare with the particle density of the atoms of an ideal gas at standard conditions? (c) If a next-generation tokamak could achieve igni tion, with a plasma temperature of 10 keV and confinement time of 1 s, what would the particle density of its plasma have to be? Solution
(a) Using Lawson’s criterion (Eq. 9), we must have 1020 s-m -^ = 2.5X 1020 m -^ n=■ 0.40 s
(b) The number density of atoms in an ideal gas at standard conditions is given by where is the Avogadro constant and (= 2.24 X 10” 2 mVniol) is the molar volume of an ideal gas at standard conditions, which gives ,_ N ^ _ "
6.02 X 1Q23 m o l-» = 2.7 X 10^5 m -^ 2.24 X 10-^ mVmol
The particle density of the plasma we found in part (a) is smaller than that of an ideal gas by a factor of about 10^. Figure 12 A worker inside the toroidal chamber of the Tokamak Fusion Test Reactor at Princeton University.
(c) From Fig. 13 we see that the 10-keV temperature line intersects the curve marked “ignition” at a value of the Lawson number of about 1 X 102's* m“ ^. (In making this last estimate.
Section 55-10
Inertial Confinement
1181
bear in mind that the scale is logarithmic.) The necessary particle density is then 1 X 1Q2' s * m -5
n=■
1s
= 1 X 1 Q 2 'm - ^
55-10 INERTIAL CONFINEM ENT A second technique for confining plasm a so th at th erm o nuclear fusion can take place is called inertial confine m ent. In term s o f Law son’s criterion (Eq. 9), it involves w orking with extrem ely high particle densities n for ex trem ely short confinem ent tim es t . These tim es are arranged to be so short th at the fusion episode is over before the particles o f the plasm a have tim e to m ove ap preciably from the positions they occupy at the onset o f fusion. T he interacting particles are confined by their own inertia. L aser fusion, which relies on the inertial-confinem ent principle, is being investigated in laboratories throughout the world. At the Law rence L iverm ore Laboratory, for exam ple, in the NO V A laser fusion project (see Fig. 14) d e u te riu m -tritiu m fuel pellets, each sm aller th an a grain o f sand (see Fig. 15), are to be “ zapped” by 10 synchro nized, high-powered laser pulses, sym m etrically arranged aro u n d the pellet. T he laser pulses are designed to deliver in total som e 200 kJ o f energy to each fuel pellet in less th an a nanosecond. T his is a delivered pow er o f 2 X 10‘^ W d uring the pulse, which is roughly 100 tim es the total installed electric pow er generating capacity o f the world! T he laser pulse energy serves to heat the fuel pellet, ionizing it to a plasm a a n d — it is h o p e d — raising its tem -
Figure 15 The tiny spheres, shown resting on a dime, are d eu teriu m -tritiu m fuel pellets for use in inertial confinement experiments.
perature to around 10® K. As the surface layers o f the pellet evaporate at these high therm al speeds, the reaction force o f the escaping particles com presses the core o f the pellet, increasing its density by a factor o f perhaps 10^. If all these things happened, then conditions would be right for therm onuclear fusion to occur in the core o f the highly com pressed pellet o f plasm a, the fusion reaction being the d-t reaction given in Eq. 8. In an operating therm onuclear reactor o f the laser fu sion type, it is visualized th at fuel pellets w ould be ex ploded, like m iniature hydrogen bom bs, at the rate o f perhaps 1 0 -1 0 0 per second. T he energetic em erging par ticles o f the fusion reaction C*He and n) m ight be absorbed in a “ blanket” consisting o f a m oving stream o f m olten lithium , heating it up. Internal energy w ould then be ex tracted from the lithium stream at an other location and used to generate steam , ju st as in a fission reactor or a fossil-fuel power plant. Lithium w ould be a suitable choice for a heat-transfer m edium because the energetic neutron would, with high probability, deliver up its en ergy to the “ blanket” by the reaction ^Li + n - ^ ^ H e + ^H.
Figure 14 The target cham ber of the NOVA inertial confine m ent fusion facility at the Lawrence Livermore National Lab oratory. The photo shows several of the 10 laser beam tubes.
The two charged particles would readily be brought to rest in the lithium . The tritium produced in the reaction can
1182
Chapter 55
Energy from the Nucleus
be extracted for use as fuel in the reactor. T he feasibility o f laser fusion as the basis o f a fusion reactor has not been conclusively d em onstrated as o f 1991, b u t vigorous re search is continuing.
deuterium atom, and Wj is the mass of a tritium atom. These atomic masses are related to the Avogadro constant ai'd to the corresponding molar masses (M^ and AfJ by md = A/d/^A ^I'd
m^ = M JN ^.
Combining these equations and solving for n lead to Sample Problem 8 Suppose that a fuel pellet in a laser fusion device is made of a liquid deuterium -tritium mixture contain ing equal numbers of deuterium and tritium atoms. The density d (= 200 kg/m^) of the pellet is increased by a factor of 10^ by the action of the laser pulses, {a) How many particles per unit vol ume (either deuterons or tritons) does the pellet contain in its compressed state? (b) According to Lawson’s criterion, for how long must the pellet maintain this particle density if breakeven operation is to take place? Solution pellet,
(a) We can write, for the density d' of the compressed
d '= I0^d= m a ^ - ^ m , j , in which n is the number of particles per unit volume (either deuterons or tritons) in the compressed pellet, m^ is the mass of a
"
2d'N^ A/d + A/, _ (2)(10’ X 200 kg/m»X6.02 X 10^’ m ol"') 2.0 X 10-’ kg/mol + 3.0 X 10“ ^ kg/mol = 4.8 X 10’’ m -^
(b) From Lawson’s criterion (Eq. 9), we have T>
1020s*m-3
1020s*m-3 = 2X 10-‘2s. 4.8 X 10^' m-^
The pellet must remain compressed for at least this long if break even operation is to occur. (It is also necessary for the effective temperature to be suitably high.) A comparison with Sample Problem 7 shows that, unlike tokamak operation, laser fusion seeks to operate in the realm of very high particle densities and correspondingly very short con finement times.
QUESTIONS 1. If it’s so much harder to get a nucleon out of a nucleus than to get an electron out of an atom, why try? 2. Can you say, from examining Table 1, that one source of energy, or of power, is better than another? If not, what other considerations enter? 3. To which of the processes in Table 1 does the relationship E = Am apply? 4. Of the two fission fragment tracks shown in Fig. 1, which fragment has the larger (a) momentum, (b) kinetic energy, (c) speed, (d) mass? 5. In the generalized equation for the fission of by ther mal neutrons, + n ^ A" + T + ^n, do you expect the Q of the reaction to depend on the identity of X and T? 6. Is the fission fragment curve of Fig. 2 necessarily symmetri cal about its central minimum? Explain your answer. 7. In the chain decays of the primary fission fragments (see Eq. 2) why do no P'^ decays occur? 8. The half-life of is 7.0 X 10* y. Discuss the assertion that if it had turned out to be shorter by a factor of 10 or so, there would not be any atomic bombs today. 9. ^^*U is not fissionable by thermal neutrons. What minimum neutron energy do you think would be necessary to induce fission in this nuclide? 10. The half-life for the decay of by alpha emission is 7 X 10* y; by spontaneous fission, acting alone, it would be 3 X 10'^ y. Both are barrier-tunneling processes, as Fig. 8 in Chapter 54 and Fig. 4 in Chapter 55 reveal. Why this enor mous difference in barrier-tunneling probability? 11. Compare fission with alpha decay in as many ways as possi
ble. How can a thermal neutron deliver several million electron-volts of excitation energy to a nucleus that absorbs it, as in Fig. 3fl? The neutron has essentially no energy to start with! 12. The binding energy curve of Fig. 6 in Chapter 54 tells us that any nucleus more massive than 56 can release energy by the fission process. Only very massive nuclides seem to do so, however. Why can’t lead, for example, release energy by the fission process? 13. By bombardment of heavy nuclides in the laboratory it is possible to prepare other heavy nuclides that decay, at least in part, by spontaneous fission. That is, after a certain mean life they spontaneously break up into two major fragments. Can you explain this on the basis of the theory of Bohr and Wheeler? 14. Slow neutrons are more effective than fast ones in inducing fission. Can you make that plausible? (Hint: Consider how the de Broglie wavelength of a neutron might be related to its capture cross section in 15. Compare a nuclear reactor with a coal fire. In what sense does a chain reaction occur in each? What is the energy releasing mechanism in each case? 16. Not all neutrons produced in a reactor are destined to initi ate a fission event. What happens to those that do not? 17. Explain just what is meant by the statement that in a reactor core neutron leakage is a surface effect and neutron produc tion is a volume effect. 18. Explain the purpose of the moderator in a nuclear reactor. Is it possible to design a reactor that does not need a modera
Problems
19.
20.
21.
22.
23. 24.
25.
26.
27.
28.
tor? If so, what are some of the advantages and disadvan tages of such a reactor? Describe how to operate the control rods of a nuclear reactor (a) during initial start-up, (b) to reduce the power level, and (c) on a long-term basis, as fuel is consumed. A reactor is operating at full power with its multiplication factor k adjusted to unity. If the reactor is now adjusted to operate stably at half power, what value must k now assume? Separation of the two isotopes and from natural uranium requires a physical method, such as diffusion, rather than a chemical method. Explain why. A piece of pure (or ^^’Pu) will spontaneously explode if it is larger than a certain “critical size.” A smaller piece will not explode. Explain. What can you say, if anything, about the value of the multi plication factor k in an atomic (fission) bomb? The Earth’s core is thought to be mostly iron because, dur ing the formation of the Earth, heavy elements such as iron would have sunk toward the Earth’s center and lighter ele ments, such as silicon, would have floated upward to form the Earth’s crust. However, iron is far from the heaviest element. Why isn’t the Earth’s core made of uranium? From information given in the text, collect and write down the approximate heights of the Coulomb barriers for (a) the alpha decay of (b) the fission of by thermal neu trons, and (c) the head-on collision of two deuterons. The Sun’s energy is assumed to be generated by nuclear reactions such as the proton - proton cycle. What alternative ways of generating solar energy were proposed in the past, and why were they rejected? Elements up to mass number « 56 are created by thermonu clear fusion in the cores of stars. Why are heavier elements not also created by this process? Do you think that the thermonuclear fusion reaction con
29.
30.
31.
32.
1183
trolled by the two curves plotted in Fig. 9 necessarily has its maximum effectiveness for the energy at which the two curves cross each other? Explain your answer. In Fig. 9, are you surprised that, as judged by the areas u n d ^ the curve marked n(K), the number of p r i d e s with^AT > K is smaller than the number with K < K , where K is the average thermal energy? The uranium nuclides present in the Earth today were origi nally built up and spewed into space during the explosion of stars, so-called supernova events. These explosions, which occurred before the formation of our solar system, represent the collapse of stars under their own gravity. Can you then say that the energy derived from fission was once stored in a gravitational field? Does fission energy then, in this limited sense, have something in common with energy derived from hydroelectric sources? Why does it take so long (~ 10^ y!) for gamma-ray photons generated by nuclear reactions in the Sun’s central core to diffuse to the surface? What kinds of interactions do they have with the protons, a particles, and electrons that make up the core? The primordial matter of the early universe is thought to have been largely hydrogen. Where did all the silicon in the Earth come from? All the gold?
33. Do conditions at the core of the Sun satisfy Lawson’s crite rion for a sustained thermonuclear fusion reaction? Ex plain. 34. To achieve ignition in a tokamak, why do you need a high plasma temperature? A high density of plasma particles? A long confinement time? 35. Which would generate more radioactive waste products, a fission reactor or a fusion reactor? 36. Does Lawson’s criterion hold both for tokamaks and for laser fusion devices?
PROBLEMS Section 55-2 Nuclear Fission: The Basic Process 1. You wish to produce 1.0 GJ of energy. Calculate and com pare {a) the amount of coal needed if you obtain the energy by burning coal and (b) the amount of natural uranium needed if you obtain the energy by fission in a reactor. As sume that the combustion of 1.0 kg of coal releases 2.9 X 10^ J; the fission of 1.0 kg of uranium in a reactor releases 8.2 X lO'M. 2. In the United States, coal commonly contains about 3 parts per million (3 ppm) of fissionable uranium and thorium. Calculate and compare (a) the energy derived from burning 100 kg of coal and (b) the energy that could be derived from the fission of the fissionable impurities that remain in its ashes. Assume that the combustion of 1 kg of coal releases 2.9 X 10^ J; the fission of 1 kg of uranium or thorium in a reactor releases 8.2 X 10*^ J. 3. (a) How many atoms are contained in 1.00 kg of pure ^^^U? (b) How much energy, in joules, is produced by the complete
4. 5.
6.
7.
fissioning of 1.00 kg of ^^^U? Assume Q = 200 MeV. (c) For how many years would this energy light a 100-W lamp? At what rate must nuclei undergo fission by neutrons to generate 2.00 W? Assume that Q = 200 MeV. Verify that, as reported in Table 1, the fission of the in 1.0 kg of UO 2 (enriched so that is 3.0% of the total uranium) could keep a 100-W lamp burning for 680 y. The fission properties of the plutonium isotope ^^’Pu are very similar to those of The average energy released per fission is 180 MeV. How much energy, in joules, is liberated if all the atoms in 1.00 kg of pure ^^’Pu undergo fission? Very occasionally a nucleus, having absorbed a neu tron, breaks up into three fragments. If two of these frag ments are identified chemically as isotopes of chromium and gallium and if no prompt neutrons are involved, what is at least one possibility for the identity of the fragments? Consult a nuclidic chart or table.
1184
Chapter 55
Energy from the Nucleus
8. Show that, in Sample Problem 1, there is no need to take the masses of the electrons emitted during the beta decay of the primary fission fragments explicitly into account. 9. decays by alpha emission with a half-life of 7.04 X 10* y. It also decays (rarely) by spontaneous fission, and if the alpha decay did not occur, its half-life due to this process alone would be 3.50 X 10'^ y. {a) At what rate do spontane ous fission decays occur in 1.(X) g of (b) How many alpha-decay events are there for every spontaneous fission event? Section 55-i Theory o f Nuclear Fission 10. Fill in the following table, which refers to the generalized fission reaction 255u + n —
Y+ bn.
surfaces, {a) Assuming the nuclei to be spherical, calculate the Coulomb potential energy (in MeV) of repulsion be tween the two fragments. (Hint: Use Eq. 1 in Chapter 54 to calculate the radii of the fragments.) (b) Compare this en ergy with the energy released in a typical fission process. In what form will this energy ultimately appear in the labora tory? nucleus undergoes fission and breaks up into two 18. A middle-mass fragments, and ’^Sr. (a) By what percent age does the surface area of the nucleus change during this process? (b) By what percentage does its volume change? (c) By what percentage does its electrostatic potential energy change? The potential energy of a uniformly charged sphere of radius r and charge Q is given by U
5 \4;r€or/
Section 55-4 Nuclear Reactors: The Basic Principles ■^Xe 139|
'"■Cs
loozr ’2Rb
11. Calculate the disintegration energy Q for the spontaneous fission of ^^Cr into two equal fragments. The needed masses are ^^Cr, 51.940509 u; and ^^Mg, 25.982593 u. Discuss your result. Calculate the disintegration energy Q for the fission of 12 ’*Mo into two equal parts. The needed masses are ^*Mo, 97.905406 u; and ^’Sc, 48.950022 u. If Q turns out to be positive, discuss why this process does not occur spontane ously.
13 Calculate the energy released in the fission reaction 235u + n ^ ‘^'Cs + ’2Rb + 3n. Needed atomic masses are 235U
235.043924 u
>^'Cs
140.920006 u
91.919661 u n
1.008665 u.
14. ^^*Np has a barrier energy for fission of 4.2 MeV. To remove a neutron from this nuclide requires an energy expenditure of 5.0 MeV. Is 2^^Np fissionable by thermal neutrons? 15. Consider the fission of by fast neutrons. In one fission event no neutrons were emitted and the final stable end products, after the beta decay of the primary fission frag ments, were ‘^C e and ^R u. (a) How many beta-decay events were there in the two beta-decay chains, considered together? (b) Calculate Q. The relevant atomic masses are 238U 238.050784 u n
'^C e
139.905433 u
^R u
98.905939 u.
1.008665 u
16. In a particular fission event of
by slow neutrons, it happens that no neutron is emitted and that one of the primary fission fragments is *^Ge. (a) What is the other fragment? (b) How is the disintegration energy 170 MeV split between the two fragments? (c) Calculate the ini tial speed of each fragment. according to Eq. 2, 17. Assume that just after the fission of the resulting ‘^X e and ’^Sr nuclei are just touching at their
19. Many fear that helping additional nations develop nuclear power reactor technology will increase the likelihood of nu clear war because reactors can be used not only to produce energy but, as a by-product through neutron capture with inexpensive to make ^^’Pu, which is a “fuel” for nu clear bombs (breeder reactors). What simple series of reac tions involving neutron capture and beta decay would yield this plutonium isotope? 20. A 190-MW fission reactor consumes half its fuel in 3 years. How much did it contain initially? Assume that all the energy generated arises from the fission of and that this nuclide is consumed only by the fission process. See Sample Problem 3. 21. Repeat Problem 20 taking into account nonfission neutron capture by the See Sample Problem 3. 22, (a) A neutron with initial kinetic energy K makes a head-on elastic collision with a resting atom of mass m. Show that the fractional energy loss of the neutron is given by Am^m K " (w + m j 2 ’ in which is the neutron mass, (b) Find A K/K if the resting atom is hydrogen, deuterium, carbon, or lead, (c) If = 1.00 MeV initially, how many such collisions would it take to reduce the neutron energy to thermal values (0.025 eV) if the material is deuterium, a commonly used moderator? (Note: In actual moderators, most collisions are not “headon.”) 23. The neutron generation time t^^ in a reactor is the average time between one fission and the fissions induced by the neutrons emitted in that fission. Suppose that the power output of a reactor at time t = 0 is Pq. Show that the power output a time t later is P (t\ where P(t) = P,k^/^^, where k is the multiplication factor. Note that for constant power output /c = 1. 24. The neutron generation time (see Problem 23) of a particu lar power reactor is 1.3 ms. It is generating energy at the rate of 1200 MW. To perform certain maintenance checks, the power level must be temporarily reduced to 350 MW. It is desired that the transition to the reduced power level take
Problems 2.6 s. To what (constant) value should the multiplication factor be set to effect the transition in the desired time? 25. The neutron generation time (see Problem 23) in a par ticular reactor is 1.0 ms. If the reactor is operating at a power level of 500 MW, about how many free neutrons (neutrons that will subsequently induce a fission) are present in the reactor at any moment? 26. A reactor operates at 400 MW with a neutron generation time of 30 ms. If its power increases for 5.0 min with a multiplication factor of 1.0003, find the power output at the end of the 5.0 min. See Problem 23. 27. The thermal energy generated when radiations from radio nuclides are absorbed in matter can be used as the basis for a small power source for use in satellites, remote weather sta tions, and so on. Such radionuclides are manufactured in abundance in nuclear power reactors and may be separated chemically from the spent fuel. One suitable radionuclide is ^^*Pu(/,/2 = 87.7 y) which is an alpha emitter with 0 = 5.59 MeV. At what rate is thermal energy generated in 1.(X) kg of this material? 28. Among the many fission products that may be extracted chemically from the spent fuel of a nuclear power reactor is ^S r (/,/2 = 29 y). It is produced in typical large reactors at the rate of about 18 kg/y. By its radioactivity it generates thermal energy at the rate of 2.3 W/g. {a) Calculate the effective disintegration energy associated with the decay of a ^°Sr nucleus. (0^^ includes contributions from the decay of the ^®Sr daughter products in its decay chain but not from neutrinos, which escape totally from the sample.) {b) It is desired to construct a power source generating 150 W (elec tric) to use in operating electronic equipment in an under water acoustic beacon. If the source is based on the thermal energy generated by ^S r and if the efficiency of the thermal-electric conversion process is 5.0%, how much ^S r is needed? The atomic mass o f ’^Sr is 89.9 u. 29. In an atomic bomb (A-bomb), energy release is due to the uncontrolled fission of plutonium ^^’Pu (or The mag nitude of the released energy is specified in terms of the mass of TNT required to produce the same energy release (bomb “rating”). One megaton (10^ tons) of TNT produces 2.6 X 10^* MeV of energy, (a) Calculate the rating, in tons of TNT, of an atomic bomb containing 95 kg of ^^’Pu, of which 2.5 kg actually undergoes fission. For plutonium, the aver age 0 is 180 MeV. {b) Why is the other 92.5 kg of ^^’Pu needed if it does not fission? 30. A 66-kiloton A-bomb (see Problem 29) is fueled with pure 4.0% of which actually undergoes fission, (a) How much uranium is in the bomb? (b) How many primary fission fragments are produced? (c) How many neutrons generated in the fissions are released to the environment? (On the average, each fission produces 2.47 neutrons.) 31. One possible method for revealing the presence of concealed nuclear weapons is to detect the neutrons emitted in the spontaneous fission of ^"*®Pu in the warhead. In an actual trial, a neutron detector of area 2.5 m^, carried on a helicop ter, measured a neutron flux of 4.0 s“ ' at a distance of 35 m from a missile warhead. Estimate the mass of ^^Pu in the warhead. The half-life for spontaneous fission in ^^Pu is 1.34 X 10“ y and 2.5 neutrons, on the average, are emitted in each fission.
1185
Section 55-5 A Natural Reactor 32. The natural fission reactor discussed in Section 55-5 is esti mated to have generated 15 gigawatt-years of energy during its lifetime, (a) If the reactor lasted for 2(X),000 y, at what average power level did it operate? (b) How much did it consume during its lifetime? 33. Some uranium samples from the natural reactor site de scribed in Section 55-5 were found to be slightly enriched in rather than depleted. Account for this in terms of neutron absorption by the abundant isotope and the subsequent beta and alpha decay of its products. 34. How far back in time would natural uranium have been a practical reactor fuel, with a ratio of 3.(X)%? See Sample Problem 4. Section 55-6 Thermonuclear Fusion: The Basic Process 35. Calculate the height of the Coulomb barrier for the head-on collision of two protons. The effective radius of a proton may be taken to be 0.80 fm. See Sample Problem 5. 36. The equation of the curve n(K) in Fig. 9 is K/kT
where N is the total density of particles. At the center of the Sun thMemperature is 1.5 X 10^ K and the mean proton energy is 1.9 keV. Find the ratio of the density of protons at 5.0 keV to that at the mean proton energy. 37. Methods other than heating the material have been sug gested for overcoming the Coulomb barrier for fusion. For example, one might consider using particle accelerators. If you were to use two of them to accelerate two beams of deuterons directly toward each other so as to collide “headon,” (a) what voltage would each require to overcome the Coulomb barrier? {b) Would this voltage be difficult to achieve? (c) Why do you suppose this method is not pres ently used? 38. Calculate the Coulomb barrier height for two ^Li nuclei, fired at each other with the same initial kinetic energy K. See Sample Problem 5. (Hint: Use Eq. 1 in Chapter 54 to calcu late the radii of the nuclei.) 39. For how long could the fusion of 1.00 kg of deuterium by the reaction 2H + 2h — 3He -h n
(G = + 3.27 MeV)
keep a 100-W lamp burning? The atomic mass of deuterium is 2.014 u. Section 55-7 Thermonuclear Fusion in Stars 40. We have seen that Q for the overall proton-proton cycle is 26.7 MeV. How can you relate this number to the Q values for the three reactions that make up this cycle, as displayed in Fig. 10? 41. Show that the energy released when three alpha particles fuse to form *^C is 7.27 MeV. The atomic mass or*He is 4.002603 u, and of ‘^C is 12.000000 u. 42. At the central core of the Sun the density is 1.5 X 10^ kg/m^ and the composition is essentially 35% hydrogen by mass and 65% helium, (a) What is the density of protons at the Sun’s core? (b) What is the ratio of this to the density of
1186
43.
44.
45.
46.
Chapter 55
Energy from the Nucleus
particles for an ideal gas at standard conditions of tempera ture and pressure? Calculate and compare the energy in MeV released by {a) the fusion of 1.0 kg of hydrogen deep within the Sun and {b) the fission of 1.0 kg of in a fission reactor. The Sun has a mass of 2.0 X 10^ kg and radiates energy at the rate of 3.9 X 10^^ W. (a) At what rate does the mass of the Sun decrease? (b) What fraction of its original mass has the Sun lost in this way since it began to bum hydrogen, about 4.5 X 10’ y ago? Let us assume that the core of the Sun has one-eighth the Sun’s mass and is compressed within a sphere whose radius is one-fourth of the solar radius. We assume further that the composition of the core is 35% hydrogen by mass and that essentially all of the Sun’s energy is generated there. If the Sun continues to bum hydrogen at the rate calculated in Sample Problem 6, how long will it be before the hydrogen is entirely consumed? The Sun’s mass is 2.0 X 10^ kg. Verify the Rvalues reported for the reactions in Fig. 10. The needed atomic masses are ‘H
1.007825 u
2H 2.014102 u
^He
3.016029 u
^He 4.002603 u
its overall effects to the proton-proton cycle of Fig. 10. (b) Verify that both cycles, as expected, have the same Q. 50. (a) Calculate the rate at which the Sun is generating neu trinos. Assume that energy production is entirely by the proton-proton cycle, (b) At what rate do solar neutrinos impinge on the Earth? 51. The gravitational potential energy of a uniform spherical object of mass M and radius R is U = -3 G M y 5 R , in which G is the gravitational constant, (a) Demonstrate the consistency of this expression with that of Problem 22 in Chapter 54. (b) Use this expression to find the maximum energy that could be released by a spherical object, initially of infinite radius, in shrinking to the present size of the Sun. (c) Assume that during this shrinking, the Sun radiated en ergy at its present rate and calculate the age of the Sun based on the hypothesis that the Sun derives its energy from gravi tational contraction. Section 55-8 Controlled Thermonuclear Fusion 52. Verify the Q values reported in Eqs. 6,7, and 8. The needed masses are ■H
e=^ 0.0005486 u. (Hint: Distinguish carefully between atomic and nuclear masses, and take the positrons properly into account.) 47. Coal bums according to C + O 2 — CO 2. The heat of combustion is 3.3 X 10^ J/kg of atomic carbon consumed, (a) Express this in terms of energy per carbon atom, (b) Express it in terms of energy per kilogram of the initial reactants, carbon and oxygen, (c) Suppose that the Sun (mass = 2.0 X 10^ kg) were made of carbon and oxy gen in combustible proportions and that it continued to radiate energy at its present rate of 3.9 X 10^^ W. How long would it last? 48. After converting all its hydrogen to helium, a particular star is 100% helium in composition. It now proceeds to convert the helium to carbon via the triple-alpha process ^He + ^He + ^H e ^
4-
y;
Q = 1.21 MeV. The mass of the star is 4.6 X 10^^ kg, and it generates energy at the rate of 5.3 X 10^ W. How long will it take to convert all the helium to carbon? 49. In certain stars the carbon cycle is more likely than the proton-proton cycle to be effective in generating energy. This cycle is '^ C + 'H — ‘’N + y, >’N — ’’C + e+ + V,
G, = 1.95 MeV, 0 2 = 1.19 MeV,
’H
'^O - » '*N + e+ + V, '5N + 'H - ^ '^ C + ‘'He,
04 = 7.30 MeV,
^He
3.016029 u
2.014102 u
"He
4.002603 u
3.016049 u
n
1.008665 u.
53. Suppose we had a quantity of N deuterons (^H nuclei). (a) Which of the following procedures for fusing these N nuclei releases more energy, and how much more? (A) N/2 fusion reactions of the type ^H + ^H —►^H + 'H , or (B) N/3 fusion reactions of the type ^H -I- ^H —►^He + n, using N/3 nuclei of ^H that are first made in N/3 reactions of type A. (b) List the ultimate product nuclei resulting from the two procedures and the quantity of each. 54. Ordinary water consists of roughly 0.015% by mass of “heavy water,’’ in which one of the two hydrogens is re placed with deuterium, ^H. How much average fusion power could be obtained if we “burned” all the ^H in 1 liter of water in 1 day through the reaction ^H + ^H —►^He + n + 3.27 MeV? 55. In the deuteron-triton fusion reaction of Eq. 8, how is the reaction energy Q shared between the a particle and the neutron (that is, calculate the kinetic energies and A^„)? Neglect the relatively small kinetic energies of the two com bining particles. 56. Figure 16 shows an idealized representation of a hydrogen bomb. The fusion fuel is lithium deuteride (LiD). The high temperature, particle density, and neutrons to induce fusion are provided by an atomic (fission) bomb “trigger.” The fusion reactions are ^Li + n ^ ^H + ^He + Q
03 = 7.55 MeV, '“N + 'H ^ ” 0 + y,
1.007825 u
and
0 5 = 1.73 MeV,
2H 4- ^H — ^He + n + 17.59 MeV,
06 = 4.97 MeV.
the tritium (^H) produced in the first reaction fusing with the deuterium (D) in the fuel; see Eq. 8. By calculating Q for the first reaction, find the mass of LiD required to produce a
(a) Show that this cycle of reactions is exactly equivalent in
Problems
1187
Section 55-10 Inertial Confinement A-bomb
Figure 16
Problem 56.
fusion yield of 1 megaton of TNT (=2.6 X 10^* MeV). Needed atomic masses are ^Li
6.015121 u 3.016049 u
^He 4.002603 u n
1.008665 u.
57. Assume that a plasma temperature of 1.3 X 10® K is reached in a laser-fusion device, (a) What is the most proba ble speed of a deuteron at this temperature? (b) How far would such a deuteron move in the confinement time calcu lated in Sample Problem 8? 58. The uncompressed radius of the fuel pellet of Sample Prob lem 8 is 20 /im. Suppose that the compressed fuel pellet “bums” with an efficiency of 10%. That is, only 10% of the deuterons and 10% of the tritons participate in the fusion reaction of Eq. 8. (a) How much energy is released in each such microexplosion of a pellet? (b) To how much TNT is each such pellet equivalent? The heat of combustion of TNT is 4.6 MJ/kg. (c) If a fusion reactor is constmcted on the basis of 100 microexplosions per second, what power would be generated? (Note that part of this power must be used to operate the lasers.)
CHAPTER 56 PARTICLE PHYSICS AND COSMOLOGY Research in particle physics is often done at accelerators where a beam o f particles moving at speeds close to the speed o f light (and thus having kinetic energies many times their rest energies) is incident on a target, usually consisting o f protons. In other accelerators, two high-energy particle beams moving in opposite directions m ay be brought together. Collisions o f individual particles cause reactions in which dozens or perhaps hundreds o f new particles are produced. Some o f these particles live for unimaginably short times, often less than 10~^^ s. Nevertheless, physicists can track these particles and study their properties. This is our primary means for learning about the fundamental constituents o f matter. Astrophysicists use a very different method to unlock the secrets o f the universe. From observations with telescopes and detectors that are sensitive to radiations from all parts o f the electromagnetic spectrum, they try to look backward in time to learn about the universe when it was very young, and they also project their conclusions into the future to try to understand the subsequent evolution o f the universe. These investigations are part o f cosmology, the study o f the origin and evolution o f the universe. It m ay seem surprising that we have grouped these two very different studies in a single chapter. As we shall see, measurements by particle physicists can tell us about the structure o f the universe Just after its birth, and conclusions by cosmologists can set limits on the variety o f fundamental particles and the interactions between them. Although they are at opposite ends o f the scale o f observations, particle physics and cosmology go hand-in-hand in providing an understanding o f the structure o f the universe.
56-1 PARTICLE INTERACTIONS There are tens of thousands of chemical compounds of varying degrees o f complexity. Understanding this huge number o f systems would be a hopeless task if it were not for the underlying simplicity of the 109 fundamental units (elements) o f which these compounds are made and the relatively small number of types of bonds through which they can interact. In order to understand chemis try, we need not study the properties o f tens of thousands o f compounds, but only those o f about 100 elements, along with a few basic types of bonds between them. In fact, the task is even simpler. The 109 known ele ments can be classified into groups with similar proper ties: inert gases, halogens, alkali metals, transition metals, rare earths, and so forth. If we understand the properties
of one member of a group, we can infer the properties of the other members of that group. The subatomic world can be understood in a similar way. We know that the 109 different kinds o f atoms are not fundamental units, but instead that they are in turn composed of three different particles: protons, neutrons, and electrons. When we look still further, by smashing particles together at high energy and studying the debris of the collisions (see Fig. 1), we find what appears at first glance to be a complexity approaching that of chemistry: hundreds of different particles are produced. Yet when we look carefully we find that we can classify those particles into a few groups whose members have similar properties. Eventually we find that this classification leads to clues about the underlying substructure that is based again on a small number of truly fundamental particles and a small number of possible interactions among them.
1189
1190
Chapter 56
Particle Physics and Cosmology
Figure 1 (a) The U A 1 detector at the proton-an tip ro to n collider o f the European Orga nization for Nuclear Research (CERN) accelerator near Geneva, Switzerland. Oppositely moving beams of protons and antiprotons are made to collide in the central region o f this detector, which is designed to record the trajectories o f all electromagnetically or strongly interacting particles that leave the reaction region, (b) A com puter reconstruction o f the trajectories leaving the central region after one collision. A magnetic field causes the cur vature of the paths that permits the m om entum of the particles to be determined and helps to identify the particles. Events of this type were responsible for the discovery o f the W and Z particles at CERN in 1983.
The Four Basic Forces All of the known forces in the universe can be grouped into four basic types. In order of increasing strength, these are: gravitation, the weak force, electromagnetism, and the strong force. These forces have important roles not
only in the interactions between particles, but also in the decay of one particle into other particles. 1. The gravitational force. Gravity is of course exceed ingly important in our daily lives, but on the scale of
Section 56-1
fundamental interactions between particles in the sub atomic realm, it is of no importance at all. To give a relative figure, the gravitational force between two pro tons just touching at their surfaces is about 10“^* of the strong force between them. The principal difference be tween gravitation and the other forces is that, on the prac tical scale, gravity is cumulative and infinite in range. For example, your weight is the cumulative effect of the gravi tational force exerted by each atom of the Earth on each atom of your body. 2. The weak force. The weak force is responsible for nuclear beta decay (see Section 54-5) and other similar decay processes involving fundamental particles. It does not play a major role in the binding of nuclei. The weak force between two neighboring protons is about 10“ ^ of the strong force between them, and the range of the weak force is smaller than 1 fm. That is, at separations greater than about 1 fm, the weak force between particles is negli gible. Nevertheless, the weak force is important in under standing the behavior of fundamental particles, and it is critical in understanding the evolution of the universe. 3. The electromagnetic force. Electromagnetism is im portant in the structure and the interactions of the funda mental particles. For example, some particles interact or decay primarily through this mechanism. Electromag netic forces are of infinite range, but shielding generally diminishes their effect for ordinary objects. The proper ties of atoms and molecules are determined by electro magnetic forces, and many common macroscopic forces (such as friction, air resistance, drag, and tension) are ultimately due to the electromagnetic force. The electro magnetic force between neighboring protons is about 10“^of the strong force, but within the nucleus the electro magnetic forces can act cumulatively because there is no shielding. As a result, the electromagnetic force can com pete with the strong force in determining the stability and the structure of nuclei. 4. The strong force. The strong force, which is responsi ble for the binding of nuclei, is the dominant one in the reactions and decays of most of the fundamental particles. However, as we shall see, some particles (such as the elec tron) do not feel this force at all. It has a relatively short range, on the order of 1 fm. The relative strength of a force determines the time scale over which it acts. If we bring two particles close enough together for any of these forces to act, then a
TABLE 1
Particle Interactions
1191
longer time is required for the weak force to cause a decay or reaction than for the strong force. As we shall see, the mean lifetime of a decay process is often a signal o f the type of interaction responsible for the process, with strong forces being at the shortest end of the time scale (often down to 10"^^ s). Table 1 summarizes the four forces and some of their properties. The characteristic time for each force gives a typical range of time intervals observed for systems in which each force acts. Usually this is the typical lifetime of a particle that decays through that force.
Unification o f Forces One of the landmark achievements in the history o f phys ics was the 19th century theory of electromagnetism, based on experiments by Faraday and Oersted showing that magnetic effects could produce electric fields and electrical effects could produce magnetic fields. The previously separate sciences o f electricity and magnetism became linked under the common designation o f electro magnetism. This linking was later shown to be a funda mental part of the special theory of relativity, according to which electric fields and magnetic fields can be trans formed into one another due entirely to the relative mo tion of the observer. In the 20th century, it has been attempted to carry this linking further to include other forces. First it was shown that electromagnetism and the weak force can be under stood as two different aspects of the same force, called the electroweak force. If we study particle interactions at a high enough energy, these two forces behave similarly. It is convenient for us to regard them as separate forces for many of the effects we shall discuss, just as we often find it convenient to speak separately of electric and magnetic forces when we discuss electromagnetic phenomena. The theory of the electroweak force, which was proposed inde pendently in 1967 by Stephen Weinberg and Abdus Salam (and for which they, along with Sheldon Glashow, another originator of the theory, received the 1979 Nobel prize in physics), suggested that, just as the photon is the carrier of the electromagnetic force, there should be heavy particles that carry the weak force, and these new particles should, on an energy scale of 100 GeV (about 100 times the rest energy of the proton), behave similarly to a highenergy photon. In 1983, a research team at the European Center for Nuclear Physics (CERN), led by Carlo Rubbia and using experimental techniques developed by Simon van der Meer, discovered the predicted particles, for
THE FOUR BASIC FORCES
Type
Range
Relative Strength
Characteristic Time
Strong Electromagnetic Weak Gravitational
1 fm 00 a 1 fm 00
1 10-^ I0-’ IO-»*
10-^5 s 10-10-20 s 10-*-10-'’ s Years
1192
Chapter 56
Particle Physics and Cosmology
which they were aw arded the 1984 N obel prize in physics. T he discovery o f these particles provided the evidence for the unification o f the electrom agnetic an d weak interac tions into the electrow eak interaction. N ext it was attem pted to com bine the strong and elec trow eak forces at a new higher level o f unification. T heo ries th at do so are called grand unified theories (G U Ts), and at the present tim e there are m any candidates for G U T s but none has as yet em erged as the correct one. Because the energy at which the forces m erge is im m ense, perhaps lO '^G eV (1 0 " tim es the energy o f the largest particle accelerator yet built o r even contem plated), we can n o t do experim ents to test the G U T s directly. We m ust therefore rely on tests at obtainable energies, where the effects are exceedingly small. O ne prediction o f these theories is th at the proton should not be a stable particle bu t should decay on a tim e scale greater th an 10^‘ years. (C om pare this n u m b er with the age o f the universe, about 10‘° years.) Searches for p roton decay have so far been unsuccessful and have excluded certain o f the G U T s from consideration, but as yet there has been no verification o f any o f the theories. T he final step in the unification w ould be to include gravity in the schem e to create a theory o f everything (TO E). T here is not yet a q u a n tu m theory o f gravity, so it is difficult to anticipate the form th at these theories m ight take, b ut they nevertheless provide challenges for theoreti cal speculation.
Sample Problem 1 Suppose the half-life of the proton were 10^' y. as predicted by certain GUTs. (a) On the average, how long must we observe a liter of water before we would see one of its protons decay? (b) What volume of water would be required to have a proton decay rate of one per day? Solution (a) A liter of water (approximately 1000 g) contains a number of molecules given by (1000 g)(6.02 X 10^^ molecules/mole) ^ ^ -------- -------- — ;------------------------- = 3.3 X 10” molecules. 18 g/mole Each molecule contains 10 protons (2 from the hydrogens and 8 from the oxygen), so that the number of protons in a liter of water \s N = 3.3 X 10” . The decay rate R is given by Eq. 5 of Chapter 54 as /? = /W = — N = 3 . 3 tv2 10” y = 2.3X 10-5 y -'
X 10“
1
43,000 y ■ That is, on the average we must wait for 43,000 years before a proton decay occurs in a liter of water. (Z>) If /? = 1 d“ ‘, we obtain A
1 d "' = 5.3 X 10^^ protons 0.693/(10^'y)
Figure 2 An underground chamber, lined with plastic, in the Morton salt mine near Cleveland. Its size can be inferred from the worker standing in the corner. This chamber was later filled with 10,000 tons of water in which were suspended 2048 detectors that respond to the tiny flashes of light that would be emitted in the decay of one of the protons in the water. or 5.3 X 10^^ molecules of water. This works out to be 1.6 X 10^ L, equivalent to a cube of water measuring 25 m on a side! (See Fig. 2.)
56-2 FAM ILIES OF PARTICLES We can learn a lot about things by classifying them . This is a technique used com m only by biologists; by grouping plants or anim als into categories based on certain obvious features o f their structure, a basis can be found for study ing their behavior. From the scientific standpoint, for ex am ple, it m ay be m ore enlightening to com pare one spider with an other spider th an with a fly or a m oth. Part o f the training o f a scientist is concerned with learning how to m ake and to use these classifications. The earliest classification schem e for particles was based on their masses. The lightest particles, including the electron {m^c^ = 0.511 M eV), were called leptons (from the G reek word for “ sm all” ). The heaviest particles, in cluding the proton {rripC^ = 938 M eV), were called baryons (from the G reek word for “ heavy” ). In between were particles, including the pion (m„c^ = 140 M eV), called m esons (from the G reek w ord for “ m iddle” ). T oday these classifications based on m ass are no longer valid; for ex-
S e c tio n 5 6 - 2
TABLE 2
Family
T H R E E F A M IL IE S O F P A R T IC L E S
Structure
Interactions
Spin
Examples
Leptons Fundamental Weak, electromagnetic Mesons Composite Weak, electromagnetic, strong Baryons_______ Composite__________ Weak, electromagnetic, strong
Half integral Integral Half integral
am ple, one lepton an d m any m esons are m ore massive th an the proton. However, we retain these three nam es as descriptive o f particles with sim ilar properties, even though the classification based only on m ass is no longer valid. Table 2 sum m arizes these three fam ilies o f particles an d som e o f their properties.
e“ + Ve +
Leptons T he leptons are fundam ental particles th at interact only through the weak and electrom agnetic interactions; even though the strong force can exceed the weak or electro m agnetic force in strength by m any orders o f m agnitude, the leptons do not feel this force at all. T he leptons are true fundam ental particles; they have no internal structure an d are not com posed o f o th er still sm aller particles. We can consider the leptons to be point particles with no finite dim ensions. All know n leptons have a spin o f i. Table 3 shows the six leptons, which appear as three pairs o f particles. Each pair includes a charged particle (e", //", T“ ) and an uncharged neutrin o (Vg, v^, v j . We discussed the electron n eu trin o previously in connection with beta decay (Section 54-5). Both the charged leptons and the neutrinos have antiparticles. A ccording to som e theories o f the structure o f funda m ental particles, the n eutrinos are massless (and corre spondingly travel at the speed o f light) an d stable. O ther theories predict th at the neutrinos should have a sm all but definitely nonzero m ass and should transform into one another. So far no experim ent has revealed a m ass incon sistent w ith zero, b u t only for the electron neu trino is the upper lim it very sm all (rest energy < 20 eV). T he neu trinos an d th eir possible masses have im p o rtan t im plica tions for cosm ology, as we discuss later in this chapter. T he electron is a stable particle, b u t the m u o n and tau decay to o th er leptons, according to
TABLE 3
1193
F a m i li e s o f P a r tic le s
T~ ^
p r
+
e,v
;r,K p,n
(m ean life = 2.2 X 10“ ^ s), (m ean life = 3.0 X 10"'^ s).
These decays are caused by the weak interaction, as we can conclude from the presence o f neutrinos (which always indicates a weak interaction process) am ong the decay products and as we infer from the typical decay lifetimes listed in Table 1. T he form o f these decays can be understood based on a conservation law for leptons dis cussed in Section 56-3.
Mesons M esons are strongly interacting particles having integral spin. A partial list o f som e m esons is given in Table 4. G enerally, m esons are produced in reactions by the strong interaction; they decay, usually to other m esons or lep tons, through the strong, electrom agnetic, or weak inter actions. For exam ple, pions can be produced in reactions o f nucleons, such as p + n —►P + P + tT
or
p + n —►p + n + Ti®,
and the pions can decay according to 7T —
\
(m ean life = 2.6 X 10"® s), (m ean life = 8.4 X 10"*^ s),
where the first decay occurs due to the weak interaction (indicated by the neutrinos and suggested by the m ean life) an d the second due to the electrom agnetic interaction (indicated by the photons an d suggested by the m ean life).*
* While neutrinos always indicate a weak-interaction decay, not all weak-interaction decays produce neutrinos. The same is true for photons in electromagnetic decays.
T H E L E P T O N FA M ILY
Particle
Antiparticle
Particle Charge {e)
e" Ve
e+ V.
-1 0
i
0.511 < 20 eV
Mean Life (s) 00 00
-1 0
i i
105.7 <0.3
2.2 X 10-* 00
e- +
-1 0
i i
1784 < 40
3.0 X 10-'’ 00
H~ + v^ +
IF T“ Vt
\ X* Vt
Spin {h/2n)
Rest Energy (MeV)
Typical Decay Products __ — V,
+
— V,
1194
C h a p te r 5 6
TABLE 4
Particle
P a r tic le P h y s ic s a n d C o s m o l o g y
SOM E SELEC TED M ESONS
Antiparticle
Charge^ {e)
Spin ih/2n)
-hi
0 0 0 0 0
0
TfO
-hi
0 0
n p~
n rj'
Y
1
0
0 0
-hi
DW BY
W
-hi
0
1
-hi
0
0
1
Strangeness^
0 0
Rest Energy (MeV) 140 135 494 498 549 769 958 1869 3097 5278 9460
-hi -hi
0 0 0 0 0 0 0
Mean Life (s) 2.4 X io-« 8.4 X 10-'^ 1.2 X 10-® 0.9 X 10-10 8.0 X 10-*’ 4.5 X 10-24 2.2 X 10-2* 1.1 X 10-*2 1.0 X 10-20 1.2 X i o - ‘2 1.3 X 10-20
Typical Decay Products y-hy tC
-\-n~ y-hy iC + TlP f t tC IT -h e“ D" + + tC e'^-he“
" The charge and strangeness are those of the particle. Values for the antiparticle have the opposite sign. The spin, rest energy, and mean life are the same for a particle and its antiparticle.
Baryons
Field Particles and Exchange Forces
Baryons are strongly interacting particles having half integral spins (i, 4, . . . ). A partial listing o f some baryons is given in Table 5. The familiar members o f the baryon family are the proton and neutron. Baryons have distinct antiparticles, for example, the antiproton (p) and antineutron (n). We can produce heavier baryons in reactions between nucleons, such as
There is one additional small family o f particles that can not be classified among the leptons, mesons, or baryons. These are thefield panicles, those responsible for carrying the forces with which the particles interact. Newton’s law of gravitation and Coulomb’s law o f elec trostatics were originally based on the concept o f actionat-a-distance. Later, in the nineteenth century, this con cept was replaced by the notion of a field. Two particles interact through the fields that they establish; one particle sets up a field and the other interacts with that field, rather than directly with the first particle. Quantum field theory takes this notion one step further by supposing that the fields are carried by quanta. In this view, instead o f the first particle setting up the field, we say that it emits quanta of the field. The second particle then absorbs these quanta. For example, the electromagnetic interaction be tween two particles can be viewed in terms of the emission and absorption of photons, which are quanta o f the elec-
p -l- p —^p + A® -l- K"'', which produces the A° baryon and the K"^ meson. The A“ decays according to A° ^ p -I- Tt”
(mean life = 2.6 X 10“* s).
Although there are no neutrinos produced in the decay, the mean life indicates that the decay is governed by the weak interaction. We shall learn the reason for this “slow” decay in Section 56-3.
TABLE 5
SOME SELECTED BARYONS
Particle
Antiparticle
p n A® 2+ 1° 1E®
p n A® 2+ 2® 2BO s?— A* 2* B* fi-
A* 2* "S* 5-
Charge" (e) -hi 0 0 -hi 0 -1 0 -1 -h2 ,-h i, 0 , - 1 -hi, 0 , - 1 - 1 ,0 -1
Spin {h/2it) h i i i i i \ i i i
Strangeness^
Rest Energy (MeV)
0 0 -1 -1 -1 -1 -2 -2 0 -1 -2 -3
938 940 1116 1189 1192 1197 1315 1321 1232 1385 1530 1672
Mean Life (s) 00 889 2.6 X 10-'® 0.8 X 10-'® 5.8 X 10-“ 1.5 X 10-'® 2.9 X 10-'® 1.6 X 10-'® 6 X 10-^^ 2 X 10-« 6 X 10-“ 8.2X 10-"
Typical Decay Products p -h e" -h p -h ;r P -h 71® A®-hy n -h 7T A®-h;r® A®-h7TP -h 7T A® -h 7T E -h 7T A®-hK-
“The charge and strangeness are those of the particle. Values for the antiparticle have the opposite sign. The spin, rest energy, and mean life are the same for a particle and its antiparticle.
Section 56-3 TABLE 6
1195
T H E H E L D P A R T IC L E S
Particle
Symbol
Graviton Weak boson Weak boson Photon Gluon
w+, wZ® y
g
Interaction Gravitation Weak Weak Electromagnetic Strong (color)
Charge (e)
0
A£ =
I n Ar *
Rest Energy (GeV) 80.6 91.2
0 0 0
0 0
( 1)
W e can n o t know the energy o f a system m ore precisely th an this A E unless we m easure for a tim e longer th an Ar. If we observe only for a very short tim e, the u ncertainty in the rest energy o f a p ro to n can be at least as large as the rest energy o f a pion, as the following sam ple problem d em o n strates.
Sample Problem 2 (a) What is the longest interval of time for which we can observe a proton for its rest energy to be uncertain by the pion rest energy? (b) What is the greatest distance the pion can travel in that time? Solution (a) For the proton’s rest energy to be uncertain by an amount AE = the observation time interval can, accord ing to Eq. 1, be at most 2nAE
Spin (h/2n)
0
±1
trom agnetic field. Each type o f field has its characteristic field particles. A list o f the particles associated w ith the four basic forces can be found in Table 6. A force accom plished through the exchange o f particles is called an exchange force. F o r exam ple, the force be tw een tw o nucleons in a nucleus takes place through the exchange o f pions. In this case the pions, along w ith other m esons, can act as field particles associated with the strong force betw een nucleons. H ow is it possible for a particle, such as a proton, to em it an o th er particle w ith nonzero m ass an d still rem ain a proton? T his process seem s to violate conservation o f energy. T he solution to this d ilem m a lies in the energytim e form o f the u ncertainty relationships. T he uncer tainty principle is a fundam ental lim itation on o u r ability to m easure a system. T h at is, if we observe a system for a tim e interval A/, there is a corresponding u ncertainty A £ in its energy, according to Eq. 7 o f C hapter 50, given at m in im u m by
^t =
Conservation Laws
Inmj^c^ 4.14X lO -'^eV -s = 4.7 X 10-2^ s. (271X140 MeV)
In a time interval shorter than 4.7 X 10“ ^^ s, a proton can emit and absorb a pion without our observing a violation of conserva tion of energy.
{b) If the pion travels at nearly the speed of light, the maxi mum distance d it can travel in this time interval is d = c ^ t = (3.00 X 10« m/sK4.7 X 1 0 " s) = 1.4 X 1 0 -'5 m = 1.4 fm. This distance defines the range of the nuclear force. Two nu cleons closer than about 1.4 fm can interact through the ex change of pions. If the nucleons are separated by a greater dis tance, pion exchange cannot operate, and there is no nuclear force.
56-3 CONSERVATION LAWS________ We w ould have a difficult tim e analyzing physical pro cesses w ithout the laws o f conservation o f energy and linear and angular m om entum . These conservation laws help us understand why certain outcom es occur (such as in the case o f the collisions th at we considered in C hapter 10). They also help us understand why certain processes (those th at violate the conservation laws) are never ob served. In one sense they are em pirical laws, deduced from observing physical processes and carefully tested in the laboratory. In an o th er sense they reveal to us funda m ental aspects o f the laws o f nature. An exam ple o f a conservation law is the conservation o f electric charge. By observing the outcom es o f m any pro cesses, we are led to propose this law: the net am o u n t o f electric charge m ust not change in any process. Equiva lently, we m ay say th at the net charge before a particular reaction or decay m ust equal the net charge after the reac tion or decay. N o violation o f this law has ever been ob served, even though it has been carefully tested (see Sec tion 27-6).
Conservation of Lepton Number In reactions and decays o f fundam ental particles, we often find a certain set o f outcom es but fail to observe a set o f related outcom es th at w ould otherwise be expected to occur. W hen this happens, we suspect th at there is som e unknow n conservation law at work th at perm its the first set and forbids the second. F or exam ple, we can produce an electron neutrino w hen a proton captures an electron: e“ -h p ^ n -h Ve.
1196
Chapter 56
Particle Physics and Cosmology
W e always find neutrinos in this process, b u t we never observe antineutrinos. F u rtherm ore, the reaction always produces electron n eutrinos an d never m u o n o r tau neu trinos. W e acco u n t for the failure to observe certain processes by proposing a conservation law for lepton num ber that is sim ilar to the conservation law for electric charge. To each lepton we assign a lepton n u m b er -h 1 and to each antilepton we assign a lepton n u m b er — 1. All other particles have lepton n um bers o f 0. T he law o f conserva tion o f lepton n u m b er then states: In a ny process, the lepton num ber fo r electron-type leptons, m uon-type leptons, a n d tau-type leptons m ust each rem ain constant. As far as we know, the law o f lepton conservation is strictly valid; despite precise experim ental searches for violations, none has yet been found. In the electron capture process, we assign an electron lepton n u m b er o f + 1 to the electron an d to the elec tro n neutrino, while L^ = 0 for the p ro to n an d neutron. T his process then has Lg = + 1 on both sides an d upholds the law o f conservation o f lepton num ber. If an electron ^2A2/m eu trin o were produced, the right side w ould have Le = — 1, an d the law w ould be violated. T his accounts for o u r failure to observe this process. If a n o th er type o f neutrino, for exam ple, a m u o n n eutrino, were produced, the process w ould have Le = + 1 on the left an d Le = 0 on the right. F urtherm ore, it w ould have = 0 on the left an d L^ = + \ on the right. T he process w ould therefore violate conservation o f both electron an d m u on lepton num bers, an d it has never been observed. T hrough the law o f lepton conservation, we can ac co u n t for m any experim ental observations. Like other conservation laws, this law proves to be o f great value in analyzing decays an d reactions.
Sample Problem 3 Analyze the decay of the m uon from the standpoint of conservation of lepton number.
Conservation of Baryon Number A sim ilar conservation law occurs in the case o f baryons. T o each baryon, such as the proton or neutron, we assign a baryon nu m b er 5 o f -f 1, and we assign B = — \ to antibaryons such as the antiproton. T he law o f conserva tion o f baryon nu m b er then states: In any process, the total baryon num ber m u st rem ain constant. N o violation o f this law has yet been observed. (How ever, certain speculative theories, the G U T s discussed in Sec tion 56-1, suggest th at the proton can decay into nonbaryons, which w ould violate the law o f conservation o f bar yon num ber. T his decay has never been observed; if it were observed, the law o f conservation o f baryon n u m ber w ould need to be changed accordingly.) C onsider, for exam ple, the reaction in which an tip ro tons are produced w hen a proton beam is incident on a target o f protons: p + p ^ p -h p -h p -h p B:
+1 + 1 +1 + 1 + 1 - 1
In this reaction, the net baryon n u m b er is + 2 on both the left and the right sides. C ontrary to the case o f lepton num ber, there is only one type o f baryon num ber. T he law o f conservation o f bar yon n u m b er is a m ore general version o f the rule we used in analyzing nuclear processes in C hapters 54 and 55; there we kept the total o f neutrons plus protons constant in all decays an d reactions, which, because neutrons and protons are baryons, is equivalent to conserving the total nu m b er o f baryons. Even though there are conservation laws for two types o f particles (leptons and baryons), there is no conserva tion law for m esons. F or exam ple, in a reaction o f protons on protons, any nu m b er o f m esons can be produced (as long as the incident particles have enough kinetic energy): p + p —►p + n + TT'^, p + p —►p + p + 7r‘‘‘ + TT",
Solution Let us assign lepton numbers to each particle in the decay as follows: e + V. + ll~ L.:
0
+ 1
L-
+ 1
0
-1 0
0
P + p —^ p + n + TT'*'+ 71^ + 71®. N ote the conservation o f electric charge in these pro cesses.
+ 1
Note that electron-type leptons are assigned L^ = 0 and muontype leptons are assigned = 0. We see that L^ = 0 and = + 1 both before and after the decay, so the process is allowed by conservation of lepton number. Because of this conservation law, we can understand why there m ust be an electron antineu trino and a m uon neutrino am ong the decay products, rather than, for example, an electron neutrino and a m uon antineu trino.
Strangeness T here are still other processes th at are difficult to un d er stand based only on the conservation laws we have dis cussed so far. F or exam ple, consider the group o f kaons (K m esons), w hich in m any respects are sim ilar to the pions. Because there is no conservation law for m esons, we m ight expect th at any n u m b er o f kaons can be pro duced in reactions. W hat we instead find is th a t kaons are
Section 56-4
either produced in pairs, for example, p - h p ^ p + p-h K'*’ + K“, p+ p—p+ n+
+
or if a single kaon is produced, it is always accompanied by another “strange” particle, for example, a A®, p + p ^ p + AO + We account for these processes (and the failure to observe others that appear to be permitted by the previously known conservation laws) by assigning to particles a new quantum number called strangeness, which is found to follow a new conservation law, called conservation of strangeness. Two kaons (K"^ and KP) are assigned to have strangeness 5 = + 1, and the other two (K~ and K®) are assigned 5 = — 1. All nonstrange particles (such as p, n, and e) have 5 = 0. The reactions in which two kaons are produced then have 5 = 0 on the left (only nonstrange particles) and also 5 = 0 on the right. The A° baryon is assigned 5 = — 1, so the reaction in which A° -h is produced also has 5 = 0 on both sides. When we analyze the decays of the strange particles, the conservation of strangeness appears to break down. The kaons can decay into two (nonstrange) pions, for exam ple.
1197
We can summarize these results in the law o f conserva tion of strangeness:
In processes governed by the strong or electromag netic interactions, the total strangeness must remain constant. In processes governed by the weak interac tion, the total strangeness either remains constant or changes by one unit.
Sample Problem 4 The baryon has 5 = —3. (a) It is desired to produce the Or using a beam of K“ incident on protons. What other particles are produced in this reaction? (b) How might the D"decay? Solution (a) Reactions usually proceed only through the strong interaction, which conserves strangeness. We consider the reaction K- + p — n - + ? On the left side, we have 5 = —1, B = + 1, and 0 = 0. On the right side, we have 5 = —3, B = + 1, and Q = — \. We must therefore add to the right side particles with 5 = + 2, B = 0, and Q = -\-\. Scanning through the tables of mesons and baryons, we find that we can satisfy these criteria with and K°, so the reaction is K- + p — O" -h
Here we have 5 = + 1 on the left and 5 = 0 on the right, a clear violation of the conservation of strangeness. We get a clue about how to resolve this difficulty when we mea sure the lifetime for this decay, which turns out to be about 10“®s. The kaons and pions are strongly interact ing particles, and we would expect this decay to occur with a typical strong interaction lifetime in the range of 10“^^ s (see Table 1). Instead, it is slowed by 15 orders of magni tude! What could be responsible for slowing this decay? Another clue comes from the more likely decay mode o f the K-^:
The Quark Model
+ K°.
{b) The Qr cannot decay by the strong interaction, because no 5 = —3 final states are available. It must therefore decay to particles having 5 = —2 through the weak interaction, which can change 5 by one unit. One of the product particles must be a baryon in order to conserve baryon number. Two possibilities are 0 --^ A ° + K-
and
O” — H° + 7r.
56-4 THE QUARK MODEL__________ a weak-interaction process, for which the mean life of 10“* s would not be unusual. It appears that the weak interaction can change strangeness by one unit. In either o f these kaon decay modes, we have 5 changing by one unit. Even though it does not produce the neutrinos that usually characterize a weak-interaction process, the decay K“^—► + TT®is governed by the weak interaction. In this case, the strangeness violation is a clue that it cannot be a strong-interaction process (strangeness is conserved in all strong interactions), and it must therefore be a weakinteraction decay. Does the electromagnetic interaction conserve strange ness? To answer this question, we look for strangenessviolating electromagnetic decays, such as A° n + y. This decay apparently does not occur, and so we conclude that the electromagnetic interaction conserves strange
ness.
Decays and reactions involving mesons and baryons are subject to conservation laws involving two quantities: the electric charge Q and the strangeness S. It then makes sense to ask whether there is any connection between the electric charge and the strangeness o f a particle. In a partic ular group of similar particles (the spin-0 mesons or the spin-i baryons, for example), do we find all possible com binations of Q and S or only certain ones? Finding only a restricted set o f combinations suggests that the particles are built according to a set of rules out of more fundamen tal units whose electric charge and strangeness have certain values. To answer this question, we make a plot showing elec tric charge on one axis and strangeness on another. We locate particles on this grid according to their values o f electric charge and strangeness. Figure 3 shows this kind
1198
Chapter 56
TABLE 7
Particle Physics and Cosmology
PROPERTIES OF THREE QUARKS
Quark
Symbol
Antiquark
Up Down Strange
u d s
a
Charge" {e)
Spin (h/2n)
Baryon Number^
Strangeness^
-i -i
i i i
+i +i +i
0 0 -1
u s
‘ The values for charge, baryon number, and strangeness refer to the quarks. Values for the antiquarks have opposite signs.
Q = - l
Q =0
Q= + l S= + l
s= o
S=-l
Q=-l
Q= 0
Q= + l S = 0
S = - l
S = - 2
Q=-l
Q= 0
Q= + l
Q=+2
o f plot for the spin-0 mesons, the spin-i baryons, and the spin-j baryons. The regularity of these patterns suggests that these particles are composed out of more basic units. In 1964, it was realized independently by Murray GellMann and George Zweig that these regular patterns could be explained if it were assumed that the baryons and mesons are composed of three fundamental units, which soon became known as quarks. Table 7 shows the proper ties of the three quarks, which are called up (u), down (d), and strange (s). According to this model, the mesons are composed o f a quark and an antiquark, while the baryons are^omposed of three quarks. Consider the combination ud o f an up quark and an antidown quark, such that their two spins add to 0. The charge o f the up quark (in units o f e) is -l-f, while the charge of the antidown quark is -1-^ (the charge o f an antiparticle is opposite to that of the particle). The combination ud has 0 = -t-1, 5 = 0 (because both quarks have S = 0), and B = 0 (because the quark has 5 = and the antiquark has B = —^). This combination has the same quantum numbers as the tc*' meson. Continuing in this way, we find nine possible combinations o f a quark and an antiquark, which are listed in Table 8. These nine combinations exactly reproduce the electric charge and strangeness combinations of the spin-0 mesons, as indi cated by Fig. 4a. Baryons are composed of three quarks, the simplest combination that gives B = -I- 1. There are ten different
TABLE 8
Figure 3 A chart showing (a) the spin-0 mesons, {b) the spini baryons, and (c) the spin-| baryons. Each particle is located on a grid according to its strangeness S and electric charge Q. The grid lines for electric charge have been drawn obliquely so that the patterns appear more symmetric.
QUARK-ANTIQUARK COMBINATIONS
Combination
Charge {e)
uu ud us du dd ds su sd ss
0 -hi -hi -1 0 0 -1 0 0
Spin (h/2n)
Baryon Number
Strangeness
0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1 0,1
0 0 0 0 0 0 0 0 0
0 0 -hi 0 0 -hi -1 -1 0
Section 56-4
combinations that can be made from three quarks, as listed in Table 9. Plotting the allowed spin-i and spin-^ combinations, we obtain Figs. 46 and 4c. The similarities between Figs. 3 and 4 are remarkable. Based on only three quarks, we are able to account for the Q, S, and B quantum numbers of all these particles. How ever, the quark model does far more than produce these
Q=-l
Q=0
Q=+l s= +i
s=o
S = - l
Q=-l
Q=o
Q =+\
s=o
S=-l
S=-2
Q = -l
Q=0
Q= + \
Q=+2
Figure 4 A chart showing (a) the spin-0 combinations of a quark and an antiquark, (6) the spin-J combinations of three quarks, and (c) the spin-| combinations of three quarks. Com pare with Fig. 3.
The Quark Model
1199
simple geometrical patterns. You should think of these patterns as ways of organizing particles with similar prop erties, just as the periodic table allows us to organize atoms with similar properties. Underlying the periodic table is atomic theory, which can be used to calculate properties of atoms beyond their geometrical arrange ments. In a similar way, the quark model allows us to calculate properties of particles, including masses, mag netic dipole moments, decay modes, lifetimes, and reac tion products. The agreement between the measured and calculated properties has been a spectacular success for the model. In fact, all known particles (hundreds o f them!) have been accounted for based on this model, with a few additional quarks that we describe later. The most unusual aspect of the quark model is the fractional electric charges of the quarks. All particles yet discovered have electric charges that can be expressed as integral multiples of the basic unit of charge e. No particle with a fractional electric charge has ever been seen. In fact, no one has ever seen a free quark, despite heroic experi mental efforts to search for one. It is possible that our particle accelerators do not yet have enough energy to produce a free quark. It has also been suggested that free quarks may be forbidden to exist, so that we may only observe quarks bound in mesons and baryons. Even though free quarks have never been seen, individ ual bound quarks have been observed. Scattering experi ments that probe deep inside the nucleon have revealed three pointlike objects that appear to have a spin o f i and a charge of -f ^ or —J. These experiments give direct proof of the existence o f quarklike particles within the nucleon.
The Force Between Quarks What holds the quarks together inside a meson or a nu cleon? This force is the most fundamental version o f the strong force, brought about through the exchange of par ticles called gluons. Just as the electromagnetic force be tween charged particles can be regarded as an exchange of photons, the strong force between quarks is accomplished through the exchange of gluons. We therefore picture a nucleon as composed of three quarks mutually exchang ing gluons. It is possible, through indirect means, to meas ure the fraction of the momentum of the internal struc ture of a nucleon that is due to the quarks. This fraction turns out to be only around 50%. The rest must be due to the exchanged gluons. The resulting picture o f the nu cleon is of three quarks “swimming in a sea” of exchanged gluons. The force between quarks has two unusual properties. (1) It takes a large (perhaps infinite) energy to separate two quarks to a distance greater than the size of a nucleon or a meson (about 1 fm). This may be the reason that no free quarks have yet been seen. When we try to pump energy into a nucleon to separate one of its quarks, the energy
1200
Chapter 56
TABLE 9 Combination uuu uud udd uus uss uds ddd dds dss sss
Particle Physics and Cosmology
C O M B IN A T IO N S O F T H R E E Q U A R K S Charge {e) +2 + 1
0 + 1
0 0 -1
-1 -1 -1
Spin (h/2n) U U U U U U
Baryon Number + 1 -hi -hi -hi -hi -hi -hi -hi -hi -hi
actually creates a q u a rk -a n tiq u a rk pair. T he antiquark com bines w ith one o f the quarks to form a m eson, which agrees w ith o u r observations: w hen we sm ash tw o n u cleons together at high energies, we get o u r nucleons (or oth er baryons) back, plus som e additional m esons. The m ore energy we p u t in, the m ore m esons we get out, but no free q u ark emerges. (2) Paradoxically, inside the n u cleon or the m eson, the quarks appear to m ove freely. At very short distances (less th an the size o f a nucleon), the force betw een quarks approaches zero. T his unusual behavior o f quarks an d gluons can be understood by com parison w ith electrom agnetism . Two charged particles interact w ith one a n o th er through the exchange o f photons. However, the p h o to n itself carries no electric charge, an d so the interaction between the charged particle an d the exchanged p h o to n does not re sult in the exchange o f additional photons. A quark, on the other hand, can em it a gluon an d interact with it. This force betw een the q u ark and the gluon can create addi tional gluons. W hen it interacts w ith an o th er electron, an electron can em it a photon an d still rem ain an electron. It does not sacrifice its “ electricness” (that is, its electric charge) to em it the photon. A quark, however, gives its em itted gluon a share o f its “ strongness,” w hich physicists call “ color.” In the interaction o f quarks, color plays the sam e role as electric charge in the interaction o f charged particles. A p h oton carries no electric charge, b ut a gluon carries color, and in doing so it changes the residual color left behind in the q u ark th at em itted the gluon. In effect, the q u ark is spreading its color over a sphere the size o f a nucleon (the range o f the gluons), an d as a result the interaction betw een quarks is considerably w eakened at these distances. Particle physicists have chosen am using an d whim sical nam es to describe the fundam ental particles an d their properties. N am es such as quark, strangeness, gluon, or color have m eaning only as labels. G luons do provide the “glue” th a t binds quarks together, b u t it has no sim ilarity to any other “glue” in o u r experience. T he “ color” carried by quarks an d gluons has nothing to do w ith o u r ordinary use o f color. It is sim ply easier to rem em ber and discuss these properties if we give th em fam iliar nam es.
Strangeness
0 0 0 -1
-2 -1
0 -1
-2 -3
More Quarks In sim ultaneous 1974 experim ents at the B rookhaven N ational L aboratory in New Y ork and the Stanford L in ear A ccelerator C enter in California, investigators discov ered an unusual m eson with a rest energy ab o u t three tim es th at o f the proton. T his new m eson, called ip (psi), was expected to decay into lighter m esons in a strong interaction tim e o f perhaps 10^^^ s. Instead, it was ob served to decay in a tim e o f about 10“^®s, which is m ore characteristic o f the electrom agnetic interaction (see Table 1). M oreover, its decay products were not m esons b u t an electron and a positron, an o th er signal o f an elec trom agnetic process. W hy is the rapid, strong-interaction decay path blocked for this particle, slowing its decay by three orders o f m ag nitude? We discussed a sim ilar effect in the case o f strange ness, a new q u an tu m nu m b er th at was introduced partly to explain certain slow decays. W e accounted for those decays through a violation o f the conservation o f strange ness. In a sim ilar fashion, we assum e th at the decay o f y/ is slowed by the violation o f an o th er conservation law, called charm. A ccording to this interpretation, the y/ m eson is com posed o f a new quark c (for charm ) and its an tiquark c. T he c quark has an electric charge o f + j. Just as the strange quark is assigned a strangeness q u an tu m nu m b er o f 5 = — 1, the charm ed q u ark is assigned a charm o f C = -M . The decay o f the y/ m eson is slowed, because the c quark m ust decay into o ther quarks (u, d, or s), all o f which have C = 0. The decay thus involves a violation o f the conservation o f charm and therefore can not occur through the strong interaction, which conserves charm . In 1977 a sim ilar discovery was m ade at the Ferm i N ational A ccelerator L aboratory near Chicago. Again, a heavy m eson (in this case, ten tim es the proton rest en ergy) was discovered, which was expected to decay to other m esons in a tim e characteristic o f the strong inter action, b u t instead it decayed into e ' + e"^ in about 10“ ^° s. In this case, the decay was again slowed by the violation o f yet an o th er conservation rule, involving yet
Section 56-5 an o th er new quark, called b (for botto m ) and having an electric charge o f — This new m eson, called Y (upsilon), is assum ed to be com posed o f the com b in atio n b h If we assign to the b q u ark a new q u a n tu m n u m b er th at repre sents bottom ness, th en the decay is slowed because the b quark m ust change in to lighter quarks th a t lack this prop erty; this violation o f the conservation o f bottom ness is responsible for slowing the decay.
The Big Bang Cosmology
1201
plexity? It is possible to suppose th at bigger an d bigger accelerators will reveal new generations o f ever m ore massive leptons and quarks, and the only lim it on their nu m b er m ay appear to be im posed by the am o u n t o f energy we have available. T o answ er this question, we tu rn to discoveries at the opposite end o f the scale from accelerator laboratories: we look to the earliest m om ents after the birth o f the universe, and we shall see th at the previous list o f quarks and leptons m ay be com plete.
A New Symmetry O rdinary m atter is com posed o f pro to n s and neutrons, which are in tu rn m ade up only o f u an d d quarks. O rdi nary m atter is also com posed o f electrons, an d in the conversion o f p rotons to n eu tro n s o r n eutrons to protons in the beta decay o f ordinary m atter, we find electron-type neutrinos em itted along w ith the positron or electron. W e can therefore construct o u r entire w orld and all the ph en o m en a we com m only observe o u t o f tw o pairs o f fu ndam ental particles: u an d d quarks, and e“ and leptons. W ithin each pair, the charges differ by one unit an d —i; — 1 an d 0). If we do experim ents at a som ew hat higher energy, we find new types o f particles: a new pair o f leptons (//“ and its n eu trin o v^) an d a new pair o f q uarks (c an d s). O nce again, w ithin each pair the electric charges differ by one unit. At still higher energy, we find a new p air o f leptons ( t an d ) an d a new q u ark (b). It is assum ed th at the b quark has a partner, called t (for top), and if the t q u ark has a charge o f + this latest pair o f quarks will be sim ilar to the oth er pairs. Searches for the t quark have been m ade by looking for new m esons up to ab o u t 30 tim es the p ro to n ’s rest energy, b u t as yet no evidence for this q u ark has been found. Nevertheless, physicists are sure o f its existence an d confident it will be found if enough energy is avail able. It therefore seem s th at the truly fundam ental particles, the quarks and leptons, appear in pairs, an d th a t a pair o f quarks and a pair o f leptons can be com bined into a “ gen eratio n ,” as follows: 1st generation:
an d
2nd generation:
an d
3rd generation:
an d
C) (:) C)
Properties o f these six quarks an d leptons are sum m arized in A ppendix F. It probably now occurs to you th at we m ay be headed in the sam e direction all over again. T h at is, m ight we som e day have hundreds o f “ fu n d am en tal” quarks an d leptons, so th at instead o f sim plicity we have a new layer o f com
Sample Problem 5 quark content:
Analyze these processes in terms of their
{a) p - > n + e-" + V e, {b) — A^ + K(c) K" + p ^ + K-" + K° Solution (a) Using Figs. 3 and 4 to find the quark content of each of the particles, we can rewrite the decay as uud
udd + e"^ + v^.
Canceling the common pair of ud quarks from each side, we find u —►d + e"^ +
Ve
.
The u quark changes to a d quark by beta decay. (b) The quark content is sss —►uds + su. Canceling the common pair of s quarks from each side, we find the net process to be s
u + d + u.
That is, the s quark is transformed into a d quark, and a uupair is created from the decay energy. (c) Again replacing the particles by their quark content, we can write the reaction as su + uud —►sss + us + ds, and removing the common quarks of u, d, and s from each side we are left with uu ^ ss + si. The net process consists of the annihilation of the uupair and the production of two si pairs from the reaction energy. These examples are typical of quark processes: the weak inter action can change one type of quark into another. The strong interaction can create or destroy quark-antiquark pairs, but it cannot change one type of quark into another.
56-5 THE BIG BANG COSMOLOGY Since the beginnings o f recorded history, h u m an beings have speculated about the origin and future o f the u n i verse, a branch o f science now called cosmology. U ntil the 20th century, these speculations were done m ostly by phi-
1202
Chapter 56
Particle Physics and Cosmology By eventually resolving individual stars in the nebulae, H ubble was able to show th at they are galaxies ju st like o u r M ilky Way, com posed o f hundreds o f billions o f stars. M ore startlingly, H ubble deduced th at the galaxies are m oving away from one an o th er and from us, an d th at the greater their distance from us, the greater is their reces sional speed. T h at is, if d is the distance o f the galaxy from Earth (or from any other point o f reference in the u n i verse) and V is the speed with which the galaxy appears to be m oving away from us, H ubble’s law gives v = H d,
Figure 5 Edwin Hubble (1889-1953) at the controls of the 100-in. telescope on Mount Wilson, where he did much of the research that led him to propose that the universe is expanding.
losophers an d theologians, because there was no experi m ental evidence o f any sort th at w ould form the basis o f any scientific theory. In this century, tw o m ajor experi m ental discoveries have pointed the way to a coherent theory th at is now accepted by nearly all physicists.
The Expansion of the Universe T he first o f the two great discoveries was m ade by astron om er Edwin H ubble (see Fig. 5) in the 1920s. H ubble was studying the wispy objects know n previously as nebulae.
( 2)
where / / is a proportionality constant know n as the H ub ble parameter. The H ubble param eter has the dim ensions o f inverse tim e. Its value can be learned only by experim ent: we m ust independently deduce the distance o f a galaxy from Earth and its speed relative to Earth. T he recessional speeds can be m easured in a straightforw ard way using the D oppler shift o f the light from the galaxy (see Fig. 6 o f C hapter 42), but the distance scale is difficult to deter m ine (in fact, H ubble’s early estim ates were off by a factor o f 10). Nevertheless, today we have a consistent set o f data (Fig. 6) th at confirm s H ubble’s law and gives a value o f the H ubble param eter o f about H=61
km /s M pc ’
where the M pc (m egaparsec) is a com m only used unit o f distance on the cosm ic scale: 1 M pc = 10^ pc = 3.26 X 10^ light-years = 3.084 X 10'’ km. Because o f uncertainties in the estim ates o f the cosm ic
Figure 6 The relationship between speed and distance for groups and clusters of galaxies. The straight lines show the Hub ble relationships for various values of the Hubble parameter H.
S e c tio n 5 6 -5
T h e B ig B a n g C o s m o l o g y
1203
scale o f distance, the H ubble param eter is uncertain, with possible values in the range o f 5 0 -1 0 0 (km /s)/M pc. If the universe has been expanding forever at the sam e rate, then / / “ * is the age o f the universe. U sing the ac cepted value o f the H ubble param eter, we w ould estim ate the age o f the universe as 15 X 10’ y, w ith the range o f un certainty o f H p erm itting values in the range o f 10 - 1 9 X 10’ y. However, as we shall see later, the expan sion o f the universe has n o t been constant, so the true age is less th an the currently deduced value o f H~K
The Cosmic Microwave Background Radiation A lthough there were oth er explanations o f the expansion o f the universe, the one th at gained favor was based on the assum ption that, if the galaxies are presently rushing apart, they m ust have been closer together in the distant past. If we run the cosm ic clock back far enough, we find th at in its early state the universe consisted o f u nim agin ably high densities o f m atter and radiation. As the uni verse expanded, both the m atter and the radiation cooled; you can th in k o f the w avelengths o f the rad ian t photons being stretched in the expansion. T he radiation filled the entire universe in its com pact state, and it continues to fill the entire universe in the expansion. W e should still find th a t radiation present today, cooled to the extent th at its m ost intense co m p o n en t is in the m icrow ave region o f the electrom agnetic spectrum . T his is know n as the cosm ic m icrowave background radiation. T his radiation was discovered in 1965 by A m o Penzias an d R obert W ilson o f the Bell L aboratories in New Jersey, w ho were testing a m icrow ave an ten n a used for satellite com m u n icatio n s (see Fig. 7). N o m atter where they pointed th eir anten n a, they found the sam e annoying background “ hiss.” Eventually they realized th at they were indeed seeing a rem n an t o f the early universe, and they were aw arded the 1978 N obel prize in physics for their discovery. T he m icrow ave background radiation has a true ther m al spectrum o f the type we discussed in Sections 49-1 an d 49-2. Figure 8 shows m easurem ents o f the intensity o f the background radiation at various w avelengths, and you can see how well it is fit by P lanck’s radiation law with a tem p eratu re o f 2.735 K. T he data points include recent m easurem ents m ade from a satellite in E arth orbit, thereby elim inating atm ospheric absorption. M easurem ents o f the intensity o f the m icrow ave back ground radiation in various directions show th at the radia tion has a uniform intensity in all directions; it does not appear to com e from any p articular source in the sky, but instead fills the entire universe uniform ly, as w ould be expected for radiation th at likewise filled the early uni verse. R ecent observations, however, show th at there are tem perature fluctuations o f ab o u t 10“ ^ K betw een differ en t regions o f the sky. These results have been interpreted
Figure 7 Arno Penzias (right) and Robert Wilson, standing in front of the large horn antenna with which they first de tected the microwave background radiation.
as evidence for the nonuniform distribution o f m atter in the early universe th at led ultim ately to the condensation o f stars and galaxies. T he energy density o f the radiation can be found from P lanck’s radiation law (Eq. 6 o f C hapter 49). T he n um ber density o f these background photons is about 400 per cm^, and the energy density is about 0.25 eV/cm^ (roughly corresponding to half the rest energy o f an elec tron perm ^). T he m ean energy per photon is about 0.00063 eV, which suggests why we are not ordinarily aware o f the presence o f these photons.
The Big Bang Cosmology T he cosm ological theory th at is in best agreem ent with these two experim ental findings (the H ubble law and the background radiation) is the B ig Bang cosmology. Ac cording to this theory, the universe began som e 1 0 -2 0 billion years ago in a state o f extrem e density an d tem pera ture. T here were no galaxies or even clum ped m atter as we now know it; the “ stufT’ o f the universe at early tim es was a great variety o f particles and antiparticles, plus radia tion. T he density o f radiation and m atter is related to the tem perature o f the universe. As the universe expands, it cools (just as an expanding therm odynam ic system cools). If we m ake som e reasonable assum ptions about the expansion rate, we can find a relationship between the tem perature and the tim e after the form ation o f the u n i verse: 1.5 X 10‘° s */2.K T = , 1/2 w here the tem perature T is in K and the tim e l is in seconds.
1204
Chapter 56
Particle Physics and Cosmology Figure 8 The spectrum of the cosmic microwave background radiation. The dots represent observations, and the solid line rep resents the Planck spectrum for the radiant energy corresponding to a temperature of 2.735 K. Note the excellent agreement be tween the data points and the theoretical curve. The data be tween 0.05 cm and 1.0 cm come from observations made by the CORE (COsmic Background Explorer) satellite launched in 1989.
0.01
0.1
1
10
100
Wavelength (cm)
T he radiation in the early universe consisted o f highenergy photons, whose typical energy can be roughly esti m ated as k T , where k is the B oltzm ann constan t an d T is the tem perature at a particular tim e t, d eterm ined from Eq. 3. T he d o m in an t processes in the early universe can be represented as p hotons ^ particle + antiparticle. T h at is, photo n s can engage in p a ir production and pro duce a p a rtic le -an tip artic le pair, for exam ple, an electron an d a positron or a p ro to n an d an antip ro to n . Conversely, a particle an d its antiparticle can annihilate into photons. In each case, the total energy o f the p h o to n s m ust be at least as large as the rest energy o f the particle and the antiparticle. O u r goal in describing the early universe is to un d er stand the form ation o f ordinary m atter from the particles an d radiation produced in the Big Bang. Since ordinary m atter is com posed o f nucleons, let us consider the for m ation an d annihilation o f pro to n s an d neutrons: y -I- y ±5 p -f p
an d
y + y ^ n + n,
w here we represent the p hotons as g am m a rays. F or pho tons to produce n u c le o n -a n tin u c le o n pairs, the photon energy k T m ust be at least as large as the rest energy mc^ o f a nucleon (940 M eV). T he m in im u m tem p erature o f the
universe th at will perm it production o f nucleons an d a n tinucleons is mc^ _
940 M eV = 1.1 X 10*3 K. 8.62 X 10-5 eV /K
A ccording to Eq. 3, the universe cooled below this tem per ature at the tim e _ / l . 5 X 10*0 s */2. k V _ / 1.5 X 10*®s */2.k V
V
T
/ V
1.1 X 10*3 K
/
= 2 X 10-^ s. T h at is, at tim es earlier th an 2 //s, the universe was hot enough for the photons to produce n u c le o n -a n tin u c leo n pairs, b u t after 2 /zs the photons were (on the average) not energetic enough to produce n u c le o n -a n tin u c leo n pairs. At earlier tim es (corresponding to higher tem pera tures), the radiation m ay have been able to create q u a rk an tiquark pairs. W e can regard the universe at these earli est tim es as consisting only o f fundam ental particles (quarks and leptons) and radiation. T he quarks an d an ti quarks com bined to form m esons an d baryons, which were disassociated by the radiation as rapidly as they could form . As the universe expanded and cooled, the radiation eventually becam e too feeble to blast ap art the m esons and baryons. Because the details o f the quark m odel (and the existence o f free quarks) are n o t yet con
Section 56-5 firm ed, we will begin the story at a tim e o f ab o u t 10“ ^ s, w hen we can regard the universe as being com posed o f protons, neutrons, antiprotons, antineutrons, mesons, leptons, antileptons, an d photons. T he rates o f produc tio n an d disassociation are roughly equal, so th at the n u m bers o f particles an d th eir corresponding antiparticles are roughly equal. F urtherm ore, the n u m b er o f photons is roughly equal to the n u m b er o f protons, w hich is in tu rn roughly equal to the n u m b er o f electrons. Before this tim e, the strong interaction played a p ro m in en t role in determ ining the structure an d com position o f the uni verse, through such processes as the co m binations o f quarks an d antiquarks into baryons o r m esons or colli sions o f baryons to form m esons or new baryons. After ab o u t 10"^ s (corresponding to T = 1. 5X 10*^ K or k T = 1300 M eV), the particles and radiation have too little energy to induce these reactions, an d the era o f the strong interaction ends at ab o u t this tim e. Electrom agnetic and w eak-interaction processes con tin u e to take place. Electrom agnetic processes are repre sented by the p roduction o f particles an d antiparticles (for exam ple, electrons an d positrons) by photons, while weak interactions occur through such processes as n + Vg ^ p + e“
an d
p -f Ve ^ n + e"^
an d sim ilar processes, in w hich neutrinos are being cre ated and destroyed at the sam e rate. As long as the leptons an d n eutrinos have enough energy, the forw ard and re verse reaction rates are equal, which m aintains the bal ance betw een the n u m b er o f charged leptons (e"^ and e“) an d neutrinos. Since these reactions convert neutrons to pro to n s an d p rotons to neu tro n s with equal ease, the very early universe contained roughly equal n um bers o f pro to n s and neutrons. T he difference in rest energy betw een p rotons an d neu tro n s is ab o u t A £ '= 1.3 M eV (n eutrons being slightly m ore massive). At any tem perature T, the difference be tw een the relative n um bers o f protons an d n eutrons is determ ined in p art by the B oltzm ann factor (see Section 24-4). W hen t < 10“ ^ s (corresponding to T > 1.5 X 10*^ K or / c r > 1300 M eV ), the B oltzm ann factor is very nearly equal to 1 and has a negligible influence on the relative n u m b er o f p rotons and neutrons. A t tim es after ab o u t 10“ ^ s, the radiation has on the average too little energy to create n u c le o n -a n tin u c leo n pairs (that is, we no longer have y + y n + n o r p + p), b u t n u c le o n -a n tin u c le o n annih ilatio n co ntinues to occur (n + n —►y + y an d p + p —►y + y). F rom 10“ ^ s until abo u t 10"2 s ( T = 1.5 X 10*' K o r k T = 13 M eV), w eak-interaction processes m ain tain a rough balance be tw een the num bers o f p rotons an d neutrons. Between 10“ ^ s a n d 1 s ( T = 1.5 X 10*® K o r/c T = 1.3 M eV), the B oltzm ann factor begins to upset the balance betw een proto n s an d neutrons, and at / = 1 s the ratio betw een the n u m b er o f n eutrons and the n u m b er o f pro
The Big Bang Cosmology
1205
tons is about ^“ *, so th at the nucleons consist o f about 73% protons and 27% neutrons. D uring this period, the influence o f the neutrinos has been decreasing, and by about / = 1 s the neutrinos (which are cooling along with the rest o f the universe as it expands) have too little energy to cause p ro to n -n e u tro n transform ations, w hich dim in ishes the role o f the weak interactions in the evolution o f the universe. This is know n as the tim e o f “ neutrino de coupling,” w hen the interactions between m atter and neutrinos no longer occur. From this tim e on, the neu trinos continue to fill the universe, cooling along with the expansion o f the universe. These prim ordial neutrinos today have roughly the sam e density as the m icrow ave background photons b u t a slightly lower tem perature (about 2 K). Because neutrinos interact only feebly with m atter, detection o f energetic neutrinos {E > 1 M eV ) re quires equipm ent o f great size and sophistication. D etec tion o f these prim ordial neutrinos {E < 10“ ^ eV ) seems a hopeless task, b u t observing them , m easuring th eir energy distribution, and deducing their tem perature w ould pro vide an o th er dram atic confirm ation o f the Big Bang cos mology. At a tim e o f about 6 s (T = 6 X 10’ K or / : r = 0.5 M eV), the radiation has cooled to a tem perature at which it no longer has enough energy to produce even the light est p a rticle-an tip article pairs (electrons and positrons), so no new particles are form ed by pair production. P a rticle-an tip article annihilation continues to occur, and the resulting photons slow the rate o f cooling som e what. F urtherm ore, the electrons have too little energy to cause protons to transform into neutrons (p -f e“ n + Ve no longer occurs). T he only w eak-interaction process th at continues to occur is the decay o f the n eutron (n ^ p -h e“ + \i). At this tim e the nucleons consist o f about 83% protons and 17®/o neutrons. D uring this period, p article-an tip article annihilation has continued to occur, so th at no positrons or a n tin u cleons rem ain in the universe. T he universe now consists o f a n u m ber V o f protons, an equal nu m b er V o f electrons (to m ake it electrically neutral), and ab out 0.2V neutrons. Because p article-an tip article annihilation has substan tially decreased the nu m b er o f particles while m ain tain ing the am o u n t o f radiation, there are far m ore photons (perhaps 10®-10’ V) th an nucleons or electrons. T here are about as m any neutrinos as photons. As far as we know, the present universe contains no stars or galaxies m ade o f antim atter. W hat happened to all the antim atter, w hich represented 50% o f the particles in the early universe? A ccording to the Big Bang cosmology, in an early epoch o f the evolution o f the universe one o f the forces th at acted between the particles caused a very slight im balance o f m atter over antim atter, perhaps 1 part in 10® or 10’. T he exact nature o f this force is not yet well understood, although experim ents testing this force and distinguishing m atter from an tim atter have been dupli-
1206
C h a p te r 5 6
P a r tic le P h y s ic s a n d C o s m o l o g y
cated in the laboratory. D uring the subsequent stages in the evolution o f the universe, all the an tim a tte r an n ih i lated with all b u t ab o u t 1 part in 10* o r 10’ o f the m atter. T h at is, for every 1,000,000,000 positrons there m ay origi nally have been 1,000,000,001 electrons, b u t following the annihilation o f 2,000,000,000 particles the rem ainder is ju st 1 electron. T his description o f the evolution o f the universe, illus trated in Fig. 9, has taken us from the form ation o f the universe at the Big Bang, through hot an d tu rb u len t eras dom in ated by nuclear reactions, to a tim e o f a few seconds w hen the com position becam e identical with the particles th at now m ake up ou r universe. H ow these particles com bined to form the nuclei and ato m s th at we observe today is discussed in the next section.
Sample Problem 6 When did the universe become too cool to permit the radiation to create yC pT pairs? Solution The rest energy of the muon is 105.7 MeV. Photons have this mean energy at a temperature determined by £_ 105.7 MeV = 1.23 X 10*2 K. k 8.62 X 10-5 eV/K
^
The corresponding time is found from Eq. 3: '
/ 1 . 5 X 10'Os‘/2. k V _ , cx/ ,A-4 ( 1.23 X 10'^ K
j
56-6 NUCLEOSYNTHESIS___________ A t an age o f a few seconds, th e universe consisted o f pro tons, neutrons, an d electrons. Today, the com position o f the universe is m ostly hydrogen an d helium , w ith a sm all abundance o f heavier elem ents. H ow were the present nuclei and atom s produced from th e Big Bang? T h e for m ation o f the elem ents o f the present universe is know n as nucleosynthesis. As we shall see, observing th e present abundances o f the elem ents can give us clues ab o u t the processes th a t occurred during the Big Bang.
Big Bang Nucleosynthesis T he first step in building up com plex atom s is th e form a tion o f deuterium nuclei (deuterons) from th e com bina tion o f a proton an d a neutron, according to n -I- p ^ d -I- y. T he binding energy o f the deuteron (see Section 54-2) is 2.2 M eV, which is the energy o f the y ray th a t is given off during the form ation. T he reverse reaction, d -I- y —» n -I- p, can break ap art the deuterium nuclei in to their constitu en t protons an d neutrons, if the y ray eneigy is at least 2.2 MeV.
Figure 9 The evolution of the universe according to the Big Bang cosmology. The heavy solid line shows the relationship between temperature and time according to Eq. 3. The im portant reactions in each era are shown. (Here q and q stand for quark and antiquark, respec tively.)
Section 56-6
Nucleosynthesis
1207
Finally, the ^H and ^He will also react with protons and neutrons, as given by ^H + p - ^ ^ H e - h y
Figure 10 The energy spectrum of photons at a particular time in the evolution of the universe. Photons with energy above 2.2 MeV, which constitute a tiny fraction of the total number of photons, can dissociate deuterons.
If the universe is filled with energetic photons, the two reactions will take place at the same rate, and deuterium will be disassociated as quickly as it is formed. However, if the universe is sufficiently old, the photons will not have enough energy to accomplish the disassociation reaction, and deuterium can start to build up. When we ended our story in the previous section, the universe was about 6 s old, and the mean energy of the radiation was about 0.5 MeV, which is less than what is needed to keep deuterium from forming. However, it must be remembered that the radiation has a Planck dis tribution of energies (see Fig. 10, which was discussed in Section 49-2) and that there are 10®-10’ photons for every proton or neutron. There is a high-energy tail in the Planck distribution, which suggests that no matter what the temperature of the radiation, there will always be some photons of energy above 2.2 MeV that can break apart deuterium nuclei. If, on the average, the number of these energetic photons is less than the number of protons and neutrons, deuterium can start to build up. The neutron-to-proton ratio is about 0.2 at this point in the evolution of the universe, and there are roughly 10’ photons per nucleon, so that the ratio of neutrons to pho tons is about 0.2 X 10"’. If the fraction of photons with energies above 2.2 MeV is less than 0.2 X 10"’ of the total number o f photons, there will be less than one energetic photon per neutron, and deuterium formation can pro ceed. From the expression for the Planck distribution (ob tained from Eq. 6 of Chapter 49), we find that the fraction o f photons of energy greater than 2.2 MeV will be less than 0.2 X 10"’ when the temperature has fallen to 9 X 10® K. Equation 3 shows that this temperature occurs at a time o f 250 s. At a time of 250 s, the formation of deuterium nuclei begins. Because the deuterons are less abundant than pro tons or neutrons, the deuterons will readily react with protons and neutrons, according to the reactions d -h n ^ ^H + y and
d -h p ^ ^He -h y.
and
^He + n ^ ^ H e + y.
For all four of these reactions, the binding energy o f the final particle is greater than that of the deuteron. Thus if the radiation is too feeble to prevent the formation of deuterons, it will certainly be too feeble to prevent the succeeding reactions. We can therefore assume that nearly all the deuterons are eventually converted into ^He, so that the end products of this stage of the evolution of the universe are protons and a particles. Because there are no stable nuclei with a mass number of 5, these reac tions cannot continue beyond '‘He. To find the relative number of a particles, we must find the number of available neutrons at t = 250 s, when deu terons begin to form. At / = 6 s, about 17% o f the nu cleons are neutrons, but as a result of the radioactive decay of the neutron, some neutrons will be converted into protons between / = 6 sand / = 250 s. Using the halflife of the neutron (about 11 min), we find that at t = 250 s the nucleons will consist of about 12.5% neutrons and 87.5% protons. That is, out o f every 10,000 nucleons there will be 1250 neutrons and 8750 protons. These neu trons will combine with 1250 protons to form 625 a parti cles, leaving 8750 — 1250 = 7500 protons. O f the total number of nuclei in the universe at this time, 7.7% are a particles and 92.3% are protons. In terms of mass, the a particles constitute a fraction of the total mass of the uni verse given by 4X625 = 0.25 or 25%. 7500 + 4 X 625 The abundance of^He in the present universe should equal this value, if we neglect the burning of hydrogen to helium that takes place in stars. The measured helium abundance in a variety of systems, including stars, gas eous nebulae, and planetary nebulae, turns out to be 24 ± 1%, which agrees with our estimate and indicates that our description is certainly reasonable. The final step in the production of matter in the Big Bang is the formation of neutral atoms of hydrogen and helium when the protons and a particles combine with electrons. As in the case of deuteron formation, this can not occur when there are enough photons in the highenergy tail of the Planck distribution to break apart any neutral atoms that may form. In this case, we want the relative fraction of photons with energies above 13.6 eV (the binding energy of atomic hydrogen) to be less than about 10"’. This occurs for a temperature o f about 6000 K, which corresponds to an age of the universe of around 200,000 y. (As the radiation cools, the energy density of the universe becomes less dominated by radia tion and more by matter. In this case Eq. 3, which as sumes a radiation-dominated universe, is not quite correct. Taking this effect into account, the temperature
1208
Chapter 56
Particle Physics and Cosmology
o f the universe when hydrogen atoms begin to form is closer to 3000 K, corresponding to an age of around 700,000 y.) Once neutral atoms have formed, there are essentially no free charged particles left in the universe. This is the time o f decoupling of the matter and the radiation field. The universe becomes transparent to the radiation, which can travel long distances without interacting with matter. This radiation, which has been traveling since the decou pling time, is observed today as the microwave back ground. The expansion of the universe has reduced the radiation temperature by a factor o f 1000 since the decou pling time. The story o f the evolution of the universe as described by the Big Bang cosmology is a remarkable one. It inte grates modern experiments in nuclear and particle phys ics with quantum physics and classical thermodynamics. It yields results that can be tested in the present universe, including the helium abundance, the microwave back ground radiation, and a small abundance of left-over deu terium that did not get “cooked” into mass-3 nuclei. It is a story that depends in critical ways on the strengths of nuclear or subnuclear forces and on the variety o f parti cles that took part in the early universe. For example, if there were a fourth generation of leptons, the reaction rates of weak-interaction processes would be greater, and more neutrons would be formed, thereby increasing the abundance o f “He. Although this conclusion is subject to interpretation, the observed present abundance of “He is regarded by many cosmologists as limiting the number of generations o f leptons to three.
Nucleosynthesis in Fusion Reactions After the decoupling of matter and radiation, the matter (consisting o f hydrogen and helium) was subject only to the gravitational force. Recent precise observations of the microwave background have shown that the distribution o f matter at the decoupling time was slightly nonuniform. Regions o f slightly higher density began to condense into clouds o f ever increasing density. As each cloud con tracted under its own gravity, its temperature rose until it became hot enough to initiate fusion reactions. This is how first-generation stars formed. We have seen in Chapter 55 that stars convert hydrogen into helium by means o f fusion reactions. After a star has used up its supply of hydrogen and become mostly he lium, it can again begin to contract, which increases its temperature. (This increase in temperature causes an in creased radiation pressure, which causes the radius of the star to increase. The surface area increases more rapidly than the temperature, so that the energy per unit area of the surface actually decreases, and the color of the star goes from bright yellow to red. This is the red giant phase o f the evolution of the star.) Eventually, the temperature is high enough that the Coulomb barrier between two “He
nuclei can be successfully breached by their thermal mo tion, and helium fusion can occur. The simple helium fusion reaction “He + “He ^ «Be does not contribute to the fusion in a star, because *Be is unstable and breaks apart as rapidly as it forms. Helium fusion requires a third a particle to participate, so that the net reaction is “He + “He + “He — '^C + y. Once '^C forms, we can have additional a-particle reac tions, such as '^C + “He ^ “ O + y, '*0 + “He ^ ^ojsje -|- y, 2°Ne + “He ^ ^“Mg -I- y, and so on. These reactions have increasingly high Cou lomb barriers and therefore require increasing tempera tures. When the helium fuel is exhausted, contraction sets in again to increase the temperature, so that other reactions can occur, such as carbon burning: '^C + '^C ^ ^“Mg + y. Eventually, these reactions reach the peak of the binding energy curve (Fig. 6 of Chapter 54) at mass 56. Beyond this point, energy is no longer released in fusion reactions. Figure 11 shows the abundance o f nuclei in this mass range. The relative abundances support this scenario for producing the elements in fusion reactions. Note that C is more than five orders of magnitude more abundant than Li, Be, and B, which are not produced in these processes. Also note that the even-Z nuclei are, on the average, more than an order of magnitude more abundant than their odd-Z neighbors. These fusion reactions with a particles produce only even-Z products, so the observed higher abundances of these products are consistent with this ex planation of their formation. Note also the last point in Fig. 11, which indicates that the total abundance of the 50 elements beyond the nuclei in the mass-56 range is less than the abundance o f all but one of the individual elements in the region from C to Zn. It certainly appears that most of the matter we know about was produced in fusion processes.
Nucleosynthesis by Neutron Capture The elements beyond mass 56 were produced either slowly in stars or suddenly in supernovas by neutroncapture reactions. There will be a small density o f neu trons in stars, produced through sequences such as '^C + ' H ^ ' ^ N - l - y '3-N
'^C + e+ + V,
'^C + “He ^ “ O + n.
Section 56-6
Nucleosynthesis
1209
108
Figure 11 Relative abundances (by mass) of the elements beyond helium in the solar system.
decays to ’’Co by negative beta decay with a half-life of45 days. Whether the ” Fe captures another neutron, thereby forming “ Fe, or decays to ” Co depends on the density of neutrons. If the density of neutrons is so low that the ” Fe is unlikely to encounter a neutron within a time of the order of 45 days, then it will decay to ” Co. Ifthe chance of encountering a neutron is high, the ” Fe will be converted into *°Fe and then possibly to “ Fe, “ Fe, and so forth. These nuclei are very rich in neutrons and are thus mov-
Suppose we have ’®Fe, which has been produced through fusion processes. A sequence of neutron-capture reac tions can then occur: ’‘Fe + n —»’’Fe + y ^’Fe -I- n —»’*Fe + y **Fe -I- n —»” Fe + y. Both ’’Fe and ’*Fe are stable, but ’’Fe is radioactive and
0
69Ga
►
\ 6«Zn
6 5 z n - .66zn
67Zn
85Cu
8®Cu
decay \
n capture
\ 0-^ decay
\ 63Cu
8'*Cu
----
s process
N
\
61fgi
6°Ni
62Ni
63Ni -. 64Ni
V
\
65Ni
7 l N i — 7 2 N j |— j 7 3 N j
\
zSw 8°Co
59Co
88Co
69Co
67Fe
68Fe
A s e p e — 57Fe — 58Fe
59Fe
60pe — 6ipe _ 62pe
63Fe
6^Fe
65pe _ 66pe
r process Neutron number, N
Figure 12 A section of the chart of the nuclides (Fig. 4 of Chapter 54), showing the s- and r-process paths from ^^Fe. Many r-process paths are possible, as the short-lived nuclei beta decay; only one such path is shown. All the nuclei in the r-process path are unstable and may beta decay toward the stable nuclei.
70Co-71Co
1210
Chapter 56
Particle Physics and Cosmology
NJ
s process
r process
Neutron number, N
Figure 13 The r- and s-process paths leading to Sn, Sb, and Te isotopes.
ing further from the realm of the stable nuclei shown in Fig. 4 of Chapter 54. They correspondingly have ever shorter half-lives. For example, *^Fe has a half-life of only 2 s. Eventually the half-life becomes so short that no neu tron is encountered before the decay occurs, and finally there is a beta decay to the corresponding nuclide of Co, such as is indicated in Fig. 12. The branch that proceeds through *’Co, which permits time for even long-lived beta decays to occur before the next neutron is captured, is called the s process (s for slow). The process in which the density of neutrons is so great that many captures can occur before the beta decay is called the r process (r for rapid). These two processes are indicated in Fig. 12. Of course, the highly unstable nuclei produced in the r process will eventually beta decay toward the stable nu clei, but as you can see from Fig. 12, the decays follow a path different from that of the s process. Consider, for example, the Sn nuclides illustrated in Fig. 13. The s pro cess proceeds through ‘^°Sn, '^'Sn, and '^'Sb. Because of the beta decay of '^'Sn, it cannot capture a neutron through the s process. Therefore '“ Sn, which has a natu ral abundance of 4.6%, cannot be produced in the s pro cess; it must be produced in the r process. The nuclide '“ Sn can be produced through both the s and r process, but the nuclide '^^Te can be produced only through the s process, because the r-process beta decays along the mass-124 line are stopped at stable '^^Sn. In a red giant star, the density of neutrons may be of the
order of 10'^ per m^, which is sufficient to maintain the s process. In a supernova explosion the neutron density may be 10-20 orders of magnitude larger, but the high density lasts for a time of only a fewseconds. In that time, the r-process chains occur all the way up to the heaviest nuclei, which are then hurled into space and gradually decay to form the stable r-process nuclei. The elements found on Earth beyond mass 56 were produced in firstgeneration stars, perhaps through the s process or the r process, and the planets of our solar system (and in fact we ourselves) are made of the recycled ashes of bumt-out stars.
56-7 THE AGE OF THE UNIVERSE In Section 54-7, we discussed the use of radioactive dating methods to determine the age of the Earth. By examining the relative amounts of parent and daughter isotopes in certain radioactive decay processes having half-lives in the range 10*-10’ y (for example, ^**U ^°*Pb, *’Rb *’Sr, and “ K —»“ Ar), it has been determined that the age of the oldest rocks on Earth is about 4.5 X 10’ y. An iden tical value is obtained for meteorites and for rocks from the Moon. We can therefore be fairly certain that this value represents the time since the condensation of the solar system. We know that the universe must be much older than
Section 56-7
this value, because the solar system formed out of ele ments that were created in the interiors of stars or in supernovas. The present chemical composition of the solar system was determined during a previous era of nucleosynthesis, which occurred in a previous generation o f stars. To find the true age of the universe, we must determine the time interval needed for the elements to be produced. The total time from the Big Bang to the present can be divided into four periods: (1) from the Big Bang until the formation o f neutral H and He atoms (/,); (2) the conden sation o f galaxies and the formation of first-generation stars (/2 ); (3) nucleosynthesis in stars and supernovas, leading to the present chemical elements (/3 ); and (4) for mation and evolution of the solar system from the debris o f earlier stars (^4 ). The present age of the universe is just the sum of these four terms:
t = tx A-t2-\~h^U.
(4)
We know from our discussion of the Big Bang cosmology that the time from the Big Bang until neutral atoms formed is no more than 10^ y. The time ^2 for galaxies to condense from hydrogen and helium produced in the Big Bang is not precisely known but has been estimated to be in the range 1- 2 X 10’ y. Since U is known to be 4.5 X 1 0 ’ y, the age of the universe can be determined if we can find the time t^ associated with nucleosynthesis. This time must be estimated from the relative abun dances of the products that remain at the end of nucleo synthesis. For example, consider the isotopes and which at present have a relative abundance of 0.72% (see the discussion in connection with the natural fission reactor in Section 55-5). Both and have been decaying during the interval since the formation of the solar system. Their ratio 4.5 X 10’ years ago is (see Sam ple Problem 4 o f Chapter 55) = (0.0072)^^®’®^“
^
= 0.30.
During the interval ty, both isotop)es were being formed
The Age o f the Universe
1211
more-or-less continuously through the r process, while the relative decay of course also took place. Because o f the production of both isotopes during this time, the ratio of their abundances in this period did not change as rapidly as it did during the free decay in the interval t^; see Fig. 14. Evidence from the uranium abundance suggests that t-^is in the range 4 - 9 X 10’ y; similar values result from the analysis of the abundances of other r-process nuclei. An independent estimate of t^ comes from the r- and s-process production of the isotope **^Os, which is illus trated in Fig. 15. The isotope *®^Re is formed only in the r process, and it decays to *®^Os with a half-life of 40 X 1 0 ’ y, which is in the proper range to serve as a measure of the age of the universe. Comparing the relative amounts of parent *®^Re and daughter *®^Os should give a measure of the duration of r-process nucleosynthesis. However, ‘®^Os can also be formed through the s process, as shown in Fig. 15. Correcting its observed abundance for the frac tion produced in the s process (which can be determined from the abundance of *®^Os), we find from the relative amounts of *®^Re and '*^Os that ^3 is in the range 9 -1 2 X 10’ y. The lower end of this range, 9 X 10’ y, is consistent with the upper end of the range for t^determined from the uranium abundances, so we choose this value as our esti mate for ^3 . Combining these results, we have as our estimate for the age of the universe / = /i -h ^2 + ^3 + ^4
= 10^ y + 1- 2 X 10’ y + 9 X 10’ y + 4.5 X 10’ y = 15 X 10’ y. This number is somewhat uncertain as a result o f the range of values in the estimate for Taking this uncer tainty into account, we obtain / = 10-18 X 10’ y. Consider the enormous amount of physics contained in this simple result. To determine t ^, we used the cumula tive knowledge of particle physics, electromagnetism.
Figure 14 The change in the ratio with time. D ur ing the life of the solar system (the tim e t^\ the ratio changes due only to the relative decays, eventually reaching the present value of 0.0072. During the interval t-^, production by the r process occurs along with the decay. The duration de duced for the interval ty depends on the value that we take for the initial ratio, which must be determined from calculation.
Time before present (lO^y)
1212
C h a p te r 5 6
P a r ti c le P h y s ic s a n d C o s m o l o g y
r process
Figure 15 The r- and s-process formation of Re and Os isotopes.
thermal physics, and atomic and nuclear physics to trace the formation o f matter as we know it. The interval ti is determined from calculations using thermodynamics and gravitational theory to analyze the condensation of cold matter into hot stars. Our estimate for tj is based on our knowledge o f r-process and s-process nucleosynthesis based on nuclear physics studies in laboratories on Earth, and the interval 14 is based on further experiments in nuclear physics and research in geochemistry.
/ = 10-19 X 10’ y, which corresponds with the range determined from nu cleosynthesis. Our assumption about the constant separation speed o f the galaxies is almost certainly incorrect. The mutual grav itational attraction o f the galaxies has been slowing their separation since the Big Bang, so that at earlier times the speed of separation was greater than it is at present. Figure 16 shows a representation o f a typical intergalactic separa-
Cosmological Determination of the Age If we make the rough but not quite correct assumption that the universe has been expanding at the same rate since its formation, than the separation (/between typical galaxies should be related to the age o f the universe roughly according to
d = vt, where v is the (assumed constant) speed of separation. Comparing this result with Eq. 2 shows that the age t of the universe is just the inverse of the Hubble parameter:
t = H-
(5)
The present best estimate for the Hubble parameter, H = 67 (km/s)/Mpc, gives a value for the age o f the universe of / = 15X
10’
y,
in remarkably good agreement with the value obtained from the nucleosynthesis calculation. However, the range o f uncertainty o f the Hubble parameter, 5 0 -1 0 0 (km/s)/ Mpc, is very large and gives a corresponding range in ages of
Figure 16 The dependence of a typical galactic separation distance on time during the evolution of the universe accord ing to different models. If the universe has been expanding at a constant rate (straight line), we can extrapolate backward to zero separation (the Big Bang) at a time of H ~' before the present. If the expansion has been slowing due to the gravita tional interaction (a more reasonable scenario), the Big Bang occurred at a time less than H ~' before the present. If the gravitational interaction is strong enough, the expansion may eventually become a contraction.
Q u e s tio n s
tion distance as a function of the time. If the “constantspeed” model were valid, the age of the universe would be t = H~K If the speed has been decreasing since the Big Bang, the deduced age depends on the rate o f decelera tion. Since humans have not been observing long enough to detect any change in the separation rate, we must rely on two indirect methods to determine the deceleration: ( 1 ) we can measure the red shifts and thereby deduce the speeds o f the most distant (and therefore the oldest) ob jects we can observe with telescopes, or ( 2 ) we can calcu late the deceleration based on the gravitational effects of the total amount of matter in the universe. If the galaxies are decelerating, the most distant objects would be observed to have larger recessional speeds than we would deduce from the Hubble relationship. Unfortu nately, our observations of these distant objects are not yet precise enough to indicate any definite deceleration. The second method is also unsuccessful: the amount of matter observed with telescopes (that is, matter that emits some type of electromagnetic radiation) is not even enough to account for the gravitational attraction within galaxies and clusters of galaxies. It is possible that as much as 90% o f the matter in the universe is in a form unknown to us, possibly neutrinos (if they have nonzero mass) or other elementary particles left over from the Big Bang or perhaps bumt-out stars. Because of this unknown “dark
1213
matter,” we cannot make a reliable calculation o f the deceleration of the universe. Various cosmological models have been proposed that give curves in Fig. 16 of differing curvatures and therefore different ages of the universe. For example, some of these give an age that is one-half or two-thirds o f H~\ or 5 -1 2 X 10’ y. Although we don’t know which (if any) of these models is correct, it seems clear that both the nu cleosynthesis and cosmological estimates for the age of the universe are consistent with values in the range 10-15 X 10’ y. It is a source of great frustration for physicists not to be able to view the history of the universe with more cer tainty, because our ability to look forward is similarly limited. Will the expansion continue forever, or is there enough matter present to reverse the expansion? Figure 16 shows several possible outcomes. Perhaps cosmologists of later eras will observe the galaxies rushing together as the universe “heats up” and the galaxies come together, eventually reaching a single point (a “Big Crunch”) that may be followed by another Big Bang. Or perhaps the expansion continues forever, until the universe is cold and dark. If the solution to this fundamental problem is to be found, it will require vigorous investigations at the forefronts of astrophysics, nuclear physics, and particle physics.
QUESTIONS 1. The ratio of the gravitational force between the electron and the proton in the hydrogen atom to the magnitude of the electromagnetic force of attraction between them is about 10“ ^ . If the gravitational force is so very much weaker than the electromagnetic force, how was it that the gravitational force was discovered first and is so much more apparent to us? 2. What is really meant by an elementary particle? In arriving at an answer, consider such properties as lifetime, mass, size, decays into other particles, fusion to make other particles, and reactions. 3. Why do particle physicists want to accelerate particles to higher and higher energies? 4. Name two particles that have neither mass nor charge. What properties do these particles have? 5. Why do neutrinos leave no tracks in detecting chambers? 6. Neutrinos have no mass (presumably) and travel with the speed of light. How, then, can they carry varying amounts of energy? 7. Do all particles have antiparticles? What about the photon? 8. In the beta decay of an antineutron to an antiproton, is a neutrino or an antineutrino emitted? 9. Photons and neutrinos are alike in that they have zero charge, zero mass, and travel with the speed of light. What are the differences between these particles? How would you produce them? How do you detect them?
10. Explain why we say that the tt®meson is its own antiparticle. 11. Why can’t an electron decay by disintegrating into two neu trinos? 12. Why is the electron stable? That is, why does it not decay spontaneously into other particles? 13. Why cannot a resting electron emit a single gamma-ray photon and disappear? Could a moving electron do so? 14. A neutron is massive enough to decay by the emission of a proton and two neutrinos. Why does it not do so? 15. A positron invariably finds an electron and they annihilate each other. How then can we call a positron a stable particle? 16. What is the mechanism by which two electrons exert forces on each other? 17. Why are particles not grouped into families on the basis of their mass? 18. A particle that responds to the strong force is either a meson of baryon. You can tell which it is by allowing the particle to decay until only stable end products remain. If there is a proton among these products, the original particle was a baryon. If there is no proton, the original particle was a meson. Explain this classification rule. 19. How many kinds of stable leptons are there? Stable mesons? Stable baryons? In each case, name them. 20. Most particle physics reactions are endothermic, rather than exothermic. Why?
1214
C h a p te r 5 6
P a r tic le P h y s ic s a n d C o s m o l o g y
21. What is the lightest strongly interacting particle? What is the heaviest particle unaffected by the strong interaction? 22. For each of the following particles, state which of the four basic forces are influential: {a) electron; {b) neutrinos; (c) neutron; {d) pion. 23. Just as Xrays are used to discover internal imperfections in a metal casting caused by gas bubbles, so cosmic-ray muons have been used in an attempt to discover hidden burial chambers in Egyptian pyramids. Why were muons used? 24. Are strongly-interacting particles affected by the weak inter action? 25. Do all weak-interaction decays produce neutrinos? 26. Mesons and baryons are each sensitive to the strong force. In what ways are they different? 27. By comparing Tables 3 and 7, point out as many similarities between leptons and quarks as you can and also as many differences. 28. What is the experimental evidence for the existence of quarks? 29. We can explain the “ordinary” world around us with two leptons and two quarks. Name them. 30. The neutral pion has the quark structure uu and decays with a mean life of only 8.3 X 10“ *^ s. T l^chaigedpion, on the other hand, has a quark structure of udand decays with a mean life of 2.6 X 10“ ®s. Explain, in terms of their quark structure, why the mean life of the neutral pion should be so much shorter than that of the charged pion. {Hint: Think of annihilation.)
31. Do leptons contain quarks? Do mesons? Do photons? Do baryons? 32. The A* baryon can have an electric charge of + 2e (see Table 5). Based on the quark model, do we expect to find mesons with charge + 2^? Baryons with charge —2^? 33. The baryon decays with a mean life characteristic of the weak interaction (see Table 5). It should be able to decay to the A° baryon by the strong interaction without changing strangeness. Why doesn’t it? 34. Why can’t we find the center of the expanding universe? Are we looking for it? 35. Due to the effect of gravity, the rate of expansion of the universe must have decreased in time following the Big Bang. Show that this implies that the age of the universe is less than 1///. 36. It is not possible, using telescopes that are sensitive in any part of the electromagnetic spectrum, to “look back” any farther than about 500,(X)0 y from the Big Bang. Why? 37. How does one arrive at the conclusion that visible matter may account for only about 10% of the matter in the uni verse? 38. Are we always looking back in time as we observe a distant galaxy? Does the direction in which we look make a differ ence? 39. Can you think of any possible explanation for the expanding universe other than a Big Bang?
PROBLEMS Section 56-1 Particle Interactions 1. (a) An electron and a positron are separated by a distance r. Find the ratio of the gravitational force to the electrostatic force between them. What do you conclude from the result concerning the forces acting between particles detected in a bubble chamber or similar detector? (b) Repeat for a proton-antiproton pair. 2. Some of the GUTs predict the following possible decay schemes for the proton: p -►e+ + 7, p — e+ + wO {a) Calculate the Q-values for these decays, (b) Show that the decays do not violate the conservation laws of charge, rela tivistic energy, or linear momentum. The rest energy of a proton is 938.27 MeV, of an electron is 0.511 MeV, and of a neutral pion is 135 MeV. 3. An electron and a proton are placed a distance apart equal to one Bohr radius Oq, Find the radius R of a lead sphere that must be placed directly behind the electron so that the gravi tational force on the electron just overcomes the electro static attraction between the proton and the electron; see Fig. 17. Assume that Newton’s law of gravitation holds, and that the density of the sphere equals the density of lead on the Earth.
Figure 17
Problem 3.
Section 56-2 Families o f Particles 4. A neutral pion decays into two gamma rays: Calculate the wavelengths of the gamma rays produced by the decay of a neutral pion at rest. 5. The rest energy of many short-lived particles cannot be measured directly, but must be inferred from the measured momenta and known rest energies of the decay products. Consider the meson, which decays by tT. Cal culate the rest energy of the meson given that each of the oppositely directed momenta of the created pions has mag nitude 358.3 MeV/c. See Table 4 for the rest energies of the pions. 6. Observations of neutrinos emitted by the supernova SN 1987a in the Large Magellanic Cloud, see Fig. 18, place an upper limit on the rest energy of the electron neutrino of
Problems
1215
14. What conservation law is violated in these proposed reac tions and decays? (a) A® p + K“; {b) K“ + p —►A® + n'*’. 15. Use the conservation laws to identify the particle labeled x in the following reactions, which proceed by means of the strong interaction, {a) p + p —►p + A® 4- x; (^) p + p ^ n + X; (c) 7T- + p — E® + K® + X. Section 56-4 The Quark Model
Figure 18
7.
8.
9.
10.
Problem 6.
20 eV. Suppose that the rest energy of the neutrino, rather than being zero, is in fact equal to 20 eV. How much slower than light is a 1.5-MeV neutrino, emitted in a )?-decay, mov ing? A neutral pion has a rest energy of 135 MeV and a mean life of 8.4 X 10“ '^ s. If it is produced with an initial kinetic en ergy of 80 MeV and it decays after one mean lifetime, what is the longest possible track that this particle could leave in a bubble chamber? Take relativistic time dilation into ac count. A positive tau (t"^, rest energy = 1784 MeV) is moving with 2200 MeV of kinetic energy in a circular path perpendicular to a uniform 1.2-T magnetic field, (a) Calculate the mo mentum of the tau in kg • m/s. Relativistic effects must be considered, {b) Find the radius of the circular path. {Hint: See Section 34-3.) Calculate the range of the weak force between two neighbor ing protons. Assume that the Z° boson is the field particle; see Table 6. Identify the interaction responsible for each of the following decays: { a ) t] ^ y - \- y; {b) ^ v/, {c) rf r j 7 t ^ -\7C\ (d) K° —► + tC.
Section 56-3 Conservation Laws 11. What conservation law is violated in each of these proposed decays? (a) //" —►e"^ + (b) . 12. The reaction Tr'^ + p —►p + p + n proceeds by the strong interaction. By applying the conservation laws, deduce the charge, baryon number, and strangeness of the antineutron. 13. By examining strangeness, determine which of the following decays or reactions proceed via the strong interaction. (a) + n-\ (b) A° + p — -h n; (c) A° — p + 7r\ (d) K“ + p —►A® + 7T°. See Tables 4 and 5 for values of S.
16. Show that, if instead of plotting S versus Q for the spin-^ baryons in Fig. 3b and for the spin-0 mesons in Fig. 3a, the quantity Y = B-\- S is plotted against the quantity = Q — {B, then the hexagonal patterns emerge with the use of nonsloping (perpendicular) axes. (The quantity Y is called hypercharge and is related to a quantity called isospin.) 17. The quark composition of the proton and the neutron are uud and udd, respectively. What are the quark composi tions of (a) the antiproton and (b) the antineutron? 18. From Tables 5 and 7, determine the identity of the baryons formed from the following combinations of quarks. Check your answers with the baryon octet shown in Fig. 3b. (a) ddu; {b) uus; (c) ssd. 19. What quark combinations form {a) A°; {b) E°? 20. Using the up, down, and strange quarks only, construct, if possible, a baryon (a) with Q = -\-\ and S = —2. {b) With 0 = + 2 and 5 = 0. 21. There is no known meson with Q = -\-\ and 5 = —1 or with Q = — \ and 5 = + 1. Explain why, in terms of the quark model. 22. Analyze the following decays or reactions in terms of the quark content of the particles: (a) >n 4- n~; (b) K® ^ n'^ H- tT; (c) + p —> (t/) y + n —►tt” + p. Section 56-5 The Big Bang Cosmology 23. By choosing two points on each line of Fig. 6 and calculating the slopes, verify the given numerical values of the Hubble parameter. 24. If Hubble’s law can be extrapolated to very large distances, at what distance would the recessional speed become equal to the speed of light? 25. What is the observed wavelength of the 656.3-nm H^ line of hydrogen emitted by a galaxy at a distance of 2.4 X 10* pc? 26. In the laboratory, one of the lines of sodium is emitted at a wavelength of 590.0 nm. When observing the light from a particular galaxy, however, this line is seen at a wavelength o f602.0 nm. Calculate the distance to the galaxy, assuming that Hubble’s law holds. 27. The wavelength of the photons at which a radiation field of temperature T radiates most intensely is given by = (2898 pm • K )/^ (see Eq. 4 in Chapter 49). (a) Show that the energy E in MeV of such a photon can be computed from £ = (4.28X 10"‘0 MeV/K)T. {b) At what minimum temperature can this photon create an electron-positron pair? 28. The recessional speeds of galaxies and quasars at great dis tances are close to the speed of light, so that the relativistic Doppler shift formula (see Eq. 10 in Chapter 42) must be
1216
Chapter 56
Particle Physics and Cosmology
used. The redshift is reported as z, where z = AA/Ao is the (fractional) red shift, (a) Show that, in terms of z, the reces sional speed parameter p = v/c is given by ^+ 2z z2 + 2z + 2 {b) The most distant quasar detected (as of 1990) has z = 4.43. Calculate its speed parameter, (c) Find the distance to the quasar, assuming that Hubble’s law is valid to these distances. 29. Due to the presence everywhere of the microwave radiation background, the minimum temperature possible of a gas in interstellar or intergalactic space is not 0 K but 2.7 K. This implies that a significant fraction of the molecules in space that possess excited states of low excitation energy may, in fact, be in those excited states. Subsequent de-excitation leads to the emission of radiation that could be detected. Consider a (hypothetical) molecule with just one excited state, {a) What would the excitation energy have to be in order that 23% of the molecules be in the excited state? {Hint: see Section 52-6.) {b) Find the wavelength of the pho ton emitted in a transition to the ground state. 30. Will the universe continue to expand forever? To attack this question, make the (reasonable?) assumption that the reces sional speed y of a galaxy a distance r from us is determined only by the matter that lies inside a sphere of radius r cen tered on us; see Fig. 19. If the total mass inside this sphere is A/, the escape speed is given by = yflGM/r (see Sample
Problem 6 in Chapter 16). (a) Show that the average density p inside the sphere must be at least equal to the value given by p = 3H ySnG to prevent unlimited expansion, (b) Evaluate this “critical density” numerically; express your answer in terms of Hatoms/m^. Measurements of the actual density are difficult and complicated by the presence of dark matter. 31. (a) What is the minimum temperature of the universe neces sary for the photons to produce 7t^ - n~ pairs? (b) At what age did the universe have this temperature? Section 56-7 The Age o f the Universe 32. The existence of dark (i.e., nonluminous) matter in a galaxy (such as our own) can be inferred by determining through observation the variation with distance in the orbital period of revolution of stars about the galactic center. This is then compared with the variation derived on the basis of the distribution of matter as indicated by the luminous material (mostly stars). Any significant deviation implies the exis tence of dark matter. For example, suppose that the matter (stars, gas, dust) of a particular galaxy, total mass A/, is distributed uniformly throughout a sphere of radius R. A star, mass m, is revolving about the center of the galaxy in a circular orbit of radius r < R . (a) Show that the orbital speed V of the star is given by v = H G M /R \ and therefore that the period T of revolution is T = 2n^R V G M , independent of r. {b) What is the corresponding formula for the orbital period assuming that the mass of the galaxy is strongly concentrated toward the center of the galaxy, so that essentially all of the mass is at distances from the center less than r? These considerations applied to our own Milky Way galaxy indicate that substantial quantities of dark mat ter are present. 33. (a) Show that the number N of photons radiated, per unit area per unit time, by a cavity radiator at temperature T is given by
Figure 19
Problem 30.
(Hint: In evaluating the integral, ignore the “ 1” in the de nominator of R{k)\ see Eq. 6 in Chapter 49. Use the change of variables given in Problem \l{a) in Chapter 49.) {b) To the same approximation, show that the fraction of photons, by number, with energies greater than 2.2 MeV at a tempera ture of 9 X 10*Kis 2.1 X 10"'0.
APPENDIX A THE INTERNATIONAL SYSTEM OF UNITS (SI)*
T H E SI BASE U N IT S Quantity
Name
length
meter
Symbol m
mass
kilogram
kg
time
second
s
electric current
ampere
A
thermodynamic temperature
kelvin
K
amount of substance
mole
mol
luminous intensity
candela
cd
Definition “ . . . the length of the path traveled by light in vacuum in 1/299,792,458 of a second.” (1983) “. . . this prototype [a certain platinum-iridium cylinder] shall henceforth be considered to be the unit of mass.” (1889) “ . . . the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom.” (1967) “ . . . that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to 2 X 10“^ newton per meter of length.” (1946) “ . . . the fraction 1/273.16 of the thermodynamic temperature of the triple point of water.” (1967) “ . . . the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12.” (1971) “ . . . the luminous intensity, in the perpendicular direction, of a surface of 1/600,000 square meter of a blackbody at the temperature of freezing platinum under a pressure of 101.325 newton per square meter.” (1967)
• Adapted from “The International System of Units (SI),” National Bureau of Standards Special Publication 330, 1972 edition. The definitions above were adopted by the General Conference of Weights and Measures, an international body, on the dates shown. In this book we do not use the candela.
A-l
A-2
Appendix A
The International System o f Units (SI)
SO M E SI D E R IV E D U N IT S Quantity
Name o f Unit
Symbol
area volume frequency mass density (density) speed, velocity angular velocity acceleration angular acceleration force pressure work, energy, quantity of heat power quantity of electricity potential difference, electromotive force electric field electric resistance capacitance magnetic flux inductance magnetic field entropy specific heat capacity thermal conductivity radiant intensity
square meter cubic meter hertz kilogram per cubic meter meter per second radian per second meter per second squared radian per second squared newton pascal joule watt coulomb volt volt per meter ohm farad weber henry tesla joule per kelvin joule per kilogram kelvin watt per meter kelvin watt per steradian
m^ m^ Hz kg/m^ m/s rad/s m/s^ rad/s^ N Pa J W C V V/m Q. F Wb H T J/K J/(kg*K) W /(m-K) W/sr
Quantity
Name o f Unit
Symbol
plane angle solid angle
radian steradian
rad sr
T H E SI S U P P L E M E N T A R Y U N IT S
Equivalent
s"‘
kg-m/s^ N/m2 N -m J/s A*s N -m /C N/C V/A A-s/V Vs V-s/A Wb/m^, N/A • m
APPENDIX B SOME FUNDAM ENTAL CONSTANTS OF PHYSICS
Constant
Symbol
Speed of light in a vacuum Elementary charge Electron rest mass Permittivity constant Permeability constant Electron rest mass‘d Neutron rest mass‘d Hydrogen atom rest mass‘d Deuterium atom rest mass‘d Helium atom rest mass‘d Electron charge-to-mass ratio Proton rest mass Proton-to-electron mass ratio Neutron rest mass Muon rest mass Planck constant Electron Compton wavelength Universal gas constant Avogadro constant Boltzmann constant Molar volume of ideal gas at STP*^ Faraday constant Stefan-Boltzmann constant Rydberg constant Gravitational constant Bohr radius Electron magnetic moment Proton magnetic moment Bohr magneton Nuclear magneton Fine structure constant Magnetic flux quantum Quantized Hall resistance
c e m. €o /^o w. "Jn m ('H ) m(^H) m(“He) e/m. Wp Wp/m.
h K R k F o R G ^0 /^c
a o Rh
Best (1986) value
Computational Value
Value^
Uncertainty^
3.00 X 10* m/s 1.60 X 10-” C 9.11 X 10-^' kg 8.85 X 10-'^ F/m 1.26 X 10-
2.99792458 1.60217733 9.1093897 8.85418781762 1.25663706143 5.48579902 1.008664904 1.007825035 2.014101779 4.00260324 1.75881962 1.6726231 1836.152701 1.6749286 1.8835327 6.6260755 2.42631058 8.314510 6.0221367 1.3806513 2.2413992 9.6485309 5.670399 1.0973731571 6.67259 5.29177249 9.2847700 1.41060761 9.2740154 5.0507865 1/137.0359895 2.06783461 25812.8056
exact 0.30 0.59 exact exact 0.023 0.014 0.011 0.012 0.012 0.30 0.59 0.020 0.59 0.61 0.60 0.089 8.4 0.59 1.8 1.7 0.30 6.8 0.00036 128 0.045 0.34 0.34 0.34 0.34 0.045 0.30 0.045
Same unit and power of ten as the computational value. * Parts per million. Mass given in unified atomic mass units, where 1 u = 1.6605402 X 10“^^ kg. d STP— standard temperature and pressure = 0*C and 1.0 bar.
A -3
APPENDIX C SOME ASTRONOMICAL DATA
THE SUN, THE EARTH, AND THE MOON Sun"
Earth
Moon
1.99 X 10^® 6.96 X 10» 1410 274 618 26-37* 2.6 X lO'"-' 2.4 X 10* y '
5.98 X 102* 6.37 X 10* 5520 9.81 11.2 0.997 1.50 X 10** 1.00 y*
7.36 X 1022 1.74 X 10* 3340 1.67 2.38 27.3 3.82 X lOV 27.3 d /
Property Mass (kg) Mean radius (m) Mean density (kg/m^) Surface gravity (m/s^) Escape velocity (km/s) Period of rotation'" (d) Mean orbital radius (km) Orbital period
The Sun radiates energy at the rate of 3.90 X 10^^ W; just outside the Earth’s atmosphere solar energy is received, assuming normal incidence, at the rate of 1380 W/m^. ^ The Sun—a ball of gas—does not rotate as a rigid body. Its rotational period varies between 26 d at the equator and 37 d at the poles. Measured with respect to the distant stars. ^ About the galactic center. ^ About the Sun. ^ About the Earth.
SOME PROPERTIES OF THE PLANETS Mercury
Venus
Earth
Mars
150
228
Jupiter
Neptune
Pluto
2,870
4,500
5,900 248
57.9
108
Period of revolution (y)
0.241
0.615
1.00
1.88
11.9
29.5
84.0
165
Period of rotation'* (d)
58.7
243*
0.997
1.03
0.409
0.426
0.451*
0.658
6.39
Orbital speed (km/s)
47.9
35.0
29.8
24.1
13.1
9.64
6.81
5.43
4.74
Inclination of axis to orbit
O.O^’
2.6**
23.5*^
24.0®
3.08®
26.7®
82.1“
28.8®
65®
Inclination of orbit to Earth’s orbit
7.00
3.39*’
—
1.85®
1.30“
2.49®
0.77®
1.77®
17.2®
Eccentricity of orbit
0.206
0.0068
0.0167
0.0934
0.0485
0.0556
0.0472
0.0086
0.250
Equatorial diameter (km)
4,880
12,100
12,800
6,790
143,000
120,000
51,800
49,500
3,400
Mass (Earth = 1)
0.0558
0.815
1.000
0.107
318
95.1
14.5
17.2
0.002
Mean density (g/cm^)
5.60
5.20
5.52
3.95
1.31
0.704
1.21
1.67
0.5(?)
Surface gravity*" (m/s^)
3.78
8.60
9.78
3.72
22.9
9.05
7.77
11.0
0.03
Escape speed (km/s)
4.3
10.3
11.2
5.0
59.5
35.6
21.2
23.6
1.3
Known satellites
0
0
1
2
16 + rings
19 + rings
15 + rings
8 + rings
1
A -4
1,430
Uranus
Mean distance from Sun (10^ km)
Measured with respect to the distant stars. ^ The sense of rotation is opposite to that of the orbital motion. Measured at the planet’s equator.
778
Saturn
APPENDIX D PROPERTIES OF THE ELEMENTS
Element Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Boron Bromine Cadmium Calcium Californium Carbon Cerium Cesium Chlorine Chromium Cobalt Copper Curium Dysprosium Einsteinium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold Hafnium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium
Symbol Ac A1 Am Sb Ar As At Ba Bk Be Bi B Br Cd Ca Cf C Ce Cs Cl Cr Co Cu Cm Dy Es Er Eu Fm F Fr Gd Ga Ge Au Hf He Ho H In I Ir Fe Kr La Lr
Atomic number, Z 89 13 95 51 18 33 85 56 97 4 83 5 35 48 20
98
6 58 55 17 24 27 29 96 66 99 68 63 100 9 87 64 31 32 79 72 2 67 1 49 53 77 26 36 57 103
Molar mass (g/mol)
Density (g/cm^) at 2 0 X
(227) 26.9815 (243) 121.75 39.948 74.9216
2.699 13.7 6.69 1.6626 X 105.72
( 210)
137.33 (247) 9.0122 208.980 10.811 79.909 112.41 40.08 (251)
12.011 140.12 132.905 35.453 51.996 58.9332 63.54 (247) 162.50 (252) 167.26 151.96 (257) 18.9984 (223) 157.25 69.72 72.61 196.967 178.49 4.0026 164.930 1.00797 114.82 126.9044 192.2 55.847 83.80 138.91 (260)
3.5 1.848 9.75 2.34 3.12 (liquid) 8.65 1.55 2.25 6.768 1.873 3.214 X 10"M 0X) 7.19 8.85 8.96 —
8.55 —
9.07 5.245 —
1.696 X 10-3 (OX) —
7.90 5.907 5.323 19.32 13.31 0.1664 X 10-3 8.79 0.08375 X 10-3 7.31 4.94 22.5 7.87 3.488 X 10-3 6.145 —
Melting point rc) 1050 660 994 630.5 -1 8 9 .2 817(28 at.) 302 725 12.78 271.3 20.79 -1 2 320.9 839 3550 798 28.40 -1 0 1 1857 1495 1083.4 1340 1412
Boiling point rc) 3200 2467 2607 1750 -1 8 5 .7 613 337 1640
0.092 0.900
2970 1560 2550 58 765 1484
1.83 0.122 1.11 0.293 0.226 0.624
__
0.691 0.188 0.243 0.486 0.448 0.423 0.385
3443 6.69 -3 4 .6 2672 2870 2567 —
2567
—
—
1529 822
2868 1527
—
-2 1 9 .6 (27) 1313 29.78 937.4 1064.43 2227 -2 7 2 .2 1474 -259.34 156.6 113.5 2410 1535 -1 5 6 .6 918 —
Specific heat (J/g-C “) at 25 X
—
-1 8 8 .2 (677) 3273 2403 2830 2808 4602 -2 6 8 .9 2700 -252.87 2080 184.35 4130 2750 -1 5 2 .3 3464 —
—
0.205 0.523 0.331 —
0.205
—
0.172 —
0.167 0.163 —
0.753 —
0.234 0.377 0.322 0.131 0.144 5.23 0.165 14.4 0.233 0.218 0.130 0.447 0.247 0.195 — (Continued)
A-5
A-6
Appendix D
Element Lead Lithium Lutetium Magnesium Manganese Mendelevium Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Samarium Scandium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium
Properties o f the Elements
Symbol Pb Li Lu Mg Mn Md Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Pr Pm Pa Ra Rn Re Rh Rb Ru Sm Sc Se Si Ag Na Sr S Ta Tc Te Tb T1 Th Tm Sn Ti W U V Xe Yb Y Zn Zr
Atomic number, Z 82 3 71 12
25 101
80 42 60 10
93 28 41 7 102 76 8
46 15 78 94 84 19 59 61 91 88 86 75 45 37 44 62 21 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40
Molar mass (g/mol) 207.19 6.939 174.97 24.305 54.9380 (258) 200.59 95.94 144.24 20.180 (237) 58.69 92.906 14.0067 (259) 190.2 15.9994 106.4 30.9738 195.09 (244) (209) 39.098 140.907 (145) (231) (226) (222) 186.2 102.905 85.47 101.107 150.35 44.956 78.96 28.086 107.68 22.9898 87.62 32.066 180.948 (98) 127.60 158.924 204.38 (232) 168.934 118.71 4788 183.85 (238) 50.942 131.30 173.04 88.905 65.37 91.22
Density (g/cm’) at 20‘C 11.36 0.534 9.84 1.74 7.43 —
13.55 10.22 7.00 0.8387 X 10-3 20.25 8.902 L 1 ^ 9 X 10-3 22.57 1.3318 X 10-3 12.02 1.83 21.45 19.84 9.24
0.86 6.773 7.264 5.0 9.96 X 10-^ (0“C) 21.04 12.44 1.53 12.2 7.49 2.99 4.79 2.33 10.49 0.9712 2.54 2.07 16.6 11.46 6.24 8.25 11.85 11.72 9.31 7.31 4.54 19.3 19.07 6.1 5.495 X 10-3 6.966 4.469 7.133 6.506
Melting point VC)
327.50 180.54 1663 649 1244 —
-3 8 .8 7 2617 1021 -248.67 640 1453 2468
-210 3045 -2 1 8 .4 1554 44.25 1772 641 254 63.25 931 1042 1600 700 -7 1 3180 1965 38.89 2310 1074 1541 217 1410 961.9 97.81 769 112.8 2996 2172 449.5 1357 304 1750 1545 231.97 1660 3410 1132 1890 -111.79 819 1552 419.58 1852
Boiling point VC)
Specific heat (J/g*C^) a t2 5 “C
1740 1342 3402 1090 1962
0.129 3.58 0.155 1.03 0.481
—
—
357 4612 3074 -2 4 6 .0 3902 2732 4742 -1 9 5 .8
0.138 0.251 0.188 1.03 1.26 0.444 0.264 1.03
5027 -1 8 3 .0 3140 280 3827 3232 962 760 3520 (3000)
0.130 0.913 0.243 0.741 0.134 0.130
1140 -6 1 .8 5627 3727 686 3900 1794 2836 685 2355 2212 882.9 1384 444.6 5425 4877 990 3230 1457 (3850) 1950 2270 3287 5660 3818 3380 -1 0 8 1196 5338 907 4377
0.758 0.197
—
0.092 0.134 0.243 0.364 0.239 0.197 0.569 0.318 0.712 0.234 1.23 0.737 0.707 0.138 0.209 0.201 0.180 0.130 0.117 0.159 0.226 0.523 0.134 0.117 0.490 0.159 0.155 0.297 0.389 0.276
The values in parentheses in the column of atomic masses are the mass numbers of the longest-lived isotopes of those elements that are radioactive. Melting points and boiling points in parentheses are uncertain. All the physical properties are given for a pressure of one atmosphere except where otherwise specified. The data for gases are valid only when these are in their usual molecular state, such as Hj, He, O2, Ne, etc. The specific heats of the gases are the values at constant pressure. Source: Handbook of Chemistry and Physics, 71st edition (CRC Press, 1990).
APPENDIX E PERIODIC TABLE OF THE ELEMENTS ALKALI METALS (including hydrogen)
NOBLE GASES
1
2
H
He
3
4
5
6
7
8
9
10
Li
Be
B
c
N
O
F
Ne
11
12
13
14
15
16
17
18
Na Mg
A1
Si
P
s
Cl Ar
31
32
33
34
35
19
20
21
22
23
24
Cr Mn Fe Co Ni Cu Zn Ga Ge As Se
Br Kr
42
52
53
54
Sn Sb Te
I
Xe
85
86
25
26
27
28
29
30
K
Ca Sc
Ti
V
37
38
39
40
41
Rb Sr
Y
Zr Nb Mo Tc Ru Rh Pd Ag Cd In
55
56
57-71
Cs Ba 87
88
Fr Ra
89-103
43
44
46
45
47
48
49
50
51
36
73
74
75
76
77
78
81
82
83
H f Ta
w
Re Os
Ir
Pt Au Hg Tl
Pb
Bi Po At Rn
104
106
107
108
109
♦♦
**
**
*♦
72
105
Rf* Ha*
57 Lanthanide series
59
60
80
84
J 61
62
63
64
65
66
67
68
69
70
71
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 91
92
Ac Th Pa
U
89 Actinide series
58
79
90
93
95
96
97
98
99
100
101
102
103
Np Pu Am Cm Bk Cf Es Fm Md No Lr
* The names of these elements (Rutherfordium and Hahnium) have not been accepted because of conflicting claims of discovery. A group in the USSR has proposed the names Kurchatovium and Nielsbohrium.
94
** Discovery of these elements has been reported but names for them have not yet been adopted.
APPENDIX F ELEMENTARY PARTICLES
1.
THE FUNDAMENTAL PARTICLES
LEPTONS
Particle
Symbol
Anti particle
e“ Vc
Ve
Electron Electron neutrino Muon Muon neutrino Tau Tau neutrino
T
Vt
Vt
Charge (e)
Spin {h/2n)
Rest energy (MeV)
-1 0 -1 0 -1 0
1/2 1/2 1/2 1/2 1/2 1/2
0.511 <0.00002 105.7 <0.3 1784 < 40
Mean life (s) 00 00 2.2 X 10-* 00 3.0 X 10-‘» 00
Typical decay products
e" + V, + //- +
+ V,
QUARKS Flavor
Symbol
Antiparticle
Charge (e)
Spin (h/lii)
Rest energy^ (MeV)
Other property
Up Down Charm Strange Top^ Bottom
u d c s t b
u d c s t b
+ 2/3 -1 /3 + 2/3 -1 /3 + 2/3 -1 /3
1/2 1/2 1/2 1/2 1/2 1/2
300 300 1500 500 >40,000 4700
C = S= T= B= 0 C = S^T = B = 0 Charm (C) = + 1 Strangeness (5) = —1 Topness (T) = + 1 Bottomness (^) = —1
FIELD PARTICLES Particle Graviton^ Weak boson Weak boson Photon Gluon
A -8
Symbol
w+, w z® y
g
Interaction
Charge {e)
Spin {h/2n)
Rest energy (GeV)
Gravity Weak Weak Electromagnetic Strong (color)
0 ±1 0 0 0
2
0 80.6 91.2 0 0
1 1 1 1
Appendix F
2.
A -9
SO M E C O M PO SITE PARTICLES
BARYONS
Particle Proton Neutron Lambda Omega Delta Charmed lambda
Symbol p n A® QA++ A?
Spin {h/ln)
Rest energy (MeV)
Mean life
p n A® £2A++ A?
Charge {e) + 1 0 0 -1 +2 + 1
1/2 1/2 1/2 3/2 3/2 1/2
938 940 1116 1673 1232 2285
>10“ 889 2.6 X 10-'® 8.2X 10-" 5.7 X 10-^“ 1.9X 10-'»
Quark content
Anti particle
Charge {e)
Spin (h/2n)
Rest energy (MeV)
Mean life
ud uu H- dd us di ud cd cc ub bb
n~ 7^ KK^ P~ D¥ B-
+ 1 0 + 1 0 + 1 + 1 0 + 1 0
Quark content uud udd uds sss uuu udc
Anti particle
Typical decay
(s)
;t® + e+(?) p + e- + V. p + n~ A® + K p + jf*' A° + n*
MESONS
Particle Pion Pion Kaon Kaon Rho D-meson Psi B-meson Upsilon
Symbol tC
K-^ YsP D-^ ¥
Y
Y
0 0 0 0 1 0 1 0 1
140 135 494 498 768 1869 3097 5278 9460
Typical decay
(s) 2.6 X 8.4 X 1.2 X 0.9 X 4.5 X 1.1 X 1.0 X 1.2 X 1.3 X
10-* 10-” 10-* 10-'® 1 0 -2 4
10-'2 10-2® 10-'2 10-2®
y+ y tC tC
+ n~ H- 7l~ K - + TT-^ + 7t^ e"*" + e“ D - + 71+ + 7T+ e+ + e“
^ The rest energies listed for the quarks are not those associated with free quarks; since no free quarks have yet been observed, measuring their rest energies in the free state has not yet been possible. The tabulated values are effective rest energies corresponding to constituent quarks, those bound in composite particles. ^ Particles expected to exist but not yet observed. Source: “Review of Particle Properties,” Physics Letters B, vol. 239 (April 1990).
APPENDIX G CONVERSION FACTORS
Conversion factors may be read directly from the tables. For example, 1 degree = 2.778 X 10“^ revolutions, so 16.7® = 16.7 X 2.778 X 10“^ rev. The SI quantities are
capitalized. Adapted in part from G. Shortley and D. Wil liams, Elements of Physics, Prentice-Hall, Englewood Cliffs, NJ, 1971.
PLANE ANGLE 0
//
'
RADIAN
rev
1 degree =
1
60
3600
1.745 X 10-2
2.778 X 10-2
1 minute =
1.667 X 10-2
1
60
2.909 X 10-“
4.630 X 10-’
1 second =
2.778 X 10-“
1.667 X 10-2
1
4.848 X 10-*
7.716 X 10“^
1 RADIAN =
57.30
3438
2.063 X 10’
1
0.1592
1 revolution =
360
2.16 X 10“
1.296 X 10‘
6.283
1
SOLID ANGLE 1 sphere = An steradians = 12.57 steradians
LENGTH
ft
mi
1 centimeter =
1
10-2
10-’
0.3937
3.281 X 10-2
6.214 X 10-*
1 METER =
100
1
10-2
39.37
3.281
6.214 X 10“^
1 kilometer =
10’
1000
1
3.937 X 10“
3281
0.6214
1 inch =
2.540
2.540 X 10-2
2.540 X 10-’
1
8.333 X 10-2
1.578 X 10“5
1 foot =
30.48
0.3048
3.048 X 10-“
12
1
1.894 X 10“^
1 mile =
1.609 X 10’
1609
1.609
6.336 X 10“
5280
1
cm
METER
km
in.
1 light-year = 9.460 X 10'^ km 1 parsec = 3.084 X 10'^ km 1 fathom = 6 ft 1 Bohr radius = 5.292 X 10“ “ m
1 angstrom = 10“ ‘° m 1 nautical mile = 1852 m = 1.151 miles = 6076 ft 1 fermi = 10“ *^ m
1 yard = 3 ft 1 ro d = 16.5 ft 1 mil = 10“^ in. 1 nm = 10“’ m
AREA METER2
cm^
ft2
in.^
1 SQUARE METER =
1
10“
10.76
1550
1 square centimeter =
10-“
1
1.076 X 10-2
0.1550
1 square foot =
9.290 X 10-2
929.0
1
144
1 square inch =
6.452 X 10-“
6.452
6.944 X 10-2
1
) acres 1 barn = ■10“^* m^
A-IO
1 acre = 43,560 ft^ 1 hectare = 10"* m^ = 2.471 acre
Appendix G
A -1 1
VOLUME cm^
METERS
ft2
L
in.^
1 CUBIC METER =
1
10*
1000
35.31
6.102 X 10“
1 cubic centimeter =
10-*
1
1.000 X 10-2
3.531 X 10-2
6.102 X 10-2
1 liter =
1.000 X 10"’
1000
1
3.531 X 10-2
61.02
1 cubic foot =
2.832 X 10-2
2.832 X 10“
28.32
1
1728
1 cubic inch =
1.639 X 10-*
16.39
1.639 X 10-2
5.787 X 10-“
1
1 U.S. fluid gallon = 4 U.S. fluid quarts = 8 U.S. pints = 128 U.S. fluid ounces = 231 in.^ 1 British imperial gallon = 277.4 in^ = 1.201 U.S. fluid gallons
MASS KILOGRAM
slug
u
oz
lb
ton
1 gram =
1
0.001
6.852 X 10-2
6.022 X 1022
3.527 X 10-2
2.205 X 10-2
1.102 X 10-*
1 KILOGRAM =
1000
1
6.852 X 10-2
6.022 X 102*
35.27
2.205
1.102 X 10-2
14.59
1
8.786 X 1022
514.8
32.17
1.609 X 10-2
g
1 slug =
1.459 X 10“
1u =
1.661 X 10-2“ 1.661 X 10-22
1 ounce =
28.35
2.835 X 10-2
1 pound =
453.6
1 ton =
9.072 X 10*
1.138 X 10-2* 1
5.857 X 10-2* 3.662 X 10-22
1.830 X 10-20
1
6.250 X 10-2
3.125 X 10-2
1.943 X 10-2
1.718 X 1022
0.4536
3.108 X 10-2
2.732 X 102*
16
1
0.0005
907.2
62.16
5.463 X 1025
3.2 X 10“
2000
1
1 metric ton = 1000 kg Quantities in the colored areas are not mass units but are often used as such. When we write, for example, 1 kg 2.205 lb this means that a kilogram is a mass that weighs 2.205 pounds under standard condition of gravity (g = 9.80665 m/s^).
DENSITY slug/ft^
KILOGRAM/METER2
1
515.4
0.5154
1 KILOGRAM per METER2 =
1.940 X 10-2
1
1 gram per cm^ =
1.940
1000
3.108 X 10-2 53.71
1 slug per ft^
1 pound per
=
1 pound per in.^ =
lb/in.2
lb/ft2
g/cm^
32.17
1.862 X 10-2
0.001
6.243 X 10-2
3.613 X 10-2
1
62.43
3.613 X 10-2
16.02
1.602 X 10-2
1
5.787 X 10-“
2.768 X 10^
27.68
1728
1
Quantities in the colored areas are weight densities and, as such, are dimensionally different from mass densities. See note for mass table.
A -1 2
Appendix G
Conversion Factors
SPEED km/h
ft/s
METER/SECOND
mi/h
cm/s
1 foot per second =
1
1.097
0.3048
0.6818
30.48
1 kilometer per hour =
0.9113
1
0.2778
0.6214
27.78
3.6
1
2.237
100
1.609
0.4470
1
44.70
0.01
2.237 X 10-2 1
1 METER per SECOND = 3.281 1 mile per hour =
1.467
1 centimeter per second =
3.281 X 10-2 3.6 X 10-2
1 knot = 1 nautical mi/h = 1.688 ft/s
1 mi/min = 88.00 ft/s = 60.00 mi/h
FO R C E dyne
NEWTON
1
1 dyne =
lb
pdl
10-5
2.248 X 10-«
7.233 X 10-5
gf 1.020 X 10-5
1.020 X 10-«
kgf
1 NEWTON =
10’
1
0.2248
7.233
102.0
0.1020
1 pound =
4.448 X 10’
4.448
1
32.17
453.6
0.4536
1 poundal =
1.383 X 10“
0.1383
3.108 X 10-2
1
14.10
1.410 X 10-2
1 gram-force =
980.7
9.807 X 10-5
2.205 X 10-5
7.093 X 10-2
1
0.001
1 kilogram-force =
9.807 X 105
9.807
2.205
70.93
1000
1
Quantities in the colored areas are not force units but are often used as such. For instance, if we write 1 gram-force “= ” 980.7 dynes, we mean that a gram-mass experiences a force of 980.7 dynes under standard conditions of gravity (g = 9.80665 m/s^)
ENERGY, W ORK, HEAT erg
Btu
ft*lb
hp-h
JOULE
cal
kW -h
eV
MeV
1 British thermal unit =
1
1.055 X 10‘®
777.9
3.929 X 10-“
1055
252.0
2.930 X 10-“
6.585 X 102'
6.585 X 10'5
1.174 X 10"'“
7.070 X 10*2
1 erg =
9.481 X 10-"
1
7.376 X 10-*
3.725 X 10-'“
10-2
2.389 X io-«
2.778 X 10-'“
6.242 X 10"
6.242 X 105
1.113 X 10-2“
670.2
1 foot-pound =
1.285 X 10-5
1.356 X 102
1
5.051 X 10-2
1.356
0.3238
3.766 X 10-2
8.464 X 10'*
8.464 X 10'2
1.509 X 10-'2
9.037 X 10’
1 horsepowerhour =
2545
2.685 X 10*5
1.980 X 10*
1
2.685 X 105
6.413 X 105
0.7457
1.676 X 1025
1.676 X 10'’
2.988 X 10-"
1.799 X 10'5
1 JOULE =
9.481 X 10-“
102
0.7376
3.725 X 10-2
1
0.2389
2.778 X 10"^
6.242 X 10'*
6.242 X 10'2
1.113 X 10-'2
6.702 X 10’
1 calorie =
3.969 X 10-5
4.186 X 102
3.088
1.560 X 10-5
4.186
1
1.163 X 10-5
2.613 X 10'’
2.613 X 10'5
4.660 X 10-'2
2.806 X lO'o
1 kilowatt-hour =
3413
3.6 X 10*5
2.655 X 10«
1.341
3.6 X 105
8.600 X 105
1
2.247 X 1025
2.247 X 10'’
4.007 X 10-"
2.413 X 10“
1 electron volt =
1.519 X 10-22
1.602 X 10-'2
1.182 X 10-“
5.967 X 10-25
1.602 X 10-'’
3.827 X 10-20
4.450 X 10-25
1
10-5
1.783 X 10-55
1.074 X 10-’
1.519 X 10-“
1.602 X 10-*
1.182 X 10-'5
5.967 X 10-20
1.602 X 10-'5
3.827 X 10'“
4.450 X 10-20
105
1
1.783 X 10-50
1.074 X 10-5
1 kilogram =
8.521 X 10*5
8.987 X 1025
6.629 X 10“
3.348 X 10'“
8.987 X 10'5
2.146 X 10“
2 .4 9 7 X lO 'o
5.610 X 10*5
5.610 X 102’
1
6.022 X 1025
1 unified atomic mass unit =
1.415 X 10-'5
1.492 X 10-5
1.101 X 10-“
5.559 X 10-'2
1.492 X lO-'o
3.564 X 10-"
4.146 X 10-'2
9.32 X 10*
932.0
1.661 X 10-22
1
1
million electron volts =
Quantities in the colored areas are not properly energy units but are included for convenience. They arise from the relativistic mass-energy equivalence formula E = mc^ and represent the energy equivalent of a mass of one kilogram or one unified atomic mass unit (u).
u
Appendix G
A-13
PRESSURE dyne/cm^
atm
cm Hg
inch of water
1.013 X 10« 406.8
1.013 X 10’
76
1 atmosphere =
1
1 dyne per cm^ =
9.869 X 10-’ 1
4.015 X 10-^ 7.501 X 10-’ 0.1
1 inch of water" at 4®C =
2.458 X 10-^ 2491
1
1 centimeter of mercury*’ 1.316 X 10-^ 1.333 X 10* 5.353 at 0°C = 1 PASCAL = 1 pound per in.^ =
6.805 X 10-2 6.895 X 1(P 27.68
1 pound per
4.725 X lO-** 478.8
=
2116
1.405 X 10-2 2.089 X 10-2
249.1
3.613 X 10-2 5.202
1
1333
0.1934
5.171
0.1922
14.70
lb/ft2
0.1868
4.015 X 10-2 7.501 X 10-“ 1
9.869 X 10-* 10
lb/in.2
PASCAL
6.895 X 102
3.591 X 10-2 47.88
" Where the acceleration of gravity has the standard value 9.80665 m/s^. 1 bar = 10^ dyne/cm^ = 0.1 MPa 1 millibar = 10^ dyne/cm^ = 10^ Pa
27.85
1.450 X 10-“ 2.089 X 10-2 144
1
6.944 X 10-2 1
1 torr = 1 millimeter of mercury
POWER Btu/h
ft*lb/s
hp
cal/s
kW
WATT
1 British thermal unit per hour =
1
0.2161
1 foot-pound per second =
4.628
1
3.929 X 10-“
6.998 X 10-2
2.930 X 10-“
0.2930
1.818 X 10-2
0.3239
1.356 X 10-2
1.356
1 horsepower =
2545
550
1
178.1
0.7457
745.7 4.186
1 calorie per second =
14.29
3.088
5.615 X 10-2
1
4.186 X 10-2
1 kilowatt =
3413
737.6
1.341
238.9
1
1000
1 WATT =
3.413
0.7376
1.341 X 10-2
0.2389
0.001
1
MAGNETIC FLUX maxwell
WEBER
1 maxwell =
1
10-*
1 WEBER =
10‘
1
gauss
MAGNETIC HELD TESLA
milligauss
1 gauss =
1
10-“
1000
1 TESLA =
10“
1
102
1 milligauss =
0.001
10-2
1
1 tesla = 1 weber/meter^
APPENDIX H MATHEMATICAL FORMULAS
GEOMETRY
MATHEMATICAL SIGNS AND SYMBOLS
Circle of radius r: circumference = 2nr\ area = nr^. Sphere of radius r: area = 47rr^; volume = ^nr^. Right circular cylinder of radius r and height h: area = Inr^ H- 2nrh\ volume = nr^h. Triangle of base a and altitude h: area = {ah.
= equals equals approximately ~ is of the order of magnitude of # is not equal to = is identical to, is defined as > is greater than ( » is much greater than) < is less than ( c is much less than) ^ is greater than or equal to (or, is no less than) ^ is less than or equal to (or, is no more than) ± plus or minus (V4 = ± 2) a is proportional to 1 the sum of X the average value of x
QUADRATIC FORMULA If
+ c = 0, then x =
—b ± y/b^ — 4ac 2a ■
TRIGONOMETRIC FUNCTIONS OF ANGLE 6 y sin 6 = -
X cos6 = -
tan ^ = -
cot 0 = —
j axis
PRODUCTS OF VECTORS Let i, j, k be unit vectors in the x, y, z directions. Then L i = j*j = k ‘k = 1,
r r sec 6 = - CSC 6 = X y
i X i = j X j = k X k = 0, i X j = k,
PYTHAGOREAN THEOREM a^-\- b^ =
i*j = j- k = k*i = 0,
j X k = i,
k X i = j.
Any vector a with components a^,, ay, a^ along the x, y, z axes can be written a= + ay} + a X Let a, b, c be arbitrary vectors with magnitudes a, b, c. Then a X (b + c) = (a X b) + (a X c) (5 a) X b = a X (5 b) = 5 (a x b)
(5 = a scalar).
Let 0 be the smaller of the two angles between a and b. Then a*b = b*a = a ^ x + i
TRIANGLES Angles A, B, C Opposite sides a, b, c A - y B - y c = 180" sin A _ sin B _ sin C a b c c^ = a^-\- b^ — 2ab cos C
A-14
j
cos 6 k
axb=-bxa= b^ by b2 (ayb^
bya^x “ 1“ {^a2b^
|a X b| = ab sin 6 a*(b X c) = b*(c X a) = c*(a x b) a X (b X c) = (a*c)b — (a*b)c
Appendix H
EXPONENTIAL EXPANSION
TRIGONOMETRIC IDENTITIES sin(90® — 6) = cos 6 cos(90® — 6) = sin 6 sin 6/cos 6 = \BXi6
1+ ^ + 2 [+ 3 [ +
sin^ 6 H- cos^ 6 = 1 sec^ 6 —tan^ 6 = 1 csc^ 6 — cot^ 6 = 1
LOGARITHMIC EXPANSION
sin 29 = 2 sin 6 cos 6 cos 26 = cos^ 6 — sin^ 6 = 2 cos^ 6 — I = I - 2 sin^ 6 sin(a ± fi) = sin a c o s f i ± cos a sin P cos(a ± p ) = cos O' cos)? + sin a sin p * / />x tan a ± tan tan(a ± P ) = - r ^ ------------^ 1 -h tan a tan sin a ± s \ n p = 2 sin ± P) cos + p)
ln (l + a:) = X —
—
%xne = e - - + -
BINOMIAL THEOREM________
------
„
e*
rx ^ 0^
26^ -jy + • • •
tan 6 = 6
DERIVATIVES AND INTEGRALS In what follows, the letters u and v stand for any functions of x, and a and m are constants. To each of the indefinite integrals should be added an arbitrary constant of integration. The Handbook o f Chemistry and Physics (CRC Press Inc.) gives a more extensive tabulation.
,
ax d , . du s '™ '- - * d , , . du
dv
dx 5. — ln x = dx X d dv
1.
dx =
2.
audx= = a | udx
3.
(u -\-v)d x
4.
x " 'd x = - ^ (m ¥ ^ -\) m -r 1
X
J
j
=
udx+
j
V dx
5. du
6.
dv , { du , u - r d x = u v — I V — dx dx j dx
7. — e^ = e^ dx
7.
e^ d x = e^
8. - r sin X = cos x dx
8.
s\n x d x = —co sx
o9. — d cosx = —sin x dx
9.
cosxdx
= sin x
10. - ^ t a n x = sec^x dx *, d , 11. — c o tx = —csc^x dx
10.
X2X i x d x
=
11.
sin^ x d x =
12.
e~^ d x = ~
13.
xe~^ dx =
14.
x^e~^ dx = — r (a^x^ + 2^zx + 2)e a^ n\ x^e~^ dx =
12. — secx = ta n x s e c x dx d cscx = —c o tx c s c x 13. —
, ^ d . du 15. — sinw = c o s w - r dx dx ,^ d . du 16. - r cos w = —sin w— dx dx
15.
10.
(W < 1 )
TRIGONOMETRIC EXPANSIONS (6 in radians)
c o s 0 = l - - + - --------
I^
A -15
e
\n
\x
a
|sec x| — isin 2x e~^
— 7
ax
{ax
+
\)e~^
V a
APPENDIX I COMPUTER PROGRAMS
netic-field program ) is small enough th at the approxim a tions used in the integrations do not introduce significant errors into the calculation. As the interval is m ade smaller, the nu m b er o f intervals becom es larger, thereby increasing both the length o f tim e it takes to run the pro gram and the q u antity o f data it produces. T o reduce the am o u n t o f output, each program has a provision for lim it ing the o u tp u t to a certain n um ber o f points. T his lim its only the o u tp u t data and does not affect the calculation. The total length o f tim e for which the program follows the m otion is equal to the product o f the n um ber o f intervals (N T) and the size o f each interval (DT).
H ere we give exam ples o f program s th at can be used to calculate the trajectories o f charged particles m oving in electric an d m agnetic fields. They are based on the pro gram s for kinem atic calculations involving n onconstant forces given in A ppendix I o f V olum e 1. T he program s are w ritten in the BASIC language and can easily be adapted for m ost personal com puters. By m aking appro priate m odifications, these program s can be used for any electric o r m agnetic field configuration. In both pro gram s, all quantities are in SI units. In using both program s, special care should be taken to determ ine th at the tim e interval D T (specified in line 130 in the electric-field program and in line 120 in the m ag
1.
ELECTRIC FIELDS
This program was used in Section 2 8 - 6 to find the m o tion o f a particle m oving along the axis o f a ring o f charge. T he program calculates m otion in the x z plane only. The X and z co m ponents o f the electric field are specified in lines 180 and 190. As given here, the electric field com po
nents are = 0, = zRX/2eo{z^ + /?^)^ (see Eq. 22 o f C hapter 28), with R = 0.02 m and A = + 2 X 10~^ C /m . T he o u tp u t o f the program is plotted in Fig. 16^ o f C hapter 28.
PR O G R A M LISTIN G
10 ■ C f l L C i j l h t i o h s o f m o t i o n 20 ' I N E L E C T R I C F I E L D I N 20 E0 = 8 , 85E-12 4 0 ■ S P E C I F Y N f l S 8 ,. C H R R G E , 50 M = 1 , 67E - 2? 60 Q = 1 , 6 E - 1 9 7 0 !-! = 0 3 0 ,il = , tr 0 90 1 0 0 '<■' Z = - 7 0 0 0 0 0 ! 1 10 ' S P E C l F t s t a r t i n g T I N E
OF KZ
CHFI RGED
PARTICLE
PLANE
INITIAL
TIME
P
08I
T I
0 N. ,
INTERVAL,
AND
INITIAL
NUMBER
OF
VELOCIT'i
INTERVALS
(C ontinued) A '1 6
Appendix I
120 130 140 150 160 170 180 190 200 210 220 230 240 250 260
270 2y0 290 300 510 520 330 340 ^30 400
T = 0 DT = 5 E - 1 0 HT=3000 ' S P E C I F Y HLI MB E R 0 F I N T E R f i L S T 0 P R I HT H=1 5 ' S P E C I F Y ELECTRI C F I E L D COMPOHENTS DEF FNEK' ::X, Z> = 0 DEF F N E 2 < X , 2 > = 2 4: , 0 3 4: .0 0 0 0 0 0 2 .••■2 .••■E 0 .•••• < i PRI NT " TIME X 2 'JX LPRI NT " TI ME X 2 I..I jxj l„l 7 P R IN T 1.1S IN G " # # . # # •••••••• •••• " T .. X ' LI'::' U 7 L P RI N T LI S I N G " # # . # # •••••••• •••• " T , FOR I = 1 TO NT T = T + DT RX = Q:t;FNEX'::X, 2; |..•••M RZ = Q:t;FNE2CX, 2 ; X = X + '...'X:t:DT + , 5:t:flX:f:DT:f:DT 7 = 2 + '...'24:DT + . 5*fi2:f:DT:f:DT I..I y = OX + flX:4DT I..17 = '.,.'2 + fi2:4DT IF '::NT..-N::'4:INT< I..-••NT4:N::' < > I P R I N T LIS I N G " # # , # # •••• ••• •••• " T . LPRI NT US I NG ' NEXT I END
: + (. , M3 .J I I
"7
2
A -1 7
■
II
SAMPLE OUTPUT 0
1 2
5 4 5 6 7 8
9 1
1 1 1 1 1
TI NE . 00E + 00 . 0 0 E “ 07 . 00E-07 . 00E-07 . 00E-07 . 00E-07 . 00E-07 . 00E-07 . 00E-07 . 00E-07 . 00E-06 . 10E-06 . 20E-06 . 50E-06 . 40E-06 . 50E-06
2.
y 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 00E+00 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 00E+00 0 . 00E+00 0 . 00E+00 0 . 00E+00 0 . 00E+00
7 5 . 0 0 E- 0 1 4 , 31E-01 3 , 63E-01 2 . 98E-01 2 , 37E-01 1 , 81E-01 1 . 35E-01 1 . 06E-01 1 . 02E-01 1 . 25E-01 1 , 66E-01 2 . 20E-01 2 , 80E-01 3 . 44E-01 4 . 1 lE-01 4 . 80E-01
i„i y
l,,l 7
0 . 00E + 00 0 . 00E + 00 0 . 00E + 00 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 00E + 00 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 0 0 E+ 0 0 0 . 00E + 00 0 . 00E + 00 0 . 00E + 00 0 . 00E + 00 0 . 00E + 00 0 . 00E + 00 0 . 0 0 E+ 0 0
- 7 . 0 0 E+ 0 5 - 6 . S5E+05 - 6 . 64E+05 - 6 . 55E+05 - 5 . 9 0 E+ 0 5 - 5 . lbE + 05 - 5 . 8 9 E+ 0 5 - 1 . 78E + 05 9 . 8 9 E+ 0 4 5 . 57E + 05 4 . 86E +05 5 . 72E+05 S . 24E+05 6 . 58E+05 6 . 81E+05 6 . 9 7 E+ 0 5
MAGNETIC FIELDS
This program, which was used in Section 34-3, calculates the motion o f a particle confined to the xy plane and subject to a magnetic field in the z direction. The zcom po
nent o f the magnetic field (in units o f tesla) is specified in line 170. As given here, the field is uniform. The output o f the program is plotted in Fig. 17a o f Chapter 34.
A -1 8
Appendix I
Computer Programs
PROGRAM LISTING
10 ' CRLCULf l TI ON OF MOTION OF CHRR GED P R R T I C L E IN XY PLRNE 20 ' WITH MfiGNETIC F I ELD IN Z DI R ECT ION 30 ' S P E C I F Y MASS, CHRRGE, IN I T I R L PO S I T I ON IN I T I R L U E L O C I T Y 40 M = 6 . 6 4 5 E - 2 7 50 Q = 3 .2 E - 1 9 60 X = 0 70 V = 0 S0 y X = 3 0 0 0 0 0 0 ! 90 y V = 0 1 0 0 ■SPE CI FY S T R RT I N G T I ME , TI ME I NT E R UR L , NUMBER OF I NT E RURL S 110 T = 0 120 DT = I E - 10 1 30 NT = 10000 140 ■SPE CI FY NUMBER OF I NTER URLS TO PRI NT 150 N = 20 1 6 0 ■SPE CI FY Z COMPCI NENT OF MRGNETIC F I E L D 170 DEF FNBZCX.. Y> = . 15 Y X UY ■ TI ME 180 PRI NT ■ UY ■ ■ Y U X ■ TI ME 190 L P R I N T ■ X UY V .! T , X , 2 0 0 PRI NT USING ■■##. •••• ■ ' T , X V UX, UY 210 L P R I N T USING '■## ## •■ FOR I = 1 TO NT 220 2 3 0 T = T + DT 240 RX = Q*'...'Y:*:FNBZCH Y>/M 250 RY = - Q * U X * F N B Z ( X Y > ■M 260 X = X + UX#DT + 5 * R X# DT * DT 270 Y = Y + UY:+:DT + 5*R^r * DT*DT 2 8 0 OX = UX + RX*DT 290 UY = UY + RY*DT ::n t ..-n > * i n t (:I..-' n t * n : < > I THEN 330 300 IF ■ JT,X, X UY PRI NT USI NG ■■##. 310 UX, UY " T X L P R I N T USING '■ ## Y 320 NEXT I 330 4 0 0 END ,
,
II
II
,
..
.
.i
,
,
,
, ,
:>
.•••.
II
,
,
..
,
SAMPLE OUTPUT
TI ME 0 . 0 0 E +0 0 5 . 00E-08 1 . 00E-07 1 . 50E-07 2 . 00E-07 2 . 50E-07 3 . 00E-07 3 . 50E-07 4 . 00E-07 4 . 50E-07 5 . 0 0 E —07 5 . 50E-07 6 . 00E-07 6 . 50E-07
0 . 00E +00 1 , 47E-01 2 , 75E-01 3 , 67E-01 4 . 12E-01 4 . 04E-01 3 , 44E-01 2 , 39E-01 1 . 03E-01 - 4 , 55E-02 - 1 . S9E-01 - 3 . 08E-01 - 3 , 87E-01 - 4 , 16E-01
0 . 00E +0 0 - 2 , 68E-02 -1 , 0 4E -01 - 2 . 21E-01 -3.63E-01 - 5 . 12E-01 - 6 . 49E-01 - 7 , 55E-01 - 8 , 18E-01 - 8 , 29E-01 - 7 . 86E-01 - 6 . 95E-01 - 5 . 69E-01 - 4 , 22E-01
3 . 0 0 E +0 6 2 . 81E +06 2 . 2 5E +06 1 . 41E +06 3 . 77E +05 - 6 . 9 9 E +0 5 - 1 . 6 9 E +0 6 - 2 . 4 6 E +0 6 - 2 . 9 1E +06 - 2 . 9 9 E +0 6 - 2 . 6 8 E +0 6 - 2 . 0 2 E +0 6 -1 . 1 l E +06 - 5 . 14E +04
0 . 00 E +0 0 ""1 1U6 E +U6 -1 . 98E +06 - 2 . 6 5 E +0 6 - 2 . 9 8E +06 - 2 . 9 2 E +0 6 - 2 . 4 8E +06 - 1 . 7 3 E +0 6 - 7 . 49 E +0 5 3 . 27E +05 1 . 36E +0 6 2 . 2 2E +06 2 . 7 9E +06 3 . 0 0 E +0 6 (Continued)
Appendix I
00E- 07 50E- 07 00E- 07 50E-07 00E- 07 50E- 07 00E- 06
92E-01 17E-01
75E-01 45E-01
0 1 E -0 1
0 8 E -0 2
94E-02 0 0E - 02 28E-01 36E-01
27E-03 86E- 03 69E-02 70E-01
1 . 01 E +0 6 1 . 95 E +0 6 63 E +0 6 9 8 E +0 6 9 4 E +0 6 52 E +0 6 1 77 E +0 6
2 . 8 3 E + 06 2 , 29E+06 1 . 45E+06 4 , 29E+05 6 . 50E+05 1 , 65E+06 2 , 43E+06
A -1 9
APPENDIX J NOBEL PRIZES IN PHYSICS*
1905 1906
Wilhelm Konrad Rdntgen Hendrik Antoon Lorentz Pieter Zeeman Antoine Henri Becquerel Pierre Curie Marie Sklowdowska-Curie Lord Rayleigh (John William Strutt) Philipp Eduard Anton von Lenard Joseph John Thomson
1907
Albert Abraham Michelson
1901 1902 1903 1904
1908 Gabriel Lippmann 1909 Guglielmo Marconi Carl Ferdinand Braun 1910 Johannes Diderik van der Waals 1911 Wilhelm Wien 1912 Nils Gustaf Dalen 1913
Heike Kamerlingh Onnes
1914 Max von Laue 1915 William Henry Bragg William Lawrence Bragg 1917 Charles Glover Barkla 1918 Max Planck 1919 Johannes Stark 1920 Charles-Edouard Guillaume 1921
Albert Einstein
1922
Niels Bohr
1923
Robert Andrews Millikan
1924 Karl Manne Georg Siegbahn 1925 James Franck Gustav Hertz 1926 Jean Baptiste Perrin
1845 -1923 for the discovery of x-rays 1853-1928 for their researches into the influence of magnetism upon radiation 1865 -1943 phenomena 1 8 5 2 - 1908 for his discovery of spontaneous radioactivity 1859 -1906 for their joint researches on the radiation phenomena discovered by 1867 -1934 Professor Henri Becquerel 1842 -1919 for his investigations of the densities of the most important gases and for his discovery of argon 1862 -1947 for his work on cathode rays 1856-1940 for his theoretical and experimental investigations on the conduction of electricity by gases 1852-1931 for his optical precision instruments and metrological investigations carried out with their aid 1845-1921 for his method of reproducing colors photographically based on the phenomena of interference 1874-1937 for their contributions to the development of wireless 1850-1918 telegraphy 1837-1932 for his work on the equation of state for gases and liquids 1864 -1928 for his discoveries regarding the laws governing the radiation of heat 1 8 6 9 - 1937 for his invention of automatic regulators for use in conjunction with gas accumulators for illuminating lighthouses and buoys 1 8 5 3 - 1926 for his investigations of the properties of matter at low temperatures which led, inter alia, to the production of liquid helium 1879-1960 for his discovery of the diffraction of Rontgen rays by crystals 1862-1942 for their services in the analysis of crystal structure by means of 1890-1971 x-rays 1877-1944 for his discovery of the characteristic x-rays of the elements 1858-1947 for his discovery of energy quanta 1874-1957 for his discovery of the Doppler effect in canal rays and the splitting of spectral lines in electric fields 1861 -1938 for the service he has rendered to precision measurements in Physics by his discovery of anomalies in nickel steel alloys 1879-1955 for his services to Theoretical Physics, and especially for his discovery of the law of the photoelectric effect 1885-1962 for the investigation of the structure of atoms, and of the radiation emanating from them 1868 -1953 for his work on the elementary charge of electricity and on the photoelectric effect 1888 - 1979 for his discoveries and research in the field of x-ray spectroscopy 1882 -1964 for their discovery of the laws governing the impact of an electron 1887 -1975 upon an atom 1 8 7 0 - 1942 for his work on the discontinuous structure of matter, and especially for his discovery of sedimentation equilibrium
* See Nobel Lectures, Physics, 1901- 1970, Elsevier Publishing Company for biographies of the awardees and for lectures given by them on receiving the prize.
A -20
Appendix J 1927
Arthur Holly Compton Charles Thomson Rees Wilson
1892-1962 1869-1959
1928
Owen Willans Richardson
1879-1959
A -2 1
for his discovery of the effect named after him for his method of making the paths of electrically charged particles visible by condensation of vapor for his work on the thermionic phenomenon and especially for the discovery of the law named after him for his discovery of the wave nature of electrons for his work on the scattering of light and for the discovery of the effect named after him for the creation of quantum mechanics, the application of which has, among other things, led to the discovery of the allotropic forms of hydrogen for the discovery of new productive forms of atomic theory
1929 Prince Louis-Victor de Broglie 1930 Sir Chandrasekhara Venkata Raman 1932 Werner Heisenberg
1892-1987 1888-1970
1933
1887-1961 1902-1984 1 8 9 1 - 1974for his discovery of the neutron 1883-1964 for the discovery of cosmic radiation 1905-1991 for his discovery of the positron 1881-1958 for their experimental discovery of the diffraction of electrons 1 8 9 2 - 1975 by crystals 1901-1954 for his demonstrations of the existence of new radioactive elements produced by neutron irradiation, and for his related discovery of nuclear reactions brought about by slow neutrons 1901-1958 for the invention and development of the cyclotron and for results obtained with it, especially for artificial radioactive elements 1888-1969 for his contribution to the development of the molecular ray method and his discovery of the magnetic moment of the proton 1898-1988 for his resonance method for recording the magnetic properties of atomic nuclei 1900-1958 for the discovery of the Exclusion Principle (Pauli Principle) 1882-1961 for the invention of an apparatus to produce extremely high pressures, and for the discoveries he made therewith in the field of highpressure physics 1892-1965 for his investigations of the physics of the upper atmosphere, especially for the discovery of the so-called Appleton layer 1897-1974 for his development of the Wilson cloud chamber method, and his discoveries therewith in nuclear physics and cosmic radiation 1907-1981 for his prediction of the existence of mesons on the basis of theoretical work on nuclear forces 1903-1969 for his development of the photographic method of studying nuclear processes and his discoveries regarding mesons made with this method 1897- ■1967 for their pioneer work on the transmutation of atomic nuclei by artificially accelerated atomic particles 19031905- •1983 for their development of new methods for nuclear magnetic precision 1912methods and discoveries in connection therewith 1888- 1966 for his demonstration of the phase-contrast method, especially for his invention of the phase-contrast microscope 1882-1970 for his fundamental research in quantum mechanics, especially for his statistical interpretation of the wave function 1891-1957 for the coincidence method and his discoveries made therewith for his discoveries concerning the fine structure of the hydrogen 1913spectrum 1911 for his precision determination of the magnetic moment of the electron 1910 •1989 for their researchers on semiconductors and their discovery of the transistor effect 1908 1991 1902 ■1987 1922 for their penetrating investigation of the parity laws which has led to 1926 important discoveries regarding the elementary particles for the discovery and the interpretation of the Cerenkov effect 1904 1908 1990 1895 1971 1905 1989 for their discovery of the antiproton 1920for the invention of the bubble chamber 19261915-1990 for his pioneering studies of electron scattering in atomic nuclei and for his thereby achieved discoveries concerning the structure of the nucleons 1929for his researches concerning the resonance absorption of y-rays and his discovery in this connection of the effect which bears his name
1938
Erwin Schrodinger Paul Adrien Maurice Dirac James Chadwick Victor Franz Hess Carl David Anderson Clinton Joseph Davisson George Paget Thomson Enrico Fermi
1939
Ernest Orlando Lawrence
1943
Otto Stem
1944
Isidor Isaac Rabi
1945 1946
Wolfgang Pauli Percy Williams Bridgman
1947
Sir Edward Victor Appleton
1948
Patrick Maynard Stuart Blackett
1949
Hideki Yukawa
1935 1936 1937
1950 Cecil Frank Powell 1951
1953
Sir John Douglas Cockcroft Ernest Thomas Sinton Walton Felix Bloch Edward Mills Purcell Frits Zemike
1954
Max Bom
1955
Walther Bothe Willis Eugene Lamb
1952
1956 1957 1958 1959 1960 1961
Polykarp Kusch William Shockley John Bardeen Walter Houser Brattain Chen Ning Yang Tsung Dao Lee Pavel Aleksejecic Cerenkov ir ja Michajlovic Frank Igor’ Evgen’ evic Tamm Emilio Gino Segre Owen Chamberlain Donald Arthur Glaser Robert Hofstadter Rudolf Ludwig MOssbauer
1901-1976
A - 2 2 1 Appendix J 1962 1963
Nobel Prizes in Physics
Lev Davidovic Landau Eugene P. Wigner
1908-1968 1902-
Maria Goeppert Mayer J. Hans D. Jensen 1964 Charles H. Townes Nikolai G. Basov Alexander M. Prochorov 1965 Sin-itiro Tomonaga Julian Schwinger Richard P. Feynman 1966 Alfred Kastler
1906-1972 1907-1973 1915192219161906-1979 19181918-1988 1902-1984
1967
Hans Albrecht Bethe
1906-
1968
Luis W. Alvarez
1911-1988
1969
Murray Gell-Mann
1929-
1970 Hannes Alven Louis Neel
19081904-
1971 Dennis Gabor 1972 John Bardeen Leon N. Cooper J. Robert Schrieffer 1973 Leo Esaki Ivar Giaever Brian D. Josephson
1900-1979 1908-1991 19301931192519291940-
1974 Antony Hewish Sir Martin Ryle 1975 Aage Bohr Ben Mottelson James Rainwater 1976 Burton Richter Samuel Chao Chung Ting 1977 Philip Warren Anderson Nevill Francis Mott John Hasbrouck Van Vleck 1978 Peter L. Kapitza Amo A. Penzias Robert Woodrow Wilson 1979 Sheldon Lee Glashow Abdus Salam Steven Weinberg 1980 James W. Cronin Val L. Fitch 1981 Nicolaas Bloembergen Arthur Leonard Schawlow Kai M. Siegbahn 1982 Kenneth Geddes Wilson
19241918-1984 192219261917-1986 19311936192319051899-1980 1894-1984 19261936193219261933193119231920192119181936-
Subrehmanyan Chandrasekhar William A. Fowler 1984 Carlo Rubbia Simon van der Meer
1910191119341925-
1983
1985 1986
Klaus von Klitzing Ernst Ruska Gerd Binnig Heinrich Rohrer
1943190619471933-
for his pioneering theories of condensed matter, especially liquid helium for his contribution to the theory of the atomic nucleus and the elementary particles, particularly through the discovery and application of fundamental symmetry principles for their discoveries concerning nuclear shell structure for fundamental work in the field of quantum electronics which has led to the construction of oscillators and amplifiers based on the maser-laser principle for their fundamental work in quantum electrodynamics, with deepploughing consequences for the physics of elementary particles for the discovery and development of optical methods for studying Hertzian resonance in atoms for his contributions to the theory of nuclear reactions, especially his discoveries concerning the energy production in stars for his decisive contribution to elementary particle physics, in particular the discovery of a large number of resonance states, made possible through his development of the technique of using hydrogen bubble chamber and data analysis for his contribution and discoveries concerning the classification of elementary particles and their interactions for fundamental work and discoveries in magneto-hydrodynamics with fruitful applications in different parts of plasma physics for fundamental work and discoveries concerning antiferromagnetism and ferrimagnetism which have led to important applications in solid state physics for his discovery of the principles of holography for their development of a theory of superconductivity for his discovery of tunneling in semiconductors for his discovery of tunneling in superconductors for his theoretical prediction of the properties of a super-current through a tunnel barrier for the discovery of pulsars for his pioneering work in radioastronomy for the discovery of the connection between collective motion and particle motion and the development of the theory of the structure of the atomic nucleus based on this connection for their (independent) discovery of an important fundamental particle for their fundamental theoretical investigations of the electronic structure of magnetic and disordered systems for his basic inventions and discoveries in low-temperature physics for their discovery of cosmic microwave background radiation for their unified model of the action of the weak and electromagnetic forces and for their prediction of the existence of neutral currents for the discovery of violations of fundamental symmetry principles in the decay of neutral K mesons for their contribution to the development of laser spectroscopy for his contribution of high-resolution electron spectroscopy for his method of analyzing the critical phenomena inherent in the changes of matter under the influence of pressure and temperature for his theoretical studies of the structure and evolution of stars for his studies of the formation of the chemical elements in the universe for their decisive contributions to the large project, which led to the discovery of the field particles W and Z, communicators of the weak interaction for his discovery of the quantized Hall resistance for his invention of the electron microscope for their invention of the scanning-tunneling electron microscope
Appendix J 1987
Karl Alex Muller J. Georg Bednorz 1988 Leon M. Lederman Melvin Schwartz Jack Steinberger 1989 Hans G. Dehmelt Wolfgang Paul Norman F. Ramsey 1990 1991
Richard E. Taylor Jerome I. Friedman Henry W. Kendall Pierre-Gilles de Gennes
1992 George Charpak
1927195019221932192119221913191519291930192619321924-
A-23
for their discovery of a new class of superconductors for experiments with neutrino beams and the discovery of the muon neutrino for their development of techniques for trapping individual atoms for his discoveries in atomic resonance spectroscopy, which led to hydrogen masers and atomic clocks for their experiments on the scattering of electrons from nuclei, which revealed the presence of quarks inside nucleons for discoveries about the ordering of molecules in substances such as liquid crystals, superconductors, and polymers for his invention of fast electronic detectors for high energy particles
ANSWERS TO ODD NUMBERED PROBLEMS
CHAPTER 27
CHAPTER 30
I. 2.74 N on each charge. 3. 0.50 C. 5. {a) 1.77 N. (^)3.07N . 7. ^, = -4 ^2 . 9. 24.5 N, along the angle bisector. II . 1.00 //C and 3.00 //C, of opposite sign. 13. (a) A charge - 4 ^ /9 must be located on the line segment joining the two positive charges, a distance L/3 from the -\-q charge. 15. q = Q /l. 17. (b) 2.96 cm. 19. a/>/2. 23. H ^m eo d ^lq Q . 25. 2.89 X 10“’ N. 27. 3.8 N. 29. 5.08 m below the electron. 31. 13.4 MC. 33. (a) 57.1 TC; no. (b) 598 metric tons. 35. (a) Boron. {b) Nitrogen, (c) Carbon.
1. (a) 484 keV. (b) Zero. 3. (a) 27.2 fj = 170 keV. (b) 3.02 X 10"^' kg, in error by a factor of about three. 5. (a) 3.0 kN. (b) 240 MeV. 7. {a) 30 GJ. (b) 7.1 km/s. (c) 9.0 X 10“ kg. 9. (a) 256 kV. (b) 0.745c. 11. 2.6 km/s. 13. — . 15. 2.17d. 17. ( a ) 24.4kV/m. 87T€q Lc, Cj J (b) 2.93 kV. 19. (a) 132 MV/m. (b) 8.43 kV/m. 21. (a) 32 MeV. 23. (a) -3 .8 5 kV. (/>) -3 .8 5 kV. 25. -1 .1 nC. 27. (a) 0.562 mm. (b)8l3 V. 29. 637 MV. 31. (a) qdl2ne.Qa(a + d). 33. (a) —5.40 nm. {b) 9.00 nm. (c) No. 37. 186 pJ. 41. (a) 4.5 m. (A) No. 45. 746 V/m. 47. - 2 .3 X lO^' V/m. 49. -3 9 .2 V/m. 51.
CHAPTER 28 1. 10.5 mN/C, westward. 3. 203 nN/C, up. 5. 144 pC. 7. 19.5 kN/C. 9.9:30. 11. (i») Parallel to p. 1 1 15. (a);.(b)-
4neo {R^ +
2n^€o {R^ +
19. To the right. 25. /?/V3. 27. (a) lO 4 n C .0 ) 1.31 X 10*’. (c) 4.96 X 10-‘. 29. (a) 6.50 cm. (b) 4.80 nC. 35. q/&n€oRK 37. (a) 6.53 cm. (b) 26.9 ns. (c) 0.121. 39. (a) 585 kN/C, toward the negative charge. (b) 93.6 fN, toward the positive charge. 41. 5e. 43. 1.64X 10-” C (=2.5% high). 45. 1.2 mm. 47. The upper plate; 4.06 cm. 49. (a) Zero. (b) 8.50 X 10“^^ N • m. (c) Zero. 51. 2pE cos 0o53. (a) iq/iKoO^.
CHAPTER 29 1. -0.0078 N • mVC. 3. (a) - nR^E. (b) nR}E. 5. 208kN-m VC. 1. q/6eo- 9. 4.6/tC. 13. (a) 22.3 N • m^C. (b) 197 pC. 15. (a) 452 nC/m^ (b) 51A kN/C. 17. (a) - Q. (b) - Q. (c) - «2 + q). (d) Yes. 19. (a) 53 MN/C. {b) 60 N/C. 21. (a) 322 nC. (*) 143 nC. 23. (a) Zero, {b) ff/Co, to the left, (c) Zero. 25. 5.11 nC/m^. 27. 5.09 29. (a) q!2n€QLr, radially inward. (b) —q on both inner and outer surfaces. (c) q/2fKoLr, radially outward. 31. —1.13 nC. 33. (a) X/2n€or. (b) Zero. 35. 270 eV. 37. (a) 2.19 MN/C, radially out. (b) 436 kN/C, radially in. 39.9 7 .9 cm. 41.0.557/?. 45. (b) pR}!2i^r.
A -24
( a ) - ^ [ V Z 7 T 7 - j.] . ( b ) ^ \ \ - - ^ = ] . 4neo 4 tco L VZT+y^J (d) 3L/4. 53. (a) F, = {b) q, = q/3; q^ = 2q/3. 55. 840 V. 57. 2.0 X 10-«. 59. {a) Zero, (b) Zero, (c) Zero. (d) Zero, (c) No. 63. (a) 1.75 kV. (b) 7.40 cm. 65. 9.65 kW.
CHAPTER 31 1. 7.5 pC. 3. 3.25 mC. 5. 0.546 pF. 7. (a) 84.5 pF. (b) 191 cm^ 11. 9090. 13. 7.17/rF. 15. (a)2.4pF. ib) = 46 = 480 pC. (c) F, = 120 V; F, = 80 V. 17. (a) d/3. (b) 3d. 19. (a) 942 pC. (b) 91.4 V. 23. (a) 45.4 V. (b) 52.7 pC. (c) 146 pC. 25. (a) 50 V. (b) Zero. 27. (a) 4| = 9.0 pC, q^ = pC, q, = 9.0pC, 4, = 16 pC. (b) 4i = 8.40 pC, 42 = 16.8 pC. 4, = 10.8 pC, q^ = 14.4 pC. 29. 200 nJ. 31. (a) 28.6 pF. (b) 17.9 nC. (c) 5.59 pJ. (d) 482 kV/m. (c) 1.03 J/m ^ 33. 74.1 m J/m \ 35. (a) 2.0 J. 37. (a) 2 V. (b) U, = €qA V^!2d\ U f = €qA V^/d. (c) €oA FV2
CHAPTER 32 1. (a) 1.33 kC. (ft) 8.31 X lO^'. 3. {a) 9.41 A/m^ north. 5. 0.400 mm. 7. 0.67 A, toward the negative terminal. 9. 7.1 ms. 11. (a) 654 nA/m^ (ft) 83.4 MA. 13. 52.5 min. 15. (a) 95.0 pC. (ft) 158 C \ 17. 0.59 Ci. 19. (a) 1.5 kA. (ft) 53 MA/m^ (c) 110 n£2-m; platinum. 23. (a) 250°C. 25. (a) 380 pV. (ft) Negative, (c) 4.3 min. 27. 54 £i. 29. 3.
Answers to Odd Numbered Problems 31. (a) 6.00 mA. (b) 15.9 nV. (c) 21.2 nil. 33. I l9 0 ( il- m ) - '. 35. (a) Cu: 55.3 A/cm^; Al: 34.0 A/cm^. (b) Cu; 1.01 kg; Al: 0.495 kg. 37. (a) Silver, (b) 60.8 nil. 39.0.036. 41. (a)8.52kil.(Z>)4.51/zA- 43. 7.16 fs. 45. 18 kC. 47. (fl) 1.03 kW.(Z») 34.5 cents. 49. (a) $4.46. (b) 144 il. (c) 833 mA. 51. (a) 2.88 X 10". (b) 24.0 M (c) 1.14 kW; 23.1 MW. 53. (a) 6.1 m. (b) 13 m. 55. 27.4 cm/s. 57. 311 nJ. 59. (a) 37.0 min. (b) 122 min. 61. (a) 1.37L. (b)0.730A.
CHAPTER 33 I. 10.6 kJ. 3. 13 h 38 min. 5 . - l O V . 7. (a) 14 il. (b) 35 mW. 9. (a) 44.2 V. (b) 21.4 V. (c) Left. I I . The cable. 13. (a) 1.5 kfl. (b) 400 mV. (c) 0.26%. 15. (a) 3.4 A. (b) 0.29 V. A nd (a) 0.59 A. (b) 1.7 V. 17. 4.0 il; 12 il. 19. 7.5 V. 21. 262 i l or 38.2 il. 23. (a) In parallel, (b) 72.0 il; 144 il. 25. (a)PA = 16.3 n il-m ; Pa = 7.48 n il-m . (b) jA =J b = 62.3 kA/cm^. (c) = 10.2 V/m; = 4.66 V/m. {d) K< = 435 V; Vb = 195 V. 27. {a)R/2.(b)5R/S. 29. (a) 3R/4. (b) 5R/6. 31. (a) R 2. (b)R t. 33. 6 P R .
i ___ £___ V
35. (50 k W )l___ _ j , X in cm. V ^22OOO+ 0 0 0 + 10 10 a:: - xV ’ 37. (a) /, = 668 mA, down; = 85.7 mA, up; I3 = 582 mA, up. ( ( ) ) - 3.60 V. 39. (a) 0.45 A. 41.0.90% . 45. (a) Top: 70.9 mA; 4.91 V; bottom: 55.2 mA; 4.86 V. {b) Top: 69.3 il; bottom: 88.0 il. 49. 4.61. 51. (a) 2.20 s. (b) 44 mV. 53. 2.35 M il. 55. (a) 955 pC/s. (b) 1.08 pW. ( c ) 2.74 p W . (d) 3.82 p W .
A-25
(d) Zero. 47. fioir^/lna^. 49. 3/o/8, into the page. 5 1.109 m. 5 3 .272 mA. 55. (a) Negative. (^) 9.7 cm.
CHAPTER 36 1. 57 /iWb. 3. (a) 31 mV. (b) Right to left. 5. (a) 1.12 mi2. (b) 1.27 T/s. 7. (^)58m A . 9. 4.97/iW. 11. (b)No. 15. (a) 28.2 fiW. (b) From c to b. 17. 80 //V; clockwise. 19. Zero. 21. iLBt/m, away from G. 23. 455 mV. 27. (b) Design it so that Nab = (5/2;:) m^. 29. 6.3 rev/s. 31. 25 liC. 33. (a) 253 //V. (b) 610 M . (c) 154 nW. (
39. {Baryaat. 41. (a) -1 .2 0 mV. 2nRD(D + b) (b) - 2 .7 9 mV. (c) 1.59 mV. 47. (a) 34 V/m. (b) 6.0 X 10" m /s’. 49. (a) 0.15°. ib)
CHAPTER 37 1. + 3 Wb. 3. (a) Stable, (b) Unstable, (c) Stable. (d) Unstable. 5 (b) 15. 23. 31.
7. (a)514G V /m .
19.0 mT. 11. 24 mJ/T. 13. (a) 0.86 pT. (b) 0.68 A/m. 0.58 K. 17. (a) 150 T. {b) 600 T. 19. Yes. (a) 3.0 pT. (b) 9.0 X 10"” J. 27. (a) 630 MA. 1660 km. 33. 61 pT; 84°.
CHAPTER 38 CHAPTER 34
1. 100 nWb. 3. 261pH/m . 5. (a) 600pH . (/>) 120.
1. 1: +; 2: —, 3: 0; 4: - . 3. (a) 3.4 km/s. 5. 8.2 X 10’. 7. 0.75k, T. 9. (a) To the East, (b) 6.27 X 10“*m /s^ (c) 2.98 mm. 11. (a) 0.34 mm. (b) 2.6 keV. 13. (a) 1.11 X 10’ m/s. {b) 0.316 mm. 15. {a) 2600 km/s. (b) 110 ns. (c) 140 keV. (d) 70 kV. 19. (a) K^. (b) K„/2. 21. (a) r^x/2. (b) r^. 23. (a) B{qml2V)'i'^6jc. (b) 7.91 mm. 25. (a) - q . (b) nm/qB. 27. (a) 0.999928c. 29. An alpha particle. 31. (a) 78.6 ns. (b) 9.16 cm. (c) 3.20 cm. 33. 240 m. 39. 37 cm/s. 41. 467 mA; left to right. 43. (a) 330 MA. (b) 1.1 X 10‘^ W. 45. 4.2 C. 47. -0.414k, N. 49. (a)0; 138 mN; 138 mN. 53. InaiB sin normal to plane of ring, up. 55. 1.63 A. 57. 2.1 GA. 59. (a) - 2.86k, A-m^. (^) 1.10k, A-m^.
7. 7.87 H. 15. ( ^ ) l n - . 17. 29.8 i2. 19. (a) 4.78 mH. \2 n / a (b) 2.42 ms. 21. 42 + 20/, V. 23. 12 A/s. 25. (a) /, = /j = 3.33 A. (b) i, = 4.55 A; = 2.73 A. (c) /, = 0; /2 = 1.82 A. ( < / ) = 0. 29. (a) 13.2 H. (/>) 124 mA. 31. 63.2 M J/m’. 33. 150 MV/m. 35. (a) 78 kJ. {b) 3.7 kg. 37. (a) 117 H. (b) 225 pJ. 39. (a) po/’/VV8/r’r ’. (6) (poV’A/V47t) In - . 41. 12 PJ. a 45. 123 mA. 47. 38 pH. 51. (a) 6.08 ps. (A) 164 kHz. (c) 3.04 ps. 53. (a) No. (b) 6.1 kHz. (c) 16 nF. 55. (a) 5800 rad/s. (()) 1.1 ms. 57. (a) qj>l3. (b) //T = 0.152. 59. (a) 6.0:1. (b) 36 pF; 220 pH. 61. (a) 180 pC. (b) 778. (c) 67 W. 63. (a) Zero, (b) 2i. 65. {L/R) In 2. 67. 8.7 m il. 69. 2.96 ii.
CHAPTER 35 1. 7.7m T. 3. 12 nT. 5. (a) 0.324 fN, parallel to current. (Z?) 0.324 fN, radially outward, (c) Zero. 7. 30.0 A, antiparallel. 9. (a) 4. (Z?) i. 11. (a) 2.43 A-m^. {b) 46 cm. 13. 2 rad. 15. 19. {b) ia \ 21.
( V
T“ “ out of figure. An \ b a)
In ^ 1 + -^ ^ ; up. 25. i?i/(a’ + /»’).
29. (c) \nia} sin(2;r//2). 31. (a) (2iioi/3nL)(2 yfl H- >/T0). (b) Greater. 35. 606 //N, toward the center of the square. 37. (b) 2.3 km/s. 39. (a) - 2 .5 /^T*m. (b) Zero. 4 1 .6 .0 p T -m . 4 5 . ( a ) | 4 . ( 6 ) f ^ . ( c ) f ^ 4 : : ^ . 2;tc’ 2nr 2nr o r — b^
CHAPTER 39 1. 377 rad/s. 3. (a) 3.75 krad/s. {b) 23.4 i2. 5. (a) 39.1 mA. (b) Zero, (c) 32.6 mA. (d) Taking energy. 7. (a) 6.73 ms. {b) 2.24 ms. (c) Capacitor, (d) 56.6 pF. 13. 1.0 kV > ) Leads, (c) Capacitive. (
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Answers to Odd Numbered Problems
CHAPTER 40 3. r = 2.5 m; 10 tn. 5. Change the potential across the plates at 1.0 kV/s. 7. (a) 1.84 A. (b) 140 GV/m •s. (c) 460 mA. (d) 578 nT •m. 9. (a) 840 mA. (b) Zero, (c) 1.3 A. 11. (a) 623 nT. (6)2.11 TV/m -s. 13. 2.27 pT. 19. 1900 km in radius, independent of its length.
CHAPTER 41 3. (a) 4.5 X lO^-* Hz. (6) 10,000 km. 5. 5.0 X 10-^' H. 7. 1.07 pT. 11. 100 kJ. 13. 4.62 X 10"” W/m^. 15. 78 cm. 17. {a) 883 m. (6) No. 19. (a) 6.53 nT. (6) 5.10 mW /m^ (c) 8.04 W. 21. (a) ±EBa^/fi(, for faces parallel to the x y plane; zero th ro u ^ each of the other four faces. (6) Zero. 23. (a) 9.14 mW/cm^ (6) 1.68 MW. 25. (a) 76.8 mV/m. (O = cB„. (6) 256 pT. (c) 12.6 kW. 29. (a) — = c; (b )S
- fWoC/ — ) sin 2(ot sin 2kx.
31. (a) E = &!r ln(6/a); B = Ho€l2nRr. (b)S=S^I2nRr^\n(bla). 33. 0.043 kg-m/s. 35. 7.7 MPa. 37. (a) 586 MN. (b) 1.66 X 10-“'. 39. (a) 94.3 MHz. (b) +z; 960 nT. (c) 1.98 m” '; 593 Mrad/s. (d) 110 W / m \ (e) 678 nN; 367 nPa. 41. 7(2 - /)/c . 45. (a) 3.60 G W /m l (b) 12.0 Pa. (c) 16.7 pN. (d) 2.78 km /s^ 47. 1.06 km^ 49. (b) 585 nm.
CHAPTER 42 1. (a) 515 nm; 610 nm. (b) 555 nm; 541 THz; 1.85 fs. 3. (a) 8.68 y. (b) 4.4 My. 5. 67 ps. 11. Yellow-orange. 13. (6) 0.80c. 15. ±0.0036 nm. 17. (a) 1.66 X 10“ ’. (b) 0.83 X 10"’. 19. (a) 6 min. (b) 12 min. (c) 6 min. 23. (a) 0.067. (6) 10°; 7.0°; 2.2°. 25. 4.43 nm. 27. 78.9°.
CHAPTER 43 1. (a) 38.0°. (6) 52.9°. 3. 1.56. 5. 1.95 X 10* m/s. 7. 1.25. 9. 1.5. 13. 74 m. 15. (6) 0.60 mm. 1 9 .4 3 mm. 2 1 .7 5 0 m. 23. 1.24 < n < 1.37. 27. (a)2c. (b) V. 33. 390 cm beneath the mirror surface. 3 5 . /„ „ = (10/9)/ow. 37. Six. 39. (a) 405 nm. (6) 2.37/zm. (c) 112°. 41. (a) 72.07°. (b) From A to B. 43. (a) 45. 187 cm. 47.(6)0.170. 49. (6) 60.2 /is. 51. (a) Yes. (6) No. (c) 43°.
CHAPTER 44 I . 11.0 cm. 3. (a ) + , + 4 0 , —2 0 ,+ 2 , no, yes. (6) Plane, <», <», —10, yes. (c) Concave, +40, + 60, —2, yes, no. (d) Concave, + 20, + 40, +30, yes, no. (e) Convex, —20, + 20, +0.50, no, yes. ( / ) Convex, —, —40, —18, +180, no, yes. (g) —20, —, —, +5.0, +0.80, no, yes. (6 ) Concave, +8.0, + 16, +12, —, yes. 9. (6) 2.0. (c) None. I I . 12 cm to the left of the lens. 13. 2.5 mm. 17. (a) 40 cm. (6) 80 cm. (c) 240 cm. (d) —40 cm. (e) —80 cm. ( / ) —240 cm. 23. 22 cm. 27. 30 cm to the left of the diverging lens; virtual; upright; m = 0.75. 29. (6) No. (c) Light passes undeviated. 31. (a) 73.6 cm on the side of the lens away from the mirror. (6) Real, (c) Upright, (d) 0.289. 35. 2.0 mm.
37. (a) 2.34 cm. (6) Smaller. 39. (a) 5.3 cm. (6) 3.0 mm. 41. 103. 43. 25 ms.
CHAPTER 45 1. (a) 0.22 rad. (6) 12°. 3. 2.3 mm. 5. 650 nm. 7.0.103 mm. 9. 600nm . 13. (a) 0.253 mm. (6) Maxima and minima are interchanged. 17. 3.2 X 10“*. 19. 0°. 23. (a) 1.21 m; 3.22 m; 8.13 m. 27. 124 nm. 29. (a) 552 nm. (6) 442 nm. 31. 215 nm. 33. 643 nm. 35. 2.4 ;zm. 37. 840 nm. 39. 141. 41. 1.89/zm. 43. (a) 34. (6) 45. 45. 1.00 m. 47. (a) 88%. (6) 95%. 49. 588 nm. 51. 1.0003.
CHAPTER 46 1. 690 nm. 3. (a) 0.430°. (6) 118/zm. 5. (a)A„ = 2Aj. (6) Minima coincide when m<, = 2wia. 7. 173/zm. 9. 1.49 mm. 11. (a) 0.186 °. (6) 0.478 rad. (c) 0.926. 13. 5.07°. 15. (6) 0; 4.493 rad; 7.725 rad; . . . (c)-0 .5 0 ; 0.93; 1.96; . . . 17. (a) 137/zrad.(6) 10.4 km. 19.51.8 m. 21. 1400 km. 23. 15 m. 25. (a) 6 .8 °. (6) No answer. 27. (a) 0.35°. (6)0.94°. 29. (6)70/zm. (c) Three times the lunar diameter. 31. XD/d. 33. (a) 3. 35. (a) 5.0 fim- (b) 20 fim-
CHAPTER 47 1. (a) 3.50/zm. (6)9.69°; 19.7°; 30.3°; 42.3°; 57.3°. 3. 523 nm. 5. (a) 6.0 /zm. (6) 1.5 /zm. (c)w = 0, 1,2, 3, 5, 6 , 7,9. 9. (6) Halfway between principal maxima, (c) /„ /9 . 13. 400 nm < A < 635 nm. 1 5 .3 . 21.491. 23.3650. 25. (a) 9.98 /zm. (6) 3.27 mm. 27. (a) 0.032 °/nm; 0.077 °/nm; 0.25 °/nm. (6) 40,000; 80,000; 120,000. 31. 2.68°. 33. 26 pm; 39 pm. 35. 49.8 pm. 39. 0.206 nm. 41. (a) ao/v/2; ao/VS; Oo/yfiO; aj4V i.
CHAPTER 48 I. (a) —y. (6) = 0,Ey — 0, Ey = —cB sin (ky + (ot). (c) Linearly polarized; z direction. 3. (a) 2.14 V/m. (6)20.3pPa. 5. i 7.27/128. 9. 15.8 W/m^. I I . (a) 0.16. (6) 0.84. 13. (a) 53.1 °. (6) Yes, slightly. 15. 55°31'to55°46'. 17. 12/zm. 21. (a) Turns plane of polarization by 90°. (6) Reverses handedness of circular polarization, (c) Light remains unpolarized. 23. (a) 2.90 X 10“ “• kg-m*/s^ (6) 2.88 h.
CHAPTER 49 1. 91 K. 3. (a) 1.06 mm; microwave. (6) 9.4/zm; infrared. (c) 1.6 /zm; infrared, (d) 500 nm; visible, (e) 0.29 nm; x ray. ( / ) 2.9 X 10"*' m; hard gamma ray. 5. 580 mW. 9. (a) 138 K. (6) 21.0/zm. 11. 1.44 W. 13. 780 K. 15. (6) 6 °C. 19. 0.796Te. 21. (a) 92.1%. (6) 58.2%. 23. (a) 1.4 X 10'^ 6.0 X 10'^ 1.4 X 10'*, Hz. (6) 5.9, 25,60, meV. (c) 27, 64, 120, N/m. 25. (a) 1110 J. (6)713J. 27. (a) 2.11 eV. 29. 1.17 eV. 31. Ultraviolet. 33. Cesium, lithium, barium. 35. (a) No. (6) 544 nm; green. 37. 172 nm. 39. (a) 1.17 V. (6) 641 km/s. 43. 2.63 m*. 45. (a) The infrared bulb. (6) 1.97 X 10*0. 47. (a) 3.10 keV. (6) 14.4 keV. 49. (a) 2.97 X 10“ s"'. (6) 48,600 km. (c) 281 m. (d) 5.91 X 10'*/m* • s; 1.97 X lO"* m"*. 51. (a) 29.8 keV.
Answers to Odd Numbered Problems {b) 7.19 X 10'* Hz. (c) 1.59 X lO” ” kg • m/s = 29.8 keV/c. 53. 2.95 cm/s. 55. (a) 2.87 pm. (b) 5.89 pm. 59. 2.64 fm. 61. (a) 4.86 pm. (b) -4 2 .1 keV. (c) 42.1 keV. 63. 42.6°. 65. (b) 1.12 keV.
CHAPTER 50 I. (a) 1.7 X 10“ ” m. 3. (a) 38.8 pm. (b) 1.24 nm. (c) 907 fm. 5. (a) 3.51 X 10‘ m/s. (b) 64.4 kV. 7. (a) 5.3 fm. 9. 3.9 X 10“ '^ m. 11. A neutron. 13. (a) 7.77 pm. (b) 7.68 pm. 15. 5.5°. 17. (a) The beams are not present, (b) 41°. 21. 690 mHz. 23. 76/teV. 25. 8.8 X lO-^"* kg • m/s. 27. A/2;t. 29. (a) 1900 MeV. (b) 1.0 MeV. 31. 88.3 eV. 33. (a) 6.2 X lO”-" J. (b) 1.0 X 10-“ . (c) 3.0 X 10"'» K. 35. (a) 8.74 keV. (b) 1.01 X 10“ “ kg • m/s. (c) 98.5 pm. 37. (a )x = N L /2 n ,N = 1,3,5, . . . , ( n - 1). (b) x = NL/n, A ^=0,l,2, . . . ,M. 39. (/>) 0.0006. (c) 0.0003. 41. (a) 9.2 X 10"‘. (b) 7.5 X 10” *. 43. 1.1 X 10'®* y.
CHAPTER 51 3. 656.3, 486.1, 434.1, 410.2, 397.0 nm. 7. 3.40 eV. 9. {a) n = 5 —* 3. (b) Paschen. 11. 66 neV; E2 = —3.4 eV. 21. (a) 54.4 eV. (b) 13.6 eV. 25. (b) n \ (c) n. (d) 1/n. (e) \ ! n \ ( / ) l/«. {g) Un*. (h) i / n \ (/) \ / n l (j) l/n^. (k) W . 31. (a) 3, 2, l , 0 , - l , - 2 , - 3 h . (Z»)- 3 , - 2 , - 1 , 0 , 1 ,2 ,3 //B_ (c) 30.0°, 54.7°, 73.2°, 90°, 107°, 125°, 150°. (d) ^ h . (e) yfUfiB. 33. (b) 0.358 meV; 1.07 meV; 2.15 meV. 37. 72 k m /sl 39. n = 4 ; l = 3; 2, 1,0 , —1, —2, —3; m^ = ±^. 41. n a 5 ; / = 4; w , = ± i . 43. 1,0, 0, i; 1 ,0 ,0 ,- f 45. All the statements are true. 47. 51 mT. 49. (a) 2150 nm "’ ; zero, (b) 291 nm"^; 10.2 nm "'. 51. 1.85. 53.5.41 X 10-’. 55. 1 .5 X 1 0 -” . 57.0.439. 59. (a) 0.764oo; 5.236oo-(^) 0.981 n m -';3 .6 1 nm "'. 61. 1.90 X 10-’. 63. (a) 11.4 meV. (b) 1.62 eV. 65. (a) 0.284 pm. (b) 2.53 keV. (c) 490 pm.
CHAPTER 52 3. 9.84 kV. 5. (a) 24.8 pm. (b) Unchanged, (c) Unchanged. 7. =2.1 keV. 9. 49.6 pm; 99.2 pm; 99.2 pm. I I . (a) 19.7 keV; 17.5 keV. (b) Zr or Mb. 13. (a) 5.72 keV. {b) 86.8 pm, 14.3 keV; 217 pm, 5.72 keV. 19. ( a ) 2 , 0 , 0 , ± i . { b ) n = 2 ; l = V,m ,= 1,0 , - 1; w, = 21. Only argon would remain an inert gas. 23. (a) 1.84; 2.26. (/>) 0.167; 0.119. 25. 2.0 X 10” s-'. 27. 3.2 X 10’. 29. 10,000 K. 31. (a) None. (6) 51.1 J. 33. 4.74 km. 35. (a) No. (b) 0.11 //m. (c) 110 km.
A-27
37. 46 nm. — 39. insulator none ext. semi. donor n — int. semi. none — conductor none — conductor none ext. semi. acceptor P 41. (a) 0.74 eV above it. (b) 5.6 X 10-’. 43. 20 Gf2; 90 il. 45. 1.1 eV;no. 47. (a) 230 nm. (^) Ultraviolet. 49. Opaque.
CHAPTER 54 I. 15.7 fm. 3. 26 MeV. I I . (a) 1.000000 u; 11.906830 u; 236.202500 u. 13. “ Mg: 10.01%; “ Mg: 11.00%. 15. (a) 19.81 MeV; 6.258 MeV; 2.224 MeV. (b) 28.30 MeV. (c) 7.075 MeV. 17. (b) 7.92 MeV. 19. (a) 2.59 fm. (b) Yes. 21. (a) 4. (b) 148 neV. (c) 8.38 m. () 0.125. (c) 0.0749. 27. (a) 7.57 X 10” s” '. (b) 4.95 X 10” s - ‘. 29. 3.84 X 10“ . 31. (a) 59.5 d. (b) 1.18. 33. 87.8 mg. 39. (a) 3.65 X 10’ s” '. (b) 3.65 X 10’ s” '. (c) 6.41 ng. 43. 03 = -9 .4 6 0 MeV; ^ = 4.679 MeV; Q, = -1 .3 2 6 MeV. 45. (a) 31.85 MeV; 5.979 MeV. (b) 73 MeV. 47. 1.17 MeV. 49. (a) 874 fm. (b) 6.4 fm. (c) No. 51. (/») 960.2 keV. 53. 596 keV. 55. 13 mJ. 57. 39.4/iCi. 59. 5.33 X 10“ 61. (a) 2.03 X 10“ (b) 2.78 X 10’ Bq. (c) 75.1 mCi. 63. 730 cm’. 65. (a) 6.3 X 10” . (b) 2.5 X 10” . (c) 200 mJ. (d) 230 mrad. (c ) 3.0rem. 67. (6) 27 TW. 69. 1.78 mg. 71. -1 .8 5 5 MeV. 77. (c) 3.9 X 10’ m/s; 8.8 X 10’ m/s; 15.6 MeV. 81. (a) 5.5 MeV. 83. (a) 5.10 X 10'* Hz. (b) 20.5 keV. 85. (a) 3.55 MeV. (b) 7.72 MeV. (c) 3.26 MeV. 87. (a) 7.19 MeV.(Z>) 12.0 MeV.(c) 8.69 MeV.
CHAPTER 55 I. (a) 34 kg. (b) 12 mg. 3. (a) 2.56 X 10“ (b) 81.9 TJ. (c) 25,900 y. 9. (a) 13.9 d“ '. (b) 4.97 X 10*. I I . -2 3 .0 MeV. 13. 174 MeV. 15. 231 MeV. 17. (a) 253 MeV. 19. “ *U + n ^ ” ’U -► ” ’Np + e; ” ’Np ^ ” ’Pu + e. 21. 548 kg. 25. 1.6X10'*. 27. 566 W. 29. (a)44kton. 31. 24 g. 35. 450 keV. 37. (a) 170 kV. 39. 24,800 y. 43. (a) 4.0 X 10“ MeV. (b) 5.1 X 10“ MeV. 45. 4.5 Gy. 47. (a) 4.1 eV/atom. (b) 9.0 MJ/kg. (c) 1500 y. 51. (b) 2.28 X lO*’ J. (c) 1.85 X 10* y. 53. (a) B; 5.19V, MeV. (b) A: i N ’H, i N n; B: j N 'H, “He, n. 55. = 3.52 MeV; K„ = 14.07 MeV. 57. (a) 1000 km/s. (b) 2.0 nm.
CHAPTER 56 CHAPTER 53 3. 5.90 X 10“ m -’. 5. (a) 0.90. (b) 0.69. (c) Sodium. 7. (a) 1.00; 0.986; 0.500; 0.014; zero, (b) 700 K. 9. 5.53eV. 11. 65.4keV. 19. 234 keV. 23. 201°C. 27. (a) 5.86 X 10“ m -’. (b) 5.51 eV. (c) 1390 km/s. (d) 524 pm. 29. (a) 52.1 nm. (b) 202. 31. (a) 1.5 X 10-‘. (b) 1.5 X 10-‘. 35. (a) 5.0 X 10“ m“ ’. (b) 1.7 X 10’.
I. (a) 2.4 X 10-*’. (b) 8.1 X 10-’’. 3. 2.84 X 10“ m. 5. 769 MeV. 7. 31 nm. 9. 2.2 X 10-'* m. I I . (a) Charge; electron lepton number. _ (b) Relativijtic energy. 13. b, d. 15. (a) K*". (b) n. (c) n®. 17. (a) uud. {b) uud. 19. (a) sud. {b) uss. 25. 690 nm. 27. W) 2.39 GK. 29. (a) 280 /zeV. (b) 4.4 mm. 31. (a) 1.6 X 10'’ K.(/>)88/ts.
PHOTO CREDITS
CHAPTER 27 Figure 2: Courtesy Xerox Corporation. Figure 9: Seattle Times. CHAPTER 28 Figure 9: Courtesy Educational Services, Inc. CHAPTER 30 Figure 23: Courtesy High Voltage Engineering Company. Fig ure 30: Courtesy NASA. CHAPTER 31 Figure 2: Courtesy Spague Electric Company. Figure 8: Cour tesy Lawrence Livermore Laboratory. Figure 19: Courtesy Pasco Scientific. CHAPTER 34 Figures 1 and 2: D. C. Heath and Company with Education Development Center. Figure 3: Courtesy Varian Associates. Figure 10: Courtesy Professor J. le P. Webb, University of Sussex, Brighton, England. Figure 11: Courtesy Argonne Na tional Laboratory. Figure 13: Courtesy Fermi National Accel erator Laboratory. CHAPTER 37 Figure 5: Courtesy GE Medical Systems. Figure 10: Courtesy R. W. De Blois. Figure 12: Dr. Syun Akasofu/Geophysical Institute, University of Alaska, Copyright © 1977. CHAPTER 40 Figure 6 : Courtesy Stanford Linear Accelerator Laboratory. CHAPTER 41 Figure la: Courtesy NASA. Figure lb: Astronomical Society of the Pacific. Figure 3: Courtesy AT&T Bell Labs. Figure 4: Courtesy NASA. Figure 5: Astronomical Society of the Pacific. Figure 17: Courtesy NASA. CHAPTER 42 Figure 2: Oregon State University. Figure 3: Copyright © Fotocentre Ltd. Oamuaru, New Zealand. Figure 6 : Courtesy M ount Wilson and M ount Palomar Observatories. CHAPTER 43 Figure 2: Education Development Center, Inc. Figure 3a: PSSC, Physics, 2nd Ed., D. C. Heath and Co. with Education Development Center, 1965, Newton, Mass. Figure 20: Science Photo Library/Photo Researchers. Figure 21: Bell System.
CHAPTER 44 Figure 27: Courtesy NASA. CHAPTER 45 Figure 2: from Atlas o f Optical Phenomena by Cagnet et al.. Springer-Verlag, Prentice-Hall, 1962. Figure 3: Education De velopment Center, Newton, Mass. Figure 16: Courtesy Bausch & Lomb Optical, Co. Figure \Sb: Courtesy Robert Guenther. CHAPTER 46 Figures 1 and 2: from Atlas o f Optical Phenomena by Cagnet et al., Springer-Verlag, Prentice-Hall, 1962. Figure 3: from Sears, Zemansky, and Young, University Physics, 5th ed., AddisonWesley, Reading, Mass., 1976. Figure 13: Atlas o f Optical Phenomena, by Cagnet et al., Springer-Verlag, Prentice-Hall, 1962. Figure 15: Courtesy Dr. G. D. Shockman from D. C. Shingo, J. B. Cornett, G. D. Shockman, J. Bacteriology, 138: 598-608, 1979. Figure 16: horn Atlas o f Optical Phenomena, by Cagnet et al., Springer-Verlag, Prentice-Hall, 1962. CHAPTER 47 Figure 2: from Atlas o f Optical Phenomena, by Cagnet et al., Springer-Verlag, Prentice-Hall, 1962. Figure 14: W. Arrington and J. L. Katz, X-Ray Laboratory, Rensselaer Polytechnic In stitute. Figures 21 and 22: Ronald R. Erickson and Museum of Holography. Figure 23: from Rigden, “Physics and the Sound of Music,” Scientific American, John Wiley & Sons, Inc., 1985. CHAPTER 48 Figure 8: Copyright © R. Mark, Experiments in Gothic Struc ture, MIT Press, Cambridge, Mass., 1982. Figure 9: Courtesy Apple Computer, Inc. and Paul Matsuda. Figure 12: from Rob ert Guenther, Modern Optics, John Wiley & Sons, Inc., 1990. CHAPTER 49 Figure 1: Courtesy Alice Halliday CHAPTER 50 Figure 1: Professor C. Jonsson, University Tubingen, Ger many. Figure 2: Courtesy G. Matteucci. Figure 24: Philippe Plailly/SPL/Photo Researchers. CHAPTER 51 Figure 1: W. Finkelnburg, Structure o f Matter, Springer-Ver lag, 1964. Figure 14: American Institute of Physics, Neils Bohr Library, Margaret Bohr Collection.
P -7
P-2
Photo Credits
CH A PTER 52
CH A PTER 55
Figure 8: Dave Roback/AP/Wide World Photos. Figure 9: Roger Ressmeyer/Starlight Pictures. Figure 10: National Bu reau of Standards.
Figure 12: Princeton University Plasma Physics Lab. Figure 14: Courtesy Lawrence Livermore Laboratory. Figure 15: Los Alamos National Laboratory.
CH A PTER 53
CH A PTER 56
Figure 18: Courtesy AT&T
Figure la: Courtesy CERN. Figure 5: American Institute of Physics, Neils Bohr Library, Margaret Bohr Collection. Figure 7: Courtesy AT&T.
INDEX
Aberration, 896, 901 chromatic, 907, 939 spherical, 939 Absorption, 1106 Accelerator, 741-742 AC circuits, 843-852 amplitude, 844 capacitive element, 845-846 inductive element, 845 phase and amplitude relations, 846 power, 849-851 resistive element, 844-845 transformer, 851 -852 transients, 844 AC generator, 843 Acoustic resonator, 864-865 Action at a distance, 606 Adaptive optics, 940 Adiabatic demagnetization, 812 Alpha decay, 1148-1149 thermonuclear fusion, 1175 Alpha particles, 1141-1143, 1207 deflecting force, 1142 Alternating current, 843-844, see also AC circuits Ammeter, 724 Ampere, 597, 698 Ampere, Andre-Marie, 762 Ampere’s law, 16%-11 A, 829, 859-861, 863 field outside solenoid. 111 line integral, 768 resonant cavity, 866 solenoids, 770-771 toroids, 11 \ -112 Amperian loop, 768-769 Amplitude reflection coefficient, 958 Amplitude transmission coefficient, 958 Analyzer, 1005-1006 Angle of incidence, 904, 908 Angle of minimum deviation, 907 Angle of reflection, 904, 909
Angle of refraction, 904, 912 Angular magnification, 938 Angular momentum: components, 808 hydrogen atom, 1076-1080 direction, 1076-1078 magnitude, 1076 orbital, magnetism and, 1078-1080 intrinsic, 1082 Angular velocity, circular motion, charged particle, 740 Antenna, electromagnetic waves trans mission, 875 Antimatter, 1205-1206 Antineutrino, 1149-1150 Antinodes, 1102 Area, conversion factors. A-10 Astronomical data, A-4 Atom: acceptor, 1124-1125 constructing, 1099-1100 donor, 1124 energy from, 1167 nuclear model, 641-643 Atomic magnetism, 807 - 809 Atomic number, 1099, 1143 Atomic oscillators, 1025-1026 Atomic particles, spins and magnetic moments, 808 Atomic physics, 1095-1111 absorption, 1106 building atoms, 1099-1100 Einstein and the laser, 1105-1107 laser light, 1104-1105 laser principles, 1107-1109 molecular structure, 1109-1111 periodic table, 1100-1104 spontaneous emission, 1106 x-ray spectrum, 1095-1097 Atomic shell structure, 1157 Atomic structure, hydrogen atom, 1088-1089 Avogadro constant, 1028
B Balmer-Rydberg formula, 1069 Balmer series, hydrogen atom, 10691070 Bands: conduction, 1123 electrical conduction, 1120-1122 valence, 1123 Bardeen-Cooper-Schrieffer theory, 1133 Barrier penetration, 916 by waves, 1060 Barrier tunneling, 1059-1062 Baryon, 1192, 1194, 1198, A-9 Baryon number, conservation, 1196 Base, 1132 BCS theory, 1133 Beam splitter, 959 Benzene, atomic structure, 1048 Beta decay, 1147 energy of emitted electrons, 1150 Betatron, 792-793 Big Bang cosmology, 1201-1206 cosmic microwave background radia tion, 1203 expansion of universe, 1202-1203 periods of time, 121 Big Bang nucleosynthesis, 1206-1208 Binary system, 874 Binding energy: curve, 1145-1146 neutron, 1170 nuclear masses, 1145-1146 Binomial theorem. A-15 Bioluminescence, 890 Biot-Savart law, 761-763 applications, 763-766 circular current loop, 763-766 long straight wire, 763 Birefringence. See Double refraction Black-body radiation, 1022-1023 Black box, 705-706 Blazed gratings, 989
7-7
1-2
In dex
Blazing, 989 Bohr magneton, 808, 1078-1079, 1146 Bohr, Niels, 1069-1070 correspondence principle, 1062 principle of complementarity, 1063 Bohr radius, 1073, 1144 Bohr theory, 1069-1074 derivation, 1072-1074 frequency postulate, 1070 Moseley plot, 1098 postulate of stationary states, 1070 Boltzmann constant, 1025 Bonding, covalent, 1110 Bradley, James, 891 Bragg, W. L., 996-997 Bragg’s law, 995-996 Branches, 722-723 Breakeven, 1179-1180 Breeder reactors, 1184 Bremsstrahlung, 1095 Brewster’s angle, 1007, 1108 Brewster’s law, 1008
Capacitance, 678 analogy with fluid flow, 678 calculation, 678-681 definition, 821 equivalent, 681 parallel-plate capacitor filled with die lectric, 686 Capacitive reactance, 846 Capacitive time constant, 726, 824 Capacitor, 677-686 AC circuits, 845-846 charged, 677 cylindrical, 680 with dielectric, 685-686 discharging, 727-728 electric field calculation, 678-679 in parallel, 681 parallel-plate, 679-680, 684, 686 potential difference calculation, 679 in series, 682 spherical, 680 Carriers, majority and minority, 1124 Cavity oscillations. Maxwell’s equa tions, 864-867 Cavity radiation, 1022-1023 spectral radiancy curves, 1022-1023 Cavity radiation problem, 1105 Center of curvature, mirror, 924 Characteristic x-ray spectrum, 10961097 Charge carriers, density, semiconduc tors, 1124 Charged bodies, 594 Charge density, 612 disk of charge, 613-614 infinite line of charge, 614-615 ring of charge, 612-613
Charge distribution. Gauss’ law, 635639 Charge-to-mass ratio, electron, 739 Charm, 1200 Chemiluminescence, 890 Chromatic aberration, 907, 939 Circuit: LR, 824-826 /?C, 725-728 see also AC circuits; CD circuits; LC circuit; R LC circuit Circular aperture, diffraction, 975-977 Circular motion, angular velocity, charged particle, 740 Circular polarization, 1012-1014 Coherence, 950-952 Coherence length, wavetrain, 963-964 Coherent waves, 947, 950-952, 1004 Collective model, 1156-1157 Collector, 1132 Collisions, electron-lattice, 704 Complementarity, principle of, 1063 Compound microscope, 938-939 Compound nucleus, 1156-1157 Compound optical systems, 936-937 Compton, Arthur H., 1032 Compton effect, 1032-1035 Compton shift, 1033-1034 Computer programs: position-dependent forces. A-18 - A-19 time-dependent forces. A-16 - A-17 velocity-dependent forces, A-17-A-18 Concave mirror, 923 Conduction band, 1123 Conduction electrons, 595, 704 drift speed, 1120 metals, 1115-1117 Conductivity, 701 Conductor, 595-596, 1121, 1123 charged isolated, 633-635 electric current, 698 electric field external, 666 inside, 697 energy bands, 707 Ohm’s law, 703 two parallel, 767-768 Confinement time, thermonuclear reac tor, 1179 Conservation of baryon number, 1196 Conservation of charge, 600-601 Conservation of lepton number, 11951196 Conservation of strangeness, 1197 Constants, fundamental, A-3, inside front cover Contact potential difference, 1029 Continuous charge distribution, 611615 electric potential, 660-662 Control rods, 1172
Converging lens, 931-933 Conversion factors, A-lO-A-13, inside front cover Convex mirror, 923 Cooper pairs, 1133-1134 Copper, electrical properties, 1123 Core electrons, 1121 Comer reflector, 906, 919 Corona discharge, 666-667 Correspondence principle, 1026, 1057, 1062 hydrogen atom, 1073 Cosmic microwave background radia tion, 1203 Cosmology: age of universe, 1210-1213 Big Bang cosmology, 1201-1206 cosmic microwave background radia tion, 1203 determination of age of universe, 1212-1213 nucleosynthesis, 1206 -1210 Coulomb, 597 Coulomb, Charles Augustin, 596 Coulomb force, 1144 Coulomb’s law, 596-599 constant in, 762 experimental tests, 639 - 641 from Gauss’ law, 632 point charge, 607 significance of, 598 vector form, 597-599 Covalent bonding, 1110 Critical temperature, superconductors, 1133 Curie, 1151 Curie’s law, 812 Curie temperature, magnetic materials, 813 Current: displacement, 862-863 eddy, 787 heat flow and, 703-704 induced. See Induced current magnetic force, 747-749 parallel and antiparallel, 767 reverse saturation, 1139 Current balance, 767-768 Current density, 699-700 Current loop: circular, Biot-Savart law, 763-766 torque, 749-751 Cutoff frequency, 1030 Cyclotron, 740-742 Cylindrical capacitor, capacitance, 680 Cylindrical symmetry, 615
D Damped oscillations, 833- 835 Dating, radioactive, 1153
In dex
Davisson-Gentler experiment, 10461047 DC circuits, 715 - 728 branches, 722-723 capacitor discharge, 727-728 current calculations, 717-718 electromotive force, 715-717 junctions, 722-723 measuring instruments, 724-725 multiloop circuits, 722-724 potential differences, 718-720 R C circuits, 725-728 resistors in series and parallel, 720-722 de Broglie’s hypothesis, testing, 10461049 de Broglie wave, 1074 de Broglie wavelength, 1045-1046, 1057 Debye-Scherrer experimental arrange ment, 999 Debye temperatures, 1028 Dees, 741 Demagnetization, adiabatic, 812 Density, conversion factors. A -11 Density of occupied states, 1117 Density of states, 1116 pairing gap, 1134 Depletion zone, 1127 Derivatives, A-15 Deuterium, nucleus, 601 Deuteron, proton-proton cycle, 1177 Diamagnetism, 809, 812-813 Dichroic material, 1009 Dielectric constant, 685 Dielectrics, 686-690 atomic view, 686-688 capacitor with, 685-686 Gauss’ law, 688-690 induced electric dipole moment, 687 induced surface charge, 688 nonpolar, 687 polar, 686-687 properties, 685 Dielectric strength, 686 Diffraction, 903, 948, 967-981 circular aperture, 975-977 double-slit analysis with phasors, 980-981 combined with interference, 977 981 electrons, 1043-1044 Fraunhofer, 969-970 Fresnel, 970 pattern, disk, 968 single-slit, 970-972 intensity, 972-975 water waves, 903-904 wave theory of light, 967-970 x-ray, 993-997 Diffraction factor, 978 Diffraction grating, 985, 989-991 Diffuse reflection, 905
Diffusion current, 1127 Diode laser, 1131-1132 Diode rectifier, 1128-1130 Dip angle, 817 Dipole: electric, 805-806 potential due to, 659-660 nonuniform field, Stem-Gerlach ex periment, 1080-1082 Dipole antenna, 875 Dipole equations, 765 Dipole moment, induced, 660 Disintegration, nuclear fission, 1169 Disintegration constant, 1147 Disintegration energy, 1149 Disk of charge, 613-614 Dispersion, 893 resolving power, 991-993 Displacement current, 862-863 Dissociation energy, 1110 Diverging lens, 931 -933 Doppler effect: light, 893-895 relativistic, 894 consequences, 897-898 derivation, 895-897 transverse, 897-898 twin paradox, 898 Dose equivalent, 1152 Double refraction, 1008-1012 definition, 1009 mechanical analogy, 1011-1012 ordinary and extraordinary waves, 1009-1010 principal indices of refraction, 10091010 Double scattering, 1016 Double-slit diffraction: analysis with phasors, 980-981 combined with interference, 977-981 Double-slit interference, 947-950, 1064-1065 analysis with phasors, 980-981 combined with diffraction, 977-981 elections, 1043-1044 intensity, 952-955 Young’s experiment, 949-950 Drain, 1132 Drift current, 1127 Drift speed, 699
Eamshaw’s theorem, 602 Earth: magnetic field, 743, 815-816 properties, A-4 Eddy currents, 787 Einstein, Albert, stimulated emission, 1105-1107 Einstein-de Haas effect, 1079 Einstein’s photon theory, 1031 -1032
1-3
Einstein’s postulates, 896 Einstein temperature, 1028-1029 Elastic scattering, 1154 Electrets, 810 Electrical conduction, 1115-1134 bands and gaps, 1120-1122 conductors, 1121, 1123 doped semiconductors, 1124-1126 Ferm i-Dirac probability function, 1118-1119 Ferm i-Dirac statistics, 1118 filling allowed states, 1117-1119 free electron gas model, 1115 insulators, 1124 metals, 1119-1120 optical electronics, 1130-1132 Pauli exclusion principle, 1116 pn junctions, 1126-1130 semiconductors, 1123-1124 superconductors, 1133-1134 transistor, 1132-1133 Electric charge, 594-595 conservation, 600-601 quantized, 599-600 Electric circuit: energy transfer, 705-706 LR, 824-826 /?C, 725-728 see also AC circuits; DC circuits; LC circuit; R LC circuit Electric current, 697-699 conductor, 698 direction, 698 drift speed, 699 lattice, 699 Electric dipole, 608-609, 660, 805-806 in electric field, 608-609, 618-620 equations, 765 lines of force, 610-611 radiation, 875-876 Electric dipole moment, 609 induced, 687 Electric field, 606-607 calculation, 678-679 from electric potential, 663-665 conservative, 791 continuous charge distribution, 611615 electric dipole, 608-609, 618-620 electric potential calculation, 655-657 electromagnetic standing wave pat terns, 1055 external, conductor in, 666 flux, 629-631,806 Gauss’ law, 634 induced, 790-792 from induced surface charges, 688 inside conductor, 697 isolated conductor, 634-635 lines of force, 609-611, 662 Lorentz force, 738-740
1-4
In d ex
Electric field (Continued) nonconservative, 791 nonuniform, 617-618 point charge, 607 - 609, 615-618 principle of superposition, 598, 608 Electric potential, 654-668 collection of point charges, 658-660 conductor in external electric field, 666 continuous charge distribution, 660662 corona discharge, 666-667 definition, 654 dipole, 659-660 electric field calculation, 655-657, 663-665 isolated conductor, 665-667 point charge, 657-658 principle of superposition, 658 Electric potential difference, 654-655 absolute value, 679 calculation, 679 point charges, 657 Electric potential energy, 652-654 energy storage, 683-684 Electric quadrupole, 660 Electromagnet, 736 Electromagnetic force, 1191 Electromagnetic interaction, strange ness, 1197 Electromagnetic oscillations: damped and forced, 833-835 qualitative, 829-831 quantitative, 831-833 Electromagnetic shielding, 798 Electromagnetic spectrum, 871-874 Electromagnetic waves, 871-883 energy transport, 880-881 generation, 874-877 incident wave vectors, 882 linearly polarized, 876-877 Maxwell’s equations, 864 Poynting vector, 880-881 propagation, 879-880 quantum and classical physics, 1017 radiation pressure, 881-883 reflection and refraction, 905-906 speed in free space, 892 standing, 1054-1055 traveling, Maxwell’s equations, 877880 Electromagnetism, 593-594 basic equations, 859-860 frames of reference, 773-774 Electromotive force, 715-717 induced, 784 internal resistance, 717-718 motional, 787-790 Electron: charge-to-mass ratio, 739 conduction, 595 metals, 1115-1117
configuration, 1102 core, 1121 diffraction, 1043-1044 drift speed, 704-705 energy from, 1167 levels, 1100 of emitted, beta decay, 1150 frequency of revolution in orbit, 1072 Maxwellian velocity distribution, 704 momentum, 1051 probability, 1085 properties, 599 quantum distribution, 704 radial probability density, 1085 reduced mass, 1088-1089 spin, 1082-1083 trapping, 1055-1057 Electron gas, 704 Electronics, 1104 Electron-lattice collisions, 704 Electron microscope, diffraction, 977 Electron number, magic, 1157 Electron-volt, 655 Electrostatic accelerator, 667-668 Electrostatics, 594, 651-652 Electroweak force, 1191 Electroweak interaction, 594 Elementary charge, 599 measurement, 617 Elements: numbering, x-rays and, 1097-1099 periodic table, A-7 properties, A-5-A-6 relative abundance in solar system, 1208-1209 Emerging nucleus, 1154 Emission: spontaneous, 1106 stimulated, 1105-1107 Emitter, 1132 Endothermic reactions, 1154 Energy: from atoms, 1167 conversion factors. A-12 density, magnetic field, 828-829 disintegration, 1149 nuclear fission, 1169 dissociation, 1110 gap, 707 ground state, well, 1057 ionization, 1102-1103 levels, allowed, 1057 nuclear fission, 1168-1171 oscillating systems, 831 pairing, 1134 photons, 1031 quantization, 1025-1027 states, 707 storage electric field, 683-685
magnetic field, 826-829 thermonuclear fusion, 1175-1176 transfer electric circuit, 705-706 reversibility, 717 transport, 880-881 zero-point, 1057 Energy theorem, classical equipartition, 1027-1028 Energy-time uncertainty relationship, 1052-1053 Equipotential surfaces, 662-^663 Equivalent capacitance, 681 Equivalent resistance, 720 Ethanol, nuclear magnetic resonance spectrum, 1083 Ether hypothesis, 960 Excess charge, isolated conductor, 665667 Exchange force, 1195 Excited states: hydrogen atom, 1086-1087 optical transitions, 1103-1104 Exothermic reactions, 1154 Expansion, multi-poles, 660 Exponential expansion. A-15 Extraordinary ray, 1009 Eye: near point, 938 sensitivity as function of wavelength, 889
Farad, 678, 783 Faraday, Michael, 593, 640, 685, 812 experiments, 783-784 Faraday’s law, resonant cavity, 866 Faraday’s law of induction, 783-795, 822, 859, 861,863 betatron, 792-793 Faraday’s experiments, 783-784 induced electric fields, 790-792 Lenz’ law, 785-787 motional electromotive force, 787790 traveling waves, 878-880 Femtometer, 1144 Fermat, Pierre, 909 Fermat’s last theorem, 909 Fermat’s Principle: law of reflection, 909 law of refraction, 913 Fermi, 1144 Ferm i-Dirac probability function, 1118-1119 Ferm i-Dirac statistics, 1118 Fermilab tunnel, 742 Fermi speed, 1119 Ferroelectrics, 810 Ferromagnetism, 809, 813-814 FET, 1132
In d ex
Field-effect transistor, 1132 Field of view, 939 Field particles, 1194-1195, A-8 Fields, 605-606 Fine structure, hydrogen atom, 1088 First focal point, thin lens, 933 Fizeau, Hippolyte Louis, 891-892 Floaters, 968 Fluid flow, capacitance analogy, 678 Fluorescence, 890 Fluorine, 1102 Flux: electric field, 629-631 magnitude, 627-628 vector field, 627-629 /n u m b er, 941 Focal length: spherical mirrors, 924 thin lens, 931 Focal point, 925 Force: basic, 1190-1191 conversion factors. A-12 Forced oscillations, resonance, 834-835 Frames of reference, electromagnetism, 773-774 Franklin, Benjamin, 595, 639-640 Fraunhofer diffraction, 969-970 Free-electron model, 704, 1115 Frequency, natural, 834 Frequency postulate, 1070 Fresnel, Augustin, 967-968 Fresnel diffraction, 970 Frustrated total internal reflection, 916, 1060 Full-angle beam divergence, 885 Fusion reactions, nucleosynthesis, 1208 Fusion reactor. See Thermonuclear re actor
Galileo, speed of light, 891 Galvanometer, 750 Gamma rays, spectrum, 874 Gaps, electrical conduction, 1120-1122 Gate, 1132 Gauss, 737 Gauss, Carl Friedrich, 631 Gaussian surface, 631-632 Gauss’ law, 631-643 applications, 635-639 infinite line of charge, 635-636 infinite sheet of charge, 636 spherically symmetric charge distri bution, 637-639 spherical shell of charge, 636-637 Coulomb’s law from, 632 dielectrics, 688-690 electric field, 634 electricity, 859, 863 experimental tests, 639-641
gravitation, 645 isolated conductor, 633-635 magnetism, 805-807, 859, 863 Geiger counter, 648 Generator, AC, 843 Geometrical optics, 903-904 see also Reflection; Refraction Geometry, mathematical formulas. A-14 Glancing angle, 996 Glashow-W einberg-Salam theory, 594 Gluons, 1199-1200 Grand unified theories, 1192 Gratings: blazed, 989 diffraction. See Diffraction gratings dispersion and resolving power, 991 993 multiple slits, 985-989 maxima width, 986-988 secondary maxima, 988-989 reflection, 989 Gravitational field, 605 Gravitational forces, 651-652, 11901191 Greek alphabet, inside back cover Ground state, 1057 hydrogen atom, 1085-1086 H Half-life, 1147 Half-wave plate, 1020 Hall, Edwin H., 745 Hall effect, 595, 745-747 quantized, 746-747 quantum, 701 Hall potential difference, 745-746 Hall voltage, 745 Halogens, 1102 Harmonic motion, analogy of oscillat ing LC circuit, 831 Heat, conversion factors. A-12 Heat capacity: quantum theory, 1028-1029 solids, 1027-1029 Heat flow, current and, 703-704 Heat radiation, 872 Heisenberg’s uncertainty principle, 1051 angular momentum vector, 1077 single-sit diffractions, 1052 Heisenberg uncertainty relationships, 1051-1052 Helium: abundance in universe, 1207 fusion reaction, 1208 Helium atoms, intensity pattern, 10441045 Helium -neon gas laser, 1107-1108 Helmholtz coil, 776 Henry, 783, 821 Henry, Joseph, 783 Holographic interferometry, 998
1-5
Holography, 997-998 Homopolar generator, 802 Hubble, Edwin, 1202 Hubble parameter, 1202-1203, 1212 Huygens, Christiaan, 907 Huygens’ Principle: law of reflection, 907-909 law of refraction, 912-913 Huygens wavelets, 1010-1011 Huygens wave surfaces, 1009-1010 Hydrogen: atomic, see Hydrogen atom molecular structure, 1109 Hydrogen atom, 1069-1089 angular momentum, 1076-1080 atomic structure, 1088-1089 Balmer series hydrogen, 1069-1070 Bohr theory, 1069-1074 correspondence principle, 1073 Einstein-de Haas effect, 1079 excited states, 1086-1087 fine structure, 1088 ground state, 1085-1086 Layman and Paschen series, 1070 « -2 , / - I subshell, 1087 potential energy function, 1075 quantum number, 1076 reduced mass, 1088-1089 Schrodinger’s equation, 1074-1076 shell, 1084 spectrum, 1071 spinning electron, 1082-1083 states, 1084 Stem-Gerlach experiment, 10801082 subshell, 1084 weighted average probability den sity, 1087 Zeeman effect, 1088 Hysteresis curve, 814 I Image: distance, 923 inverted, 932 real, 910 virtual, 910 Image formation: plane mirror, 909-912 reversal, 911-912 Impedance, R L C circuit, 847 Incandescence, 890 Inclination, 817 Incoherent waves, 947 Independent particle model, 1157 Index of refraction, 905 Induced current, 784 Joule heating, 788 Lenz’ law, 786 Induced electric dipole moment, 660 dielectrics, 687
1-6
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Induced electric fields, 790-792 Induced electromotive force, 784 Induced magnetic field, 860-863 Inductance, 821-835 calculation, 822-824 definition, 821 electromagnetic oscillations. See Elec tromagnetic oscillations LR circuit, 824-826 solenoid, 822-823 toroid, 823 Induction: Faraday’s law, 859, 861, 863 relative motion, 793-795 see also Faraday’s law of induction Induction furnace, 787 Inductive time constant, 825 Inductor, 821 AC circuits, 845 flux linkages, 822 with magnetic materials, 823 Inelastic scattering, 1154 Inert gas, 1102 Inertial confinement, thermonuclear re actor, 1179, 1181-1182 Infinite line of charge, 614-615 Gauss’ law, 635-636 Infinite sheet of charge. Gauss’ law, 636 Infrared radiation, spectrum, 872 Ink-jet printer, 616 Insulator, 595-596 energy bands, 707-708 Integrals, A-15 Intensity: diffraction gratings, pattern, 986 double-slit interference, 952-955 pattern, 953-954 single-slit diffraction, 972-975 Interference, 947 - 961 adding wave disturbances, 954-955 coherence, 950-952 constructive, 947 destructive, 947 Michelson’s interferometer, 956-961 from thin films, 955-958 see also Double-slit interference Interference factor, 978 Interference fringes, 948, 1064 circular, 957-958 of constant thickness, 957 double-slit system, 977-979 Interference pattern, 948 multiple slits, 985-986 particles, 1043-1045 Interferometer, 959 Michelson’s, 959-960 propagation, 960 - 961 Interferometry, holographic, 998 Internal reflection, frustrated total, 916, 1060 Internal resistance, single-loop circuit, 717-718
International system of units. See SI system Interplanar spacings, 996 Intrinsic angular momentum, 1082 Intrinsic magnetic moments, 808-809 Inverse square law, 640, 651 Inverted image, 932 Ionization energy, 1102-1103 Ionizing radiation, measurement, 11511152 Isolated conductor, 633-635 capacitance, 680 with cavity, 633-634 excess charge, 665-667 external electric field, 634-635 Isolated sphere, capacitance, 680 Isotopes, radioactive, 1153 Ives, H. E., 894-895
Josephson junction, 725 Joule heating, 706 induced current, 788 Joule’s law, 706 Junction rule, 723 Junctions, 722-723 Junction transistor, 1132 K Kaons, 1196-1197 Kinetic energy, 1029-1030 KirchhofTs first rule, 723 KirchhofTs second rule, 717 Klystrons, 867 K mesons, 1196-1197
Lanthanides, 1102 Laser, 952, 1104-1105 diode, 1131-1132 light characteristics, 1104 principles, 1107-1109 Laser fusion, 1179-1182 Lateral magnification: spherical mirrors, 925, 928 thin lens, 931 Lattice, 699 Laue spots, 994 LawofM alus, 1006 Law of reflection, 904 derivation, 907-909 Fermat’s principle, 909 Huygens’ principle, 907-909 Law of refraction, 904 derivation, 912-914 Fermat’s principle, 913 Huygens’ principle, 912-913 LC circuit, oscillation, 830 analogy to simple harmonic motion, 831 damped, 833-835 forced, 834-835
Lawson’s criterion, 1179 Lead, isotopes, r- and s-process paths, 1210 LEDs, 1130 Length, conversion factors. A-10 Lens. See Thin lens Lens maker’s equation, 931 Lenz, Heinrich Friedrich, 785 Lenz’ law, 785-787, 821-822 Lepton number, conservation, 11951196 Leptons, 1192-1193, A-8 pairs of, 1201 Light, 889-898 coherent, 951 Doppler effect, 893-895 double scattering, 1016 extraordinary ray, 1009-1010 incoherent, 951 intensity, polarizing sheets, 10051006 laser, characteristics, 1104 line spectra, 1035-1036 ordinary ray, 1009-1010 Planck’s radiation law, 1024-1025 propagation in matter, 893 Michelson’s interferometer, 960961 scattering, 1014-1016 spectrum, 871-872 speed of, 891-893, 960-961 in matter, 892-893 thermal radiation, 1021 -1024 visible, 889-891 wave theory, 967-970 Light-emitting diodes, 1130 Light-gathering power, 939 Linear charge density, 612 Linearly polarized wave, 1004 Line integral, 656 Ampere’s law, 768 Lines of force, 609-611 equipotential surfaces, 662-663 Line spectra, 1035-1036 Liquid crystal display, 1007 Liquid-drop fission model, 1156, 1170 Lloyd’s mirror experiment, 958-959 Logarithmic expansion. A-15 Loop rule, 717 Loop theorem, 825, 827 R LC circuit, 847 Lorentz force, 738-740 Lorentz transformation, 896 LR circuit, 824-826 Luminescence, 890-891 Luminosity, 1023-1024 Lyman series, hydrogen atom, 1070 M Macroscopic quantities, 702 Magic electron numbers, 1157
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Magic nucleon numbers, 1157 Magnet, 735 bar, magnetic field, 735, 738 poles, 738 Magnetic bottle, 743 Magnetic braking, 787 Magnetic confinement, thermonuclear reactor, 1179-1181 Magnetic deflecting force, properties, 740 Magnetic dipole, 751-752, 806-807 equations, 765 Magnetic dipole moments, 751 Magnetic domains, 814 Magnetic field, 735-752 Ampere’s law, 769 bar magnet, 735, 738 betatron, 792 circulating charges, 740-744 conversion factors. A-13 current loop torque, 749-751 cyclotron, 741-742 frequency, 740 definition, 737 earth, 743,815-816 energy density, 828-829 storage, 826-829 flux, 806 Hall effect, 745-747 induced, 860-863 lines, 766-767 Lorentz force, 738-740 magnetic braking, 787 magnetic mirror, 742-743 moving charge, 737-740 nonuniform, dipole, Stem-Gerlach experiment, 1080-1082 numerical calculation of path, 743744 poloidal, 1179 right-hand rule, 764, 766 solar system, 815-817 synchrotron, 742 toroidal, 1179-1180 values, 737 Magnetic flux, 784 conversion factors. A-13 density, 735 right-hand rule, 878 Magnetic force: current, 747-749 moving charge, 736-740 Magnetic induction, 735 Magnetic latitude, 820 Magnetic materials, 811-814 diamagnetism, 812-813 ferromagnetism, 813-814 inductors, 823 paramagnetism, 811-812 Magnetic mirror, 742-743 Magnetic moment, intrinsic, 808-809 Magnetic monopoles, 736, 807
Magnetic quantum number, 1076 Magnetism: atomic, 807-809 Einstein-de Haas effect, 1079 Gauss’ law, 805-807 nuclear, 809-810 nuclear spin, 1146-1147 orbital angular momentum, hydrogen atom, 1078-1080 planets, 815-817 Magnetization, 810-811 current, 861 saturation value, 812 Magnification: angular, 938 lateral, spherical mirrors, 925, 928 Magnifier, simple, 937-938 Majority carriers, 1124 Malus, Etienne Louis, 1006 Mass: conversion factors. A-11 reduced, electron, 1088-1089 Mass-energy relation, 1145 Mass number, 1143 Mass spectrometer, 740 Mathematical formulas, A-14-A-15 Mathematical signs and symbols. A-14, inside back cover Matter: dual wave-particle nature, 10631065 speed of light in, 892-893 Maxima, diffraction gratings, 986-988 Maxwell, James Clerk, 593-594, 863, 879-880 unification of electromagnetism, 594 Maxwell-Boltzmann statistics, 1118 Maxwell’s equations, 594, 859-867 cavity oscillations, 864-867 electromagnetic waves, 864 symmetry, 863-864 traveling waves, 877-880 Mesons, 1192-1194, 1198, A-9 Metal: conduction electrons, 1115-1117 electrical conduction, 1119-1120 resistivity, 1120 work function, 1120-1121 Metal-oxide-semiconductor FET, 11321133 Michelson, Albert A., 892 interferometer, 959-960, 1105 light propagation, 960-961 Michelson-Morley interferometer, 961 Microfarad, 678 Microscope: compound, 938-939 diffraction effects, 977 Microscopic quantities, 702 Microwaves: background radiation, 872 cosmic, 1203
1-7
spectrum, 872 transmission, 1003-1004 Millikan, Robert A., 617 oil-drop apparatus, 617 Minimum energy principle, 1099-1100 Minority carriers, 1124 Mirror: concave, 923 convex, 923 plane, image formation, 909-912 see also Spherical mirrors Mirror equation, 923-925 derivation, 927-928 Molar heat capacity, solids, 1027-1029 Molecular bonding, 1110 Momentum: electron, 1051 radiation pressure, 881-883 Monopole, magnetic, 807 Moon, properties, A-4 Moseley, Henry G. J., 1097 Moseley plot, 1097-1098 Bohr theory, 1098 MOSFET, 1132-1133 Motion: nonuniform electric fields, 617-618 relative. See Relative motion Motional electromotive force, 787-790 Moving charge, magnetic force, 736-740 Multiloop circuits, 722-724 Multi-meter, 725 Multiplication factor, 1172 Multi-poles, expansion, 660 Muons, 919 N Natural frequency, 834 Near point, 938 Negative charge, 594-595 Neutrino, 1149-1150 in beta decay, 1150 energy from Sun, 1178 Neutron: balance, 1172 nuclear reactor, 1172 binding energy, 1170 capture, nucleosynthesis, 1208 -1210 capture problem, 1171-1172 energy problem, 1171 intensity pattern, 1044-1045 leakage problem, 1171 properties, 599 quarks, 599 spin, 1083 thermal, 1048, 1168 Neutron number, 1143 Newton’s rings, 957-958 Nobel Prizes, physics, A-20-A-23 Nonpolar dielectrics, 687 North magnetic pole, 814 NOVA laser fusion project, 1181 npn junction transistor, 1132
1-8
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n-type semiconductor, 700 Nuclear fission, 1168-1171 basic process, 1168-1169 chain reaction, 1171 disintegration, 1169 distortion parameter, 1170 natural reactor, 1174-1175 potential barrier, 1170 spontaneous, 1182 test of fissionability, 1170-1171 theory, 1169- 1171 Nuclear force, 1143-1144 Nuclear fusion, 1175 see also Thermonuclear fusion Nuclear magnetic resonance, 809, 1083 Nuclear magnetism, 809-810 Nuclear magneton, 1146 Nuclear masses, binding energy, 11451146 Nuclear matter, density, 1146 Nuclear models, 1156-1158 atom, 641-643 Nuclear physics, 1141-1158 alpha decay, 1148-1149 beta decay, 1149-1151 collective model, 1156-1157 independent particle model, 1157 ionizing radiation measurement, 1151-1152 natural radioactivity, 1152-1153 nuclear force, 1143-1144 nuclear masses and binding energies, 1145-1146 nuclear models, 1156-1158 nuclear radii, 1144-1145 nuclear reactions, 1153-1156 nuclear spin and magnetism, 11461147 radioactive dating, 1153 radioactive decay, 1147-1148 terminology, 1143 Nuclear radii, 1144-1145 Nuclear reactors: basic principles, 1171-1184 chain reaction, 1171 multiplication factor, 1172 natural, 1174-1175 neutron balance, 1172 neutron capture problem, 1171-1172 neutron energy problem, 1171 neutron leakage problem, 1171 pressurized-water, 1172-1173 radioactive wastes, 1172-1173 resonance capture, 1172 response time, 1172 supercritical, 1172 see also Thermonuclear reactors Nuclear shell structure, 1157 Nuclear spin, magnetism, 1146-1147 Nucleon-antinucleon pair, 1204-1205 Nucleon number, magic, 1157
Nucleons, 1143 Nucleosynthesis, 1206 -1210 alpha particles, 1207 fusion reactions, 1208 neutron capture, 1208-1210 Nucleus: compound, 1156-1157 discovery, 1141-1143 energy from, 1167 nuclear fission, 1168-1171 secular equilibrium, 1161 thermonuclear fusion, 1175-1176 Nuclides, 1143-1144 chart, 1154 properties, 1144 s- and r-process paths, 1209-1210 test of fissionability, 1170-1171 O Object distance, 923 Occupied states, density, 1117 Oersted, Hans Christian, 593, 761 Ohm, 701 Ohmic material, 703 Ohmmeter, 724 Ohm’s law, 703-705 microscopic view, 704-705 Onnes, Kammerlingh, 708 Optical electronics, 1130-1132 Optical fibers, 914-915 Optical flat, 962 Optical instruments, 937-940 compound microscope, 938-939 refracting telescope, 939-940 simple magnifier, 937-938 Optically anisotropic, 1008 Optically isotropic, 1008 Optical path length, 913 Optical reversibility, on reflection, 958959 Optical transitions, 872 excited states, 1103-1104 Optics: adaptive, 940 compound systems, 936-937 geometrical, 903-904 wave, 903-904 Order number, 986 Ordinary ray, 1009 Oscillations: cavity. Maxwell’s equations, 864-867 energy, 831 frequency, 1026 see also Electromagnetic oscillations; LC circuit Osmium isotopes, r- and s-process for mation, 1211-1212
Pairing energy, 1134 Pairing gap, 1134
Parallax effect, 998 Parallel connections, capacitor, 681 Parallel-plate capacitor: capacitance, 679-680 energy storage, 684 filled with dielectric, 686 Paramagnetic material, 809 Paramagnetism, 811-812 Paraxial rays, 928 Parity, 911 Particle-antiparticle annihilation, 1205 Particle - antiparticle pair, 1204 Particle physics: baryons, 1192, 1194 charm, 1200 conservation of baryon number, 1196 conservation of lepton number, 1195-1196 field particles and exchange forces, 1194-1195 leptons, 1192-1193 mesons, 1192-1194 particle families, 1192-1195 particle interactions, 1189-1192 'P(PSI), 1200-1201 quark model, 1197-1201 strangeness, 1196-1197 Particles: composite, A-9 density, thermonuclear reactor, 1179 families of, 1192-1195 fundamental, A-8 interactions, 1189-1192 basic forces, 1190-1191 unification of forces, 1191-1192 localizing wave, 1049-1050 wave behavior, 1043-1045 Paschen series, hydrogen atom, 1070 Pauli exclusion principle, 1099, 1116 Periodic table, A-7 electron configurations, 1102 excited states and optical transitions, 1103-1104 ionization energy, 1102-1103 Permeability: diamagnetic materials, 813 paramagnetic materials, 812 Permeability constants, 762, 810 Permittivity constant, 597 Phase, changes on reflection, 958-959 Phasor: diagram, 844-845, 848 rotating, 954 single-slit diffraction, 973 junction diode, current-voltage plot, 703 Phosphor, 890 Phosphorescence, 890 Photocopier, 595 Photodiode, 1130 Photoelectric effect, 1029-1031
In d ex
Einstein’s photon concept, 1031 frequency problem, 1030 intensity problem, 1030 time delay problem, 1030 Photoelectrons, 1029 Photon-electron collision, 1033 Photonics, 1104 Photons, 1031 energy, 1031 energy spectrum, 1207 pair production, 1204 Phthalocyanine, structure, 995 Physical constants, table, inside front cover, A-3 Physical optics, 903-904 Physical properties, inside front cover Picofarad, 678 Pions, 919 Plank, Max, 1024 Planck constant, 1025-1026 Planck’s radiation law, 1024-1025 Plane angle, conversion factors. A-10 Plane mirror, image formation, 909-912 Plane of incidence, 904 Plane of polarization, 1004 Plane polarized wave, 1004 Planes, family of, 996 Planets: magnetism, 815-817 properties, A-4 Plasma: confinement, 743 temperature, theimonuclear reactor, 1179 Plates, 677 pw junctions: depletion zone, 1127 diffusion current, 1127 diffusion-recombination event, 1126-1127 diode rectifier, 1128-1130 drift current, 1127 LEDs, 1130 reverse-biased connection, 1129-1130 p«p junction transistor, 1132 Point charge, 596 collection of, electric potential due to, 658-660 electric field, 607 - 609, 615-618 electric potential due to, 657-658 lines of force, 610 system, potential energy, 653-654 Polar dielectrics, 686-687 Polarization, 1003 -1017 circular, 1012-1014 direction of, 1004 double refraction, 1008-1012 light scattering, 1014-1016 plane of, 1004 quantum mechanics, 1017 by reflection, 1007-1008
Polarized light, applications, 1006-1007 Polarizing angle, 1007 Polarizing sheet, 1005-1007 Polaroid, 1005 Poloidal magnetic field, 1179 Population inversion, 1107 Positive charge, 594-595 lines of force, 610-611 Positrons, kinetic energy distribution, 1150 Postulate of stationary states, 1070 Potential: differences, 718-720 contact, 1029 gradient, 663 path independence, 718 stopping, 1029-1030 Potential energy: change in, 655 curve, 1121 system of charges, 653-654 variation, 1120-1121 Potential energy function, hydrogen atom, 1075 Potentiometer, 725 Power: AC circuits, 849-851 conversion factors. A-13 Poynting vector, 880-881, 1004 Pressure, conversion factors. A-13 Pressure field, 605 Pressurized-water reactor, 1172-1173 Priestley, Joseph, 640 Principal indices of refraction, double refraction, 1009-1010 Principle of complementarity, 1063 Principle of superposition: electric field, 598, 608 electric potential, 658 Prism spectrograph, 990 Probability density, 1056, 1085 Probability function, 1117 Products of vectors, mathematical for mulas, A-14 Projectile nucleus, 1154 Proton: energy distribution, 1155 core of Sun, 1176 properties, 599 quark model, 669 quarks, 599 Proton-proton cycle. Sun, 1177 'P(PSI), 1053-1054, 1200-1201 Pulsars, 873-874 Pumping, lasers, 1107 Pythagorean theorem. A-14
Quadratic formula. A-14 Quadrupole, electric field, 609 Quadrupole moment, 622
1-9
Quality factor, 1152 Quantization, energy, 1025-1027 Quantized Hall effect, 746-747 Quantum, 874 Quantum Hall effect, 701 Quantum mechanics, polarization, 1017 Quantum number, 1025, 1056, 1071, 1074 hydrogen atom, 1076 magnetic, 1076 principle, 1099 Quantum physics: Compton effect, 1032-1035 Einstein’s photon theory, 1031 -1032 energy quantization, 1025-1027 line spectra, 1035-1036 photoelectric effect, 1029-1031 Quantum theory, heat capacity, 10281029 Quark-antiquark combinations, 1198-
1200 Quark-antiquark pairs, 1204 Quark model, 1197-1201 force between quarks, 1199-1200 fractional electric charges, 1199 new symmetry, 1201 proton, 669 Quarks, A-8, 599 force between, 1199-1200 pairs of, 1201 properties, 1198 Quarter-wave plate, 1012-1013 R Rad, 1152 Radial probability density, 1085 Radiation: cosmic microwave background, 1203 dual wave-particle nature, 10631065 Radiation field, 875-876 Radiation pressure, momentum, 881 883 Radiation problem, 1022 Radiator, ideal, 1022 Radioactive dating, 1153 Radioactive wastes, 1172-1173 Radioactivity, natural, 1152-1153 Radio astronomy, 873 Radiometers, 882 Radionuclides, 1143 Radio waves, spectrum, 872-874 Radius, Bohr, 1073 Radon gas, 1152-1153 Rare earths, 1102 Rayleigh’s criterion, 976 Ray optics, 903 Rays, paraxial, 928 Ray tracing: spherical mirrors, 926 thin lens, 933
I-IO
In dex
R C circuits, 725-728 Reactions: endothermic, 1154 exothermic, 1154 nuclear, 1153-1156 threshold energy, 1154 Real image, 925 Red giant, 1178, 1208 Reflection, 904-907 diffuse, 905 electromagnetic waves, 905-906 image formation, plane mirror, 909912 image reversal, 911-912 law of, 904-905 derivation, 907-909 optical reversibility and phase changes, 958-959 polarization by, 1007-1008 total internal, 914-916 see also Mirror Reflection coefficient, amplitude, 958 Reflection gratings, 989 Refracted wave, polarization, 10071008 Refracting surface formula, derivation, 930-931 Refracting telescope, 939-940 Refraction, 904-907 electromagnetic waves, 905-906 law of, 904 Relative motion, induction, 793-795 Relativity, electromagnetism, 864 Rem, 1152 Residual nucleus, 1154 Resistance, 700-702 in microscopic terms, 702 Ohm’s law, 703-705 Resistivity, 701 -703 metal, 1120 semiconductors, 1124 superconductors, 708 temperature coefficients, 703 conductors and semiconductors, 707, 1124 temperature variation, 702-703 Resistor, 701, 706 AC circuit, 844-845 connected in parallel, 720 energy dissipation rate, 850 potential difference across, 825 connected in series, 720-722 Resolving power, dispersion, 991-993 Resonance: capture, 1172 condition, 741, 834, 848 forced oscillations, 834-835 Reverse saturation current, 1139 Rhenium isotopes, r- and s-process for mation, 1211-1212
Right-hand rule: magnetic field, 764, 766 Poynting vector, 880 sign of flux, 878 Ring of charge, 612-613 Ritz combination principle, 1092 R LC circuit, 844 differential analysis, 849 graphical analysis, 848-849 impedance, 847 loop theorem, 847 phasor diagram, 848 single-loop, 847-849 trigonometric analysis, 847-847 Roemer, Ole, 891 Roentgen, 1152 Rotating phasor, 954 R process, 1210 Rutherford, Ernest, 641-642, 1141 Rydberg constant, 1069, 1073
Salt, ionic bonding, 1110-1111 Scalar fields, 605 Scanning tunnel microscope, 1061 Schrodinger’s equation, 1054 hydrogen atom, 1074-1076 Seat of emf, internal resistance, 717-718 Secondary maxima, diffraction gratings, 988-989 Second focal point, thin lens, 933 Selection rule, 1103 Semiconductor, 596, 706-708, 11231124 doped, 1124-1126 energy bands, 707 extrinsic, 1124 n-type, 700 properties, 1125 Series connection, capacitor, 682 Shell, 1084 Shell theorems, 636-637 Sign conventions: spherical mirrors, 925-926 thin lens, 931-933 Silicon, electrical properties, 1123 Single-loop circuit, current calculations, 717-718 Single-slit diffraction, 970-972 Heisenberg’s uncertainty principle, 1052 intensity, 972-975 SI system, A -l-A -2 capacitance, 678 coulomb, 597 electric current, 698 electric field, 606 electric potential, 655 farad, 783 henry, 783, 821
magnetic field, 737 magnetic flux, 784 prefixes, inside back cover Snell’s law, 904-905 polarization, 1008 Sodium, excited states, 1103-1104 Sodium chloride, unit cell, 994-995 Solar compass, 1015 Solar wind, 816 Solenoid: Ampere’s law, 770-771 field outside, 772 inductance, 822-823 Solid angle, conversion factors. A-10 Solids, heat capacity, 1027-1029 South magnetic pole, 815 Spectra: Bragg’s law, 995-996 holography, 997-998 x-ray diffraction, 993-997 Spectral lines, 990 Spectral radiancy, 1022-1023 Spectrographs, 990 - 991 Speed, conversion factors. A-12 Spherical aberration, 939 Spherical capacitor, capacitance, 680 Spherically symmetric charge distribu tion, Gauss’ law, 637-639 Spherical mirrors, 923-928 center of curvature, 924 focal length, 924 lateral magnification, 925, 928 mirror equation, 923-925 ray tracing, 926 sign conventions, 925-926 Spherical refracting surfaces, 928-931 refracting surface formula, 930 thin lens, 931-936 Spherical shell of charge. Gauss’ law, 636-637 Spin: electron, 1082-1083 neutron, 1083 Spontaneous emission, 1106 S process, 1210 Standard cell, 725 Stars, thermonuclear fusion, 1176-1178 States: allowed, filling, 1117-1119 density of, 1116 excited, hydrogen atom, 1086-1087 ground, 1057 hydrogen atom, 1085-1086 metastable, 1106 spectroscopic designations, 1101 Static fields, 605 Stationary states, postulate, 1070 Stefan-Boltzmann constant, 1022 Stefan-Boltzmann law, 1022 Step-down transformer, 852
Index Step-up transformer, 852 Stem -Gerlach experiment, 1080-1082 Stilwell, G. R., 894-895 Stimulated emission, 1105-1107 Stokes, G. G., 958 Stopping potential, 1029 cutoff frequency, 1030 Strangeness, 1196-1197 Strong force, 1143-1144, 1191 Subshell, 1084 Z7-2, /-1 , hydrogen atom, 1087 weighted average probability density, hydrogen atom, 1087 Subshells, 1101 Sun: barrier tunneling, 1061 neutrino energy, 1178 properties, A-4 proton energy distribution in core, 1176 thermal radiation, 890 Superconducting Supercollider, 742 Superconductivity, 708-709 Superconductors, 1133-1134 Supernova, neutrino burst, 1150 Surface charge: density, 612 induced, 688 Synchrotron, 742 System of charges, potential energy, 653-654
Target nucleus, 1154 Telescope, refracting, 939-940 Tellurium isotopes, r- and s-process paths, 1210 Temperature, superconductors, 708 critical, 1133 Temperature coefficient of resistivity, 703 conductors and semiconductors, 707, 1124 Temperature field, 605 Tesla, 737, 784 Theory of everything, 1192 Theory of relativity: Doppler effect, 894 consequences, 897-898 derivation, 895-897 Einstein’s postulates, 896 Thermal neutron, 1048, 1168 test of fissionability, 1170-1171 Thermal radiation, 872,890, 1021 -1024 cavity radiation, 1022-1023 spectral radiancy, 1022-1023 spectrum, 1021 Stefan-Boltzmann law, 1022 Wien displacement law, 1023-1024 Thermograph, 1038
Thermonuclear fusion, 1175-1182 controlled, 1178-1179 inertial confinement, 1181-1182 kinetic energy, 1176 laser fusion, 1181-1182 magnetic confinement, 1179-1181 proton-proton cycle, 1177 stars, 1176-1178 Sun, 1061 Thermonuclear reactor: inertial confinement, 1179, 11811182 laser fusion, 1181-1182 magnetic confinement, 1179-1181 requirements, 1179 tokamak, 1179-1180 Thin films, interference, 955-958 Thin lens, 931,931-936 approximation, 936 converging, 931-933 diverging, 931-933 first focal point, 933 focal length, 931 formula derivation, 935-936 lateral magnification, 931 optical path length, 933 ray tracing, 933 second focal point, 933 sign conventions, 931 -933 Thomason, J. J., 739 Thompson, George P., testing de Broglie’s hypothesis, 1047 Thomson model, 641 Threshold energy, 1154 Time, conversion factors. A-11 Time-dependent forces, computer pro grams, A-16-A-17 Time-reversal symmetry, 1013 Time-varying fields, 605 Tin isotopes, r- and s-process paths, 1210 Tokamak, 772, 1179-1180 Toner, 595 Toroid: Ampere’s law, 771 -772 inductance, 823 Toroidal magnetic field, 1179-1180 Torque, current loop, 749-751 Torsion balance, 596 Total internal reflection, 906, 914-916 frustrated, 916 Transformer, 851-852 Transistor, 1132-1133 Transmission coefficient, amplitude, 958 Transverse Doppler effect, 897-898 Traveling wave: emission, 875 Maxwell’s equations, 877-880 Triangles, mathematical formulas. A-14 Triboluminescence, 890-891 Trigonometry:
I-l 1
expansions. A-15 functions. A-14 identities. A-15 Tritium, 601 Tube length, 938 Twin paradox, 898 U alpha decay, 1149 abundance, 1211 nuclear fission, 1168-1169
238\J: abundance, 1211 alpha decay, 1148-1149 Ultraviolet spectrum, 874 Uncertainty principle, 1158 electron trapped in infinite well, 1058 Unification forces, 1191-1192 Unified atomic mass units, 1145 Uniform field, 605 Universe: age, 1210-1213 expansion of, 1202-1203 Unpolarized light, scattering, 1015 Unpolarized wave, 1004-1005
Valence band, 1123 Van Allen radiation belts, 743 Van de Graaff accelerator, 667-668 Vector field, 605 flux, 627-629 Vector products. A-14 Velocity, Fermi distribution, 1119-1120 Velocity-dependent forces, computer program. A-17 - A-18 Velocity selector, 739-740 Vibrational nodes, 1054 Virtual images, 925 Virtual object, 925-926 Visible light, 889-891 Volt, 655,716 Voltage, Hall, 745 Voltmeter, 724-725 Volume: charge density, 612 conversion factors, A-11 von Klitzing, Klaus, 746 W Water, electric dipole moment, 619-620 Wave behavior, 1043-1065 barrier tunneling, 1059-1062 complementarity, 1063 correspondence principle, 1062 Davisson-Germer experiment, 1046-1047 de Broglie wavelength, 1045-1046, 1057
1-12
In d ex
Wave behavior (Continued) double-slit interference, 1043-1044, 1064-1065 dual wave-particle nature, 10631065 energy-time uncertainty relationship, 1052-1053 frequencies, 1052 Heisenberg uncertainty relationships, 1051-1052 particles, 1043-1045 probability density, 1054, 1056 Schrodinger equation, 1054, 1056 standing electromagnetic waves, 1054-1055 testing de Broglie’s hypothesis, 10461049 G. P. Thomson’s experiment, 10471048 transmission coefficient, 1060 trapped particles and probability den sities, 1054-1059 uncertainty principle and single-slit diffraction, 1052
vibrational nodes, 1054 wave functions, 1053-1054 Wave disturbances, adding, 954-955 Wave functions, 1053-1054 Wave optics, 903-904 Wave packet, 1049-1050 Waves: coherent, 947, 950-952 incoherent, 947 localizing in space, 1049-1050 in time, 1050 path difference, 950, 973-974 phase difference, 950, 952, 973-974 Wave theory of light, diffraction, 967970 Wavetrain, 951 coherence length, 963-964 Weak force, 1191 Weak interaction, strangeness, 1197 Weber, 784 Wien displacement law, 1023-1024 Wire, long straight: Ampere’s law, 769
Biot-Savart law, 763 Wire gauge, 710 Work, conversion factors. A-12 Work function, 1031 -1032 metal, 1120-1121
X-ray astronomy, 874 X-ray diffraction, 993-997 Bragg’s law, 995-996 X rays, spectrum, 874 characteristic, 1096-1097 element numbering and, 10971099 continuous, 1095-1096
Young, Thomas, double-slit experi ment, 949-950
Zeeman effect, hydrogen atom, 1088 Zero-point energy, 1057