CHAPTER
17
Wave–Particle Duality and Quantum Physics
1* ·
The quantized character of electromagnetic radiation is revealed by ( a) the Young Young double double-slit -slit experiment. ( b )
diffraction of light by a small aperture. ( c) the photoelectric effect. (d ( d ) the J.J. Thomson cathode-ray experiment. (c ) 2
··
Two monochromatic light sources, A and B, emit the same number of photons per second. The wavelength of A is
power radiated by source source B is ( a ) equal to that of source A. (b ( b ) less λA = 400 nm, and that of B is λB = 600 nm. The power than that of source A. (c ( c) greater than that of source A. (d ( d ) cannot be compared to that from source A using the available data. (b ) 3
·
Find the photon energy in joules and in electron volts for an electromagnetic electromagn etic wave of frequency freque ncy (a ( a ) 100MHz in the
FM radio band, and (b ( b ) 900 kHz in the AM radio band. -34
-28
-9
E = = 5.96 × 10 J = 3.72 × 10 eV
(b ) E = = hf ·
-7
-15
h = 6.626 × 10 J .s = 4.136 × 10 eV .s
4
-26
E = = 6.63 × 10 J = 4.14 × 10 eV
(a ) E = = hf ;
An 80-kW 80-kW FM transmitter operates at a frequency of 101.1 101.1 MHZ. How many photons per second are emitted by
the transmitter? 4
-34
N = = 8 × 10 /(6.63 × 10
P = Nhf = Nhf ; N = P = P /hf
× 101.1 × 106) s-1
30 -1
= 1.19 × 10 s 5* ·
What are the frequencies of photons having the following energies? (a (a) 1 eV, (b (b) 1 keV, and (c (c) 1 MeV. -15
14
17
20
(a ) f = = 1/4.14 × 10 Hz = 2.42 × 10 Hz. (b ) f = = 2.24 × 10 Hz. (c) f = = 2.24 × 10 Hz. 6
·
Find the photon energy for light of wavelength (a (a) 450 nm, (b (b) 550 nm, and (c (c) 650 nm.
(a), (b (b), (c (c) E = = hc/ hc /λ = (1240 eV .nm)/λ 7
·
(a ) E = = 2.76 eV (b ) E = = 2.25 eV (c) E = = 1.91 eV
Find the photon energy if the wavelength is (a ) 0.1 nm (about 1 atomic diameter), and (b ( b) 1 fm (1 fm = 10
about 1 nuclear diameter). -6
(a ) E = = 1240/0.1 1240/0.1 eV = 12.4 12.4 keV. (b ) E = = 1240/10 eV = 1.24 GeV.
-15
m,
Chapter 17 8
··
Wave–Particle Duality and Quantum Physics
The wavelength of light emitted by a 3-mW He-Ne laser is 632 nm. If the diameter of the laser beam is 1.0 mm,
what is the density of photons in the beam? 1. Find the number of photons emitted per second
-3
-19
N = P / E = (3 × 10 /1.6 × 10 )/(1240/632) 15
= 9.56 × 10
-6
8
3
2
3
2. Find the volume containing the photons
V = Ac = (π × 10 /4) × 3 × 10 m = 2.36 × 10 m
3. Density of photons = N /V
ρ = 9.56 × 10 /2.36 × 10 m = 4.05 × 10 m
9* ·
15
2
-3
13
-3
True or false: In the photoelectric effect, (a) the current is proportional to the intensity of the incident light. (b) the
work function of a metal depends on the frequency of the incident light. ( c) the maximum kinetic energy of electrons emitted varies linearly with the frequency of the incident light. ( d ) the energy of a photon is proportional to its frequency. (a) True (b) False (c) True (d ) True 10 ·
In the photoelectric effect, the number of electrons emitted per second is (a) independent of the light intensity. (b)
proportional to the light intensity. (c) proportional to the work function of the emitting surface. ( d ) proportional to the frequency of the light. (b) The work function of a surface is φ . The threshold wavelength for emission of photoelectrons from the surface is (a) hc/ φ . (b) φ /hf . (c) hf /φ . (d ) none of the above.
11 ·
(a) When light of wavelength λ1 is incident on a certain photoelectric cathode, no electrons are emitted no matter how intense the incident light is. Yet when light of wavelength λ2 < λ1 is incident, electrons are emitted even when the
12 ··
incident light has low intensity. Explain. hc/λ must be greater than φ . Evidently, hc/λ1 < φ , but hc/λ2 > φ . 13*·
The work function for tungsten is 4.58 eV. (a) Find the threshold frequency and wavelength for the photoelectric
effect. (b) Find the maximum kinetic energy of the electrons if the wavelength of the incident light is 200 nm, and ( c) 250 nm. (a) f t = φ /h; λ = c/ f (b), (c) K m = E - φ = hc/λ - φ
-15
15
f t = 4.58/4.136 × 10 Hz = 1.11 × 10 Hz; λt = 270 nm (b) K m = (1240/200 - 4.58) eV = 1.62 eV (c) K m = 0.38 eV
Chapter 17
14 ·
Wave–Particle Duality and Quantum Physics
When light of wavelength 300 nm is incident on potassium, the emitted electrons have maximum kinetic energy of
2.03 eV. (a) What is the energy of an incident photon? ( b) What is the work function for potassium? ( c) What would be the maximum kinetic energy of the electrons if the incident light had a wavelength of 430 nm? (d ) What is the threshold wavelength for the photoelectric effect with potassium? (a) E = hc/λ
E = 1240/300 eV = 4.13 eV
(b) φ = E - K m (c) K m = E - φ
φ = 2.10 eV K m = (1240/430 - 2.10) eV = 0.784 eV
(d ) λt = hc/φ
λt = 1240/2.10 nm = 590 nm
15 ·
The threshold wavelength for the photoelectric effect for silver is 262 nm. (a) Find the work function for silver. (b)
Find the maximum kinetic energy of the electrons if the incident radiation has a wavelength of 175 nm. (a), (b) See Problem 14
(a) φ = 1240/262 eV = 4.73 eV (b) K m = (1240/175 - 4.73) eV = 2.36 eV
16 ·
The work function for cesium is 1.9 eV. (a) Find the threshold frequency and wavelength for the photoelectric
effect. Find the maximum kinetic energy of the electrons if the wavelength of the incident light is ( b) 250 nm, and (c) 350 nm. -15
14
(a) f t = φ /h; λt = hc/φ
f t = 1.9/4.136 × 10 Hz = 4.59 × 10 Hz; λt = 653 nm
(b), (c) K m = E - φ
(b) K m = (1240/250 - 1.9) eV = 3.06 eV (c) K m = 1.64 eV
17* ·· When a surface is illuminated with light of wavelength 512 nm, the maximum kinetic energy of the emitted
electrons is 0.54 eV. What is the maximum kinetic energy if the surface is illuminated with light of wavelength 365 nm? 1. Find φ = E - K m 2. Find K m for λ = 365 nm
φ = (1240/512 - 0.54) eV = 1.88 eV K m = (1240/365 - 1.88) eV = 1.52 eV o
Find the shift in wavelength of photons scattered at θ = 60 . Use Equ. 17-8; h/mec = 2.43 pm ∆λ = 2.43 × 0.5 pm = 1.215 pm
18 ·
19 ·
When photons are scattered by electrons in carbon, the shift in wavelength is 0.33 pm. Find the scattering angle. -1
Use Equ. 17-8
θ = cos (1 - 0.33/2.43) = 30.2
o
o
The wavelength of Compton-scattered photons is measured at θ = 90 . If ∆λ/λ is to be 1.5%, what should the wavelength of the incident photons be?
20 ·
Find ∆λ from Equ. 17-8; λ = ∆λ/0.015 21* ·
∆λ = 2.43 pm; λ = 162 pm = 0.162 nm
Compton used photons of wavelength 0.0711 nm. (a) What is the energy of these photons? ( b) What is the waveo
length of the photon scattered at θ = 180 ? (c) What is the energy of the photon scattered at this angle?
Chapter 17
Wave–Particle Duality and Quantum Physics
(a) E = hc/λ
E = 1240/0.0711 eV = 17.44 keV
(b) Use Equ. 17-8; λf = λi + ∆λ (c) E = hc/λ
∆λ = 2 × 2.43 pm = 0.00486 nm; λf = 0.076 nm E = 1240/0.076 eV = 16.3 keV
For the photons used by Compton, find the momentum of the incident photon and that of the photon scattered at
22 ·
o
180 , and use the conservation of momentum to find the momentum of the recoil electron in this case (see Problem 21). -24 -24 -24 p i = h/λi = 9.32 × 10 kg.m/s. p f = -8.72 × 10 kg.m/s. p e = -( p f - p i) = 18.0 × 10 kg.m/s.
23 ··
o
An X-ray photon of wavelength 6 pm that collides with an electron is scattered by an angle of 90 . (a) What is the
change in wavelength of the photon? ( b) What is the kinetic energy of the scattered electron?
∆λ = 2.43 pm = 0.00243 nm ∆ E = (1240/0.006 - 1240/0.00843) eV = 59.6 keV
(a) Use Equ. 17-8 (b) ∆ E = hc/λi - hc/λf 24 ··
How many head-on Compton scattering events are necessary to double the wavelength of a photon having initial
wavelength 200 pm? 1. Find ∆λ per collision from Equ.17-8 2. Number of collisions = ∆λ/(∆λ/collision) 25* ·
∆λ/collision = 4.86 pm N = 200/4.86 = 42
True or false: (a) The de Broglie wavelength of an electron varies inversely with its momentum. ( b) Electrons can
be diffracted. (c) Neutrons can be diffracted. (d ) An electron microscope is used to look at electrons. (a) True (b) True (c) True (d ) False 26 ·
If the de Broglie wavelength of an electron and a proton are equal, then (a) the velocity of the proton is greater
than that of the electron. (b) the velocity of the proton and electron are equal. ( c) the velocity of the proton is less than that of the electron. (d ) the energy of the proton is greater than that of the electron. ( e) both (a) and (d ) are correct. (c) p e = p p. m p > me. 27 ·
A proton and an electron have equal kinetic energies. It follows that the de Broglie wavelength of the proton is (a)
greater than that of the electron. (b) equal to that of the electron. (c) less than that of the electron. 2
2
(c) p p /2m p = p e /2me; p e < p p so λe > λ p. 28 ·
Use Equation 17-13 to calculate the de Broglie wavelength for an electron of kinetic energy (a) 2.5 eV, (b)
250 eV, (c) 2.5 keV, and (d ) 25 keV. (a) λ = 1.23/2.51/2 nm = 0.778 nm (b) λ = 0.0778 nm
(a), (b), (c), (d ) Use Equ. 17-13
(c) λ = 0.0246 nm (d ) λ = 7.78 pm 29* ·
5
An electron is moving at v = 2.5 × 10 m/s. Find its de Broglie wavelength.
Find p = mv; λ = h/ p = h/mv 30 ·
-34
-31
λ = 6.626 × 10 /9.11 × 10
× 2.5 × 105 m = 2.91 nm
An electron has a wavelength of 200 nm. Find (a) its momentum, and (b) its kinetic energy.
Chapter 17
Wave–Particle Duality and Quantum Physics -34
(a) p = h/λ 2
(b) Use Equ. 17-13; K = (1.23/λ) , λ in nm 31 ·
-27 kg.m/s = 3.31 × 10 kg.m/s
-5
K = 3.78 × 10 eV
Find the energy of an electron in electron volts if its de Broglie wavelength is (a) 5 nm, and (b) 0.01 nm. 2
-2
(a) K = 6.05 × 10 eV (b) K = 15.1 keV
(a), (b) K = (1.23/λ) , λ in nm 32 ·
-7
p = 6.626 × 10 /2 × 10
A neutron in a reactor has kinetic energy of about 0.02 eV. Calculate the de Broglie wavelength of this neutron 2
from Equation 17-12, where mc = 940 MeV is the rest energy of the neutron. Use Equ. 17-12;
λ
=
1240 1880 × 106 K
=
2.86 × 10 −2
λ = 0.202 nm
K
where λ is in nm and K in eV 33* ·
2
Use Equation 17-12 to find the de Broglie wavelength of a proton (rest energy mc = 938 MeV) that has a kinetic
energy of 2 MeV. -2 For protons, λ = 2.86 × 10 / K , λ in nm, K in eV
34 ·
A proton is moving at v = 0.003c, where c is the speed of light. Find its de Broglie wavelength.
Find p; λ = h/ p = h/m pv = (h/mec)(c/v)(me/m p) 35 ·
-2
2
(a) K = 0.818 meV (b) K = 818 MeV
Find the de Broglie wavelength of a baseball of mass 0.145 kg moving at 30 m/s. -34
-34
λ = 6.626 × 10 /4.35 m = 1.52 × 10 m
λ = h/mv 37* ·
λ = 2.43(1/0.003)(0.511/938) pm = 0.441 pm
What is the kinetic energy of a proton whose de Broglie wavelength is (a) 1 nm, and (b) 1 fm?
(a), (b) K = (2.86 × 10 /λ) , K in eV, λ in nm 36 ·
-5
λ = 2.02 × 10 nm = 20.2 fm
The energy of the electron beam in Davisson and Germer’s experiment was 54 eV. Calculate the wavelength for
these electrons. Use Equ. 17-13 38 ·
λ = 0.167 nm +
-
The distance between Li and Cl ions in a LiCl crystal is 0.257 nm. Find the energy of electrons that have a
wavelength equal to this spacing. K = (1.23/λ) 39 ·
2
An electron microscope uses electrons of energy 70 keV. Find the wavelength of these electrons.
Use Equ. 17-13 40 ·
K = 22.9 eV
-3
λ = 4.65 × 10 nm = 4.65 pm 6
What is the de Broglie wavelength of a neutron with speed 10 m/s?
Chapter 17 -34
Wave–Particle Duality and Quantum Physics -27
λ = h/ p = h/mv = 6.626 × 10 /(1.67 × 10 41* ·
× 106) m = 3.97 × 10-13 m = 0.397 pm.
Suppose you have a spherical object of mass 4 g moving at 100 m/s. What size aperture is necessary for the object
to show diffraction? Show that no common objects would be small enough to squeeze through such an aperture. -34
-3
For diffraction, the diameter of the aperture d ≈ λ. So d ≈ 6.626 × 10 /(4 × 10
× 100) = 1.66 × 10-33 m. This is many
orders of magnitude smaller than even the diameter of a proton. 42 ·
A neutron has a kinetic energy of 10 MeV. What size object is necessary to observe neutron diffraction effects?
Is there anything in nature of this size that could serve as a target to demonstrate the wave nature of 10-MeV neutrons? -2 From Problem 32, λ = 2.86 × 10 / K
43 ·
-6
d = 9.04 × 10 nm ≈ 10 fm ~ nuclear diameter
≈ d
What is the de Broglie wavelength of an electron of kinetic energy 200 eV? What are some common targets that
could demonstrate the wave nature of such an electron? From Equ. 17-13 one finds that λ = 1.23/14.1 nm = 0.0872 nm. This is of the order of the size of an atom. 44 ··
2
Sketch the wave function ψ( x) and the probability distribution ψ ( x) for the state n = 4 of a particle in a box.
45* ·· (a) Find the energy of the ground state (n = 1) and the first two excited states of a proton in a one-dimensional box -15
of length L = 10 m = 1 fm. (These are the order of magnitude of nuclear energies.) Make an energy-level diagram for this system and calculate the wavelength of electromagnetic radiation emitted when the proton makes a transition from (b) n = 2 to n = 1, (c) n = 3 to n = 2, and (d ) n = 3 to n = 1. 2
2
-11
(a) E 1 = h /8mL = 3.28 × 10
J = 205 MeV; the energy level diagram is
shown 6
(b) For n = 2 to n = 1, ∆ E = 3 E 1 so λ = 1240/615 × 10 nm = 2.02 fm (c) For n = 3 to n = 2, ∆ E = 5 E 1 and λ = 3 × 2.02/5 fm = 1.21 fm (d) For n = 3 to n = 1, ∆ E = 8 E 1 and λ = 3 × 2.02/8 fm = 0.758 fm
Chapter 17 46 ··
Wave–Particle Duality and Quantum Physics
(a) Find the energy of the ground state (n = 1) and the first two excited states of a proton in a one-dimensional box
of length 0.2 nm (about the diameter of a H 2 molecule). Calculate the wavelength of electromagnetic radiation emitted when the proton makes a transition from (b) n = 2 to n = 1, (c) n = 3 to n = 2, and (d ) n = 3 to n = 1. 2
(a) E 1 =
(hc )
8( mc2 ) L2
E 1 =
, L in nm, E in eV
E 2 = 4 E 1 ; E 3 = 9 E 1
(1240 )
2
8(9.38 × 108)(0.04)
= 5.12 meV
E 2 = 20.5 meV; E 3 = 46.1 meV 4
(b), (c), (d ) λ = 1240/∆ E
λ2-1= 8.07 × 10 nm = 80.7 µm; λ3-2 = 48.4 µm; λ3-1 = 30.3 µm
(a) Find the energy of the ground state and the first two excited states of a small particle of mass 1 µg confined to a one-dimensional box of length 1 cm. ( b) If the particle moves with a speed of 1 mm/s, calculate its
47 ··
kinetic energy and find the approximate value of the quantum number n. 2
(a) E 1 =
2
8( mc ) L
E 1 =
2
2
8(10
× 9 ×10
16
-2 2
)(10 )
-54
2
= 5.56 × 10-55 J
-54
-16
19
E = 5 × 10 J; n = 3 × 10
(b) E = 1/2mv ; n = 1/2mv / E 1 48 ··
-9
E 2 = 2.22 × 10 J; E 3 = 5.03 × 10 J
E 2 = 4 E 1; E 3 = 9 E 1 2
(6.626 ×10-34 × 3 × 108 )
-2
(hc )
A particle is in the ground state of a box of length L. Find the probability of finding the particle in the interval
∆ x = 0.002 L at (a) x = L/2, (b) x = 2 L/3, and (c) x = L. (Since ∆ x is very small, you need not do any integration because the wave function is slowly varying.) 2
2
P ( x)∆ x = ψ ( x)∆ x; ψ( x) = ψ1( x) = (2/ L)1/2 sin (π x/ L) (a) Evaluate P ( x) at x = L/2; P = P ( x)∆ x
P ( x) = (2/ L) sin (π x/ L) P = (2/ L)(0.002 L) = 0.004
(b) Repeat as in (a) for x = 2 L/3
P = (2/ L)(0.75)(0.002 L) = 0.003
(c) Repeat as in (a) for x = L
P = 0
49* ·· Do Problem 48 for a particle in the first excited state (n = 2). 2
2
2
Repeat procedure of Problem 47 with P ( x) = ψ ( x) = ψ2 ( x) = (2/ L) sin (2π x/ L). The results are: (a) P = 0. (b) P = 0.003. (c) P = 0. 50 ··
Do Problem 48 for a particle in the second excited state (n = 3). 2
2
2
Proceed as in Problem 48 with P ( x) = ψ ( x) = ψ3 ( x) = (2/ L) sin (3π x/ L). The results are: (a) P = 0.004. (b) P = 0. (c) P = 0. 51 ··
The classical probability distribution function for a particle in a box of length L is given by P ( x) = 1/ L. Use this 2
to find 〈 x〉 and 〈 x 〉 for a classical particle in such a box. L
〈 x〉
=
L
∫ (x/L) dx = L/2 ; 〈 x 〉 = ∫ ( x /L) dx = L / 3 . 2
0
2
0
2
Chapter 17 52 ··
Wave–Particle Duality and Quantum Physics 2
(a) Find 〈 x〉 for the first excited state (n = 2) for a particle in a box of length L, and (b) find 〈 x 〉.
Proceed as in Example 17-8, replacing ψ1( x) with ψ2( x). L
(a) ( x ) =
2π
2 x 2 L 2 2π x 2 sin dx = 2 θ sin θd θ L L π 0 0
∫
∫
L
(b) ( x ) = 2
∫ 0
∫ θ sin
2
2 x 2 L
sin
2
2π x L
dx
=
2 L2 π3
= L . 2
2π
∫
2
2
θ sin θd θ
0
θ 2 θ sin 2θ cos 2θ θd θ = − − ; 4 8 4
∫ θ
2
sin
2
1 1 = − 2 L2 = 0.321 L2 . We have used 3 8π
θ 3 θ 2 1 θ cos 2θ θd θ = − − sin 2θ − . 4 6 4 8 2
53* ·· (a) Find 〈 x〉 for the second excited state (n = 3) for a particle in a box of length L, and (b) find 〈 x 〉.
We proceed as in the preceding problem. Now the integrals over θ extend from 0 to 3π . 2 2 2 2 2 (a) 〈 x〉 = L/2. (b) 〈 x 〉 = (1/3 - 1/18 π ) L = 0.328 L . (Note that 〈 x 〉 approaches the classical value 1/3 as the quantum number n increases.) 2
A particle in a one-dimensional box is in the first excited state (n = 2). (a) Sketch ψ ( x) versus x for this state. (b) What is the expectation value 〈 x〉 for this state? (c) What is the probability of finding the particle in
54 ··
some small region dx centered at x = 1/2 L? (d ) Are your answers for (b) and (c) contradictory? If not, explain. (a) The probability density is shown (b) 〈x〉 = L/2 as found in Problem 52 (c) Since P ( L/2) = 0, P ( L/2) dx = 0 (d ) Parts (b) and (c) are not contradictory. (b) states that the average value of measurements of the position of the particle will yield L/2, even though the probability that any one measurement of the position will yield L/2 is zero. - x/a A particle of mass m has a wave function given by ψ( x) = A e , where A and a are constants. (a) Find the normalization constant A. (b) Calculate the probability of finding the particle in the region - a ≤ x ≤ a.
55 ··
∞
(a) We evaluate
∞
∫ Ψ ( x)dx = 1 = 2∫ A e 2
−∞
2
0
− 2 x / a
∞
dx
= 2 A
2
∫ e 0
− 2 x / a
dx = aA ; A = 1 a . 2
Chapter 17 a
(b) P =
∫
a
Ψ
( x ) dx = 2
2
−a
56 ··
∫ 0
Wave–Particle Duality and Quantum Physics
1 −2 x / a 2 e dx = 1 − e a
= 0 .865 .
A particle in a one-dimensional box of length L is in its ground state. Calculate the probability that the particle
will be found in the region (a) 0 < x < 1/2 L, (b) 0 < x < L/3, and (c) 0 < x < 3 L/4. 2
The probability density is given by P ( x) = (2/ L) sin (π x/ L). We must now evaluate the integral of P ( x) between the limits specified in (a), (b), and (c). Changing the variable from x to θ as in Example 17-8, we have
(a) P =
∫
L π
2 (b) P = π
(c) P =
π /2
2 L
2 π
2
sin θd θ
= 1. 2
0
π /3
∫ sin θd θ = 13 − 4π3 = 0.196 . 2
0
3π / 4
∫
2
sin θd θ
=3+ 4
0
1 2π
= 0.909 .
57* ·· Repeat Problem 56 for a particle in the first excited state of the box. 2
2
For the first excited state, i.e., for ψ ( x) = (2/ L) sin (2π x/ L), the integrals over θ go from 0 to π , 0 to 2π /3, and 0 to 3π /2 for parts (a), (b), and (c), respectively. The other change is that the factor ( L/π ) is replaced by ( L/2π ). π
2 L (a) P = L 2π
(b)
1 P = π
1 (c) P = π
58 ··
∫ sin
θd θ
=1.
θd θ
=1+
2
2
0
2π / 3
∫ sin
2
3
0
3π / 2
∫ 0
2
sin θ d θ
3 8π
= 0.402 .
= 3 = 0.75 . 4
(a) For the wave functions ψ n (x) =
2/L sin (n π x /L), n = 1, 2, 3, ... corresponding to a particle in 2
2
2
2 2
the nth state of a one-dimensional box of length L, show that 〈 x 〉 = ( L /3) - [ L /(2n π )]. (b) Compare this result for n >> 1 with your answer for the classical distribution of Problem 51.
Chapter 17
Wave–Particle Duality and Quantum Physics
We proceed as in Problem 52( b), now replacing the argument (n π x/ L) by θ. 3
2 L (a) ( x ) = L (nπ )3 2
nπ
∫
2
= L2 1 − 12 2 . 3 2n π
2
θ sin θd θ
0
2
(b) For large values of n, the result agrees with the classical value of L /3 given in Problem 51. The wave functions for a particle of mass m in a one-dimensional box of length L centered at the origin (so
59 ··
that the ends are at x = ± L/2) are given by ψ n (x) =
2/L cos(n π x /L), n = 1, 3, 5, 7, ... and
2/L sin (n π x /L), n = 2, 4, 6, 8, ... . Calculate 〈 x〉 and 〈 x2〉 for the ground state.
ψ n (x) = 2
2
1. Since ψ1 ( x) is an even function of x, xψ1 ( x) is an odd function of x. It follows that the integral between - L/2 and L/2 is zero. Thus 〈 x〉 = 0 for all values of n.
2. ( x ) = 2
2
L / 2
∫
2
2
π x
x cos dx L − L / 2 L
2 π /2
2 L
=
π
3
∫ θ
2
(1 − sin θ ) d θ = 2
−π / 2
2 L2 π 3 π
3
π 3 π 2 1 1 − + = L − 2 . 12 2π 12 24 4
Note: The result differs from that of Example 17-8. Since we have shifted the origin by ∆ x = L/2, we could have 2
2
2
arrived at the above result, without performing the integration, by subtracting ( ∆ x) = L /4 from 〈 x 〉 as given in Example 17-8. 2
Calculate 〈 x〉 and 〈 x 〉 for the first excited state of the box described in Problem 59.
60 ··
〈 x〉 = 0. (see Problem 59)
〈 x 〉 2
61* ·
= L
2
1 3
-
8 p 2 1
2
L 4
1 12
= L2
-
1 8π
. (see Note of Problem 59) 2
Can the expectation value of x ever equal a value that has zero probability of being measured?
Yes 62 ·
Explain why the maximum kinetic energy of ele ctrons emitted in the photoelectric effect does not depend on
the intensity of the incident light, but the total number of electrons emitted does. In the photoelectric effect, an electron absorbs the energy of a single photon. Therefore, K max = hf -φ , independent of the number of photons incident on the surface. However, the number of photons incident on the surface determines the number of electrons that are emitted. 63 ··
A six-sided die has the number 1 painted on three sides and the number 2 painted on the other three sides. ( a)
What is the probability of a 1 coming up when the die is thrown? ( b) What is the expectation value of the number that comes up when the die is thrown? (a) P (1) = 1/2. (b) 〈n〉 = (3 × 1 + 3 × 2)/6 = 1.5. 64 ··
True or false: (a) It is impossible in principle to know precisely the position of an electron. ( b) A particle that
is confined to some region of space cannot have zero energy. ( c) All phenomena in nature are adequately described by classical wave theory. (d ) The expectation value of a quantity is the value that you expect to
Chapter 17
Wave–Particle Duality and Quantum Physics
measure. (a) False (b) True (c) False (d ) False; it is the most probable value of the measurement. 65* ·· It was once believed that if two identical experiments are done on identical systems under the same
conditions, the results must be identical. Explain why this is not true, and how it can be modified so that it is consistent with quantum physics. According to quantum theory, the average value of many measurements of the same quantity will yield the expectation value of that quantity. However, any single measurement may differ from the expectation value. 66 ·
2
A light beam of wavelength 400 nm has an intensity of 100 W/m . (a) What is the energy of each photon in 2
the beam? (b) How much energy strikes an area of 1 cm perpendicular to the beam in 1 s? (c) How many photons strike this area in 1 s? (a) E ph = hc/λ
E ph = 1240/400 eV = 3.1 eV
(b) E = IAt
E = 100 × 10
-4
× 1 = 0.01 J = 6.25 × 1016 eV 16 16 N = 6.25 × 10 /3.1 = 2.02 × 10
(c) N = E / E ph 67 ·
-6
-1
A mass of 10 g is moving with a speed of about 10 cm/s in a box of length 1 cm. Treating this as a one-
dimensional particle in a box, calculate the approximate value of the quantum number n. 2
1. Write the energy of the particle
E = 1/2mv
2. Write the expression for E n
E n = n h /8mL
3. Solve for n
n = 2mvL/h = 3.02 × 10
68 ·
2 2
2 19
(a) For the classical particle of Problem 67, find ∆ x and ∆ p, assuming that these uncertainties are given by _
∆ x/ L = 0.01% and ∆ p/p = 0.01%. (b) What is (∆ x∆ p)/h ? -4 -2 -6 -4 -4 -9 -3 -16 (a) ∆ x = 10 × 10 m = 10 m; ∆ p = 10 (mv) = 10 × 10 × 10 kg.m/s = 10 kg.m/s. _ -22 -34 12 (b) ∆ x∆ p/h = 10 /1.05 × 10 = 0.948 × 10 . 69* ·
-12
In 1987, a laser at Los Alamos National Laboratory produced a flash that lasted 1 × 10 s and had a power 15
of 5.0 × 10 W. Estimate the number of emitted photons if their wavelength was 400 nm. 3
N = E / E ph = ( P ∆t )/(hc/λ)
N = (5 × 10
× 1.6 × 10-19 eV)/3.1 eV = 1022
You can’t see anything smaller than the wavelength λ used. What is the minimum energy of an electron needed in an electron microscope to see an atom, which has a diameter of about 0.1 nm?
70 ·
2
2
Use Equ. 17-13; K = 1.23 /λ eV, λ in nm 71 ·
K = 151 eV
A common flea that has a mass of 0.008 g can jump vertically as high as 20 cm. Estimate the de Broglie
wavelength for the flea immediately after takeoff. 1. p = m 2gh 2. λ = h/ p 72 ·
-5 p = 1.584 × 10 kg.m/s -34
-5
-29
λ = 6.626 × 10 /1.584 × 10 m = 4.2 × 10 m
The work function for sodium is φ = 2.3 eV. Find the minimum de Broglie wavelength for the electrons
Chapter 17
Wave–Particle Duality and Quantum Physics
emitted by a sodium cathode illuminated by violet light with a wavelength of 420 nm. 1. Use Equ. 17-3 to find K max
K max = (1240/420 - 2.3) eV = 0.652 eV
2. Use Equ. 17-13 to find λ
λ = 1.23/ 0.652 = 1.52 nm
73* ·· Suppose that a 100-W source radiates light of wavelength 600 nm uniformly in all directions and that the eye
can detect this light if only 20 photons per second enter a dark-adapted eye having a pupil 7 mm in diameter. How far from the source can the light be detected under these rather extreme conditions? 2
2
2
1. At a distance R from the source, the fraction of the light energy entering the eye is Aeye/4π R = r /4 R . -19
N = P / E ph = 100/[(1240/600) × 1.6 × 10 ]
2. Find the number of photons emitted per second
20
= 3.02 × 10 /s 20
3. Solve for R from 20 = 3.02 × 10 74 ··
× r 2/4 R2
R = 6800 km (neglects absorption by atmosphere)
Data for maximum kinetic energy of the electrons versus wavelength for the photoelectric effect using
sodium are λ, nm K max, eV
200
300 4.20
400 2.06
500 1.05
600 0.41
0.03
Plot these data so as to obtain a straight line and from your plot find (a) the work function, (b) the threshold frequency, and (c) the ratio h/e. We plot K max versus frequency, f . The plot is shown below. 15
(a) φ = hf th, where f th is the threshold frequency for emission of electrons. Here f th = 0.5 × 10 Hz; φ = 2.07 eV. (b) See (a). f th = 0.5 PHz. (c) h/e is the slope of the straight line divided by e. The slope is 4.2/10
15
-15
eV/Hz; so h/e = 4.2 × 10
-15
V/Hz, which
-15
gives the value of h as 4.2 × 10 eV .s, in fair agreement with the exact value of 4.136 × 10 eV .s.
75 ··
The diameter of the pupil of the eye under room-light conditions is about 5 mm. (It can vary from about 1 to 8
mm.) Find the intensity of light of wavelength 600 nm such that 1 photon per second passes through the pupil. I = P / A = E / At = hc/λ At ; evaluate I
I = 1240/(600 -14
× π × 2.52 × 10-6) eV/m2.s 2
= 1.68 × 10 W/m
Chapter 17 76 ··
Wave–Particle Duality and Quantum Physics
A light bulb radiates 90 W uniformly in all directions. (a) Find the intensity at a distance of 1.5 m. ( b) If the 2
wavelength is 650 nm, find the number of photons per second that strike a surface of area 1 cm oriented so that the line to the bulb is perpendicular to the surface. 2
2
2
(a) I = P / A = P /4π R
I = 90/4π × 2.25 W/m = 3.18 W/m
(b) N = IA/ E ph
N = [3.18 × 10 /1.6 × 10 (1240/650)] = 1.04 × 10
-4
-19
15
77* ·· When light of wavelength λ1 is incident on the cathode of a photoelectric tube, the maximum kinetic energy of the emitted electrons is 1.8 eV. If the wavelength is reduced to λ1/2, the maximum kinetic energy of the emitted
electrons is 5.5 eV. Find the work function φ of the cathode material. 1. Use Equ. 17-3 for λ1 and λ1/2 1.8 eV = 1240/ λ1 - φ ; 5.5 eV = 2480/λ1 - φ 2. Solve for φ φ = 1.9 eV 78 ··
A photon of energy E is scattered at an angle ′ of θ. Show that the energy E ′ of the scattered photon is given
by E ′ =
E (E/ me c2)(1 - cos θ ) + 1 h
From Equ. 17-8, λ ′ =
E ′ =
79 ··
hc λ′
me c
(1 - cos θ ) + λ , and
hc
= λ +
h me c
hc λ
=
(1 - cos θ )
1+
hc 2
me c λ
E
=
( 1 - cos θ )
E
1+
me c
2
.
(1 - cos θ )
A particle is confined to a one-dimensional box. In making a transition from the state n to the state n - 1,
radiation of 114.8 nm is emitted; in the transition from the state n - 1 to the state n - 2, radiation of wavelength 147 nm is emitted. The ground-state energy of the particle is 1.2 eV. Determine n. 2
2
1. ∆ E = n E 1 - (n - 1) E 1 = (2n - 1) E 1 = hc/λ
2n - 1 = 1240/1.2 × 114.8 = 9
2. Solve for n
n = 5
80 ··
A particle confined to a one-dimensional box has a ground-state energy of 0.4 eV. When irradiated with light
of 206.7 nm it makes a transition to an excited state. When decaying from this excited state to the next lower state it emits radiation of 442.9 nm. What is the quantum number of the state to which the particle has decayed? 1. Find E f, energy of the final state 2
2. E f = n E 1; solve for n
E f = (0.4 + 1240/206.7 - 1240/442.9) eV = 3.6 eV 2
n = 9; n = 3
81* ·· When a surface is illuminated with light of wavelength λ the maximum kinetic energy of the emitted electrons is 1.2 eV. If the wavelength λ′ = 0.8λ is used the maximum kinetic energy increases to 1.76 eV, and for
wavelength λ″ = 0.6λ the maximum kinetic energy of the emitted electrons is 2.676 eV. Determine the work
Chapter 17
Wave–Particle Duality and Quantum Physics
function of the surface and the wavelength λ. 1. Use Equ. 17-3
1240/λ = 1.2 eV + φ ; 1240/0.8λ = 1.76 eV + φ (1550 - 1240)/ λ = 310/λ = 0.56 eV; λ = 553.6 nm
2. Solve for λ 3. Evaluate φ 82 ··
φ = 1.04 eV
A simple pendulum of length 1 m has a bob of mass 0.3 kg. The energy of this oscillator is quantized to the
values E n = (n + 1/2)hf 0, where n is an integer and f 0 is the frequency of the pendulum. ( a) Find n if the angular o
amplitude is 10 . (b) Find ∆n if the energy changes by 0.01%. o
E = 0.3 × 9.81 × 1(1 - cos 10 ) = 0.0447 J
(a) Find E of pendulum; E = mgL(1 - cos θ)
32
n = 1.35 × 10
h g Set E = (n + 2 ) and solve for n 2π L 1
∆n = 1.35 × 1028 -4
-4
(b) For ∆ E = 10 E , ∆n = 10 n 83 ··
(a) Show that for large n, the fractional difference in energy between state n and state n + 1 for a particle in
a box is given approximately by ( E n + 1 - E n)/ E n ≈ 2/n. (b) What is the approximate percentage energy difference between the states n1 = 1000 and n2 = 1001? (c) Comment on how this result is related to Bohr’s correspondence principle.
(a)
E n +1 - E n E n
2
=
(n + 1 ) - n2 2
n
=
2n + 1 2
n
≈
2 n
.
(b) Using the above, the percentage difference is 0.2%. (c) Classically, the energy is continuous. For very large values of n (see, e.g., Problem 67) the energy difference between adjacent levels is infinitesimal. 84 ··
12
In 1985, a light pulse of 1.8 × 10 photons was produced in an AT&T laboratory during a time interval of -15
8 × 10 s. The wavelength of the produced light was λ = 2400 nm. Suppose all of the light was absorbed by the black surface of a screen. Estimate the force exerted by the photons on the screen. p = E /c = Nhf /c = Nh/λ; then F = ∆ p/∆t = Nh/λ∆t
12
F = (1.8 × 10
× 6.626 × 10-34/2.4 × 10-6 × 8 × 10-15) N =
0.0621 N 85* ·· This problem is one of estimating the time lag (expected classically but not observed) in the photoelectric 2
2
effect. Let the intensity of the incident radiation be 0.01 W/m . (a) If the area of the atom is 0.01 nm , find the energy per second falling on an atom. ( b) If the work function is 2 eV, how long would it take classically for this much energy to fall on one atom? (a) P = IA (b) t = E / P
-2
× 10-20 J/s = 6.25 × 10-4 eV/s -4 t = 2/6.25 × 10 s = 3200 s = 53.3 min P = 10
