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0
(6.626 × 10 )(1 − cos 35 ) (9.109 × 10 31 )(2.998 × 108 ) 13 ∆λ = 4.39 ×10 m = 0.439 pm λ ' = λ + ∆λ = 2.17 + 0.439 = 2.607 pm (6.626 × 10 34 )(1 − cos1150 ) ∆λ = (9.109 × 10 31 )(2.998 × 10 8 ) 12 ∆λ = 3.45 × 10 m = 3.45 pm λ ' = λ + ∆λ = 2.17 + 3.45 = 5.62 pm
∆λ =
(a)
−
−
−
(b)
−
−
Problem 1-53
An X-ray photon of energy ' E ' , moving along the x-axis, undergoes Compton scattering with an electron which initially at rest. The photon is scattered at angle 'θ ' with the x-axis. Show that the energy ' E ′ ' of the scattered photon is given by E ′ =
E
1 + (2 E / m0 c 2 ) sin 2 (θ / 2)
Solution
The Compton shift in wavelength is given by ∆λ =
h m0 c
(1 − cos θ )
The energy of the scattered photon is given by E ′ = E ′ = E ′ =
h c
λ ′
=
h c
λ + ∆λ
=
h c
λ + (h / m0 c )(1 − cos θ )
E (h c / λ ) = 1 + {(h c / λ ) /(m0c 2 )}(1 − cos θ ) 1 + ( E / m0 c 2 )(1 − cos θ ) E 2
2
1 + (2 E / m0 c ) sin (θ / 2)
Q
(1 − cos θ ) = 2 sin 2 (θ / 2)
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Problem 1-54
A photon of 1 MeV collides with a free electron and scatters through 900. What are the energy of the scattered photon and the kinetic energy of the recoiling electron? Solution
The energy of the scattered photon is given by E ′ =
E
1 + (2 E / m0 c 2 ) sin 2 (θ / 2) 1 E ′ = = 0.338 MeV 1 + {2(1) / 0.511} sin 2 (45 0 ) The kinetic energy of the recoiling electron is given by K = E − E ′ = 1 − 0.338 = 0.662 MeV Problem 1-55
Gamma rays of energy 0.662 MeV are Compton scattered. (a) What is the energy of the scattered photon observed at a scattering angle of 600? (b) What is the kinetic energy of the scattered electrons? Solution
(a) The energy of the scattered photon is given by E ′ =
E
1 + ( E / m0c 2 ) (1 − cosθ ) 0.662 E ′ = = 0.402 MeV 1 + {(0.662) / 0.511}(1 − cos 600 ) 2 Q m0c = 0.511 MeV (b) The kinetic energy of the recoiling electron is given by K = E − E ′ = 0.662 − 0.402 = 0.260 MeV Problem 1-56
At what scattering angle will incident 100 keV X-rays leave a target with energy of 90 keV?
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Solution
The energy of the scattered photon in terms of energy of the incident photon and scattering angle is given by E
E ′ =
1 + ( E / m0 c 2 )(1 − cosθ ) Rearranging the above equation we get cosθ = 1 −
E − E ′
2 m0 c E E ′
(100 − 90)(511) = 0.43222 (100)(90) 3 = 0.511 MeV = 0.511 × 10 keV = 511 θ = cos 1 (0.43222) = 64.4 0
cosθ = 1 − Q
m0 c
2
keV
−
Problem 1-57
If the Compton shift in an experiment is found to be 0.0121 Ǻ, then calculate the scattering angle. Solution
The Compton shift in wavelength is given by ∆λ =
m0 c( ∆λ ) h
=
h m0 c
(1 − cos θ )
(1 − cosθ )
cosθ = 1 −
m0 c (∆λ ) h
(9.109 ×10 31 )(2.998 ×108 )(0.0121×10 10 ) cosθ = 1 − 6.626 ×10 34 cosθ = 0.50130 or θ = 59 0 54′50′′ or 59.9 0 −
−
−
Problem 1-58
In a Compton experiment, the wavelength of the incident Xrays is 7.078 × 10 11 m while the wavelength of the outgoing X-rays is 7.314 × 10 11 m . At what angle the scattered radiation was measured? B.U. B.Sc. 2007S −
−
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Solution
Now
h
∆λ =
m0 c ( ∆λ )
=
h
m0 c
(1 − cos θ )
(1 − cosθ )
cos θ = 1 −
m 0 c(∆λ ) h
(9.109 × 10 31 )(2.998 × 10 8 ){(7.314 − 7.078) × 10 10 } cosθ = 1 − 6.626 × 10 34 cosθ = 0.02734 or θ = cos 1 (0.02734) = 88.4 0 or 88 0 24′ −
−
−
−
Problem 1-59
On scattering via the Compton Effect, a photon undergoes a fractional wavelength change [(λ ′ − λ ) / λ ] equal to 6 %. If the incident photon has a wavelength of 0.020 nm, at what angle is the detector to the incident beam? Solution
The Compton shift in wavelength is given by λ ′ − λ = ∆λ =
h m0 c
(1 − cos θ )
λ ′ − λ ∆λ h = = (1 − cos θ ) λ λ m0 c λ m0 c λ ∆λ = 1 − cosθ h λ
cos θ = 1 −
m0 c λ ∆λ h
λ
(9.109 × 10 19 )(2.998 × 10 8 )(0.020 × 10 9 ) 6 cosθ = 1 − 6.636 × 10 34 100 cosθ = 0.50543 or θ = cos 1 (0.50543) = 59 0 38′ 24′′ −
−
−
−
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Problem 1-60
In Compton scattering, calculate the maximum kinetic energy given to the scattered electron for a given photon energy. Solution
The kinetic energy given to the recoiled electron is given by K E . . = E − E ′ The above energy will have a maximum value if the energy of the scattered photon i.e. E ′ is a minimum. Now E ′ =
h c
λ ′
=
h c
λ + ∆λ
=
h c
λ + (h / m0 c )(1 − cos θ )
It is clear that E ′ will be a minimum for θ = 180 0 i.e. h c
′ = E min
1 + (2 h / m0 c) (h c / λ ) (h c / λ ) = = 1 + (2 h / λ m0 c) 1 + 2(h c / λ )(1 / m0 c 2 )
′ = E min
E
1 + (2 E / m0c 2 )
=
m0 c 2 E m0 c 2 + 2 E
Hence 2
( K . E .) max ( K E . .) max
= E −
=
2
m0 c E 2
m0 c + 2 E
=
2
2
m0 c E + 2 E − m0 c E 2
m0 c + 2 E
2 E 2 2 E + m0 c 2
Problem 1-61
If the maximum kinetic energy given to the electrons in a Compton scattering experiment is 10 keV, what is the wavelength of the incident X-rays? Solution
Now
( K E . .) max =
2 E 2 2 E + m0 c 2
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2
2 E 2 E + 511 20 E + 5110 = 2 E 2 2 E 2 − 20 E − 5100 = 0 2 E − 10 E − 2555 = 0 Using quadratic formula for positive root we get 2 − ( −10) + ( −10) − 4(1)(−2555) E = = 55.8 keV 2(1) The wavelength of the incident X-rays is calculated as 10 =
E =
h c
λ
or λ =
h c E
−34
(6.626 ×10 )(2.998 ×108 ) 11 λ = = 2.2 × 10 m = 0.022 3 19 (55.8 ×10 ×1.602 ×10 ) −
−
nm
Problem 1-62
What percentage increase in wavelength leads to 75 percent loss of photon energy in a collision of photon with a stationary electron? B.U. B.Sc. 2004A Solution
Let E and E´ be the energies of incident and scattered photons respectively, then fractional loss of photon energy is given by E − E ′ E ′ (hc / λ ′) λ λ ′ − λ ∆λ f = =1− =1− =1− = = E E (hc / λ ) λ ′ λ ′ λ + ∆λ f ∆λ 0.75 = = =3 λ 1 − f 1 − 0.75 Hence a 300 percent increase in wavelength will lead to 75 percent loss of photon energy. Problem 1-63
When photons of wavelength 0.024 Ǻ are incident on a target, scattered photons are detected at an angle of 60o. Calculate (a) the wavelength of the scattered photon. (b) the angle at which the electron is scattered. B.U. B.Sc. (Hons.) 1991A
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Solution
(a) The wavelength of the scattered photon is given by λ ′ = λ +
h mo c
(1 − cosθ )
(6.626 × 10−34 )(1 − cos 60o ) λ ′ = (0.024 × 10 ) + − 31 8 (9.108 × 10 )(2.998 × 10 ) = 3.6 × 10−12 m = 0.036 Ǻ −10
(b) The scattering angle φ of the electron is given by υ ′ sin θ λ sin θ = tan φ = υ − υ ′ cosθ λ ′ − λ cosθ (0.024) sin 60o = = 0.866 0.036 − (0.024) cos 60o φ = 400 53′ 33′′ or 40.9o Problem 1-64
X-ray photons of wavelength 0.220 nm are Compton scattered at 450. Calculate the energy of the scattered photon in eV. Solution
The Compton shift in wavelength is given by ∆λ =
h m0 c
(1 − cos θ )
(6.626 × 10 34 )(1 − cos 45 0 ) ∆λ = (9.109 × 10 31 )(2.998 × 10 8 ) 13 ∆λ = 7.11 × 10 m The wavelength of the scattered photon is λ ′ = λ + ∆λ = (0.220 × 10 9 ) + (7.11 × 10 13 ) λ ′ = 2.20711×10 10 m The energy of the scattered photon will be −
−
−
−
−
−
E ′ =
h c
λ
in J
or
E ′ =
hc
λ e
in eV
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(6.626 × 10 34 )(2.998 × 10 8 ) E ′ = 5.62 × 10 3 eV = 5.62 10 19 = (2.20711 × 10 )(1.602 × 10 )
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−
−
−
keV
Problem 1-65
X-rays of wavelength 0.040 Å are scattered from a carbon block. Determine (a) the momentum of a photon scattered at angle of 300 and (b) the kinetic energy of the recoil electron. Solution
(a) The Compton shift in wavelength is given by ∆λ =
h
(1 − cos θ )
m0 c
(6.626 × 10 34 )(1 − cos 30 0 ) 13 ∆λ = = 3.25 × 10 m 31 8 (9.109 × 10 )(2.998 × 10 ) λ ′ = λ + ∆λ = (0.040 × 10 10 ) + (3.25 × 10 13 ) λ ′ = 4.325 ×10 12 m The momentum of the scattered photon is given by h 6.626 × 10 34 22 p ′ = = = 1.532 × 10 J • s / m 12 λ 4.325 × 10 (b) The kinetic energy of the recoil electron is given by h c h c h c (λ ′ − λ ) −
−
−
−
−
−
−
−
−
K =
λ
−
=
λ ′
λ λ ′
−34
(6.626 × 10 )(2.998 × 10 8 )(3.25 × 10 13 ) K = (0.040 × 10 10 )(4.325 × 10 12 ) K = 3.732 × 10 15 J 3.372 × 10 15 3 K = eV = 2.33 × 10 eV = 23.3 keV 19 1.602 × 10 −
−
−
−
−
−
Problem 1-66
How many head-on Compton scattering events are necessary to double the wavelength of a photon having initial wavelength 200 pm?
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Solution
The Compton shift in wavelength ∆λ is independent of the wavelength λ of the incident photon and will have maximum value for head-on collision i.e. 180 0. Now ∆λ =
h m0c
(1 − cosθ )
(6.626 ×10 34 )(1 − cos1800 ) 4.853 ×10 12 m ∆λ = 31 8 = (9.109 ×10 )(2.998 ×10 ) λ = 200 pm = 200 × 10 12 m = 2 × 10 10 m . Hence and Number of Compton Scattering events necessary to double λ ∆λ 2 × 10 10 = = = 41 λ 4.853 × 10 12 −
−
−
−
−
−
−
Problem 1-67
An X-ray photon of frequency 1019 Hz is scattered through an angle of 450 with the stationary electron. (a) What is its new frequency? (b) What is the kinetic energy of electron after collision? Solution
(a) The Compton shift in wavelength is given by ∆λ =
λ ′ − λ = c
or
−
c
ν ′ ν
1 ν ′
=
1 ν
+
=
h m0 c h m0 c
h m0 c
2
(1 − cos θ ) (1 − cos θ )
(1 − cos θ )
1 h ν ′ = + θ ( 1 cos ) − 2 ν m0 c
−1
1 (6.626 × 10 −34 )(1 − cos 45 0 ) ′ ν = + 19 (9.109 × 10 −31 )(2.998 × 10 8 ) 2 1 × 10 ν ′ = 9.768 ×1018 Hz
−1
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the desired frequency of the scattered photon. (b) The kinetic energy of electron after collision is given by K = hν − hν ′ = h(ν − ν ′) K = (6.626 ×10 34 ){(1×1019 ) − (9.768 ×1018 )} 16 = 1.537 × 10 J = 959 eV ≅ 1 keV Q 1 eV = 1.602 × 10 19 J −
−
−
Problem 1-68
A 300 keV photon undergoes a Compton Scattering. The kinetic energy of recoil electron is 250 keV. Calculate the wavelength of the scattered photon. Solution
Energy of the scattered photon E ′ = Energy of the incident photon – Energy of recoil electron = (300 – 250) = 50 keV 6 19 15 = (50 × 10 )(1.602 × 10 ) J = 8.01 × 10 J −
Now
E ′ =
−
hc
λ ′
(6.626 × 10 34 )(2.998 × 10 8 ) = E ′ (8.01 × 10 15 ) = 24.8 pm
λ ′ =
hc
−
−
=
2.480 × 10
−11
m
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CONCEPTUAL QUESTIONS (1) How does the total intensity of thermal radiation vary when temperature of an object is doubled? Answer: - The total intensity of thermal radiation is given by Stephen’s law 4
I = σ T
If the temperature of an object is doubled, then new intensity will be 24 = 16 times the initial intensity. (2) If all objects radiate energy, then why we are unable to see in the dark? Answer: - As the radiation emitted by the object is in the infrared region which is not visible to eye, therefore we are unable to see the objects in the dark. (3) What is ‘ultraviolet catastrophe’ of classical physics? Answer: - The discrepancy between theoretical (Rayleigh-Jeans formula) and experimental spectral energy density towards the ultraviolet end of spectrum of black body is termed as ‘ultraviolet catastrophe’. (4) How ‘ultraviolet catastrophe’ was resolved by quantum action? Answer: - It was resolved by Planck’s law for black body radiation. (5) What is Planck’s concept of energy quantization? Answer: - The energy of a photon is given by E = n h ν n = 0,1,2.3........ where h is Planck’s constant and ν is the frequency of the photon. (6) UV light causes sunburn whereas visible light does not. Explain. Answer: -As the energy of ultraviolet (UV) photon is greater than that of a visible photon, therefore it is capable of causing sunburn. (7) Which photon is more energetic- violet or red? Answer: - Violet because the frequency of violet photon is greater than the red photon.
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(8) Do we get quantization of energy in classical mechanics? Answer: - No. One can assign any value of energy from zero to maximum to the classical system instead of certain discrete values. (9) What is the rest mass of a photon? Answer: - The rest mass of a photon is zero. (10) If an X-ray photon is scattered by an electron, does its wavelength change? If so, does it increase or decrease? Answer: - Yes. The wavelength of the scattered photon is greater than that of the incident photon. (11) At what angle maximum shift in wavelength will be observed in Compton Effect experiment? Answer:- θ = 1800 (12) What is the physical significance of Compton Effect? Answer: - The Compton Effect confirms the particle nature of electromagnetic radiation. (13) Can we observe Compton Effect if electron is replaced by a proton? Answer: - No. A proton is much heavier than electron (1836 times), therefore the Compton shift in wavelength will be too small to be observed experimentally. (14) Why is Compton Effect not observed with visible light? Answer: - The energy of the visible photon is far too sufficient to eject even the most loosely bound electron. (15) Why does not the photoelectric effect work for free electrons? Answer: - The rest mass of photon is zero. Since any material particle has nonzero rest mass, the photoelectric cannot take place with a free electron. (16) Why the alkali metals i.e. Sodium, Potassium and Calcium etc. are most suitable for studying the photoelectric emission? Answer: - The alkali metals have a low value of work function and show the photoelectric effect for visible light ( λ = 4000 Å to 7500 Å).
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(17) Is it possible to observe photoelectric emission at all frequencies? Answer: - No. In order to observe photoelectric emission, the frequency of the incident photon must be greater or equal to the threshold frequency of the given material i.e. ν ≥ ν 0 . (18) What will be the velocity of emitted photoelectrons if the frequency of incident photon is just equal to threshold frequency? Answer: - Zero. (19) Which parameter governs the kinetic energy of the photoelectrons emitted from a material? Answer: - The kinetic energy of photoelectrons is governed by the frequency of the incident radiation provided that it is greater than threshold frequency of the given material. (20) What will happen to the velocity of the photoelectrons if the wavelength of incident radiation is increased? Answer: - The velocity of the photoelectrons will decrease with increase in wavelength of the incident photon. (21) Name two effects which support the photon theory of light? Answer: - Compton Effect, Photoelectric effect, Pair production etc.
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ADDITIONAL PROBLEMS (1) Taking the sun as a 5800 K blackbody how does its ultraviolet radiance at 200 nm compare with its visible radiance at its peak wavelength of 500 nm? (2) The photoelectric threshold of tungsten is 2300 Ǻ. Determine the energy of electrons ejected from the surface by ultraviolet radiation of wavelength 1800 Å. {B.U. B.Sc.(Hons.) 1983S} (3) The photoelectric threshold of tungsten is 2300 Ǻ. Determine the energy of electrons ejected from the surface by ultraviolet radiation of wavelength 1500 Å. Express your answer in eV. (B.U. B.Sc. 1990S) (4) Calculate the value of Planck’s constant when a light of frequency 11.5 × 10 14 Hertz is incident upon a material of work function 4.4 × 10 14 Hertz and the stopping potential is 3 volts. (B.U. B.Sc. 1992A) (5) If the photoelectric threshold wavelength of sodium is 542 nm, calculate the maximum velocity of photo electrons ejected by photons of wavelength 400 nm. (B.U. B.Sc. 2004S) (6) Photoelectrons emitted from a photocell by light with a wavelength of 2500 Ǻ can be stopped by applying a potential of 2 volts to the collector. Calculate the work function (in eV) for the surface. {B.U. B.Sc.(Hons.) 1988S} (7) Monochromatic X-rays of wavelength λ = 0.124 Ǻ are scattered from a carbon block. Determine the wavelength of the X-rays scattered through 180 o. (B.U. B.Sc. 1986S) (8) X-rays with λ = 1.0 Ǻ are scattered from a carbon block. The scattered radiation is viewed at 90 o to the incident beam. What is the Compton shift in wavelength? (B.U. B.Sc. 1987A)
