Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 40, Problem 66AP
To determine
To prove that the energy of the scattered photon is
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A photon with wavelength X scatters off an electron at rest, at an angle with the incident direction.
The Compton wavelength of the electron Ac = 0.0024 nm.
a)
For λ = 0.0006 nm and 0 = 53 degrees, find the wavelength X' of the scattered photon in
nanometres.
b)
Obtain a formula for the energy of the electron Ee after collision, in terms of the universal constants
h, c and the variables X, X' and Ac. The answer must be expressed in terms of these variables
only. (Please enter an algebraic expression using latex format; do not input any numerical values)
c)
Using the energy conservation condition, find the value of the electron energy Ee after scattering in
units of keV.
d)
Write an algebraic expression for the electron's momentum pe in
terms of its energy Ee, its mass me and the speed of light c.
e)
What is the de Broglie wavelength of the scattered electron ? Express your answer in terms of Ee,
me, and X and c.
f)
Find the value of the de Broglie wavelength of the scattered electron…
X-ray photons of wavelength 0.0248 nm are incident on a target and the Compton-scattered photons are observed at 80.0° above the photons' incident line of travel. [Use relativistic units for this problem!](a) What is the momentum of the incident photons? eV/c(b) What is the momentum (magnitude and angle) of the scattered electrons? eV/c°magnitude=61802.35
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In the Compton scattering, the photon of energy 9,2 MeV is scattered from a free electron of mass 9.1×10−319.1×10−31 kg, What is the kinetic energy acquired by the electron (in MeV), if the scattering angle is 15∘15∘?
Chapter 40 Solutions
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Ch. 40.1 - Prob. 40.1QQCh. 40.2 - Prob. 40.2QQCh. 40.2 - Prob. 40.3QQCh. 40.2 - Prob. 40.4QQCh. 40.3 - Prob. 40.5QQCh. 40.5 - Prob. 40.6QQCh. 40.6 - Prob. 40.7QQCh. 40 - Prob. 1OQCh. 40 - Prob. 2OQCh. 40 - Prob. 3OQ
Ch. 40 - Prob. 4OQCh. 40 - Prob. 5OQCh. 40 - Prob. 6OQCh. 40 - Prob. 7OQCh. 40 - Prob. 8OQCh. 40 - Prob. 9OQCh. 40 - Prob. 10OQCh. 40 - Prob. 11OQCh. 40 - Prob. 12OQCh. 40 - Prob. 13OQCh. 40 - Prob. 14OQCh. 40 - Prob. 1CQCh. 40 - Prob. 2CQCh. 40 - Prob. 3CQCh. 40 - Prob. 4CQCh. 40 - Prob. 5CQCh. 40 - Prob. 6CQCh. 40 - Prob. 7CQCh. 40 - Prob. 8CQCh. 40 - Prob. 9CQCh. 40 - Prob. 10CQCh. 40 - Prob. 11CQCh. 40 - Prob. 12CQCh. 40 - Prob. 13CQCh. 40 - Prob. 14CQCh. 40 - Prob. 15CQCh. 40 - Prob. 16CQCh. 40 - Prob. 17CQCh. 40 - The temperature of an electric heating element is...Ch. 40 - Prob. 2PCh. 40 - Prob. 3PCh. 40 - Prob. 4PCh. 40 - Prob. 5PCh. 40 - Prob. 6PCh. 40 - Prob. 7PCh. 40 - Prob. 8PCh. 40 - Prob. 9PCh. 40 - Prob. 10PCh. 40 - Prob. 11PCh. 40 - Prob. 12PCh. 40 - Prob. 14PCh. 40 - Prob. 15PCh. 40 - Prob. 16PCh. 40 - Prob. 17PCh. 40 - Prob. 18PCh. 40 - Prob. 19PCh. 40 - Prob. 20PCh. 40 - Prob. 21PCh. 40 - Prob. 22PCh. 40 - Prob. 23PCh. 40 - Prob. 25PCh. 40 - Prob. 26PCh. 40 - Prob. 27PCh. 40 - Prob. 28PCh. 40 - Prob. 29PCh. 40 - Prob. 30PCh. 40 - Prob. 31PCh. 40 - Prob. 32PCh. 40 - Prob. 33PCh. 40 - Prob. 34PCh. 40 - Prob. 36PCh. 40 - Prob. 37PCh. 40 - Prob. 38PCh. 40 - Prob. 39PCh. 40 - Prob. 40PCh. 40 - Prob. 41PCh. 40 - Prob. 42PCh. 40 - Prob. 43PCh. 40 - Prob. 45PCh. 40 - Prob. 46PCh. 40 - Prob. 47PCh. 40 - Prob. 48PCh. 40 - Prob. 49PCh. 40 - Prob. 50PCh. 40 - Prob. 51PCh. 40 - Prob. 52PCh. 40 - Prob. 53PCh. 40 - Prob. 54PCh. 40 - Prob. 55PCh. 40 - Prob. 56PCh. 40 - Prob. 57PCh. 40 - Prob. 58PCh. 40 - Prob. 59PCh. 40 - Prob. 60APCh. 40 - Prob. 61APCh. 40 - Prob. 62APCh. 40 - Prob. 63APCh. 40 - Prob. 64APCh. 40 - Prob. 65APCh. 40 - Prob. 66APCh. 40 - Prob. 67APCh. 40 - Prob. 68APCh. 40 - Prob. 69APCh. 40 - Prob. 70APCh. 40 - Prob. 71APCh. 40 - Prob. 72CPCh. 40 - Prob. 73CPCh. 40 - Prob. 74CPCh. 40 - Prob. 75CPCh. 40 - Prob. 76CP
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- Consider the Compton scattering of a photon of wavelength Ao by a free electron moving with a momentum of magnitude P in the same direction as that of the incident photon. (a) Show that in this case the Compton equation (1.42) becomes (ро + P) с sin? AX = 240 E – Pc 2 h/Ao is the magnitude of the incident photon momentum, 0 is the photon scattering angle where po 1/2 and E = (m²c* + P²c²)*/² is the initial electron energy.arrow_forwardA photon of frequency v is scattered by an electron initially at rest. Verify that the maximum kinetic energy of the recoil electron is KEmax = (2h² v/mc²)/(1 + 2 hv/mc²).arrow_forwardAssume that the total energy E of an electron greatly exceeds its rest energy E0. If a photon has a wavelength equal to the de Broglie wavelength of the electron, what is the photon’s energy? Repeat the problem assuming E= 2E0 for the electron.arrow_forward
- Problem 4: A photon originally traveling along the x axis, with wavelength λ = 0.100 nm is incident on an electron (m = 9.109 x 10-31 kg) that is initially at rest. The x-component of the momentum of the electron after the collision is 5.0 x 10-24 kg m/s and the y-component of the momentum of the electron after the collision is -6.0 x 10-24 kg m/s. If the photon scatters at an angle + from its original direction, what is wavelength of the photon after the collision. h= 6.626 x 10:34 J·s and c = 3.0 x 108 m/s.arrow_forwardCan you please help with the following question? Thanks!arrow_forwardA) Calculate the de Broglie wavelength of a neutron (mn = 1.67493×10-27 kg) moving at one six hundredth of the speed of light (c/600). (Enter at least 4 significant figures.) B) Calculate the velocity of an electron (me = 9.10939×10-31 kg) having a de Broglie wavelength of 230.1 pm.arrow_forward
- Help Mearrow_forwardAn x-ray photon is scattered from a free electron (mass m) at rest. The wavelength of the scattered photon is I′, and the final speed of the struck electron is v. (a) What was the initial wavelength I of the photon? Express your answer in terms of I′, v, and m. (Hint: Use the relativistic expression for the electron kinetic energy.) (b) Through what angle f is the photon scattered? Express your answer in terms of I,I ′, and m. (c) Evaluate your results in parts (a) and (b) for a wavelength of 5.10 x 10-3 nm for the scattered photon and a final electron speed of 1.80 x 108 m/s. Give f in degrees.arrow_forwardAn electron has kinetic energy E = 295KeV which is equal to the energy of a photon. Let λ1 be the de-Broglie wavelength of the electron and λ2 be the wavelength of the photon. What would be the ratio of λ1/λ2?arrow_forward
- In the Compton scattering, the photon of energy 8.7 MeV is scattered from a free electron of mass 9.1 × 10 31 the electron (in MeV), if the scattering angle is 164°? kg, What is the kinetic energy acquired by Answer:arrow_forwardWhen an electron is accelerated through a potential difference Δφ it acquires a kinetic energy e Δφ. Calculate the momentum, and hence the de Broglie wavelength, of an electron accelerated from rest through (a) 1.00V, (b) 1.00 kV, (c) 100 kV.arrow_forwardAn electron has a de Broglie wavelength λ = 4.5×10−10 m . h=6.626×10−34 J⋅s, e=1.602×10−19 C, me=9.109×10−31 kg. What is its momentum? (p =h/)arrow_forward
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