FUND. OF PHYSICS FOR LSU WILEY+ NEXT GEN
11th Edition
ISBN: 9781119749295
Author: Halliday
Publisher: WILEY
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Question
Chapter 38, Problem 58P
To determine
To find:
a) The energy of photon corresponding to wavelength
b) The kinetic energy of an electron with de Broglie wavelength
c) The energy of photon corresponding to wavelength
d) The kinetic energy of an electron with de Broglie wavelength
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What are (a) the energy of a photon corresponding to wavelength 1.00 nm, (b) the kinetic energy of an electron with de Broglie wavelength 1.00 nm, (c) the energy of a photon corresponding to wavelength 1.00 fm, and (d) the kinetic energy of an electron with de Broglie wavelength 1.00 fm?
(a) What is the de Broglie wavelength (in m) of a proton moving at a speed of 3.30 × 104 m/s?
m
(b) What is the de Broglie wavelength (in m) of a proton moving at a speed of 1.92 × 108 m/s?
m
m
(c) What is the de Broglie wavelength for an electron having a kinetic energy of 3.15 MeV?
What is the wavelength of (a) a photon with energy 1.00 eV, (b) an electron with energy 1.00 eV, (c) a photon of energy 1.00 GeV, and (d) an electron with energy 1.00 GeV?
Chapter 38 Solutions
FUND. OF PHYSICS FOR LSU WILEY+ NEXT GEN
Ch. 38 - Prob. 1QCh. 38 - Prob. 2QCh. 38 - Prob. 3QCh. 38 - Prob. 4QCh. 38 - Prob. 5QCh. 38 - Prob. 6QCh. 38 - Prob. 7QCh. 38 - Prob. 8QCh. 38 - Prob. 9QCh. 38 - Prob. 10Q
Ch. 38 - Prob. 11QCh. 38 - Prob. 12QCh. 38 - Prob. 13QCh. 38 - Prob. 14QCh. 38 - Prob. 15QCh. 38 - Prob. 16QCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Prob. 12PCh. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - Prob. 19PCh. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Prob. 27PCh. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - Prob. 30PCh. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Prob. 33PCh. 38 - Prob. 34PCh. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53PCh. 38 - Prob. 54PCh. 38 - Prob. 55PCh. 38 - Prob. 56PCh. 38 - Prob. 57PCh. 38 - Prob. 58PCh. 38 - Prob. 59PCh. 38 - Prob. 60PCh. 38 - Prob. 61PCh. 38 - Prob. 62PCh. 38 - Prob. 63PCh. 38 - Prob. 64PCh. 38 - Prob. 65PCh. 38 - Prob. 66PCh. 38 - Prob. 67PCh. 38 - Prob. 68PCh. 38 - Prob. 69PCh. 38 - Prob. 70PCh. 38 - Prob. 71PCh. 38 - Prob. 72PCh. 38 - Prob. 73PCh. 38 - Prob. 74PCh. 38 - Prob. 75PCh. 38 - Prob. 76PCh. 38 - Prob. 77PCh. 38 - Prob. 78PCh. 38 - Prob. 79PCh. 38 - Prob. 80PCh. 38 - Prob. 81PCh. 38 - Prob. 82PCh. 38 - Prob. 83PCh. 38 - Prob. 84PCh. 38 - Prob. 85PCh. 38 - Prob. 86PCh. 38 - Prob. 87PCh. 38 - Prob. 88PCh. 38 - Prob. 89PCh. 38 - Prob. 90P
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- An 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_forwardWhat is the de Broglie wavelength for an electron with speed (a) v = 0.480c and (b) v = 0.960c?arrow_forwardWhat is the de Broglie wavelength for an electron with speed (a) v = 0.469c and (b) v = 0.958c? (Hint: Use the correct relativistic expression for linear momentum if necessary.)arrow_forward
- An electron has a kinetic energy of 4.1 × 10–¹³ J. 46. What is the electron's speed? (a) 0.960c (b) 0.968c (c) 0.972c (d) 0.980c (e) 0.986c 47. What is the de Broglie wavelength of the electron described? (a) 4.09 × 10-13 m (b) 4.94 × 10-13m (c) 5.42 × 10-13 m (d) 6.25 × 10-13 m (e) 7.81 x 10-13marrow_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_forwardA particle has a de Broglie wavelength of 2.80×10−102.80×10−10 m. Then its kinetic energy doubles. What is the particle's new de Broglie wavelength, assuming that relativistic effects can be ignored?arrow_forward
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