College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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Chapter 27, Problem 31P
To determine
The de Broglie wavelength for the relativistic electron.
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De Broglie postulated that the relationship λ =h/p is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 3.00 MeV?
De Broglie postulated that the relationship
? = h/p
is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 2.98 MeV?
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?
Chapter 27 Solutions
College Physics
Ch. 27.5 - Prob. 27.1QQCh. 27.5 - Prob. 27.2QQCh. 27.5 - Prob. 27.3QQCh. 27.6 - Prob. 27.4QQCh. 27.6 - Prob. 27.5QQCh. 27 - Prob. 1CQCh. 27 - Prob. 2CQCh. 27 - Prob. 3CQCh. 27 - Prob. 4CQCh. 27 - Prob. 5CQ
Ch. 27 - Prob. 6CQCh. 27 - Prob. 7CQCh. 27 - Prob. 8CQCh. 27 - Prob. 9CQCh. 27 - Prob. 10CQCh. 27 - Prob. 11CQCh. 27 - Prob. 12CQCh. 27 - Prob. 13CQCh. 27 - Prob. 14CQCh. 27 - Prob. 15CQCh. 27 - Prob. 16CQCh. 27 - Prob. 1PCh. 27 - Prob. 2PCh. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6PCh. 27 - Prob. 7PCh. 27 - Prob. 8PCh. 27 - Prob. 9PCh. 27 - Prob. 10PCh. 27 - Prob. 11PCh. 27 - Prob. 12PCh. 27 - Prob. 13PCh. 27 - Prob. 14PCh. 27 - Prob. 15PCh. 27 - Prob. 16PCh. 27 - Prob. 17PCh. 27 - Prob. 18PCh. 27 - Prob. 19PCh. 27 - Prob. 20PCh. 27 - Prob. 21PCh. 27 - Prob. 22PCh. 27 - Prob. 23PCh. 27 - Prob. 24PCh. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - Prob. 28PCh. 27 - Prob. 29PCh. 27 - Prob. 30PCh. 27 - Prob. 31PCh. 27 - Prob. 32PCh. 27 - Prob. 33PCh. 27 - Prob. 34PCh. 27 - Prob. 35PCh. 27 - Prob. 36PCh. 27 - Prob. 37PCh. 27 - Prob. 38PCh. 27 - Prob. 39PCh. 27 - Prob. 40PCh. 27 - Prob. 41APCh. 27 - Prob. 42APCh. 27 - Prob. 43APCh. 27 - Prob. 44APCh. 27 - Prob. 45APCh. 27 - Prob. 46APCh. 27 - Prob. 47APCh. 27 - Prob. 48APCh. 27 - Prob. 49APCh. 27 - Prob. 50APCh. 27 - Prob. 51APCh. 27 - Prob. 52AP
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- A neutron of mass 1.675 × 10-27 kg has a de Broglie wavelength of 7.8x10-12 m. What is the kinetic energy (in eV) of this non-relativistic neutron? Please give your answer with two decimal places. 1 eV = 1.60 × 10-19 J, h = 6.626 × 10-34 J ∙ s.arrow_forwardAn electron is accelerated through a potential difference of 26 000 V. What is the de Broglie wavelength of the electron (in m)?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
- De Broglie postulated that the relationship ? = h/p is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 3.31 MeV? answer in marrow_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(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?arrow_forward
- What is the de Broglie wavelength of a proton whose kinetic energy is 2.0 MeV? 10.0 MeV?arrow_forwardA particle of matter is moving with a kinetic energy of 7.53 eV. Its de Broglie wavelength is 2.85 x 10^-12 m. What is the mass of the particle?arrow_forwardHow fast does a proton have to be moving in order to have the same de Broglie wavelength as an electron that is moving at 4.15×1064.15×106 m/s?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_forwardThe electron beam in a scanning electron microscope (SEM) can have an energy ranging from 0.2 keV to 40 keV. (Note: an eV is an electronvolt, the measure of an amount of kinetic energy for a single electron. 1 eV = 1.60218×10−19 J and 1 J = 1 kg m2 s−2). What is the de Broglie wavelength of the electron beam in the SEM when the electrons are accelerated to kinetic energy of 6.6 keV?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
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