Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 44AP
To determine
Show that the ratio of Compton wavelength to the de Broglie wavelength is
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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/)
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?
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.
Chapter 39 Solutions
Physics for Scientists and Engineers with Modern Physics
Ch. 39.1 - Prob. 39.1QQCh. 39.2 - Prob. 39.2QQCh. 39.2 - Prob. 39.3QQCh. 39.2 - Prob. 39.4QQCh. 39.3 - Prob. 39.5QQCh. 39.5 - Prob. 39.6QQCh. 39.6 - Prob. 39.7QQCh. 39 - Prob. 1PCh. 39 - Prob. 2PCh. 39 - Prob. 3P
Ch. 39 - Prob. 4PCh. 39 - Prob. 5PCh. 39 - Prob. 6PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - Prob. 10PCh. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - Prob. 18PCh. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 22PCh. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 35PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40APCh. 39 - Prob. 41APCh. 39 - Prob. 43APCh. 39 - Prob. 44APCh. 39 - Prob. 45APCh. 39 - Prob. 46APCh. 39 - Prob. 47CPCh. 39 - Prob. 48CPCh. 39 - Prob. 49CPCh. 39 - Prob. 50CP
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- Assume 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_forwardAn alpha particle (m = 6.64 × 10−27 kg) emitted in the radioactive decay of Uranium-238 has an energy of 4.20 MeV. What is its de Broglie wavelength?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
- An electron has a non-relativistic speed of v=1.5 x 105 m/s. Determine: a) Its De Broglie wave length, its kinetic energy and the rest energy of the electron in joules and electron volts b) What is the speed of the electron if the relativistic kinetic energy is equal to the rest energy? c) The momentum of the electron at a speed of 0.80c and the De Broglie wavelength for an electron with a speed of 0.80c.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_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. (I got the answer 949.4 pm but it is wrong, please help) B) Calculate the velocity of an electron (me = 9.10939×10-31 kg) having a de Broglie wavelength of 230.1 pm.arrow_forward
- The diameter of an atomic nucleus is about 10 fm (1 fm = 10-15 m). What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of 10 fm?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_forwardWhat 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?arrow_forward
- In an experiment on the photoelectric effect, a metal is illuminated by visible light of different wavelengths. A photoelectron has a maximum kinetic energy of 0.9 eV when red light of wavelength 640 nm is used. With blue light of wavelength 420 nm, the maximum kinetic energy of the photoelectron is 1.9 eV. Use this information to calculate an experimental value for the Planck constant h. [arrow_forwardA relativistic electron has a de Broglie wavelength of 2.45 pm2.45 pm (1 pm=10−12 m).(1 pm=10−12 m). Determine its velocity, expressed as a fraction of the speed of light ?.arrow_forwardDe 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?arrow_forward
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