Explanation Given info: The photon energy emits by the F 57 e is 14.0 keV and the rest energy is M c 2 = 8.60 × 10 − 9 J . Since the total momentum is zero before decays. Then it is necessary after the decay that the momentum of the nucleus is equal to the momentum of the photon. Then, p nucleus = p photon = E γ c = 14.0 keV c Here, E γ is the energy emits by the F 57 e . c is the speed of light. The equation for the energy of the object is, E 2 = p 2 c 2 + ( m c 2 ) 2 Here, m is the mass of the F 57 e . For the nucleus recoiling, E 2 = p 2 c 2 + ( m c 2 ) 2 with M c 2 = 8.60 × 10 − 9 J . Thus, ( M c 2 + K ) 2 = ( 14.0 keV ) 2 + ( M c 2 ) 2 ( 1 + K M c 2 ) 2 = ( 14.0 keV M c 2 ) 2 + 1 Here, K is the kinetic energy. The term ( 14.0 keV M c 2 ) 2 is very much less than 1 . ( 14.0 keV M c 2 ) 2 < < 1 Then, ( 14

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN: 9781305116399

Author: Raymond A. Serway, John W. Jewett

Publisher: Cengage Learning

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Question

Chapter 39, Problem 39.87AP

To determine

To show: The kinetic energy of the recoiling nucleus in electron volts.

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Chapter 39 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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The kinetic energy of the recoiling nucleus in electron volts.

Solution Summary: The author explains the kinetic energy of the recoiling nucleus in electron volts.

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To show: The kinetic energy of the recoiling nucleus in electron volts.

Chapter 39 Solutions