University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 37 - Relativity - Problems - Exercises - Page 1250: 37.41


At such a high speed, the electrons are relativistic. a. $2.06 \times10^6 volts$. b. $2.06 \times10^6 eV=3.3\times10^{-13}J$.

Work Step by Step

a. $\gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}}=5.025$ The relativistic kinetic energy is $(\gamma-1)mc^2=4.025mc^2=3.3\times10^{-13}J$. Convert that energy to electron volts, $K=2.06 MeV$. The potential difference required is $2.06 \times10^6 volts$. b. From part A, $K = 3.3\times10^{-13}J = 2.06 MeV$.
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