College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 26 - Problems - Page 1011: 38


$\gamma = 1.08$

Work Step by Step

We can find the kinetic energy of the protons: $K = q~\Delta V$ $K = (1.6\times 10^{-19}~C)(75\times 10^6~V)$ $K = 1.2\times 10^{-11}~J$ We can find the Lorentz factor $\gamma$: $K = (\gamma-1)~m_0~c^2$ $\gamma = 1+\frac{K}{m_0~c^2}$ $\gamma = 1+\frac{ 1.2\times 10^{-11}~J}{(1.67\times 10^{-27}~kg)(3.0\times 10^8~m/s)^2}$ $\gamma = 1.08$
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