College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1012: 52


$E = 0.64~MeV$

Work Step by Step

We can find the total energy: $E = \gamma~m~c^2$ $E = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m~c^2$ $E = \frac{1}{\sqrt{1-\frac{(0.60~c)^2}{c^2}}}~\times (9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $E = (1.25)(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $E = 1.024\times 10^{-13}~J$ $E = (1.024\times 10^{-13}~J)\times (\frac{1~MeV}{1.60\times 10^{-13}~J})$ $E = 0.64~MeV$
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