College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1011: 33


$v = 2.8\times 10^8~m/s$

Work Step by Step

We can find the electron's speed: $p = \gamma~m~v$ $v = \frac{p}{\gamma~m}$ $v = \frac{p}{\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~m}$ $v = \frac{p}{m}~\sqrt{1-\frac{v^2}{c^2}}$ $v^2 = \frac{p^2}{m^2}~(1-\frac{v^2}{c^2})$ $1 = \frac{p^2}{v^2~m^2}~-\frac{p^2}{m^2~c^2}$ $\frac{p^2}{v^2~m^2} = 1+\frac{p^2}{m^2~c^2}$ $\frac{p^2}{v^2~m^2} = \frac{m^2~c^2}{m^2~c^2}+\frac{p^2}{m^2~c^2}$ $\frac{v^2~m^2}{p^2} = \frac{m^2~c^2}{m^2~c^2+p^2}$ $v^2 = \frac{p^2~c^2}{m^2~c^2+p^2}$ $v = \frac{p~c}{\sqrt{m^2~c^2+p^2}}$ $v = \frac{(2.4\times 10^{-22}~kg~m/s)(3.0\times 10^8~m/s)}{\sqrt{(9.1\times 10^{-31}~kg)^2~(3.0\times 10^8~m/s)^2+(2.4\times 10^{-22}~kg~m/s)^2}}$ $v = 2.8\times 10^8~m/s$
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