College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1011: 37


(a) $v = 0.9998~c$ (b) $t = 1.0\times 10^{-11}~s$

Work Step by Step

(a) We can find the speed: $K \approx \frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}$ $\sqrt{1-\frac{v^2}{c^2}} = \frac{mc^2}{K}$ $1-\frac{v^2}{c^2} = (\frac{mc^2}{K})^2$ $\frac{v^2}{c^2} = 1-(\frac{mc^2}{K})^2$ $\frac{v^2}{c^2} = 1-(\frac{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2}{25\times 10^6~eV})^2$ $\frac{v^2}{c^2} = 1-(\frac{(8.19\times 10^{-14}~N)(6.24\times 10^{18}~eV/N)}{25\times 10^6~eV})^2$ $\frac{v^2}{c^2} = 1-(\frac{5.11\times 10^5~eV}{25\times 10^6~eV})^2$ $\frac{v^2}{c^2} = 0.99958$ $v = \sqrt{0.99958}~c$ $v = 0.9998~c$ (b) We can find $L$, the distance in the electron frame: $L = L_0~\sqrt{1-\frac{v^2}{c^2}}$ $L = (15~cm)~\sqrt{1-\frac{(0.9998~c)^2}{c^2}}$ $L = (15~cm)~\sqrt{1-(0.9998)^2}$ $L =0.30~cm$ We can find the time it takes the electron to move this distance: $t = \frac{3.0\times 10^{-3}~m}{(0.9998)(3.0\times 10^8~m/s)}$ $t = 1.0\times 10^{-11}~s$
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