College Physics (4th Edition)

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

Chapter 26 - Problems - Page 1013: 70


$L = 0.111~m$

Work Step by Step

We can find the kinetic energy $K$: $E = K+E_0$ $K = E - E_0$ $K = 46\times 10^9~eV - 0.511\times 10^6~eV$ $K = 45.999489\times 10^9~eV$ We can find $\gamma$: $K = (\gamma-1)mc^2$ $\gamma = 1+\frac{K}{mc^2}$ $\gamma = 1+\frac{(45.999489\times 10^9~eV)(1.6\times 10^{-19}~J/eV)}{(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2}$ $\gamma = 89866$ Let $L_0 = 10.0~km$. We can find the length $L$ in the reference frame of the electrons: $L = \frac{L_0}{\gamma}$ $L = \frac{10000~m}{89866}$ $L = 0.111~m$
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