Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 36 - Relativity - Exercises and Problems - Page 1062: 73


$\lambda = 1.06\times 10^{-12}~m$

Work Step by Step

We can find $\gamma$ for the electron and positron: $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-\frac{(0.9~c)^2}{c^2}}}$ $\gamma = \frac{1}{\sqrt{1-0.81}}$ $\gamma = 2.294$ We can find the total energy for the electron: $E = \gamma~mc^2$ $E = (2.294)(9.109\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2$ $E = 1.88\times 10^{-13}~J$ Note that the positron also has the same total energy. We can find the wavelength of each photon with this energy: $E = \frac{hc}{\lambda}$ $\lambda = \frac{hc}{E}$ $\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{1.88\times 10^{-13}~J}$ $\lambda = 1.06\times 10^{-12}~m$
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