Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 26 - The Special Theory of Relativity - Problems - Page 768: 42

Answer

260 MeV

Work Step by Step

The muons’ total energy (rest energy plus kinetic energy) becomes electromagnetic energy. Use equation 26–6b. $$E_{photon}=E_1+E_2=\gamma_1mc^2+\gamma_2mc^2=(\gamma_1+\gamma_2)mc^2$$ $$ =(\frac{1}{\sqrt{1-0.53^2}}+\frac{1}{\sqrt{1-0.65^2}})(105.7MeV)$$ $$ =263.7MeV\approx260MeV$$
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