Physics: Principles with Applications (7th Edition)

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

Chapter 32 - Elementary Particles - General Problems - Page 945: 43

Answer

a. 1.022 MeV b. 1876.6 MeV

Work Step by Step

The initial kinetic energy is approximately 0. Therefore, the total energy released is the rest mass energy of the pair of particles. a. $E_t=2mc^2=2(0.511MeV)=1.022MeV$ b. $E_t=2mc^2=2(938.3MeV)=1876.6MeV$
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