College Physics (4th Edition)

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

Chapter 27 - Problems - Page 1043: 58


$E = 0.78~MeV$

Work Step by Step

We can find the energy of the photon: $E = m_e~c^2+m_p~c^2+2K$ $E = (9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2+(9.1\times 10^{-31}~kg)(3.0\times 10^8~m/s)^2+(2)(0.22~MeV)$ $E = 5.46\times 10^{-14}~J+0.44~MeV$ $E = (5.46\times 10^{-14}~J)(\frac{1~eV}{1.6\times 10^{-19}~J})+0.44~MeV$ $E = 3.4\times 10^5~eV+0.44~MeV$ $E = 0.34\times 10^6~eV+0.44~MeV$ $E = 0.34~MeV+0.44~MeV$ $E = 0.78~MeV$
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