Answer
$\gamma=2.9569514$
Work Step by Step
We know that
$\gamma=\frac{K}{m_ec^2}+1$
We plug in the known values to obtain:
$\gamma=\frac{1MeV}{0.51099MeV}+1=2.9569514$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 37 - Relativity - Problems - Page 1149: 58c
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