Answer
$140MeV$
Work Step by Step
We can find the required rest energy as
$m_{\circ}c^2=\frac{(pc)^2-K^2}{2K}$
We plug in the known values to obtain:
$m_{\circ}c^2=\frac{[105(MeV/c)(c)]^2-(35.0MeV)^2}{2(35.0MeV)}$
$m_{\circ}c^2=140MeV$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 29 - Relativity - Problems and Conceptual Exercises - Page 1044: 90
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