Answer
$1.02MeV$
Work Step by Step
We can find the required energy as follows:
$E=m_{electron}c^2+m_{anti-electron}c^2$
$E=2m_e c^2$
We plug in the known values to obtain:
$E=2(0.51MeV)$
$E=1.02MeV$
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 1043: 59
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