Answer
$2.21\times10^{-11}J$
Work Step by Step
$\Delta E=m_{m}c^{2}$
$m_{m}=270m_{e}=270\times 9.11\times10^{-31}kg$
Also, $c=3\times10^{8}$
Therefore, $\Delta E=270 \times9.11\times10^{-31}\times(3\times10^{8})^{2}$= $2.21\times10^{-11}J$
College Physics (7th Edition)
by Wilson, Jerry D.; Buffa, Anthony J.; Lou, Bo
Published by
Pearson
ISBN 10:
0-32160-183-1
ISBN 13:
978-0-32160-183-4
Chapter 30 - Nuclear Reactions and Elementary Particles - Learning Path Questions and Exercises - Exercises - Page 1031: 28
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