Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 32 - Nuclear Physics and Nuclear Radiation - Problems and Conceptual Exercises - Page 1153: 48

Answer

$171.1$ MeV

Work Step by Step

We know that $m_i=1.008665u+235.043925u=236.052590u$ and $m_f=132.915237u+97.910331u+5(1.008665u)=235.868893u$ Now, $\Delta m=m_f-m_i=235.868893u-236.052590u=-0.183697u$ We can find the required energy as $E=\Delta mc^2$ We plug in the known values to obtain: $E=0.183697u(\frac{931.5MeV/c^2}{1u})c^2=171.1$ MeV
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