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: 45

Answer

$15.66MeV$

Work Step by Step

We can find the required energy as follows: $m_f=15.003065+1.008665u=16.011730u$ Now $\Delta m=m_f-m_i=16.011730-15.994915u=0.016815u$ We know that $E=\Delta mc^2$ We plug in the known values to obtain: $E=0.016815u(\frac{931.5MeV/c^2}{1u})c^2=15.66MeV$
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