Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 31 - Nuclear Energy; Effects and Uses of Radiation - Problems - Page 911: 11

Answer

0.626 MeV.

Work Step by Step

The reaction is $$n + \;^{14}_{7}N\rightarrow\;^{14}_{6}C +?$$ Conserve nucleon number: 1 + 14 = 14 + A. We see that A = 1. Conserve charge: 0 + 7 = 6 + Z. We see that Z = 1. The unknown product is a proton. $$n + \;^{14}_{7}N\rightarrow\;^{14}_{6}C +p$$ Calculate the Q-value, using data from Appendix B. We use atomic masses. $$Q=\left( m_{n}+m(^{14}_{7}N)-m(^{14}_{6}C)-m_{p} \right)c^2$$ $$ =\left(1.008665u+14.003074u-14.003242u-1.007825u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$ $$=0.626MeV$$
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