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: 6

Answer

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Work Step by Step

a. Check the sign of the Q-value, using data from Appendix B. $$Q=\left( m_n+m(^{24}_{12}Mg)-m(^{23}_{11}Na)-m_d \right)c^2$$ $$ =\left(1.008665u+23.985042u-22.989769u-2.014102u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$ $$=-9.468MeV$$ This is negative, so it requires energy to proceed. Yes, the 18.00 MeV of kinetic energy is enough to make the reaction happen. b. The energy released is 18.00MeV - 9.468 MeV = 8.53MeV.
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