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: 14

Answer

17.35 MeV is released.

Work Step by Step

Calculate the Q-value, using data from Appendix B. We use atomic masses. $$Q=\left(m_p+m(^{7}_{3}Li)-m(^{4}_{2}He)-m_{\alpha} \right)c^2$$ $$ =\left(1.007825u+7.016003u-4.002603u-4.002603u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$ $$=17.35MeV$$ The Q-value is positive. The reaction is exothermic, and releases 17.35 MeV of energy.
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