Physics: Principles with Applications (7th Edition)

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

Chapter 30 - Nuclear Physics and Radioactivity - Problems - Page 882: 31

Answer

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Work Step by Step

Start with the $\alpha$ emission. Calculate the energy release from the difference in the masses, using data from Appendix B. $$E_{release}=\left( m(^{218}_{84}Po) -m(^{214}_{82}Pb) -m(^{4}_{2}He) \right)c^2$$ $$ =\left(218.008973u-213.999806u-4.002603u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$ $$ E_{release}=6.114MeV$$ Next analyze the $\beta^-$ emission. Calculate the energy release from the difference in the masses, using data from Appendix B. Assume that the neutrino mass is negligible. To balance the electrons, use the mass of the At atom. $$E_{release}=\left( m(^{218}_{84}Po) -m(^{218}_{85}At) \right)c^2$$ $$ =\left(218.008973u-218.008695u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$ $$ E_{release}=0.259MeV$$
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