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 881: 27

Answer

5.515 MeV.

Work Step by Step

a. Column 6 of Appendix B shows that $^{24}_{11}Na$ is a $\beta^-$ emitter. b. The decay described is $^{24}_{11}Na\rightarrow\;^{24}_{12}Mg +\;^{0}_{-1}e+\overline{\nu}$. Calculate the energy release from the difference in the masses, using data from Appendix B. The mass of the emitted $\beta^-$, which is an electron, is accounted for by adding 11 electrons to each side of the equation, then using the atomic mass of $^{24}_{12}Mg $ on the right hand side. $$E_{release}=\left( m(^{24}_{11}Na)-m(^{24}_{12}Mg) \right)c^2$$ $$ =\left(23.990963u-23.985042u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$ $$ E_{release}=5.515MeV$$ If we assume that the magnesium and the neutrino have negligible KE, and assume zero mass for the neutrino, the energy release is the maximum kinetic energy of the emitted $\beta^-$. $$ KE_{max}=5.515MeV$$
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