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: 29

Answer

2.822 MeV.

Work Step by Step

The decay described is $^{60}_{27}Co\rightarrow\;^{60}_{28}Ni +\;^{0}_{-1}e+\overline{\nu}$. 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 27 electrons to each side of the equation, then using the atomic mass of $^{60}_{28}Ni $ on the right hand side. $$E_{release}=\left( m(^{60}_{27}Co)-m(^{60}_{28}Ni) \right)c^2$$ $$ =\left(59.933816u-59.930786u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$ $$ E_{release}=2.822MeV$$ If we assume that the nickel and the neutrino have negligible KE, and assume zero mass for the neutrino, the energy released is the maximum kinetic energy of the emitted $\beta$.
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