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# Consider the beta decay ${ }^{10} \mathrm{Be} \rightarrow{ }^{10} \mathrm{~B}+\beta^{-}$, with the beryllium nucleus initially at rest. (c) What are the possible energies of that other particle?

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As we have shown in the a) part of the problem the energy released from the reaction is $1.35 \times 10^{-16} \text{ J}$ and this energy can go into the speed of the electron ( $\beta ^{-}$ ) and antineutrino. So we can see two edge cases, first one is that the neutrino gets all of the available energy and that the electron is stationary while the second one is that the electron gets almost all of the available energy and the neutrino gets a negligible bit of it, and of course all of the distributions in between those edge cases. So in conclusion, the antineutrino can take energy values from the maximum ( $1.35 \times 10^{-16} \text{ J}$ ) all the way to close to zero.

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