Answer
See answers.
Work Step by Step
The decay described is $^{23}_{10}Ne\rightarrow\;^{23}_{11}Na +\;^{0}_{-1}e+\overline{\nu}$.
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 10 electrons to each side of the equation, then using the atomic mass of $^{23}_{11}Na $ on the right hand side.
$$E_{release}=\left( m(^{23}_{10}Ne)-m(^{23}_{11}Na) \right)c^2$$
$$ =\left(22.9947u-22.989769u\right)c^2\left( \frac{931.49MeV/c^2}{u}\right)$$
$$ E_{release}=4.59MeV$$
The sum of the kinetic energy of the electron and the neutrino’s energy is 4.59 MeV.
If we assume that the sodium and the neutrino have negligible KE, and assume zero mass for the neutrino, then 4.59 MeV is the maximum kinetic energy of the emitted $\beta$. The neutrino has zero KE then.
The minimum KE of the emitted $\beta$ is zero, and occurs when the neutrino has 4.59 MeV of energy.
