The nucleus $_{10} Ne ^{23}$ decays by $\beta ^{-}$ emission. write down the $\beta^{-}$ decay equation and determine the maximum kinetic energy of the electrons emitted from the following data. $m (_{10} Ne ^{23} ) = 22.994466\;amu \;\qquad m (_{11} Na ^{23} ) = 22.989770\;amu$

Question

$\begin{array}{1 1} 0.04372\;Mev \\ 0.4372\;Mev \\ 4.372\;Mev \\ 43.72\;Mev \end{array}$