The $\beta^{-}$ decay of $^{23}_{10}\mathrm{Ne}$ may be explained as $^{23}_{10}\mathrm{Ne}$ → $^{23}_{11}\mathrm{Na} + ^{0}_{-1}\mathrm{e} + \bar{\nu} + Q$ The expression or the kinetic energy released may be written as Q = $\left[ m(^{23}_{10}\mathrm{Ne}) - m(^{23}_{11}\mathrm{Na}) - m_e \right] c^2$ = $\left[ m(^{23}_{10}\mathrm{Ne}) - m(^{23}_{11}\mathrm{Na}) \right] c^2$ ≈ [22.994466 – 22.989770] × 931.5 MeV ≈ 0.004696 × 931.5 MeV = 4.374 MeV As $^{23}_{11}\mathrm{Na}$ is massive, the kinetic energy released is mainly shared by electron-positron pair. When the neutrino carries no energy, the electron has a maximum kinetic energy equal to 4.374 MeV.