The nucleus $_{10}^{23} Ne$ decays by $\beta^{-}$ emission. Write down the $\beta^{-}$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$m(_{10}^{23} Ne) = 22.994466 \; u$
$m(_{11}^{23} Na) = 22.989770 \; u$

(N/A) In $\beta^{-}$ emission,a neutron in the nucleus converts into a proton,emitting an electron $(e^{-})$ and an antineutrino $(\bar{\nu})$. The decay equation is:
$_{10}^{23} Ne \rightarrow _{11}^{23} Na + e^{-} + \bar{\nu} + Q$
The $Q$-value of the reaction is given by the mass defect:
$Q = [m(_{10}^{23} Ne) - m(_{11}^{23} Na)] c^{2}$
Note: The mass of the emitted electron is accounted for by the difference in atomic masses (since the daughter nucleus has one extra orbital electron compared to the parent).
$Q = (22.994466 \; u - 22.989770 \; u) c^{2}$
$Q = 0.004696 \; u \times 931.5 \; MeV/u = 4.374 \; MeV$
Since the daughter nucleus is much heavier than the electron and antineutrino,it carries away negligible kinetic energy. The maximum kinetic energy of the emitted electron is approximately equal to the $Q$-value,which is $4.374 \; MeV$.