Neon-23 decays in the following way, ${}_{10}{}^{23}Ne\to {}_{11}{}^{23}Na+{}_{-1}{}^{0}e+\overline{v}$

Find the minimum and maximum kinetic energy that the beta particle $\left({}_{-1}{}^{0}e\right)$ can have. The atomic masses of ${}^{23}Ne$ and ${}^{23}Na$ are 22.9945 u and 22.9898 u, respectively.

Here, atomic masses are given (not the nuclear masses), but still we can use that for calculating the mass defect because mass of electrons get cancelled both sides. Thus, Mass defect $∆m=\left(22.9945-22.9898\right)=0.0047 u$

$$\begin{gathered} \therefore Q=\left(0.0047 u\right)\left(931.5 MeV\right) \\ =4.4 MeV \end{gathered}$$

Hence, the energy of beta particles can range from 0 to 4.4 $MeV$.