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

$m\left(_{10}^{23} Ne \right)=22.994466 \;u$

$m\left(_{11}^{23} Na\right) =22.089770\; u$

In $\beta^{-}$ emission, the number of protons increases by $1,$ and one electron and an antineutrino are emitted from the parent nucleus. $\beta^{-}$ emission of the nucleus

$_{10}^{23} Ne _{10}^{23} Ne \rightarrow_{11}^{23} Na +e^{-}+\bar{v}+Q$

It is given that:

Atomic mass $m\left(_{10}^{23} Ne \right) o f=22.994466 u$

Atomic mass $m\left(_{11}^{23} N a\right) o f=22.989770 u$

Mass of an electron, $m_{e}=0.000548 u$

$Q$ - value of the given reaction is given as:

$Q=\left[m\left(^{23}_{10} N e\right)-\left[m\left(_{11}^{23} Na\right)+m_{e}\right]\right] c^{2}$

There are 10 electrons in $_{10} N e^{23}$ and 11 electrons in $_{11}^{23} N a$. Hence, the mass of the electron is cancelled in the $Q$ - value equation. $\therefore Q=[22.994466-22.9897770] c^{2}$

$=\left(0.004696 c^{2}\right) u$

But $1 u=931.5 MeV / c ^{2}$

$\therefore Q=0.004696 \times 931.5=4.374 MeV$

The daughter nucleus is too heavy as compared to $e^{-}$ and $v .$ Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., $4.374 MeV$.