The nucleus ${^{23}_{10}Ne}$ decays by $β^-$ emission. The maximum kinetic energy of the electrons emitted is:

Given that: $m({^{23}_{10}Ne}) = 22.994466 u, m({^{23}_{11}Na}) = 22.989770 u$ and $1u ≡ 931 MeV/c^2$

4.372 MeV

The correct answer is Option (4) → 4.372 MeV

For $\beta^-$ decay:

$Q = [m(^{23}_{10}Ne) - m(^{23}_{11}Na)]c^2$

Given: $m(^{23}_{10}Ne) = 22.994466 \, u$, $m(^{23}_{11}Na) = 22.989770 \, u$

Mass difference $= 22.994466 - 22.989770 = 0.004696 \, u$

$Q = 0.004696 \times 931 \, \text{MeV} = 4.37 \, \text{MeV}$

In $\beta^-$ decay, the maximum kinetic energy of the emitted electron $\approx Q$ (neglecting neutrino rest mass).

Final Answer: $K_{max} = 4.37 \, \text{MeV}$