**Ans:**

We know that the number of protons gets increased by 1, and one electron and an antineutrino is emitted from the parent nucleus during *β***− **emission.

The reaction equation is given by

We are given that –

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

Therefore, Q-value of the given reaction is given as:

There are 10 electrons and 11 electrons in given Ne and Na isotopes respectively. Hence, the mass of the electron gets cancelled out in the Q-value equation. We get –

Q = [ 22.994466 – 22.989770 ] c^{2}

= (0.004696 c^{2}) u

We know that 1 u = 931.5 MeV/c^{2}. So,

Q = 0.004696 x 931.5 = 4.374 MeV

In comparison to e^{–} and v^{_}, the daughter nucleus is too heavy. As a result, it carries very little energy. The antineutrino’s kinetic energy is nearly zero. As a result, the released electrons’ maximal kinetic energy is nearly equal to the Q-value, i.e. 4.374 MeV.