The nucleus decays
β by emission. Write down the β  decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
The nucleus decays
β by emission. Write down the β  decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:  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 βemission.

The reaction equation is given by We are given that –  Mass of an electron, me = 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 ] c2

= (0.004696 c2) u

We know that 1 u = 931.5 MeV/c2. 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.