NCERT Solutions
Class 12 Physics
Nuclei

Q.14
The nucleus 2310Ne decays by β–emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted.
Given that: m (2310Ne) = 22.994466 um (2311Na) = 22.089770 u.
In β- emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
β- Emission of the nucleus (2310Ne)
(2310Ne) à (2311Na) + e- + ̅ν + Q
It is given that:
Atomic mass m (2310Ne) of = 22.994466 u
Atomic mass m (2311Na) of = 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
Q= [m (2310Ne) + [m (2311Na) +me]] c2
There are 10 electrons in and 11 electrons in (2311Na). Hence, the mass of the electron is cancelled in the Q-value equation.
Therefore Q = [22.994466 - 22.989770] c2
= (0.004696 c2) u.
But 1u = 9.31
Therefore Q= (0.004696 x 931.5) =4.374MeV.
The daughter nucleus is too heavy as compared to e- and ̅ν. 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.