NEET-XII-Physics
13: Nuclei
- #14The nucleus
decays by
emission. Write down the
decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
= 22.994466 u
= 22.989770 u.
Ans : In emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
emission of the nucleus is given as:
It is given that:
Atomic mass of= 22.994466 u
Atomic mass of= 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
There are 10 electrons in and 11 electrons in
. Hence, the mass of the electron is cancelled in the Q-value equation.
The daughter nucleus is too heavy as compared to
and 
. 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.