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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.
Asked by Pralhad Last Modified
Answer 
Shivam Sharma
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.
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