Question

The nucleus `.^(23)Ne` deacays by `beta`-emission into the nucleus `.^(23)Na`. Write down the `beta`-decay equation and determine the maximum kinetic energy of the electrons emitted. Given,`(m(._(11)^(23)Ne) =22.994466 am u` and `m (._(11)^(23)Na =22.989770 am u`. Ignore the mass of antineuttino `(bar(v))`.

Answer 1

In `beta^(-)` emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
`beta^(-)`emission of the nucleas `""_(10)^(23)Ne` is given as:d
`""_(10)^(23)Neto""_(11)^(23)Na+e^(-)+overset(-)(v)+Q`
It is given that
Atomic mass of `m(""_(10)^(23)Ne)` = 22.994466 u
Atomic mass of `m(""_(11)^(23)Na)`= 22.989770 u
Mass of an electron, `m_(e)`= 0.000548 u
Q- value of the given reaction is given as:
`Q=[m(""_(10)^(23)Ne)-[m(""_(11)^(23)Na)+m_(e)]]c^(2)`
Therse are 10 electron, `""_(10)^(23)Ne` and 11 electron in `""_(11)^(23)Na`. Hence, the mass of the electron is cancelled in the Q-value equation.
`thereforeQ=[22.994466-22.989770]c^(2)`
`=(0.004696^(2))` u
But 1 u = 931.5 `MeV//c^(2)`
`thereforeQ=0.004696xx931.5=4.374MeV`
The daughter nucleus is too heavy as compared to `e^(-)and overset(-)(v)`.
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.