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))`.

Explanation

the `beta`-decay of `""_(10)Ne^(23)` is `""_(10)Ne^(23) to ""_(11)Na^(23) + e^(-) + barv +Q`
where Q is the Q-value of equation and is given by
`Q=(m(""_(10)Ne^(23)) -m_(N)(""_(11)Na^(23))-m_(e))c^(2)`
Where the neutrino.s rest mass has been ignored. The masses given here are atomic. We need to use nuclear masses.
`Q=[m(""_(10)Ne^(23))-10m_(e)-m(""_(11)Na^(23)) + 11m_(e) -m_(e)]c^(2)`
`=[m(""_(10)Ne^(23))-m(""_(11)Na^(23))] xx c^(2)`
`=(22.994466 - 22.989770)u xx c^(2)`
or Q = 4.374 MeV
This kinetic energy is mainly shared jointly by the `e^(-) - barv` pair since `Na^(23)` is much more massive than this pair and its recoil energy is therefore negligible.
The maximum energy of emission is equal to total kinetic energy of the `e^(-) - barv` . (When the electron has maximum kinetic energy, the neutrino carries no energy). Thus, the maximum kinetic energy of the e~ emitted is 4.37 MeV.