The nucleus `.^(23)Ne` decays 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 antineutrino `(bar(v))`.

Answer

`4.374 MeV`

Explanation

The `beta`-decay of `.^(23)Ne` is represented by the equation
.`_(10)^(23)Ne rarr. _(10)^(23)Na + bar( e) +bar (v) +Q`
Mass defect for the above reaction is
`Deltam=[(._(10)^(23)Ne)-M(._11)^(23)Na)]`
`=[22.994466-22.9897770] a.m.u`
`=0.004696 a.m.u.`
Now, `Q` value of the above reaction is given as
`Q =Delta m xx 931.5MeV =4.374 MeV`
The energy `Q` released is shared by `.^(23)Na` nucleus and the electron - antineutrino pair. Since `.^(23)Na` is massive, most of the kinetic energy if the antineutrino carries zero energy. Thus, the maximum energy of the electrons emitted is` .4.374 MeV`.