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

The equation representing `beta`-decay of `._(10)^(23)Ne` is
`._(10)^(23)Nerarr._(11)^(23)Na+beta+overset(-)v` where `Q` is the kinetic shared by `._(10)^(23)Ne` and `._(11)^(23)Na`. Ignoring the rest mass of antineutrino `(overset(-)v)` and electron.
Mass defect `Delta m`
`=m(._(10)^(23)Ne)-m(._(11)^(23)Ne)(._(11)^(23)Na)-m(beta^(-))`
`22.994466-22.989770=0.004696u`
`:. Q = 0.004696xx931MeV=4.372 MeV`
`:.` Max. `K.E` of `beta^(-)=4.372 MeV` when energy carried by is zero.