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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))`. |
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Answer» The `beta^(-)` decay of `._10Ne^(23)` may be represented as `._10Ne^(23)to ._11Na^(23) +._(-1)e^0+barv+Q` Ignoring the rest mass antineutrino `(barv)` and electron Mass defect, `Deltam=m(._10Na^(23))-m(._11Na^(23)) =22.994466-22.989770-0.004696a.m.u` `:. Q= 0.004696xx931MeV=4.372MeV` As `._11Na^(23)` is very massive, this energy of `4.3792MeV`, is shared by `e^(-) and barv` pair. The max. K.E. of `e^(-)=4.372 MeV`, when energy carried by `barv` is zero. |
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