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The nucleus .^(23)Ne deacays by beta-em...

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))`.

Text Solution

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The equation representing `beta^(-)` decay of `._(10)^(23)Ne` is `._(10)^(23)Ne rarr ._(11)^(23)Na + beta^(-)+bar(v)+Q`
where, Q is the kinetic energy shared by `._(10)^(23)Ne and ._(11)^(23)Na`.
Ignoring the rest mass of antineutrino `(bar(v))` and electron.
Mass defect `Deltam=m (._(10)^(23)Ne)-m(._(11)^(23)Na)-m(beta^(-))=22.994466-22.989770=0.004696 u`
`:. Q=0.004696xx931 MeV = 4.372 MeV`
`:. ` Max. K.E. of `beta^(-)=4.372` MeV when energy carried by `bar(v)` is zero.
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