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.^(23)Ne decays to .^(23)Na by negative ...

`.^(23)Ne` decays to `.^(23)Na` by negative beta emission. What is the maximum kinetic enerfy of the emitter electron ?

Text Solution

Verified by Experts

This decay is of the form given by equation
`._(10)^(23)Ne_(13) rarr ._(11)^(23)Na_(12)+ e^(-) + barv`
and the Q value is found from equation, using atomic masses :
`Q = [m(.^(23)Ne) - m(.^(23)Na)]c^(2)`
`= (22.994465u - 22.989768u) (931.5 MeV//u)`
`= 4.375 MeV`
Except for a small corection for the kinetic energy of the recoiling nucleus, the maximum kinetic energy of the electron is equal to this value. (This occurs when the neutrino has a negligible energy. Similarly the maximum neutrino energy occurs when the electron has a negligibly small kinetic energy)
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