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Find the maximum energy that a beta part...

Find the maximum energy that a beta particle can have in the following decay
`^176 Lu rarr ^176 Hf + e + vec v`.
Alomic mass of `^176 Lu` is `175.942694 u` and that of `^176 Hf` is `175.941420 u`.

Text Solution

Verified by Experts

The kinetic energy available for the beta particle and the antineutrione is
`Q = [m(^176 Lu) - m (^176 Hf)] c^2`
`= (175.942694 u - 175.941420 u) (931 MeVu^(-1))`
`= 1.182 MeV`.
This energy is shared by the beta particle and the antineutrino. The maximum kinetic energy of a beta particle in this decay is, therefore, `1.182 MeV` when the antineutriono practically does not get any share.
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