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Consider the beta decay ^198 Au rarr ^...

Consider the beta decay
`^198 Au rarr ^198 Hg ** + Beta^(-1) + vec v`.
where `^198 Hg^**` represents a mercury nucleus in an excited state at energy `1.088 MeV` above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass of `^198 Au` is `197.968233 u` and that of `^198 Hg` is `197.966760 u`.

Text Solution

Verified by Experts

If the product nucleus `^198 Hg` is formed in its ground state, the kinetic energy available to the electron and the antineutrino is
`Q = [m(^198 Au) - m (^198 Hg)]c^2`.
As `^198 Hg^**` has energy `1.088 MeV` more than `^198 Hg` in ground state, the kinetic energy actually available is
`Q = [m(^198 Au) -m (^198 Hg)] c^2 - 1.088 MeV`
`= (197.968233 u - 197.966760 u) (931 MeV u^(-1)`
`-1.088 MeV`
`=1.3686 MeV - 1.088 MeV = 0.2806 MeV`
This is also the maximum possible kinetic energy of the electron emitted.
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