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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

Mass defect in the process
`Deltam=197.968233-197.966760`
`=0.001473 a.m.u.`
`:.` Total B.E.`=0.001473xx931MeV`
`=1.3786MeV.`
As Hg is in excited state with an energy of 1.088MeV, therefore, max. possible K.E. of electron emitted is `=1.3786-1.088=0.2806MeV`
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