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`.

Explanation

If the product nucleus `^.(198)Hg` is formed in its ground state, the kinetic energy avaialbe to the electron and the antineutino is
`Q=[m(.^(198)Au)-m(.^(198)Hg)]c^(2)`
As `.^(198)Hg^(ast)` 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`
`=(197.968223 u -197.966760 u)(931(MeV)/(u)) - 1.088 MeV`
`=1.3686 MeV - 1.088 MeV=0.2806 MeV`
This is also the maximum possible kinetic energy of the electron emitted.