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 necleus `.^(198)Hg` is formed in its ground state, the kinetic energy available to the electron and the antineutrino is
`Q = [m(.^(18)Au)-m(.^(198)Hg)]c^(2)`.
As `.^(198)Hg^(**)` has energy `1.088 MeV` more than `.^(198)Hg` in gorund state, the kinetic energy actually available is
`Q = [m(.^(198)Au)-m(.^(198)Hg)]c^(2)-1.088MeV`.
`= (197.968233 u - 197.966760 u)(931(Me V)/(u)) - 1.088 MeV`
`= 1.3686 Me V - 1.088 MeV = 0.2806 Me V`.
This is also the maximum possible kinetic energy of the electron emitted.