Gold `._(79)^(198)Au` undergoes `beta^(-)` decay to an excited state of `._(80)^(198)Hg`. If the excited state decays by emission of a `gamma`-photon with energy `0.412 MeV`, the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligible energy. The recoil energy of the `._(80)^(198)Hg` nucleus can be ignored. The masses of the neutral atoms in their ground states are `197.968255 u` for `._(79)^(198)Hg`).

Options

A. `0.412 MeV`
B. `1.371 MeV`
C. `0.959 MeV`
D. `1.473 MeV`

Answer

C

Explanation

`m(.^(198)Au_(79))=197.968225 u`
`m(._(80)^(198)Hg) =197.966752 u`
Mass defect,
`Delta m=1.473 xx 10^(-3) u=1.371MeV`
Energy of `gamma`-photon `=0.412 MeV`
Maximum kinetic energy of the electron emitted in the decay is
`E_(e)=1.371 MeV-0.412 MeV=0.959 MeV`.