Gold, ${?}_{79}^{198}Au,$ undergoes $-$ decay to an excited state of ${?}_{80}^{198}Hg.$ If the excited state decays by emission of a $$ photon with energy $0\mathrm{.}412$ MeV $,$ what is the maximum kinetic energy of the electron emitted in the decay. 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 $197\mathrm{.}968225$ u for ${?}_{79}^{198}Au$ and $197\mathrm{.}966752$ u for ${?}_{80}^{198}Hg.)$ solve step by step and explain everything please