Consider the following decay schemes:
${ }_{79}^{198} \mathrm{Au} \rightarrow X+\beta_1^{-} \rightarrow{ }_{80}^{198} \mathrm{Hg}+\gamma_1 \quad ; \quad{ }_{79}^{198} \mathrm{Au} \rightarrow Y+\beta_2^{-} \rightarrow{ }_{80}^{198} \mathrm{Hg}+\gamma_2$
${ }_{79}^{199} \mathrm{Au} \rightarrow X+\beta_1^{-} \rightarrow Y+\gamma_3 \rightarrow{ }_{80}^{198} \mathrm{Hg}+\gamma_2$
The energies corresponding to X and Y with respect to ${ }_{80}^{198} \mathrm{Hg}$ are 1.088 MeV and 0.412 MeV respectively [Note the energy of ${ }_{80}^{198} \mathrm{Hg}$ is taken to be 0.0 MeV ]. The atomic masses of ${ }_{79}^{198} \mathrm{Au}$ and ${ }_{80}^{198} \mathrm{Hg}$ are 197.968 u and 197.966 u , respectively, where 1 u should be taken as $931 \mathrm{MeV} / \mathrm{c}^2$. Then,

• Correct answer
  A

  the energies corresponding to $\gamma_1$ and $\gamma_2$ are 1.088 MeV and 0.412 MeV respectively.
• B

  the energies corresponding to $\gamma_1$ and $\gamma_3$ are 1.088 MeV and 0.412 MeV respectively.
• Correct answer
  C

  the maximum kinetic energies of $\beta_1^{-}$and $\beta_2^{-}$are about 0.77 MeV and 1.45 MeV respectively.
• D

  the energy corresponding to ${ }_{79}^{198} \mathrm{Au}$ is 1.50 MeV .

Hence, the correct answer is options: 'A', 'C'.