Obtain the maximum kinetic energy of $\beta$-particles and the radiation frequencies of $\gamma$-decays in the decay scheme shown in the figure. You are given that:
$m(^{198}Au) = 197.968233 \; u$
$m(^{198}Hg) = 197.966760 \; u$

(N/A) From the given $\gamma$-decay diagram,we can determine the energy of the emitted photons.
$1$. For $\gamma_1$ decay (from $1.088 \; MeV$ to $0 \; MeV$):
$E_1 = 1.088 \; MeV = 1.088 \times 1.6 \times 10^{-13} \; J$
Frequency $\nu_1 = E_1 / h = (1.088 \times 1.6 \times 10^{-13}) / (6.626 \times 10^{-34}) \approx 2.628 \times 10^{20} \; Hz$.
$2$. For $\gamma_2$ decay (from $0.412 \; MeV$ to $0 \; MeV$):
$E_2 = 0.412 \; MeV = 0.412 \times 1.6 \times 10^{-13} \; J$
Frequency $\nu_2 = E_2 / h = (0.412 \times 1.6 \times 10^{-13}) / (6.626 \times 10^{-34}) \approx 9.949 \times 10^{19} \; Hz$.
$3$. For $\gamma_3$ decay (from $1.088 \; MeV$ to $0.412 \; MeV$):
$E_3 = 1.088 - 0.412 = 0.676 \; MeV = 0.676 \times 1.6 \times 10^{-13} \; J$
Frequency $\nu_3 = E_3 / h = (0.676 \times 1.6 \times 10^{-13}) / (6.626 \times 10^{-34}) \approx 1.633 \times 10^{20} \; Hz$.
$4$. Maximum kinetic energy of $\beta$-particles:
The total energy available from the decay $^{198}Au \rightarrow ^{198}Hg$ is:
$Q = [m(^{198}Au) - m(^{198}Hg)] \times 931.5 \; MeV/u$
$Q = [197.968233 - 197.966760] \times 931.5 = 0.001473 \times 931.5 \approx 1.372 \; MeV$.
Maximum kinetic energy of $\beta_1$ (decaying to $1.088 \; MeV$ level):
$K_{\beta 1} = 1.372 - 1.088 = 0.284 \; MeV$.
Maximum kinetic energy of $\beta_2$ (decaying to $0.412 \; MeV$ level):
$K_{\beta 2} = 1.372 - 0.412 = 0.960 \; MeV$.