Obtain the maximum kinetic energy of β-particles, and the radiation frequencies of γ decays in the decay scheme shown in Fig. You are given that
m (¹⁹⁸Au) = 197.968233 u
m (¹⁹⁸Hg) = 197.966760 u

To obtain the maximum kinetic energy of $\mathrm{\beta }$- particles = ? 

Radiation frequency, $\mathrm{v} = \frac{({\mathrm{E}}_2 -{\mathrm{E}}_1)}{\mathrm{h}}$

Frequency, $\mathrm{v}\left({\mathrm{\gamma }}_1\right) = \frac{(1.088-0) \times 1.6\times {10}^{-13}}{6.62 \times {10}^{-34}}$ 

                         $= 2.63 \times {10}^{20} {\mathrm{s}}^{-1}$  

$\left[\begin{array}{c}\because 1 \mathrm{MeV} = {10}^6 \times 1.6 \times {10}^{-19}\mathrm{J}\\ 1 \mathrm{MeV} = 1.6 \times {10}^{-13}\mathrm{J}\end{array}\right]$


$$\begin{gathered} \mathrm{Frequency}, \mathrm{v}\left({\mathrm{\gamma }}_2\right) = \frac{(0.412-0) \times 1.6 \times {10}^{-13}}{6.62 \times {10}^{-34}} \\ = 9.96 \times {10}^{20}{\mathrm{s}}^{-1} \\ Frequency, \mathrm{v}\left({\mathrm{\gamma }}_3\right) = \frac{(1.088-0.412) \times 1.6\times {10}^{-13}}{6.62\times {10}^{-34}} \\ = 1.63\times {10}^{20}{\mathrm{s}}^{-1} \end{gathered}$$ 

The emission process of ${\mathrm{\beta }}_1^{-}$ decay may be represented as:
             ${}_{79}{}^{198}\mathrm{Au} \to {}_{80}{}^{198}\mathrm{Hg} + {}_{-1}{}^0\mathrm{e} + \mathrm{E}\left({\mathrm{\beta }}_1^{-}\right) + \mathrm{E}\left({\mathrm{\gamma }}_1\right)$

where,               $\mathrm{E}\left({\mathrm{\gamma }}_1\right) = 1.088 \mathrm{MeV}$ ( as per the fig.)

Now,
      $\mathrm{E}\left({\mathrm{\beta }}_1^{-}\right) = \left[\mathrm{m} \left({}_{79}{}^{198}\mathrm{Au}\right) - \mathrm{m}\left({}_{80}{}^{198}\mathrm{Hg}\right) - {\mathrm{m}}_{\mathrm{e}}\right] \times 931.5 - \mathrm{E}\left({\mathrm{\gamma }}_1\right)$

where,

$\mathrm{m}\left({}_{79}{}^{198} \mathrm{Au}\right) \mathrm{and} \mathrm{m} \left({}_{80}{}^{198} \mathrm{Hg}\right)$ are masses of the ${}_{79}{}^{198} \mathrm{Au} \mathrm{and} {}_{80}{}^{198}\mathrm{Hg}$ nuclei. 

$\therefore$ $\mathrm{E}\left({\mathrm{\beta }}_1^{-}\right) = \left[\left\{\mathrm{M} \left({}_{79}{}^{198} \mathrm{Au}\right) -79 {\mathrm{m}}_{\mathrm{e}}\right\} - \left\{\mathrm{M} \left({}_{80}{}^{198}\mathrm{Hg}\right) - 80 {\mathrm{m}}_{\mathrm{e}}\right\}-{\mathrm{m}}_{\mathrm{e}}\right] \times 931.5-1.088$

         $= \left[\mathrm{M} \left({}_{79}{}^{198} \mathrm{Au}\right) - \mathrm{M}\left({}_{80}{}^{198} \mathrm{Hg}\right)\right] \times 931.5 - 1.088$ 

          = (197.968233 - 197.966760) x 931.5 - 1.088
          $= 1.372 -1.088 = 0.284 \mathrm{MeV}$ 

The emission process of ${\mathrm{\beta }}_2^{-}$ decay may be represented as: 

            ${}_{79}{}^{198}\mathrm{Au} \to {}_{80}{}^{198}\mathrm{Hg} + {}_{-1}{}^0\mathrm{e} + \mathrm{E}\left({\mathrm{\beta }}_2^{-}\right) + \mathrm{E}\left({\mathrm{\gamma }}_2\right)$ 

As in case of ${\mathrm{\beta }}_1^{-}$ decay, it can be deduced in a similar manner.  

 $\therefore$ $\mathrm{E}\left({\mathrm{\beta }}_2^{-}\right) = \left[\mathrm{M} \left({}_{79}{}^{198}\mathrm{Au}\right) -\mathrm{M}\left({}_{80}{}^{198}\mathrm{Hg}\right)\right] \times 931.5 - \mathrm{E} \left({\mathrm{\gamma }}_2\right)$

             $$\begin{gathered} = 1.372 - 0.412 \\ = 0.960 \mathrm{MeV}. \end{gathered}$$