Obtain the maximum kinetic energy of $\beta$ -particles, and the radiation frequencies of $\gamma$ decays in the decay scheme shown.

You are given that
$m{(}^{198}Au)=197.968233u$
$m{(}^{198}Hg)=197.966760u$

Where, 

h= Plank's constant and $v_1$ = frequency of radiation radiated by ${\gamma }_1$ decay.

It can be observed the given gamma decay diagram that ${\gamma }_2$ decays from the 0.412 MeV energy level to the 0 MeV energy level. 

Hence, the energy corresponding to ${\gamma }_2$ decay is given as 

Where $v_2$ is the frequency of the radiation radiated by ${\gamma }_2$ decay. 

It can be observed from the given gamma decay diagram that ${\gamma }_3$ decays from the 1.088 MeV energy level to the 0.412 MeV energy level. Hence, the energy corresponding to ${\gamma }_3$ is given as 

Where $v_3=$ frequency of the radiation radiated by ${\gamma }_3$ decay

Now mass of $Au$ is 197.968 u and mass of $Hg$ = 197.9667 u

Energy of the highest level is given as :

${\beta }_1$ decays from the 1.3720995 MeV level to the 1.088 MeV level

$\therefore$ maximum kinetic energy of the ${\beta }_1$ particle $=1.3720995-1.088=0.2840995MeV$

${\beta }_2$ decays from the 1.3720995 MeV level to the 0.412 MeV level 

$\therefore$ Maximum kinetic energy of the ${\beta }_2$ particle = $1.3720995-0.412=0.9600995$