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Q. 13.29 Obtain the maximum kinetic energy of $\beta$ - particles, and the radiation frequencies of $\gamma$ decays in the decay                    scheme shown in Fig. 13.6. You are given that

                   $m(^{198}Au)=197.968233\; u$

                   $m(^{198}Hg)=197.966760 \; u$

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$\gamma _{1}$ decays from 1.088 MeV to 0 V

Frequency of $\gamma _{1}$ is

$\\\nu _{1}=\frac{1.088\times 10^{6}\times 1.6\times 10^{-19}}{6.62\times 10^{-34}}\\ \nu _{1}=2.637\times 10^{20}\ Hz$ Plank's constant, h=6.62 $\times$ 10^-34 Js $E=h\nu$

Similarly, we can calculate frequencies of $\gamma _{2}$ and $\gamma _{3}$

$\\\nu _{2}=9.988\times 10^{19}\ Hz\\ \nu _{3}=1.639\times 10^{20}\ Hz$

The energy of the highest level would be equal to the energy released after the decay,

Mass defect is

$\\\Delta m=m(_{79}^{196}\textrm{U})-m(_{80}^{196}\textrm{Hg})\\ \Delta m=197.968233-197.966760\\\Delta m=0.001473u$

We know 1u = 931.5 MeV/c²

Q value= 0.001473 $\times$ 931.5=1.3721 MeV

The maximum Kinetic energy of $\beta _{1}^{-}$ would be 1.3721-1.088=0.2841 MeV

The maximum Kinetic energy of $\beta _{2}^{-}$ would be 1.3721-0.412=0.9601 MeV

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