Obtain the maximum kinetic energy of `beta`-particles, and the radiation frequencies of `gamma` decays in the decay scheme shown in Fig. `14.6`. You are given that `m(.^(198)Au)=197.968233 u, m(.^(198)Hg)=197.966760 u`

Explanation

It can be observed from the given Y-decy diagram that `y_(1)` decays from the 1.088 MeV energy level to the 0 MeV energy level.
Hence, the energy corresponding to `y_(1)`-decay is given as :
`E_(1)`=1.088-0=1.088 MeV
`hv_(1)=1.088xx1.6xx10^(-19)xx10^(6)J`
Where, h=Planck's constant `=6.6xx10^(-34)Js`
`v_(1)`= Frequency of radiation radiated by `y_(1)`-decay
`thereforev_(1)=(E_(1))/(h)`
` =(1.088xx1.6xx10^(-19)xx10^(6))/(6.6xx10^(-34))=2.637xx10^(20)Hz`
It can be observed form the given y-decay diagram that `y_(2)` decays from the 0.412 MeV energy level to the 0MeV energy level.
Hence, the energy corresponding to `y_(2)`-decay is given as:
`E_(2)=0.412-0=0.412 MeV`
`hv_(2)=0.412xx1.6xx10^(-19)xx10^(6)J`
Where,
`v_(2)=` Frequency of radation radiated by `y_(2)`- decay
`thereforev_(2)=(E_(2))/(h)`
`=(0.412xx1.6xx140^(-19)xx10^(6))/(6.6xx10^(-34))=9.988xx10^(19)Hz`
It can be observed form the given y-decay diagram that `y_(3)` decays from the 1.088 MeV energy level to the 0.412 MeV energy level.
Hence, the energy corresponding to `y_(3)`-decay is given as :
`E_(3)=1.088-0.412=0.676 MeV`
`hv_(3)=0.676xx10^(-19)xx10^(6)J`
Where,
`v_(3)`=Fequency of radiation radiated by `y_(3)`-decay
`therefore v_(3)=(E_(3))/(h)`
`=(0.676xx1.6xx10^(-19)xx10^(6))/(6.6xx10^(-34))=1.639xx10^(20)Hz`
Mass of `m(""_(78)^(198)Au)=197.968233u`
Mass of `m(""_(80)^(198)Hg)=197.966760ub `
`1u=931.5 MeV//c^(2)`
Energy of the highest level is given as :
`E=[m(""_(78)^(198)Au)-m(""_(80)^(190)Hg)]`
`=197.968233-197.966760=0.001473u`
`=0.001473xx931.5=1.372099MeV`
`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 =0.2840995 MeV
`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 MeV