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

Beta decay equation for `""_(79)Au^(198)` is
`""_(79)Au^(198) to ""_(80)Hg^(198) + _(-1)e^(0) + barv +Q`
But `Q=(m_(N)(""_(79)Au^(198))-[m_(N)(""_(80)Hg^(198))+m_(e)]c^(2)`
(Ignoring rest mass of antineutrino) where mN represents nuclear mass. In terms of atomic masses
`Q=(m(""_(79)Au^(198))-79m_(e))-(m(""_(80)Hg^(198))-80m_(e))+m_(e))c^(2)`
`=(m(""_(79)Au^(198))-m(""_(80)Hg^(198))c^(2)`
Hence, K.E. of particles.
`Q =(m(""_(79)Au^(198))-(m(""_(80)Hg^(198))u xx c^(2)`
`=0.001473 xx 931.5 MeV = 1.3721` MeV
In `beta_(1)^(-)` transition, energy shared by
`""_(80)Hg^(198) = 1.088 MeV` (from the given figure)
Max kinetic energy of `beta_(1)` (particle ignoring kinetic energy of `barv`)
`=1.3721-1.088`
`=0.2841` MeV
In `beta_(2)^(-)` transition, K.E. shared by
`""_(80)Hg^(198)=0.412` MeV (from the given figure)
Mass kinetic energy of `beta_(2)^(-)` particle.
`=1.3721 - 0.412 = 0.9601` MeV
For `lambda_(1)` transition, energy is given as follows:
`hv_(1)=1.088 MeV`
`=1.088 xx 1.6 xx 10^(-13)` J
Corresponding frequency is calculated as follows:
`v_(1) = (1.088 xx 1.6 xx 10^(-13))/h`
`=2.6273 xx 10^(20)` Hz
For `gamma_(2)` transition, energy is given as follows:
`hv_(2)=0.412` MeV
`=0.412 xx 1.6 xx 10^(-13)` J
Corresponding frequency is calculated as follows:
`v_(2)=(0.412 xx 1.6 xx 10^(-13))/h = (0.412 xx 1.6 xx 10^(-12))/(6.626 xx 10^(-34))`
`=0.99488 xx 10^(20)` Hz
For `gamma_(2)`
`hv_(3)=hv_(1)-hv_(2)=((hv_(1)-hv_(2))/h) = v_(1)-v_(2)`
`=2.62730 xx 10^(20) - 0.99488 xx 10^(20)`
`=1.63242 xx 10^(20)` Hz.