Obtain the maximum kinetic energy of *β*-particles, and the radiation frequencies of *γ*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

#### Solution

It can be observed from the given *γ*-decay diagram that *γ*_{1} decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to *γ*_{1}-decay is given as:

*E*_{1} = 1.088 − 0 = 1.088 MeV

*h**ν*_{1}= 1.088 × 1.6 × 10^{−19} × 10^{6} J

Where,

*h* = Planck’s constant = 6.6 × 10^{−34} Js

ν_{1 }= Frequency of radiation radiated by *γ*_{1}-decay

`:. v_1 = E_1/h`

`= (1.088 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 2.637 xx 10^20 Hz`

It can be observed from the given *γ*-decay diagram that *γ*_{2} decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to *γ*_{2}-decay is given as:

*E*_{2} = 0.412 − 0 = 0.412 MeV

*h**ν*_{2}= 0.412 × 1.6 × 10^{−19} × 10^{6} J

Where,

ν_{2 }= Frequency of radiation radiated by *γ*_{2}-decay

`:. v_2 = E_2/h`

`= (0.412 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 9.988 xx 10^(19) Hz`

It can be observed from the given *γ*-decay diagram that *γ*_{3} decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence, the energy corresponding to *γ*_{3}-decay is given as:

*E*_{3} = 1.088 − 0.412 = 0.676 MeV

*h**ν*_{3}= 0.676 × 10^{−19} × 10^{6} J

Where,

ν_{3 }= Frequency of radiation radiated by *γ*_{3}-decay

`:. v_3 = E_3/h`

`= (0.676 xx 1.6 xx 10^(-19) xx 10^6)/(6.6 xx 10^(-34)) = 1.639 xx 10^20 Hz`

Mass of `m(""_78^198Au)`= 197.968233 u

Mass of `m(""_80^198 Hg)`= 197.966760 u

1 u = 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 = 0.001473 u`

`= 0.001473 xx 931.5 = 1.3720955 MeV`

β_{1 }decays from the 1.3720995 MeV level to the 1.088 MeV level

∴Maximum kinetic energy of the *β*_{1 }particle = 1.3720995 − 1.088

= 0.2840995 MeV

β_{2 }decays from the 1.3720995 MeV level to the 0.412 MeV level

∴Maximum kinetic energy of the *β*_{2 }particle = 1.3720995 − 0.412

= 0.9600995 MeV