Question -
Answer -
It can be observed from the given γ-decaydiagram that γ1 decays from the 1.088 MeV energylevel to the 0 MeV energy level.
Hence, the energy corresponding to γ1-decayis given as:
E1 = 1.088 − 0 =1.088 MeV
hν1= 1.088 × 1.6 × 10−19 ×106 J
Where,
h = Planck’s constant = 6.6 × 10−34 Js
ν1 = Frequency of radiationradiated by γ1-decay
It can be observed from the given γ-decaydiagram that γ2 decays from the 0.412 MeV energylevel to the 0 MeV energy level.
Hence, the energy corresponding to γ2-decayis given as:
E2 = 0.412 − 0 =0.412 MeV
hν2= 0.412 × 1.6 × 10−19 ×106 J
Where,
ν2 = Frequency of radiationradiated by γ2-decay
It can be observed from the given γ-decaydiagram that γ3 decays from the 1.088 MeV energylevel to the 0.412 MeV energy level.
Hence, the energy corresponding to γ3-decayis given as:
E3 = 1.088 − 0.412= 0.676 MeV
hν3= 0.676 × 10−19 ×106 J
Where,
ν3 =Frequency of radiation radiated by γ3-decayMass of 
= 197.968233 u Mass of 
= 197.966760 u 1 u = 931.5 MeV/c2
Energy of the highest level is given as:
β1 decays from the 1.3720995MeV level to the 1.088 MeV level
∴Maximum kineticenergy of the β1 particle = 1.3720995 − 1.088
= 0.2840995 MeV
β2 decays from the 1.3720995MeV level to the 0.412 MeV level
∴Maximum kineticenergy of the β2 particle = 1.3720995 − 0.412
= 0.9600995 MeV