Question

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 (¹⁹⁸Au) = 197.968233 u

m (¹⁹⁸Hg) =197.966760 u

Answer 1

It can be observed from the given γ-decay diagram that γ₁ decays from the 1.088 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ₁-decay is 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 radiation radiated by γ₁-decay

It can be observed from the given γ-decay diagram that γ₂ decays from the 0.412 MeV energy level to the 0 MeV energy level.

Hence, the energy corresponding to γ₂-decay is given as:

E₂ = 0.412 − 0 = 0.412 MeV

hν₂= 0.412 × 1.6 × 10⁻¹⁹ × 10⁶ J

Where,

ν₂ = Frequency of radiation radiated by γ₂-decay

It can be observed from the given γ-decay diagram that γ₃ decays from the 1.088 MeV energy level to the 0.412 MeV energy level.

Hence, the energy corresponding to γ₃-decay is 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-decay

Mass 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 1.3720995 MeV level to the 1.088 MeV level

∴Maximum kinetic energy of the β₁ particle = 1.3720995 − 1.088

= 0.2840995 MeV

β₂ decays from the 1.3720995 MeV level to the 0.412 MeV level

∴Maximum kinetic energy of the β₂ particle = 1.3720995 − 0.412

= 0.9600995 MeV