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 m¹⁹⁸Hg) =197.966760 u

Answer 1

Refer to the attachment

Frequency of γ₁  = (E₂ - E₁) /h

= (1.088) × (1.6 × 10⁻¹³) / 6.626 × ⁻³⁴

= 2.63 × 10²⁰ Hz

Frequency of γ₂ = (0.412 -0 )× (1.6 × 10⁻¹³) / 6.626 × ⁻³⁴

= 9.96 × 10²⁰ Hz

Frequency of γ₃  =  (1.088- 0.412) × (1.6 × 10⁻¹³) / 6.626 × ⁻³⁴

= 1.63 × 10²⁰ Hz.

Now for β₁⁻ emission reaction

¹⁹⁸₇₉Au -----→  ¹⁹⁸₈₀Hg + ₋₁⁰e + E ( β₁⁻) + E( γ₁)

Now , E( γ₁) = 1.088 MeV

E(β₁⁻) = [ m( ¹⁹⁸₇₉Au) - m( ¹⁹⁸₈₀Hg) - $m_e$ ] × 931.5 - E( γ₁)

= (197.968233 -197.966760)× 931.5 -1.088 MeV

E(β₁⁻) = (1.372 -1.088) MeV

E(β₁⁻) = 0.284 MeV

Now for β₂⁻ emission reaction ,

¹⁹⁸₇₉Au -----→  ¹⁹⁸₈₀Hg + ₋₁⁰e + E ( β₂⁻) + E( γ₂)

E(β₂⁻) = [ m( ¹⁹⁸₇₉Au) - m( ¹⁹⁸₈₀Hg) - $m_e$ ] × 931.5 - 0.412

E(β₂⁻) = (1.372 - 0.412) MeV

E(β₂⁻)  = 0.960 MeV

Hope it helps :-)