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NCERT Class XII Chapter
Nuclei
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Question : 29 of 31
Marks: +1, -0
Obtain the maximum kinetic energy of β-particles and the radiation frequencies of γ-decays in the decay scheme shown in figure. You are given that m (79198Au)({}^{198}_{79}\mathrm{Au}) = 197.968233 u, m (80198Hg)({}^{198}_{80}\mathrm{Hg}) = 197.966760 u
Solution:  
Energy corresponding to γ1\gamma_1
E1E_1 = 1.088 – 0 = 1.088 MeV = 1.088 × 1.6 × 10−1310^{-13} joule
∴ Frequency, υ (γ1)(\gamma_1) = E1h\frac{E_1}{h} = 1.088×1.6×10−136.63×10−34\frac{1.088 \times 1.6 \times 10^{-13}}{6.63 \times 10^{-34}} = 2.626 × 102010^{20} H
Similarly, υ (γ2)(\gamma_2) = E2h\frac{E_2}{h} = 0.412×1.6×10−136.63×10−34\frac{0.412 \times 1.6 \times 10^{-13}}{6.63 \times 10^{-34}} = 9.95 × 101910^{19} Hz
and υ (γ3)(\gamma_3) = E3h\frac{E_3}{h} =
(1.088−0.412)×1.6×10−136.63×10−34\frac{(1.088-0.412) \times 1.6 \times 10^{-13}}{6.63 \times 10^{-34}}
= 1.631 × 102010^{20} Hz
Maximum K.E. of β1\beta_1 particle
Kmax(β1)K_{\text{max}} (\beta_1) = [m (79198Au)({}^{198}_{79}\mathrm{Au}) - mass of second excited state of 80198Hg{}^{198}_{80}\mathrm{Hg}] × 931 MeV
=
[m(79198Au)−m(80198Hg)−1.088931]\left[ m({}^{198}_{79}\mathrm{Au}) - m({}^{198}_{80}\mathrm{Hg}) - \frac{1.088}{931} \right]
× 931 MeV
= 931 [197.968233 – 197.966760] – 1.088 MeV
= 1.371 – 1.088 = 0.283 MeV
Similarly, Kmax(β2)K_{\text{max}} (\beta_2) = 0.957 MeV.
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