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NCERT Class XII Chapter
Nuclei
Questions With Solutions

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Question : 29
Total: 31
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 (
198
‌
79
Au) = 197.968233 u, m (
198
‌
80
Hg) = 197.966760 u

Solution:  
Energy corresponding to γ1
E1 = 1.088 – 0 = 1.088 MeV = 1.088 × 1.6 × 10–13 joule
∴ Frequency, υ (γ1) =
E1
h
 =
1.088×1.6×10−13
6.63×10−34
 = 2.626 × 1020 H
Similarly, υ (γ2) =
E2
h
 =
0.412×1.6×10−13
6.63×10−34
 = 9.95 × 1019 Hz
and υ (γ3) =
E3
h
 =
(1.088−0.412)×1.6×10−13
6.63×10−34
= 1.631 × 1020 Hz
Maximum K.E. of β1 particle
Kmax(β1) = [m (
198
‌
79
Au) - mass of second excited state of
198
‌
80
Hg] × 931 MeV
=
[m(
198
‌
79
Au)−m(
198
‌
80
Hg)−
1.088
931
]
× 931 MeV
= 931 [197.968233 – 197.966760] – 1.088 MeV
= 1.371 – 1.088 = 0.283 MeV
Similarly, Kmax(β2) = 0.957 MeV.
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