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NCERT Class XII Chapter
Nuclei
Questions With Solutions

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Question : 27
Total: 31
Consider the fission of
238
‌
92
U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments are
140
‌
58
Ce and
99
‌
44
Ru. Calculate Q for this fission process. The relevant atomic and particle masses are:
m (
238
‌
92
U) = 238.05079 u, m (
140
‌
58
Ce) = 139.90543 u , m (
99
‌
44
Ru) = 98.90594 u
Solution:  
The fission of U-238 by fast neutrons into fragments Ce-140 and Ru- 99 with energy released Q,
The Q value,
Q = [m (U-238) + mn – m (Ce-140) – m (Ru-99) c2
Q = [238.05079 + 1.00867 – 139.90543 – 98.90594] amu × c2
Q = 0.24809 × 931.5 MeV = 231.09 MeV = 231.1 MeV.
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