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NCERT Class XII Chapter
Nuclei
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Question : 27 of 31
Marks: +1, -0
Consider the fission of 92238U{}^{238}_{92}\mathrm{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 58140Ce{}^{140}_{58}\mathrm{Ce} and 4499Ru{}^{99}_{44}\mathrm{Ru}. Calculate Q for this fission process. The relevant atomic and particle masses are:
m (92238U)({}^{238}_{92}\mathrm{U}) = 238.05079 u, m (58140Ce)({}^{140}_{58}\mathrm{Ce}) = 139.90543 u , m (4499Ru)({}^{99}_{44}\mathrm{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) + mnm_n – m (Ce-140) – m (Ru-99) c2c^2
Q = [238.05079 + 1.00867 – 139.90543 – 98.90594] amu × c2c^2
Q = 0.24809 × 931.5 MeV = 231.09 MeV = 231.1 MeV.
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