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NCERT Class XII Chapter
Nuclei
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Question : 18 of 31
Marks: +1, -0
A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much  92235U\,{}^{235}_{92}U did it contain initially? Assume that the reactor operates 80% of the time and that all the energy generated arises from the fission of  92235U\,{}^{235}_{92}U and that this nuclide is consumed by the fission process.
Solution:  
In the fission of one nucleus of  92235U\,{}^{235}_{92}U , energy generated is 200 MeV.
Energy generated in fission of 1 kg of  92235U\,{}^{235}_{92}U
= 200 × 6.023×1023235\frac{6.023\times10^{23}}{235} × 1000 N = 5.106 × 102610^{26} MeV
= 5.106 × 102610^{26} × 1.6 × 10−1310^{-13} J = 8.17 × 101310^{13} J
Time for which reactor operates = 80100\frac{80}{100} × 5 yr 4 yr.
Total energy generated in 5 year = 1000 × 10610^{6} × 60 × 60 × 24 × 365 × 4 J
∴ Amount of  92235U\,{}^{235}_{92}U consumed in 5 years
=
1000×106×60×60×24×365×48.17×1013\frac{1000\times10^{6}\times60\times60\times24\times365\times4}{8.17\times10^{13}}
Kg = 1544 kg
∴ Initial amount of  92235U\,{}^{235}_{92}U = 2 × 1544 kg = 3088 kg
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