Q. 13.18 A 1000\; MW fission reactor consumes half of its fuel in5.00 \; y. How much _{92}^{235}\textrm{U} did it contain initially? Assume that the reactor operates 80\; ^{}o/_{0} of the time, that all the energy generated arises from the fission of _{92}^{235}\textrm{U} and that this nuclide is consumed only by the fission process.

Answers (1)
S Sayak

The amount of energy liberated on fission of 1  _{92}^{235}\textrm{U} atom is 200 MeV.

The amount of energy liberated on fission of 1g  _{92}^{235}\textrm{U} 

\\=\frac{200\times 10^{6} \times 1.6\times 10^{-19}\times 6.023\times 10^{23}}{235}\\=8.2\times 10^{10}\ Jg^{-1}

Total Energy produced in the reactor in 5 years

\\=1000\times 10^{6}\times 0.8\times 5\times 365\times 24\times 3600\\ =1.261\times 10^{17}\ J

Mass  of _{92}^{235}\textrm{U} which underwent fission, m

=\frac{1.261\times 10^{17}}{8.2\times 10^{10}}

=1537.8 kg

The amount present initially in the reactor = 2m

=2\times1537.8

=3075.6 kg

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