# Q. 13.18 A $1000\; MW$ fission reactor consumes half of its fuel in$5.00 \; y$. 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.

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$\times$1537.8

=3075.6 kg

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