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A $1000 \;MW$ fission reactor consumes half of its fuel in $5.00\; y$. How much $_{92}^{235} U$ (in $kg$) did it contain initially? Assume that the reactor operates $80 \%$ of the time, that all the energy generated arises from the fission of $_{92}^{235} U$ and that this nuclide is consumed only by the fission process.
$3076$
$1538$
$2446$
$3486$
Solution
Half life of the fuel of the fission reactor, $t_{\frac{1}{2}}=5$ years $=5 \times 365 \times 24 \times 60 \times 60 s$
We know that in the fission of $1\; g$ of $_{92}^{235} U$ nucleus, the energy released is equal to $200 MeV$.
$1$ mole, i.e., $235 \;g$ of $_{92}^{235} U$ contains $6.023 \times 10^{23}$ atoms.
$\therefore 1 \;g\;\; _{92}^{235} U \quad \frac{6.023 \times 10^{23}}{235}$ atoms contains
The total energy generated per gram of $_{92}^{235} U$ is calculated as:
$E=\frac{6.023 \times 10^{23}}{235} \times 200 MeV / g$
$=\frac{200 \times 6.023 \times 10^{23} \times 1.6 \times 10^{-19} \times 10^{6}}{235}=8.20 \times 10^{10} J / g$
The reactor operates only $80 \%$ of the time. Hence, the amount of $_{92}^{235} U$ consumed in $5$ years by the $1000 MW$ fission reactor is calculated as
$\frac{5 \times 80 \times 60 \times 60 \times 365 \times 24 \times 1000 \times 10^{6}}{100 \times 8.20 \times 10^{10}}\, g$
$\approx 1538 \,kg$
Initial amount of $_{92}^{235} U=2 \times 1538=3076\, kg$
