# A 1000 MW fission reactor consumes half of its fuel in 5.00 y.

Question:

A $1000 \mathrm{MW}$ fission reactor consumes half of its fuel in $5.00 \mathrm{y}$. How much ${ }_{92}^{235} \mathrm{U}$ did it contain initially? Assume that the reactor operates $80 \%$ of the time, that all the energy generated arises from the fission of ${ }_{92}^{235} \mathrm{U}$ and that this nuclide is consumed only by the fission process.

Solution:

Half life of the fuel of the fission reactor, $t_{\frac{1}{2}}=5$ years

= 5 × 365 × 24 × 60 × 60 s

We know that in the fission of $1 \mathrm{~g}$ of ${ }_{92}^{235} \mathrm{U}$ nucleus, the energy released is equal to $200 \mathrm{MeV}$.

1 mole, i.e., $235 \mathrm{~g}$ of ${ }_{92}^{235} \mathrm{U}$ contains $6.023 \times 10^{23}$ atoms.

$\therefore 1 \mathrm{~g}{ }_{92}^{235} \mathrm{U}$ contains $\frac{6.023 \times 10^{23}}{235}$ atoms

The total energy generated per gram of ${ }_{92}^{235} \mathrm{U}$ is calculated as:

$E=\frac{6.023 \times 10^{23}}{235} \times 200 \mathrm{MeV} / \mathrm{g}$

$=\frac{200 \times 6.023 \times 10^{23} \times 1.6 \times 10^{-19} \times 10^{6}}{235}=8.20 \times 10^{10} \mathrm{~J} / \mathrm{g}$

The reactor operates only 80% of the time.

Hence, the amount of ${ }_{92}^{235} \mathrm{U}$ consumed in 5 years by the $1000 \mathrm{MW}$ fission reactor is calculated as:

$=\frac{5 \times 80 \times 60 \times 60 \times 365 \times 24 \times 1000 \times 10^{6}}{100 \times 8.20 \times 10^{10}} \mathrm{~g}$

$\approx 1538 \mathrm{~kg}$

$\therefore$ Initial amount of ${ }_{92}^{235} \mathrm{U}=2 \times 1538=3076 \mathrm{~kg}$