Question

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

Answer 1

Half life of the fuel of the fission reactor,  ${\displaystyle {\text{t}}_{\frac{1}{2}}}$ =  years

= 5 × 365 × 24 × 60 × 60 s

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

1 mole, i.e., 235 g of ${\displaystyle {}_{92}^{235}\text{U}}$  contains 6.023 × 10²³ atoms.

∴1 g ${\displaystyle {}_{92}^{235}\text{U}}$ contains ${\displaystyle \frac{6.023\times {10}^{23}}{235}}$

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

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

${\displaystyle =\frac{200\times 6.023\times {10}^{23}\times 1.6\times {10}^{-19}\times {10}^6}{235}}$

= 8.20 × 10¹⁰ J/g

The reactor operates only 80% of the time.

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

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

${\displaystyle \approx 1538\text{kg}}$

∴ Initial amount of ${\displaystyle {}_{92}^{235}\text{U}}$ = 2 × 1538 = 3076 kg