A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much `""_92^235"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"U"` and that this nuclide is consumed only by the fission process.
Solution
Half life of the fuel of the fission reactor, `"t"_(1/2)` = years
= 5 × 365 × 24 × 60 × 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 × 1023 atoms.
∴1 g `""_92^235 "U"` contains `(6.023 xx 10^23)/235`
The total energy generated per gram of `""_92^235 "U"` is calculated as:
`"E" = (6.023 xx 10^23)/235 xx 200 " MeV/g"`
`= (200 xx 6.023 xx 10^23 xx 1.6 xx 10^(-19) xx 10^6)/235`
= 8.20 × 1010 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:
`= (5 xx 80 xx 60 xx 60 xx 365 xx 24 xx 1000 xx 10^6)/(100 xx 8.20 xx 10^10)`
`~~ 1538 "kg"`
∴ Initial amount of `""_92^235"U"` = 2 × 1538 = 3076 kg
