A 1000 MW fission reactor consumes half of its fuel in 5 yr. 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.

Given power of reactor P = 1000MW

We will use concept that the energy generated in one fission of ${}*{92}^{235}U$ is 200 MeV.
Number of ${}*{92}^{235}U$ attom in 1 g = 1/235 x 6.023 x ${{10}^{23}}$

Energy generated per gram of ${}

*{92}^{235}U$ =(1/235 x 6.023x${{10}^{23}}$ x 200 x 1.6x${{10}^{-13}}$*

Total energy generated in 5 year with 80% of the time = 1000 x ${{10}^{6}}$x5x365x24x60x60x80/100

Mass of ${}{92}^{235}U$ consumed in 5 yr,

Total energy generated in 5 year with 80% of the time = 1000 x ${{10}^{6}}$x5x365x24x60x60x80/100

Mass of ${}

m = Total energy/Energy consumed per gram

= 1.538 x ${{10}^{6}}$ g = 1538 kg

Initial amount of ${}_{92}^{235}U$ = 1544 x 2 kg = 3076 kg