A 1000 MW fission reactor consumes half of its fuel in 5 y. How much `._92^235U` did it contain initially ? Assume that the reactor was active 80% of the time and all the energy generated arises from the fission of `._92^235U` and that this nuclide is consumed by the fission process.

Explanation

Energy generated per gram of `._92^235U=(6.023xx10^23)/235 xx 200xx1.6xx10^(-13) J .g^(-1)`
Energy generated in 5 y = power x time x 80%
`=(1000xx10^6) xx (5xx365xx24xx60xx60)xx80%` J
`therefore` Amount of spent `._92^235U=((1000xx10^6)xx(5xx365xx24xx60xx60)xx0.8xx235)/(6.023xx10^23xx200xx1.6xx10^(-13))g`
=1538 kg
`therefore` Initial mass of `._92^235U` = 2 x 1538 kg = 3076 kg