A 1000 MW fission reactor consumes half of its fuel in 5.00y. How much `._92U^(235)` did it contain initally? Assume that the reactor operates `80%` of the time and that all the energy generated arises form the fission of `._92U^(235)` and that this nuclide is consumed by the fission process.

Explanation

In the fission of one nucleus of `._92U^(235)` , energy generated is 200 MeV.
`:.` Energy generated in fission of 1kg of `._92U^(235)=200xx(6xx10^(23))/235xx1000MeV`
`= 5.106xx10^(26)MeV=5.106xx10^(26)xx1.6xx10^(-13)J=8.17xx10^(13)J`
Time for which reactor operates `=80/100xx5yr=4yr`
Total energy generated in 5 years by 1000MW generator `=1000xx10^6xx60xx60xx24xx365xx4J`
`:.` Amount of `U._(92)^(235)` consumed in 5 years `=(1000xx10^6xx60xx60xx24xx365xx4)/(8.17xx10^(13))kg=1544kg`
`:.` Initial amount of `._92U^(235)=2xx1544kg=3088kg`