A 1000 MW fission reactor consumes half of its fuel in 5.00yr. How much `._92U^(235)` did it contain initially? 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

Half life of the fission reactor `t_((1)/(2))=5`
`=5xx365xx24xx60xx60s`
we know that fission of 1 g of `""_(92)^(235)U` uncleus , the energy relased is equal to 200 MeV .
1 mole i.e ., 235 g of `""_(92)^(235)U` contains `5.023xx10^(23)` atoms .
` therefore 1 g ""_(92 )^(235)U` contains `(6.023xx10^(23))/(235)` atoms
the total energy generated per gram of `""_(92)^(235)U` is calcularted as :
`E=(6.023xx10^(23))/(235) xx200MeV//g`
`=(200xx6.023xx10^(23)xx1.6xx10^(-19)xx10^(6))/(235)=8.20xx10^(10) j/g`
the reactor opertes only 80 % of the time .
Hence , the amount of `""_(92)^(235)U` consume d in 5 years by the 1000 MW fission reactor is calculated as :
`=(5xx80xx60xx60xx365xx24xx1000 xx10^(6))/(100xx8.20xx10^(10) )g`
`~~1538 kg`
`therefore amount of `""_(92)^(235)U=2xx1538=3076 kg`