Suppose India has a target of producing by 2020AD, 200 GW of electric power, ten precent of which is to be obtained from nuclear power plants. Suppose the efficiency remained at `25%`, what amount of fuel may be required per year by 2020? Take the heat energy per fission of `.^(235)U` to be about 200 MeV.

Explanation

Nuclear power to be generated `=10% "of 200 hw=20 GW"`
At `25%` efficiency actual required `=(100)/(25)xx20GW=80GW`
Total energy required for 1 year production of power `=80xx10^(9)xx365xx8.64xx10^(4)J`
`"i.e., " E=2.523xx10^(18)J`
Energy per fission `=200 MeV`
`=200xx1.6xx10^(-13)J`
`=320.0xx10^(-13)J`
`E=3.2xx10^(-11)J`
Total number of fission atoms required `=(2.523xx10^(18))/(3.2xx10^(-11))=7.884xx10^(28)`
But `6.023xx10^(23)` atoms of Uranium have a mass of `235g`.
`therefore 7.884xx10^(30)` atoms of Uranium have a mass of xg.
`"i.e., " x=(235xx7.884xx10^(28))/(6.023xx10^(23))`
`x=307.6xx10^(5)`
Hence, `m=307.6xx10^(2) kg`
`=3.076xx10^(4)kg`
or `m=30.76` tons.