Suppose India has a target of producing by 2020 A.D., `2xx10^(5)MW` of electric power, ten percent of which is to be obtained form nuclear power plants. Suppose we are given that on an average, the efficiency of utilization (i.e., conservation to electrical energy) of thermal energy produced in a reactor is `25%`. How much amount of fissionable uranium will our country need per year? Take the heat energy per fission `U^(235)` to be about 200MeV.

Explanation

Total energy power =`2xx10^(5)MW`, Total Nuclear Power `=10%` of `2xx10^5 MW2xx10^4 MW`.
Energy produced/fission `=200MeV`, Efficiency of power plant`=25%`
`:.` Energy converted into electrical energy per fission `=25/100xx200=50MeV=50xx1.6xx10^(-13)"joule"`
Total electrical energy to be produced `=2xx10^4 MW =2xx10^4xx10^6 "watt"=2xx10^(10) "joule"//sec`
`=2xx10^(10)xx60xx60xx24xx365 "joule"//"years"`
No. of fission in one year `=(2xx10^(10)xx60xx60xx24xx365)/(50xx1.6xx10^(-13))=2xx(36xx24xx365)/8xx10^(24)`
Mass of `6.023xx10^(23)` atoms of `U^(235)=235gm.=235xx10^(-3)kg`
Mass of `(2xx36xx24xx365)/8xx10^(24)"atoms"=(235xx10^(-3))/(0.023xx10^(23))xx(2xx36xx24xx365xx20^(24))/8=3.08xx10^(4) kg`.
Hence, mass of uranium needed per year=`3.08xx10^(4) kg`