Question

Suppose India has a target of producing by `2020 AD, 200,000 MW` of electric power, ten percent of which was to be obtained from nuclear power plants. Suppose we are given that, on an avedrage, the efficiency of utilization(i.e conversion to electric energy) of thermal energy produced in a reactor was `25%`. How much amount of fissionable uranium would our country need per year by `2020`? Take the heat energy per fission of `.^(235)U` to be about `200 MeV`.

Answer 1

Amount of electric power to generated , `P=2xx10^(5)MW`
10% of this amount has to be obtained from nuclear power plants.
`therefore` Amount of nuclear power, `P_(1)=(10)/(100)xx2xx10^(5)`
`=2xx10^(4)MW`
`=2xx10^(4)xx10^(6)J//s`
`2xx10^(10)xx60xx60xx24xx365J//y`
Heat energy released per fission of a `""^(235)U` nucleus, E=200 MeV
Efficienty of a reactor `=25 %`
Hence, the amount of energy converted into the electrical energy per fission is calculated as:
`(25)/(100)xx200=-50MeV`
`=50xx1.6xx10^(-19)xx10^(6)=8xx10^(-12)J`
Number of atoms required for fission per year:
`(2xx10^(10)xx60xx60xx24xx365)/(8xx10^(-12))=78840xx10^(24) " atoms"`
1 mole, i.e, 235 g of `U^(235)` contains `6.023xx10^(23)` atoms.
`therefore` Mass of `6.023xx10^(23)` atoms of `U^(235)=235g=235xx10^(-3)kg`
`therefore` Mass of `78840xx10^(24)` atoms of `U^(235)`
`=(235xx10^(-3))/(6.023xx10^(23))xx78840xx10^(24)`
`=3.076xx10^(4)kg`
Hence, the mass of uranium needed per year is `3.076xx10^(4) kg`.