Suppose India had a target of producing by 2020 AD, 200,000 MW of electric power , 10% of which was to be obtained from nuclear power plant. Suppose we are given that, on an average, the efficiency of utilisation (i.e., conversion to electric energy ) of thermal energy produced in a reactor was 25% . How much amount of fissionable uranium did our country need per year by 2020 ? Take the heat per fission of `.^235U` to be about 200 MeV. Avogadro's number `=6.023xx10^(-23) mol^(-1)` .

Explanation

Nuclear power target = 10% of total generation
=10% x 2 x `10^11` W
`=2xx10^10` W
Efficiency ,`eta`=25%
`therefore` Total power generated by the nuclear reactor `=(2xx10^(-10))/(25%)W=8xx10^10` W
`therefore` Generated heat by the reactor in 2020,
`H=8xx10^10xx366xx24xx60xx60J`
`therefore` Number of fission required for generation of this heat,
`N=H/(200xx1.6xx10^(-13))`
If m g of `._92^235U` contains this number of nucleus , then ,
`m=(Nxx235)/(6.023xx10^23)g`
`=(8xx10^10xx366xx24xx60xx60xx235)/(200xx1.6xx10^(-13)xx6.023xx10^23)` g
`=3.084 xx10^4` kg