In a nuclear reactor, `""^(235)U` undergoes fission liberating 200 MeV of energy. The reactor has a 10% efficiency and produces 1000 MW power. If the reactor is to function for 10 years, find the total mass of uranium required.

Explanation

The reactor produces 1000 MW power or `10^(9)` W power or `10^(9)Js^(-1)` of power. The reactor is to function for 10 years. Therefore, total energy which the reactor will supply in 10 years is
E = (Power) (time)
= `(10^(9)Js^(-1))(10xx365xx24xx3600s)`
`=3.1536xx10^(17)J`
But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or `3.1536xx10^(18)J`. One uranium atom liberates 200 MeV of energy or `200xx1.6xx10^(-13)Jor3.2xx10^(-11)J` of energy. So number of uranium atoms needed are
`(3.1536xx10^(18))/(3.2xx10^(-11))=0.9855xx10^(29)`
or number of kg-moles of uranium needed are
`n=(0.9855xx10^(29))/(6.02xx10^(26))=163.7`
Hence total mass of uranium required is
m = (n) M = (163.7) (235) kg
or m = 38470 kg
or m = `3.847xx10^(4)` kg