In a nuclear reactor `^235U` 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 yr, find the total mass of uranium required.

Answer

A, C, D

Explanation

The reactor produces 1000 MW power or `10^9J//s`. The reactor is to function for 10 yr. Therefore, total energy which the reactor will supply in 10 yr is
`E=(power)(time)`
`=(10^9J//s)(10xx365xx24xx3600s)`
`=3.1536xx10^17J`
But since the efficiency of the reactor is only 10%, therefore actual energy needed is 10 times of it or `3.1536xx10^18J`. One uranium atom liberates 200 MeV of energy or `200xx1.6xx10^-13J` or
`3.2xx10^-11J` 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~~38470kg`
or `m=3.847xx10^4kg`