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

`3.847 xx 10^(4) kg`

Explanation

The reactor produces 1000 MW power or `10^(9) W` power of `10^(9) J//s` of power. The reactor is to function for `10 yr`. Therefore, total energy which the reactor will supply in `10 yr` is `E=("power") ("time")`
`=(10^(9) J//s)(10xx365xx24xx3600 s)=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` energy or `200xx1.6xx10^(-13)` or `3.2xx10^(-11)J` of energy. So, number of uranium atoms needed are
`n=(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`
`implies m ~~38470 kg implies m=3.847xx10^(4) kg`