What is the power output of a `._(92) U^(235)` reactor if it is takes `30` days to use up `2 kg` of fuel, and if each fission gives `185 MeV` of usable energy ?.

Options

A. `45` megawatt
B. `58.46` megawatt
C. `72` megawatt
D. `92` megawatt

Answer

b

Explanation

Number of atoms in `2 kg` fuel
`(2)/(235) =6.02 xx 10^(26) =5.12 xx 10^(24)`
Fissions rate =Number of atoms fissioned in one second
`(5.12 xx 10^(24))/(30 xx 24 xx 60 xx 60)`
`=1.975 xx 10^(18)s^(-1)`
Each fission gives `185 MeV`. Hence, energy obatined in one second,
`P=185xx 1.975 xx 10^(18)MeV s^(-1)`
`=185 xx 1.975 xx 10^(18) xx 1.6 xx 10^(-19) J s^(-1)` .