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)`
Fission rate = Number of atoms fissioned in one second = `(5.12 xx 10^(24))/(30 xx 24 xx 60 xx 60) = 1.975 xx 1018 s^(-1)`
Each fission gives 185 MeV . Hence , energy obtained in one second .
`P = 185 xx 1.975 xx 10^(18) xx Me V s^(-1)`
`= 185 xx 1.975 xx 10^(18) xx 1.6 xx 10^(-19) J s^(-1)`