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. 56.3 MW
B. 60.3 MW
C. 58.3 MW
D. 54.3 MW

Answer

C

Explanation

Mass of `._(92)^(235)U` per second in the reactor is
`m=(2xx10^(3))/(30xx24xx60xx60)=7.72xx10^(-4)` g/sec
`therefore` Number of fissions reaction per second
`=(6xx10^(23))/(235)xx m = (6xx10^(23)xx7.72xx10^(-4))/(235)`
`=1.97xx10^(18)//sec`
Power of nuclear reactor `= 1.97xx10^(18)xx185` MeV/s
`= 1.97xx10^(18)xx185xx10^(6)xx1.6xx10^(-19) J//s`
`= 58.3 MW`.