In a nuclear reactor `U^(235)` undergoes fission releasing energy 100 MeV. The reactor has 20% efficiency and the power produces is 2000 MW. If the reactor is to function for 5 yr, find the total mass of uranium required.

Explanation

Given, output power, `P_0 = 2000 MW = 2 xx 10^9 W `
Efficiency, `eta = ("Output power"P_o)/("Input power " P_i)`
`implies (20^3)/100 = (2 xx 10^9)/(P_i)`
`implies P_9 = 10^10 W`
Consider the mass of uranium be m.
Number of fissions = Number of atoms
`= ("Given mass ")/("Atomic mass") xx` Avogadro number
Energy produced,
E = Number of fissions `xx` Energy per fission
`= m/235 xx 6.023 xx 10^23 xx 2000 MeV`
`= m/(235) xx 6.023 xx 10^(23) xx 2000 xx 1.6 xx 10^(-13) J`
`(because 1MeV = 16 xx 10^(-13)J)`
`= m xx 0.041 xx 10^(12) J`
`= m xx 4.1 xx 10^10 J`
Power = `("Energy")/("Time")`
`implies 10^10 = (m xx 4.1 xx 10^10)/(5 xx 365 xx 24 xx 3600)`
`m = 3.84 xx 10^7 g = 38.4 xx 10^3 kg`