Question

A nuclear reactor using `.^(235)U` generates `250 MW` of electric power. The efficiency of the reactor (i.e., efficiency of conversion of thermal energy into electrical energy) is `25%`. What is the amount of `.^(235)U` used in the reactor per year? The thermal energy released per fission of `.^(235)U` is `200 MeV`.

Answer 1

Rate of electrical energy generration is `250 MW =250 xx 10^(6) W(or J s^(-1))`. So, electrical energy generation is `250 MW =250 xx 10^(6) W(or J s^(-1))`. Therefore, electrical energy generated in `1 year` is
`(250 xx 10^(6) J s^(-1)) xx (365 xx 24 xx 60 xx 60 s)=7.884 xx 10^(15) J` Thermal eneregy from fission of one `.^(235)U` nuvleus is `200 MeV =200 xx 1.6 xx 10^(-13) = 3.2 xx 10^(-11J` . Since the efficiency is `25%`, the electrical energy obtained from the fission of one `.^(235)U` nucleus is
`E_(1)=3.2 xx 10^(-11) xx(25)/(100) =8.0 xx 10^(-12) J`
Therefore, the number of fission of `.^(235)U` required in one year will be
`N=(7.884 xx10^(15))/(8.0 xx 10^(-12) =9.855 xx 10^(26))`
Number of moles of `.^(235)U` required per year is
`n=(9.855 xx 10^(26))/(6.023 xx 10^(23)) =1.636 xx10^(3)`
Therefore, mass of `.^(235)U` required per year is
`m=1.636 xx 10^(3) xx 235`
`=3.844 xx 10^(5) g=384.4 kg`.