Question

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How much `._(92)^(235)U` did it contain initially ? Assume that the reacotr operates 80% of the time that all the energy generated arises from the fission of `._(92)^(235)U` and that this nuclide is consumed only by the fission process.

Answer 1

In the fission of one nucles of `._(92)U^(235)` energy generated is 200 MeV
`therefore` Energy generated in fission of 1 kg of
`._(92)U^(235)=200xx(6xx10^(23))/(235)xx1000` MeV
`= 5.106xx10^(26)MeV = 5.106xx10^(26)xx1.6xx10^(-13)J`
`= 8.17xx10^(3)J`
Time for which reactor operates `(80)/(100)xx5`
years = 4 years.
Total energy generated in 5 years.
`= 1000xx10^(6)xx60xx60xx24xx365xx4J`
`therefore` Amount of `U_(92)^(235)` consumed in 5 years
`= (1000xx10^(6)xx60xx60xx24xx365xx4)/(8.17xx10^(13))kg`
= 1544 kg
`therefore` Initial amount of `._(92)U^(235)=2xx1544 kg`
= 3088 kg