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.

A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How muc

1.	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» Solution :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

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