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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 reactor operaters 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. |
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Answer» Solution :In one act of fission, the ENERGY liberated is 200 MEV `therefore` NUMBER of atoms in 1g of `" "_(92)^(235)U = N_(A)/A= (6.023xx10^(23))/235` `therefore` Energy generated by 1g of `" "^(235)U =(200MeV xx6.023xx10^(23))/235=(200 xx 1.6xx 10^(-13) xx 6.023xx10^(23))/235 Jg^(-1)= 8.20 xx 10^(10) J g^(-1) = 8.20 xx 10^(13) Jkg^(-1)` As the generator operates only 80% of the time, in 5 years it will effectively operate for a time `f=5 xx80/100 year= 4 year = 4 xx 3.154 xx 10^(7)s` `therefore` Total energy generated by the reactor in 5 years E=P.t=`1000MWxx4xx3.154xx10^(7)s=1000xx10^(6)Wxx4xx3.154xx10^(7)J=12.616xx10^(16)J` `therefore` AMOUNT of `" "_(92)^(235)U` consumed`=(12.616xx10^(16))/(8.20xx10^(13))kg=1538kg` As the reactor consumes half of the fuel in 5 years, hence amount of fuel present in the reactor initially`=2xx1538=3076kg`. |
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