InterviewSolution
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InterviewSolution
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In a nuclear reactor `.^235U` undergoes fission liberating `200 MeV` of energy. The reactor has a `10%` efficiency and produces `1000 MW` power. If the reactor is to function for `10 yr`, find the total mass of uranium required. |
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Answer» Total energy produced by the reactor in time `t = 10 years` `E =1000 xx 10^(6) xx 10 xx 3.15 xx 10^(7) J` `=3.15 xx 10^(7) J` `Efficiency = ("output energy")/("input energy")` ` rarr` Input energy caused by fission `=(output energy)/(efficiency)` `=(3.15 xx10^(17))/((10//100))=3.15xx10^(18) J` Energy produced by `1` fission of `.^(135)U=200 MeV` `=200 xx1.6xx10^(-13) J` ` =3.2 xx 10^(-11) J` Therefore, Number of fissions required `=(Total energy )/(Energy per fission)` `=(3.15 xx 10^(18))/(3.2 xx10^(28))` `~~9.8 xx 10^(28)` Hence, mass of nranium required is given by ` m=(N)/(N_(a)) xx 235 kg =(9.8xx 10^(28))/(6.02 xx 10^(26))` `=38.2 xx 10^(3) kg`. |
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