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Assuming that energy released by the fission of a single ""_(92)^(235)U nucleus is 200MeV, calculate the number of fissions per second required to produce 1 kilowatt power. |
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Answer» <P> Solution :The fission of a single `""_(92)^(235)"U"` nucleus RELEASES 200 MEV of energyEnergy released in the fission is given by the formula, `E = (Pt)/(n) Rightarrow (n)/(t) = (P)/(E)` `E = 200 MeV = 200 xx 10^(6) xx 1.6 xx 10^(-19)` `E = 320 xx 10^(-13)` `E = 3.2 xx 10^(-11)`J `(n)/(t) = (P)/(E) = (1)/(3.2 xx 10^(-11)) = 0.3125 xx 10^(11) = 3.125 xx 10^(10)` `(n)/(t) = 3.125 xx 10^(10)` |
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