InterviewSolution
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InterviewSolution
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A nuclear reactor generates power at 50% efficiency by fission of `._(92)^(235)U` into two equal fragments of `._(46)^(116)U` into two equal fragments of `._(46)^(116)Pd` with the emission of two gamma rays of 5.2 MeV each and three neutrons. The average binding energies per particle of `._(92)^(235)U` and `._(46)^(116)Pd` are 7.2 MeV and 8.2MeV respectiveley. Calculate the energy released in one fission event. Also-estimate the amount to `.^(235)U` consumed per hour to produce 1600 megawatt power. |
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Answer» `._(92)^(235)Urarr2._(46)^(116)Pd +3._(0)^(1)n+2gamma` `B.E.` of `._(92)^(235)U = 235xx7.2 = 1692MeV` `B.E.` of `._(46)^(116)Pd = 2[116 xx 8.2] = 1902.4MeV` Energy of `2gamma` photons `= 2xx5.2 = 10.4MeV` Enegry released per fission `= 1692-[1902.4+10.4]` `= 200MeV` NUmber of atoms in `m gram` of `._(92)^(235)U` is `(m)/(235) xx6.023xx10^(23)` Energy produced due to fission of `m` gram of `._(92)^(235)U` is `(m)/(235)xx6.023xx10^(23)xx200xx1.6xx10^(-13) J` Output power `= 1600MW = 1600 xx 10^(6) = 1.6 xx 10^(9)J//sec` Input power `= (1.6xx10^(9))/((50//100)) = 3.2 xx 10^(9)J//sec` Input energy `= 3.2 xx 10^(9)xx60xx 60J` `(m)/(235)xx6.023xx106(23)xx200xx1.6xx10^(-13)` `= 3.2 xx 10^(9) xx 3600` `m = 140.5 g` |
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