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.

Explanation

`._(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`