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

Binding energy for the free particles neutron is assumed to be zero. Binding energy of `""_(92)U^(235)` nucleus is `7.2 xx 235 =1,692` MeV.
Binding energy of two `""_(46)Pd^(126)` nuclei is `2 xx 8.2 xx 116 = 1902.4` MeV.
Energy emitted as `gamma`-rays is `2 xx 5.2 = 10.4` MeV
Hence, energy released in one fission of `""_(92)U^(235)` can be written as follows:
Number of fission per second `=(1,600 xx 10^(8))/(3.2 xx 10^(-11)) = 5 xx 10^(19)`
Hence, we requires `5 xx 10^(19) U^(235)` nuclei per second. So required mass per second can be written as follows:
`M.. = (5 xx 10^(19))/(6.02 xx 10^(23)) xx 235 = 195.18 xx 10^(-4)`g
Required mass per hour can be written as follows:
`M. = 195.18 xx 10^(-4) xx 3,600 = 70.26`g
Since reactor efficiency is 50%, required amount of uranium will be double to the above calculated value. Mass of uranium required can be written as follows:
`M = 2 xx 70.26 = 140.52`g