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Consider one of the fission reactions of...

Consider one of the fission reactions of `U^(235)` by thernmal neutrons :
`._(92)U^(235) + n rarr ._(38)Sr^(94) + ._(54)Ce^(140) + 2n`
The fission fragments are, however, not stable. They unaergo successive `beta-`decays unit `._(38)Sr^(94)` becomes `._(40)Zr^(94)` and `._(54)Xe^(140)` becomes `._(58)Ce^(140)`. Estimate the total energy released in the process. Is all that energy available as kinetic energy of the fission products (Zr and Ce)? You are given that
`m (U^(235)) = 255.0439 am u," " m_(n) = 1.00866 am u " " m(Zr^(94)) = 93.9065 am u, " " m(Ce^(140)) = 139.9055 am u`

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Consider one of fission reactions of ^(235)U by thermal neutrons ._(92)^(235)U +n rarr ._(38)^(94)Sr +._(54)^(140)Xe+2n . The fission fragments are however unstable and they undergo successive beta -decay until ._(38)^(94)Sr becomes ._(40)^(94)Zr and ._(54)^(140)Xe becomes ._(58)^(140)Ce . The energy released in this process is Given: m(.^(235)U) =235.439u,m(n)=1.00866 u, m(.^(94)Zr)=93.9064 u, m(.^(140)Ce) =139.9055 u,1u=931 MeV] .

Calculate the energy released by fission from 2 gm of ._92U^235 in KWH. Given that the energy released per fission is 200 Mev.

A thermal neutron strikes U_(92)^(235) nucleus to produce fission. The nuclear reaction is as given below : n_(0)^(1) + U_(92)^(235) to Ba_(56)^(141) + Kr_(36)^(92) +3n_(0)^(1) + E Calculate the energy released in MeV. Hence calculate the total energy released in the fission of 1 Kg of U_(92)^(235) . Given mass of U_(92)^(235) = 235.043933 amu Mass of neutron n_(0)^(1) =1.008665 amu Mass of Ba_(56)^(141)=140.917700 amu Mass of Kr_(36)^(92)=91.895400 amu