A slow neutron strikes a nucleus of `._(92)U^(235)` splitting it into lighter nuclei of barium and krypton and releasing three neutrons. Write the corresponding nuclear reaction. Also calculate the energy released on this reaction
Given `m(._(92)U^(235))=235.043933 "amu" m(._(0)n^(1))=1.008665` amu
`m(._(56)Ba^(141))=140.917700 "amu" m(._(36)Kr^(92))=91.895400` amu


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Correct Answer - `198.77` MeV