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

Explanation

The nuclear reaction is
`""_(3)Li^(6) + ""_(0)n^(1) to ""_(2)He^(4) + ""_(1)H^(3) + DeltaE`
Initial masses :
`(m""_(3)Li^(6)) = 6.015126` amu
`(""_(0)n^(1)m) = 1.0086654` amu
------------------------------------
Total = 7.0237914 amu
From masses:
`(""_(2)He^(4)m) = 4.0026044 amu`
`m(""_(1)H^(3))= 3.016049` amu
...........................................
Mass defect,
`Deltam = (7.0237914 - 7.0186534)` atm
= 0.005138 amu
Energy released,
`E = Deltam xx 931 MeV`
`= 0.005138 xx 931` MeV = 4.78 MeV