Calcualte the excitation energy of the compound nuclei produced when
Given:
`{:(M(.^(235)U) = 235.0439 "amu"",",M(n) = 1.0087 "amu"","),(M(.^(238)U)238.0508 "amu,",M(.^(236)U) = 236.0456 "amu,"),(M(.^(239)U) = 239.0543 "amu",):}`.

Answer

`6.5MeV, 4.8 MeV`

Explanation

The two reaction are
`{:(n+.^(235)U rarr.^(236)Uast),(n+.^(238)U rarr.^(239)Uast):} `
We find the excitataion energy from the atomic masses. A thermal neutron has a negigible kinetic energy `(about 0.03 eV)`.
`E(.^(236)U^(ast)) =[m(n)+M(.^(235)U) -M(.^(236)U)]c^(2)`
`=[1.0087 a.m.u. +235.0439 a.m.u.`
`-236.0456 a.m.u.]c^(2)`
`=0.0070 xx 931.5 = 6.5 MeV`
`E(.^(239)Uast) =[m(n)+M(.^(238)U) -M(.^(239)U)]c^(2)`
`=[1.0087 a.m.u. +238.0508 a.m.u.`
`-239.0543 a.m.u.]c^(2)`
`=0.0052 xx 931.5 = 4.8 MeV`
Thus, `(.^(236)Uas)` has more excitation energy than `(.^(239)Uast)` wehn both are proeduced by thermal neutron absorption. This is why `(.^(235)U)` more easily undergoes thermal neutron fisssion.