Consider the fission of `""_(92)^(238)U` by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are `""_(58)^(140)Ce` and `""_(44)^(99)Ru`. Calculate Q for this fission process. The relevant atomic and particle masses are
`m(""_(92)^(238)U) =238.05079 u`
`m( ""_(58)^(140)Ce ) =139.90543 u`
`m(""_(44)^(99)Ru ) = 98.90594 u`

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

We are given the following atomic masses: ""_(92)^(238) U = 238.05079 u " " _(2)^(4)He = 4.00260 u ""_(90)^(234)Th = 234.04363 u" "_(1)^(1)H= 1.00783 u ""_(91)^(237)Pa = 237.05121 u Here the symbol Pa is for the element protactinium (Z = 91). (a) Calculate the energy released during the alpha decay of ""_(92)^(238)U . (b) Show that ""_(92)^(238)U can not spontaneously emit a proton.

Calculate the energy released in the following nuclear reaction and hence calculate the energy released when 235 gram of uranium-235 undergoes fission. U_(92)^(235) + n_(0)^(1) to Kr_(36)^(92) + Ba_(56)^(141) + 3n_(0)^(1) Rest masses of U^(235), Ba^(141) ,Kr^(92) and neutron are 235.04390 amu, 140.91390 amu, 91.89730 amu and 1.00867 amu respectively. Avogadro number = 6.023 xx 10^(23) .