Calculate the ground state `Q` value of the induced fission raction in the equation
`n+._(92)^(235) Urarr._(92)^(236)Uastrarr._(40)^(99)Zr+._(52)^(134)Te+2n`
If the neutron is thermal. A thermal neutorn is in thermal equilibrium with its envitronmnet, it has an avergae kinetic energy given by `(3//2) kT`. Given :
`m(n) =1.0087 am u, M(.^(235)U)=235.0439) am u`,
`M(.^(99)Zr)=98.916 am u, M(.^(134)Te)=133.9115 am u`.

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] .

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",):} .

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",):} .

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

Show that ._(92)^(230)U does not decay by emitting a neutron or proton. Given: M(._(92)^(230)U)=230.033927 am u, M(._(92)^(230)U)=229.033496 am u , M(._(92)^(229)Pa)=229.032089 am u, M(n)=1.008665 am u m(p)=1.007825 am u .

Show that ._(92)^(230)U does not decay by emitting a neutron or proton. Given: M(._(92)^(230)U)=230.033927 am u, M(._(92)^(230)U)=229.033496 am u , M(._(92)^(229)Pa)=229.032089 am u, M(n)=1.008665 am u m(p)=1.007825 am 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

U-235 is decayed by bombardment by neutron as according to the equation: ._(92)U^(235) + ._(0)n^(1) rarr ._(42)Mo^(98) + ._(54)Xe^(136) + x ._(-1)e^(0) + y ._(0)n^(1) Calculate the value of x and y and the energy released per uranium atom fragmented (neglect the mass of electron). Given masses (amu) U-235 = 235.044 , Xe = 135.907, Mo = 97.90, e = 5.5 xx 10^(-4), n = 1.0086 .

U-235 is decayed by bombardment by neutron as according to the equation: ._(92)U^(235) + ._(0)n^(1) rarr ._(42)Mo^(98) + ._(54)Xe^(136) + x ._(-1)e^(0) + y ._(0)n^(1) Calculate the value of x and y and the energy released per uranium atom fragmented (neglect the mass of electron). Given masses (amu) U-235 = 235.044 , Xe = 135.907, Mo = 97.90, e = 5.5 xx 10^(-4), n = 1.0086 .