Calculate the ground state Q value of th...

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`.

184.84 MeV

200 MeV

130 MeV

300 MeV

Text Solution

Verified by Experts

The correct Answer is:

`Q=Deltamxx931`
= [(235.0439 + 1.0087) – (98.916 + 133.9115+3.0261] × 931 Q = 184.84 MeV

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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

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