1.

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`.A. 184.84 MeVB. 200 MeVC. 130 MeVD. 300 MeV

Answer» Correct Answer - A
`Q=Deltamxx931`
= [(235.0439 + 1.0087) – (98.916 + 133.9115+3.0261] × 931 Q = 184.84 MeV


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