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
$m ({}_{92}^{238} U) = 238.05079 u m ({}_{58}^{140} Ce) = 139.90543 u m ({}_{44}^{99}{Ru}) = 98.90594 u .$

Solution

Fission reaction is given by,

{}_{92}^{238}U + {}_{0}^{1}n \to {}_{58}^{140}{Ce} + {}_{44}^{99}{Ru} + Q

Q - value = (mass of U^{238} + mass of {}_{0}^{1}n - mass of {Ce}^{140} - mass of {Ru}^{99})

x 931.5 MeV

= (238.05079 + 1.00867 - 139.90543 - 98.90594) \times 931.5 MeV = 231.1 MeV .

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