Question

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

Answer 1

The reaction for the fission of ${}_{93}^{238}U$ by fast neutrons is as shown
${}_{93}^{238}U{+}_0^{1}n{\to }_{58}^{140}Ce{+}_{44}^{99}U$
$Q=\left[m{(}_{92}^{238}U\right)+m_N-m{(}_{58}^{140}Ce)-m{(}_{44}^{99}Ru)]\times$ 931 .5 MeV
$Q=[238.05079+1.00893-139.90543-98.90594]\times$ 931 .5 MeV
$Q=0.24835\times$ 931 .5 MeV
$Q=231.1$ MeV