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 the $Q$-value for this fission process. The relevant atomic and particle masses are:
$m(_{92}^{238} U) = 238.05079 \; u$
$m(_{58}^{140} Ce) = 139.90543 \; u$
$m(_{44}^{99} Ru) = 98.90594 \; u$
$m(_{0}^{1} n) = 1.008665 \; u$

(A) The nuclear reaction for the fission of $_{92}^{238} U$ by a neutron,resulting in $_{58}^{140} Ce$ and $_{44}^{99} Ru$ with the emission of $10$ $\beta$-particles,is:
$_{92}^{238} U + _{0}^{1} n \rightarrow _{58}^{140} Ce + _{44}^{99} Ru + 10 \; _{-1}^{0} e$
The $Q$-value is calculated using the mass defect $\Delta m$:
$Q = [m(_{92}^{238} U) + m(_{0}^{1} n) - m(_{58}^{140} Ce) - m(_{44}^{99} Ru)] c^2$
Note: The mass of the $10$ electrons emitted in $\beta$-decay is accounted for by using atomic masses,as the total number of electrons remains conserved in the process $(92 = 58 + 44 - 10)$.
Substituting the given values:
$Q = [238.05079 + 1.008665 - 139.90543 - 98.90594] \; u \times c^2$
$Q = [239.059455 - 238.81137] \; u \times c^2$
$Q = 0.248085 \; u \times c^2$
Using $1 \; u = 931.5 \; MeV/c^2$:
$Q = 0.248085 \times 931.5 \; MeV \approx 231.09 \; MeV$
Thus,the $Q$-value of the fission process is approximately $231.09 \; MeV$.