In a fission event of U(238,92) by fast moving neutrons . . .

Question : In a fission event of U(238,92) by fast moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are Ce(140,58) and Ru(99,44). Calculate Q for this process. Neglect the masses of electrons/positrons emitted during the intermediate steps. 

Given 

m[U(238,92)]=238.05079 u

m[Ce(140,58)]=139.90543 u

m[Ru(99,44)]=98.90594 u

m[n(1,0)]=1.008665 u

Doubt by Dev & Aishwarya 

Solution : 

U(238,92) + n(1,0) → Ce(140,58) + Ru(99,44) + electrons / positrons

ATQ
Mass of Electrons and Positrons has to be neglected. 

Mass Defect = [Mass of Reactants] - [Mass of Products]

Δm = m[U(238,92)]+m[n(1,0)] - {m[Ce(140,58)]+m[Ru(99,44)]}
Δm = m[U(238,92)]+m[n(1,0)] - m[Ce(140,58)]-m[Ru(99,44)]
Δm = 238.05079 + 1.008665 - 139.90543 - 98.90594

Δm = 239.059455 - 238.81137
Δm = 0.248085 u

We know, 
1 u = 931 MeV

So
0.248085 u will have [0.248085 × 931] MeV of energy.

Q = 0.248085 × 931
Q = 230.967135
Q = 230.97 MeV

Required Q Value of the reaction is 230.97 MeV.