Calculate the energy released nuclear reaction

₉₂U²³⁸ ——> ₉₀Th²³⁴ + ₂He⁴ + Q

The exact mass of ₉₂U²³⁸ isotope = 238.1249 a.m.u. and that of ₉₀Th²³⁴ = 234.1165 a.m.u.

The exact mass of ₂He⁴ = 4.0039 a.m.u.
Where, a.m.u. is Atomic Mass Unit.

The value of Q can be calculated from the masses of the products and reactants of the equation.

(Given) The exact mass of ₉₂U²³⁸ isotope = 238.1249 a.m.u. and that of ₉₀Th²³⁴ = 234.1165 a.m.u.

The exact mass of ₂He⁴ = 4.0039 a.m.u.
Where, a.m.u. is Atomic Mass Unit.

Hence
Sum of the masses of reactants = 238.1249 a.m.u.
Sum of the masses of products =234.1165 + 4.0039 = 238.1204 a.m.u.
Therefore, ΔM = Sum of the masses of products – Sum of the masses of reactants

= 238.1204 – 238.1249 = – 0.0045 a.m.u.

As there is decrease of mass, hence energy is being released.
The value of Q will be therefore Positive i.e.
Q = 0.0045 a.m.u.X 931.5 MeV/a.m.u = + 4.19175MeV = 4.2MeV
Hence the above reaction is exoergic.