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.

Category: Nuclear And Radiation Chemistry

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