Calculate Q-Value of the following nuclear reaction

₃Li⁷ + ₁H¹ ——–> 2₂He⁴

Check whether the reaction is exoergic or endoergic

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

The exact mass of ₃Li⁷ isotope = 7.01601 a.m.u. and that of ₁H¹ = 1.00738 a.m.u.
The exact mass of ₂He⁴ = 4.00260 a.m.u.
Where, a.m.u. is Atomic Mass Unit.

Hence
Sum of the masses of reactants = 7.01601 + 1.00738 = 8.02384 a.m.u.
Sum of the masses of products = 2 X 4.00260 = 8.00520 a.m.u.
Therefore, ΔM = Sum of the masses of products – Sum of the masses of reactants

= 8.00520 – 8.02384 = – 0.01864 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 = + 0.01864 a.m.u. X 931.5 MeV/a.m.u = + 17.36 MeV
Hence the above reaction is exoergic.