2 _{1}H^{1} + 2 _{0}n^{1} —–>_{2}He^{4}

**Check whether the reaction is exoergic or endoergic?**

The exact mass of _{1}H^{1} isotope = 1.0081 a.m.u. and that of _{0}n^{1} = 1.0086 a.m.u.

The exact mass of _{2}He^{4} = 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 _{1}H^{1} isotope = 1.0081 a.m.u. and that of _{0}n^{1} = 1.0086 a.m.u.

The exact mass of _{2}He^{4} = 4.0039 a.m.u.

Where, a.m.u. is Atomic Mass Unit.

Hence

Sum of the masses of reactants = 2 X 1.0081 + 2 X 1.0086 = 2.0162a.m.u. + 2.0172a.m.u. = 4.0334a.m.u.

Sum of the masses of products = 4.0039 a.m.u.

Therefore, ΔM = Sum of the masses of products – Sum of the masses of reactants

= 4.0039 – 4.0334= – 0.0295a.m.u.

As there is decrease of mass,hence energy is being released.

The value of Q will be therefore Positive i.e.

Q = 0.0295a.m.u.X 931.5 MeV/a.m.u = + 27.47925 MeV

Hence the above reaction is exoergic.