Problem 6: Find the energy released in the fusion reaction; 1H2 + 1H3 ----------→ 2He4 + 0n1

Solution

Theory

Fusion reaction is one in which two atoms of hydrogen combine together (fuse) to form one atom of helium. Part of the mass of the hydrogen is converted into energy. The easiest fusion is one given in the problem, i-e, combining deuterium with tritium to make helium.

Now part of the hydrogen is converted to energy. Therefore, there must be a difference in the sums of the masses on LHS and RHS. If we calculate that difference in masses (Δm), it could be easily converted to energy by multiplying with 931.5 (as 1 u = 931.5 MeV).

Solution

Consider the given equation.

Sum of the masses of the reactants (LHS) = Mass of 1H2 + Mass of 1H3

= 2.014102 + 3.01604 = 5.030151 u

Sum of the masses of the products (RHS) = Mass of 2He4 + Mass of 0n1

= 4.00388 + 1.00866 = 5.01254 u

Difference in masses = Δm = 5.030151 u – 5.01254 u = 0.17611 = 0.018 u

This much mass is converted to energy. Therefore,

Energy released ≈ Δ m × 931.5 = 0.018 × 931.5 = 16.8 MeV