Energy released when two deuterons ( ₁ H²) fuse to form a helium nucleus ( ₂ He⁴) is : (Given : Binding energy per nucleon of ₁ H²=1.1 MeV and binding energy per nucleon of ₂ He⁴=7.0 MeV )

Solution

Physics sir · 3 steps

Do not subtract the per-nucleon numbers. Multiply each by its nucleon count first, then compare totals.

1. Total binding energies. Binding energy per nucleon times the number of nucleons gives the total binding energy. A deuteron has 2 nucleons, so its total is 2 × 1.1 = 2.2 MeV. A helium nucleus has 4 nucleons, so its total is 4 × 7.0 = 28 MeV.
2. Before and after. Before fusion there are two deuterons, with a total of 2 × 2.2 = 4.4 MeV. After fusion there is one helium nucleus, with 28 MeV. The product is more tightly bound, so the difference is released as energy.

   Q = 28 - 4.4 = 23.6 MeV

3. So. The energy released is 23.6 MeV, which is option (c).

The idea to keep. Energy released in a nuclear reaction is the total binding energy of the products minus that of the reactants. Always convert per-nucleon values to totals before you subtract.

Answer from the NTA final answer key. Solution by PrepWiser.