Energy released when two deuterons $$(_{1}H^2)$$ fuse to form a helium nucleus $$(_{2}He^4)$$ is :
(Given : Binding energy per nucleon of $$_{1}H^2 = 1.1$$ MeV and binding energy per nucleon of $$_{2}He^4 = 7.0$$ MeV)

The energy released in a nuclear reaction equals the increase in total binding energy of the nuclei involved.
$$Q = \text{(total BE of products)} - \text{(total BE of reactants)}$$

Step 1: Binding energy of the reactants
Each deuteron $$_{1}H^{2}$$ contains $$A = 2$$ nucleons.
Given binding energy per nucleon = $$1.1$$ MeV.
Total binding energy of one deuteron = $$1.1 \times 2 = 2.2$$ MeV.
Since two deuterons take part, total BE of reactants = $$2 \times 2.2 = 4.4$$ MeV.

Step 2: Binding energy of the product
Helium-4 nucleus $$_{2}He^{4}$$ has $$A = 4$$ nucleons.
Given binding energy per nucleon = $$7.0$$ MeV.
Total binding energy of product = $$7.0 \times 4 = 28.0$$ MeV.

Step 3: Energy released
$$Q = 28.0 \text{ MeV} - 4.4 \text{ MeV} = 23.6 \text{ MeV}.$$

Therefore, the energy liberated when two deuterons fuse into a helium-4 nucleus is $$23.6$$ MeV.

Option C is correct.