Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV):

We first recall the definition: the total binding energy $$B$$ of a nucleus equals its binding energy per nucleon multiplied by the number of nucleons in that nucleus.

For one deuteron (symbolically $$^2_1H$$) we have two nucleons, so

$$B_{\text{deuteron}} = \bigl(1.1\ \text{MeV per nucleon}\bigr)\times(2\ \text{nucleons}) = 2.2\ \text{MeV}.$$

Because the reaction involves two deuterons, the total initial binding energy becomes

$$B_{\text{initial}} = 2 \times 2.2\ \text{MeV} = 4.4\ \text{MeV}.$$

Next we look at the product nucleus, helium-4 (symbolically $$^4_2He$$). It contains four nucleons, and the given binding energy per nucleon is $$7.0\ \text{MeV}$$, so

$$B_{\text{helium}} = \bigl(7.0\ \text{MeV per nucleon}\bigr)\times(4\ \text{nucleons}) = 28.0\ \text{MeV}.$$

The energy released in a nuclear reaction equals the difference between the total binding energy of the products and that of the reactants. Stating this as a formula,

$$Q = B_{\text{products}} - B_{\text{reactants}}.$$

Substituting the values just obtained, we get

$$Q = 28.0\ \text{MeV} - 4.4\ \text{MeV} = 23.6\ \text{MeV}.$$

So the fusion of two deuterons into one helium-4 nucleus liberates $$23.6\ \text{MeV}$$ of energy.

Hence, the correct answer is Option A.