To calculate the energy released when two deuterons (${}_1 \mathrm{H}^2$) fuse to form a helium nucleus (${}_2 \mathrm{He}^4$), consider the following:

The reaction is represented as:

$$ {}_1{H^2} + {}_1{H^2} \to {}_2{He^4} $$

Given the binding energy per nucleon:

For ${}_1 \mathrm{H}^2$: 1.1 MeV

For ${}_2 \mathrm{He}^4$: 7.0 MeV

Calculating Binding Energy

Binding Energy for Reactants:

Each deuteron (${}_1 \mathrm{H}^2$) has a binding energy of $1.1 \, \text{MeV}$ per nucleon.

Since there are two nucleons in a deuteron, the total binding energy for one deuteron is $1.1 \times 2 = 2.2 \, \text{MeV}$.

Therefore, for two deuterons: $2.2 \times 2 = 4.4 \, \text{MeV}$.

Binding Energy for Product:

For ${}_2 \mathrm{He}^4$: Each of the four nucleons has a binding energy of $7.0 \, \text{MeV}$.

Total binding energy for the helium nucleus: $7.0 \times 4 = 28.0 \, \text{MeV}$.

Energy Released (Q)

The energy released, $ \mathrm{Q} $, is the difference between the binding energy of the products and the reactants:

$ \mathrm{Q} = \mathrm{BE}_{\text{product}} - \mathrm{BE}_{\text{reactants}} = 28.0 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV} $

Thus, the energy released when two deuterons fuse to form a helium nucleus is 23.6 MeV.