Energy released when two deuterons $\left({ }_1 \mathrm{H}^2\right)$ fuse to form a helium nucleus $\left({ }_2 \mathrm{He}^4\right)$ is : (Given : Binding energy per nucleon of ${ }_1 \mathrm{H}^2=1.1 \mathrm{MeV}$ and binding energy per nucleon of ${ }_2 \mathrm{He}^4=7.0 \mathrm{MeV}$ )

<p>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:</p> <p>The reaction is represented as:</p> <p>${}_1{H^2} + {}_1{H^2} \to {}_2{He^4}$</p> <p>Given the binding energy per nucleon:</p> <p><p>For ${}_1 \mathrm{H}^2$: 1.1 MeV</p></p> <p><p>For ${}_2 \mathrm{He}^4$: 7.0 MeV</p></p> <h3>Calculating Binding Energy</h3> <p><strong>Binding Energy for Reactants</strong>:</p> <p><p>Each deuteron (${}_1 \mathrm{H}^2$) has a binding energy of $1.1 \, \text{MeV}$ per nucleon.</p></p> <p><p>Since there are two nucleons in a deuteron, the total binding energy for one deuteron is $1.1 \times 2 = 2.2 \, \text{MeV}$.</p></p> <p><p>Therefore, for two deuterons: $2.2 \times 2 = 4.4 \, \text{MeV}$.</p></p> <p><strong>Binding Energy for Product</strong>:</p> <p><p>For ${}_2 \mathrm{He}^4$: Each of the four nucleons has a binding energy of $7.0 \, \text{MeV}$.</p></p> <p><p>Total binding energy for the helium nucleus: $7.0 \times 4 = 28.0 \, \text{MeV}$.</p></p> <h3>Energy Released (Q)</h3> <p>The energy released, $ \mathrm{Q} $, is the difference between the binding energy of the products and the reactants:</p> <p>$ \mathrm{Q} = \mathrm{BE}_{\text{product}} - \mathrm{BE}_{\text{reactants}} = 28.0 \, \text{MeV} - 4.4 \, \text{MeV} = 23.6 \, \text{MeV} $</p> <p>Thus, the energy released when two deuterons fuse to form a helium nucleus is <strong>23.6 MeV</strong>.</p>