Question 1[Maximum number: 2]In one type of nuclear process, deuterium (12H)\left({ }_{1}^{2} \mathrm{H}\right)(12H) undergoes the reaction12H+12H⟶23He+01n.{ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n} .12H+12H⟶23He+01n.Explain, with reference to your line in (a)(ii), why this reaction results in the release of energy.Show Answerboth particles have low A valuesorboth particles are at left-hand end of graphHe-3 has higher binding energy (per nucleon) than H-2Add to Test