The easiest fusion reaction to initiate is $${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}$$ Calculate the energy released per \({ }_{2}^{4} \mathrm{He}\) nucleus produced and per mole of \({ }_{2}^{4}\) He produced. The atomic masses are \({ }_{1}^{2} \mathrm{H}, 2.01410 ;{ }_{1}^{3} \mathrm{H}\), \(3.01605\); and \({ }_{2}^{4}\) He, \(4.00260\). The masses of the electron and neutron are \(5.4858 \times 10^{-4}\) amu and \(1.00866\) amu, respectively.

The mass defect of the reaction is calculated as: Mass defect = (2.01410 + 3.01605) - (4.00260 + 1.00866) = 0.01889 amu Converting the mass defect to kg: Mass defect (in kg) = 0.01889 amu x 1.66054 x 10^-27 kg/amu = 3.1378 x 10^-29 kg Calculating the energy released using E=mc^2: Energy released (in Joules) = (3.1378 x 10^-29 kg) x (2.998 x 10^8 m/s)^2 ≈ 2.818 x 10^-12 J Now converting the energy released to per mole basis: Energy released per mole of Helium-4 = (2.818 x 10^-12 J) x (6.022 x 10^23 mol^-1) ≈ 1.696 x 10^12 J/mol Therefore, the energy released per \(_{2}^{4}\) Helium nucleus produced is approximately \(2.818 x 10^{-12}\) Joules, and the energy released per mole of \(_{2}^{4}\) Helium nuclei produced is approximately \(1.696 x 10^{12}\) Joules per mole.