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

mass loss = 5.03015 u - 5.01126 u mass loss = 0.01889 u #tag_title#Calculate the energy released per Helium-4 nucleus#tag_content# To calculate the energy released, we will use the mass-energy equivalence formula: \(E = mc^2\), where \(E\) is the energy, \(m\) is the mass, and \(c\) is the speed of light in a vacuum. We will convert the mass loss from atomic units (u) to kilograms (kg) using the conversion factor: 1 u = 1.66054 × 10^{-27} kg. The speed of light is approximately 3 × 10^8 m/s. Energy = (0.01889 u) (1.66054 × 10^{-27} kg/u) (3 × 10^8 m/s)^2 Energy ≈ 1.6904 × 10^{-12} J #tag_title#Calculate the energy released per mole of Helium-4#tag_content# Now that we have the energy released per Helium-4 nucleus, we can calculate the energy released per mole of Helium-4. Since there are \(6.022 \times 10^{23}\) particles in one mole, we simply multiply the energy per nucleus by Avogadro's number to obtain the energy per mole. Energy per mole = (1.6904 × 10^{-12} J/nucleus) (6.022 × 10^23 nuclei/mole) Energy per mole ≈ 1.017 × 10^12 J/mole Thus, the energy released per Helium-4 nucleus produced is approximately \(1.6904 \times 10^{-12}\) joules, and the energy released per mole of Helium-4 produced is approximately \(1.017 \times 10^{12}\) joules.