Based on the following atomic mass values \(-1 \mathrm{H}\), 1.00782 amu; \({ }^{2} \mathrm{H}, 2.01410 \mathrm{amu} ;{ }^{3} \mathrm{H}, 3.01605 \mathrm{amu} ;{ }^{3} \mathrm{He}\) 3.01603 amu; \({ }^{4}\) He, 4.00260 amu- and the mass of the neutron given in the text, calculate the energy released per mole in each of the following nuclear reactions, all of which are possibilities for a controlled fusion process: (a) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (b) \({ }_{1}^{2} \mathrm{H}+{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{3} \mathrm{He}+{ }_{0}^{1} \mathrm{n}\) (c) \({ }_{1}^{2} \mathrm{H}+{ }_{2}^{3} \mathrm{He} \longrightarrow{ }_{2}^{4} \mathrm{He}+{ }_{1}^{1} \mathrm{H}\)

For each reaction, we will find the mass difference between the reactants and products. The mass difference will be used to calculate the energy released per mole in the next step. (a) \[ \begin{split} \Delta m_a &= (m(^2H) + m(^3H)) - (m(^4He) + m(^1n)) \\ &= (2.01410 + 3.01605) - (4.00260 + 1.00866) \\ &= 5.03015 - 5.01126 \\ &= 0.01889 \text{ amu} \end{split} \] (b) \[ \begin{split} \Delta m_b &= (m(^2H) + m(^2H)) - (m(^3He) + m(^1n)) \\ &= (2.01410 + 2.01410) - (3.01603 + 1.00866) \\ &= 4.02820 - 4.02469 \\ &= 0.00351 \text{ amu} \end{split} \] (c) \[ \begin{split} \Delta m_c &= (m(^2H) + m(^3He)) - (m(^4He) + m(^1H)) \\ &= (2.01410 + 3.01603) - (4.00260 + 1.00782) \\ &= 5.03013 - 5.01042 \\ &= 0.01971 \text{ amu} \end{split} \]

Now that we have the mass difference for each reaction, we can convert it to energy released per mole using the mass-energy equivalence relationship and the conversion factor from amu to kg. Mass of 1 amu = 1.66054 × 10^-27 kg Speed of light, c = 2.998 × 10^8 m/s (a) \[ \begin{split} E_a &= \Delta m_a \times (1.66054 \times 10^{-27} \text{ kg/amu}) \times (2.998 \times 10^8 \text{ m/s})^2 \\ &= 0.01889 \times 1.66054 \times 10^{-27} \text{ kg} \times (2.998 \times 10^8 \text{ m/s})^2 \\ &= 2.799 \times 10^{-12} \text{ J} \end{split} \] (b) \[ \begin{split} E_b &= \Delta m_b \times (1.66054 \times 10^{-27} \text{ kg/amu}) \times (2.998 \times 10^8 \text{ m/s})^2 \\ &= 0.00351 \times 1.66054 \times 10^{-27} \text{ kg} \times (2.998 \times 10^8 \text{ m/s})^2 \\ &= 5.204 \times 10^{-13} \text{ J} \end{split} \] (c) \[ \begin{split} E_c &= \Delta m_c \times (1.66054 \times 10^{-27} \text{ kg/amu}) \times (2.998 \times 10^8 \text{ m/s})^2 \\ &= 0.01971 \times 1.66054 \times 10^{-27} \text{ kg} \times (2.998 \times 10^8 \text{ m/s})^2 \\ &= 2.922 \times 10^{-12} \text{ J} \end{split} \]

Finally, we will convert the energy released for each reaction to energy per mole by multiplying by Avogadro's number (6.022 × 10^23 mol^-1). (a) \[ E_a \text{ per mole} = 2.799 \times 10^{-12} \text{ J} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 1.684 \times 10^{12} \text{ J/mol} \] (b) \[ E_b \text{ per mole} = 5.204 \times 10^{-13} \text{ J} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 3.131 \times 10^{11} \text{ J/mol} \] (c) \[ E_c \text{ per mole} = 2.922 \times 10^{-12} \text{ J} \times 6.022 \times 10^{23} \text{ mol}^{-1} = 1.759 \times 10^{12} \text{ J/mol} \] So the energy released per mole for each fusion reaction is: (a) 1.684 × 10^12 J/mol (b) 3.131 × 10^11 J/mol (c) 1.759 × 10^12 J/mol