Calculate the amount of energy released per gram of hydrogen nuclei reacted for the following reaction. The atomic masses are \({ }^{1} \mathrm{H}, 1.00782 \mathrm{amu} ;{ }_{1}^{2} \mathrm{H}, 2.01410\) amu; and an electron, \(5.4858 \times\) \(10^{-4}\) amu. (Hint: Think carefully about how to account for the electron mass.) $${ }_{1}^{1} \mathrm{H}+{ }_{1}^{1} \mathrm{H} \longrightarrow{ }_{1}^{2} \mathrm{H}+{ }_{+1}^{0} \mathrm{e}$$

The mass defect for the given reaction is calculated as: Mass defect = (2 * 1.00782 - 2.01410) amu = 0.00154 amu Converting the mass defect to energy using Einstein's mass-energy equivalence formula: \(E = mc^2 = (0.00154 \times 1.6605 \times 10^{-27} \,\text{kg}) (3.0 \times 10^8\, \text{m/s})^2 = 7.641 \times 10^{-13}\, \text{J}\) Finally, calculating the energy released per gram of hydrogen nuclei reacted: Energy per gram = \(\frac{7.641 \times 10^{-13}\, \text{J}}{2 \times 1.00782 \times 1.6605 \times 10^{-24} \,\text{kg}} = 2.27 \times 10^{11}\, \frac{\text{J}}{\text{g}}\) Thus, the energy released per gram of hydrogen nuclei reacted is approximately \(2.27 \times 10^{11} \, \frac{\text{J}}{\text{g}}\).