Calculate the energy released (per mole of deuterium consumed) for the following fusion reaction, ${ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} n$ given the following molar masses of nucleons and nuclei.
(c = 2.998 × 10⁸ m/s) particle
$\begin{array}{ll}\text { particle } & \text { mass }(\mathrm{g} / \mathrm{mol}) \\\hline \text { proton } & 1.007825 \\\text { neutron } & 1.008665 \\\text { deuterium } & 2.0140 \\\text { tritium } & 3.01605 \\\text { helium-4 } & 4.00260\end{array}$

Question

Calculate the energy released (per mole of deuterium consumed) for the following fusion reaction,

{ }_{1}^{2} \mathrm{H}+{ }_{1}^{3} \mathrm{H} \rightarrow{ }_{2}^{4} \mathrm{He}+{ }_{0}^{1} n

given the following molar masses of nucleons and nuclei.
(c = 2.998 × 10⁸ m/s) particle

\begin{array}{ll}\text { particle } & \text { mass }(\mathrm{g} / \mathrm{mol}) \\\hline \text { proton } & 1.007825 \\\text { neutron } & 1.008665 \\\text { deuterium } & 2.0140 \\\text { tritium } & 3.01605 \\\text { helium-4 } & 4.00260\end{array}

Quizplus · Accepted Answer

The energy released is calculated using the mass defect and Einstein's equation $E=mc^2$. The mass defect is the difference in mass between the reactants and products: $(2.0140 + 3.01605) - (4.00260 + 1.008665) = 0.018785 \, \text{g/mol}$. Converting this to kg/mol and using $c = 2.998 \times 10^8 \, \text{m/s}$, the energy is $0.018785 \times 10^{-3} \, \text{kg/mol} \times (2.998 \times 10^8 \, \text{m/s})^2 = 1.69 \times 10^9 \, \text{kJ/mol}$.

Calculate the Energy Released (Per Mole of Deuterium Consumed)for the Following