Question

Consider the reaction $$ 2{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He} $$ (a) Calculate \(\Delta E\) in kilojoules per gram of deuterium fused. (b) How much energy is potentially available from the fusion of all the deuterium in seawater? The percentage of deuterium in water is about \(0.0017 \%\). The total mass of water in the oceans is \(1.3 \times 10^{24} \mathrm{~g}\). (c) What fraction of the deuterium in the oceans would have to be consumed to supply the annual energy requirements of the world \(\left(2.3 \times 10^{17} \mathrm{~kJ}\right) ?\)

Short answer

Answer: About \(0.000174\%\) of the deuterium in the oceans would have to be consumed to supply the annual energy requirements of the world.

Step-by-step solution

Calculate the mass difference and \(\Delta E\) per gram of deuterium fused

We will begin by calculating the mass difference between the reactants and products in the nuclear reaction. Then, we will use the relation between energy and mass, given by Einstein's equation, \(E=mc^{2}\), to calculate the energy released per gram of deuterium fused. Consider the reaction: $$ 2{ }_{1}^{2} \mathrm{H} \longrightarrow{ }_{2}^{4} \mathrm{He} $$ The mass of one deuterium atom \((^2H)\) is approximately \(2.014 \,\text{u}\), and the mass of a helium atom \((^4He)\) is approximately \(4.002 \,\text{u}\). In this reaction, 2 deuterium atoms fuse to form 1 helium atom. The mass difference \(\Delta m\) is: $$ \Delta m = 2 \times 2.014\,\text{u} - 4.002\,\text{u} = 0.026\,\text{u} $$ Convert the mass difference to kilograms: $$ \Delta m = 0.026\,\text{u} \times \frac{1.66 \times 10^{-27}\,\text{kg}}{1\,\text{u}} = 4.316 \times 10^{-29}\,\text{kg} $$ Now, we can use Einstein's equation to calculate the energy released per fusion reaction, \(\Delta E\): $$ \Delta E = \Delta m \times c^2 = 4.316 \times 10^{-29}\,\text{kg} \times (3 \times 10^8\,\text{m/s})^2 = 3.877 \times 10^{-12}\,\text{J} $$ Finally, calculate the energy released per gram of deuterium fused: $$ \Delta E_\text{per gram} = \frac{3.877 \times 10^{-12}\,\text{J}}{2 \times (2.014\,\text{u} \times 1.66 \times 10^{-27}\,\text{kg/u})} \times \frac{1\,\text{kJ}}{1000\,\text{J}} \approx 59.67\,\text{kJ/gram} $$

Calculate the energy available from the fusion of all the deuterium in seawater

The percentage of deuterium in water is \(0.0017\%\). To find the mass of deuterium in seawater, we first need to know the mass of water in the oceans. We are given that the total mass of water in the oceans is \(1.3 \times 10^{24}\,\text{g}\). $$ \text{Deuterium mass} = \frac{0.0017}{100} \times 1.3 \times 10^{24}\,\text{g} \approx 2.21 \times 10^{21}\,\text{g} $$ Now, we can calculate the total energy available from the fusion of all the deuterium in seawater using the energy released per gram of deuterium fused: $$ \text{Total energy} = \Delta E_\text{per gram} \times \text{Deuterium mass} = 59.67\,\text{kJ/gram} \times 2.21 \times 10^{21}\,\text{g} \approx 1.32 \times 10^{23}\,\text{kJ} $$

Calculate the fraction of deuterium required to meet the annual energy requirements of the world

We are given that the annual energy requirements of the world are \(2.3 \times 10^{17}\,\text{kJ}\). To calculate the fraction of deuterium required to meet these energy demands, we will divide the required energy by the total energy available per gram of deuterium fused: $$ \text{Fraction of deuterium required} = \frac{\text{Required energy}}{\text{Total energy available}} = \frac{2.3 \times 10^{17}\,\text{kJ}}{1.32 \times 10^{23}\,\text{kJ}} \approx 1.74 \times 10^{-6} $$ Thus, about \(1.74 \times 10^{-6}\) or \(0.000174\%\) of the deuterium in the oceans would have to be consumed to supply the annual energy requirements of the world.