91Ó°ÊÓ

The barrier energy is the electrostatic potential energy when the two nuclei are close enough to touch. Given that each deuterium nucleus has one proton, the potential energy is given by the formula:\[ E_{r} = \frac{k \cdot e^{2}}{r} \]Where:- \( k = 8.99 \times 10^{9} \text{ N m}^2/ ext{C}^2 \) is Coulomb's constant,- \( e = 1.6 \times 10^{-19} \text{ C} \) is the elementary charge,- \( r \approx 2 \times 10^{-15} \text{ m} \) is the approximate nuclear radius for deuterium.Plugging in these values:\[ E_{r} = \frac{8.99 \times 10^{9} \times (1.6 \times 10^{-19})^2}{2 \times 10^{-15}} \approx 3.84 \times 10^{-14} \text{ J} \]To convert this energy to MeV:\[ E_{r} = \frac{3.84 \times 10^{-14}}{1.6 \times 10^{-13}} \approx 0.24 \text{ MeV} \]

The energy liberated in the fusion reaction can be calculated using the mass defect:\[ \Delta m = (2m_{D}) - (m_{He} + m_{n}) \]Where:- \( m_{D} = 2.0141 \text{ u} \),- \( m_{He} = 3.0160 \text{ u} \),- \( m_{n} = 1.0087 \text{ u} \).Calculate \( \Delta m \):\[ \Delta m = (2 \times 2.0141) - (3.0160 + 1.0087) = 0.0045 \text{ u} \]Convert to energy using \( E = \Delta m \cdot c^2 \) (1 u = 931.5 MeV):\[ E = 0.0045 \times 931.5 = 4.19 \text{ MeV} \]Convert MeV to Joules:\[ E = 4.19 \times 1.6 \times 10^{-13} \approx 6.7 \times 10^{-13} \text{ J} \]

Since deuterium gas is diatomic (\(D_2\)), we have Avogadro's number of molecules (\(N_A = 6.022 \times 10^{23} \text{ mol}^{-1}\)) per mole. Each molecule contains two deuterium atoms, so the energy per mole is:\[ E_{\text{per mole}} = 2 \times 4.19 \text{ MeV/molecule} \times 6.022 \times 10^{23} \text{ molecules/mol} \]Convert to Joules:\[ E_{\text{per mole}} = 2 \times 4.19 \times 1.6 \times 10^{-13} \times 6.022 \times 10^{23} \approx 8.05 \times 10^{11} \text{ J/mol} \]This is much greater than the heat of combustion of hydrogen, \(2.9 \times 10^{5} \text{ J/mol} \).