91Ó°ÊÓ

The characteristic rotational temperature \(\Theta_{\text{rot}}\) is given by the formula:\[ \Theta_{\text{rot}} = \frac{h^2}{8\pi^2Ik} \]where \(h\) is Planck's constant, \(I\) is the moment of inertia of the molecule, and \(k\) is Boltzmann’s constant.

The moment of inertia for a diatomic molecule is given by:\[ I = \mu \cdot r^2 \]where \(\mu\) is the reduced mass of the molecule and \(r\) is the bond length.For \(\mathrm{H}_2\), each hydrogen atom has a mass \(m_H\) of approximately 1 u (atomic mass unit).Reduced mass \(\mu_{\mathrm{H}_2}\) is:\[ \mu_{\mathrm{H}_2} = \frac{m_H \cdot m_H}{m_H + m_H} = \frac{1 \cdot 1}{1 + 1} = 0.5 \; \text{u} \]For \(\mathrm{D}_2\), each deuterium atom has a mass \(m_D\) of approximately 2.014 u.Reduced mass \(\mu_{\mathrm{D}_2}\) is:\[ \mu_{\mathrm{D}_2} = \frac{m_D \cdot m_D}{m_D + m_D} = \frac{2.014 \cdot 2.014}{2.014 + 2.014} = 1.007 \; \text{u} \]Convert the bond length to meters: \(r = 74.16 \; \text{pm} = 74.16 \times 10^{-12} \; \text{m}\).Calculate \(I\) for each gas:\[ I_{\mathrm{H}_2} = \mu_{\mathrm{H}_2} \cdot r^2 = 0.5 \times (74.16 \times 10^{-12})^2 \; \text{kg} \cdot \text{m}^2 \]\[ I_{\mathrm{D}_2} = \mu_{\mathrm{D}_2} \cdot r^2 = 1.007 \times (74.16 \times 10^{-12})^2 \; \text{kg} \cdot \text{m}^2 \]

The atomic mass unit is defined to be 1/12th of the mass of a carbon-12 atom, or approximately \(1.66053906660 \times 10^{-27} \; \text{kg}\).For \(\mathrm{H}_2\):\[ \mu_{\mathrm{H}_2} = 0.5 \; \text{u} = 0.5 \times 1.66053906660 \times 10^{-27} \; \text{kg} = 8.302695333 \times 10^{-28} \; \text{kg} \]For \(\mathrm{D}_2\):\[ \mu_{\mathrm{D}_2} = 1.007 \; \text{u} = 1.007 \times 1.66053906660 \times 10^{-27} \; \text{kg} = 1.672612743 \times 10^{-27} \; \text{kg} \]

Substitute the reduced masses into the moment of inertia calculations:For \(\mathrm{H}_2\):\[ I_{\mathrm{H}_2} = 8.302695333 \times 10^{-28} \cdot (74.16 \times 10^{-12})^2 \; \text{kg} \cdot \text{m}^2 = 4.60315 \times 10^{-48} \; \text{kg} \cdot \text{m}^2 \]For \(\mathrm{D}_2\):\[ I_{\mathrm{D}_2} = 1.672612743 \times 10^{-27} \cdot (74.16 \times 10^{-12})^2 \; \text{kg} \cdot \text{m}^2 = 9.05448 \times 10^{-48} \; \text{kg} \cdot \text{m}^2 \]

Recall the formula for \(\Theta_{\text{rot}}\):\[ \Theta_{\text{rot}} = \frac{h^2}{8\pi^2I k} \]Using \(h = 6.62607015 \times 10^{-34} \; \text{J} \cdot \text{s}\) and \(k = 1.380649 \times 10^{-23} \; \text{J/K}\), calculate:For \(\mathrm{H}_2\):\[ \Theta_{\text{rot}}(\mathrm{H}_2) = \frac{(6.62607015 \times 10^{-34})^2}{8 \times \pi^2 \times 4.60315 \times 10^{-48} \times 1.380649 \times 10^{-23}} = 87.595 \; \text{K} \]For \(\mathrm{D}_2\):\[ \Theta_{\text{rot}}(\mathrm{D}_2) = \frac{(6.62607015 \times 10^{-34})^2}{8 \times \pi^2 \times 9.05448 \times 10^{-48} \times 1.380649 \times 10^{-23}} = 44.385 \; \text{K} \]