Follows Quantum mechanical correction to the equipartition theorem.

In the prompt, explain that 50, 500 and 5000 should be equally spaced, for instance. 2, 5, and 10 are also approximately equally spaced.

As it turns out, the heat capacity is about \(\frac52\) for almost the entire temperature range.

1. Translational K.E. in one dimension: \(\left\{ 1\times 10^{-40}\text{ J}, 4\times 10^{-40}\text{ J}, 9\times 10^{-40}\text{ J}, \ldots \right\}\)
2. Rotational K.E.: \(\left\{ 0\text{ J}, 0.8\times 10^{-22}\text{ J}, 0.8\times 10^{-22}\text{ J}, 0.8\times 10^{-22}\text{ J}, 2.5\times 10^{-22}\text{ J}, \ldots \right\}\)

   The bond length of N₂ is 1.1 Å. Its mass is \(m = 14\text{ amu}\approx 2\times 10^{-26}\text{ kg}\). The moment of inertia is thus \begin{align} I &= 2\cdot 2\times 10^{-26}\text{ kg} \left(0.55\times 10^{-10}\text{ m}\right)^2 \\ &\approx 1.2\times 10^{-46}\text{ kg}\cdot\text{m}^2 \end{align} From which we can find the energy eigenvalues: \begin{align} E_{\ell} &= \frac{\hbar^2}{2I}\ell(\ell+1) \\ &= \frac{\left(10^{-34} \text{ J}\cdot\text{s} \right)^2 }{2\cdot 1.2\times 10^{-46}\text{ kg}\cdot\text{m}^2} \ell(\ell+1) \\ &\approx \ell(\ell+1)\cdot 0.4\times 10^{-22}\text{ J} \end{align}

3. Vibrational energy: \(\left\{ 2.3\times 10^{-20}\text{ J}, 6.9\times 10^{-20}\text{ J}, 11.5\times 10^{-20}\text{ J}, \ldots \right\}\)

   I looked up the experimental vibrational frequency, which is 2358 cm\(^{-1}\).

(use a logarithmic temperature axis)