I have a doubt regarding energy levels. I saw that translational energy levels are quasi-continuous, rotational energy levels are discrete and vibrational energy levels are more discrete. I want to ask what does this energy level describe. This question is related to specific heat of gases.

My thought:
For example, if we take a diatomic molecule under consideration, only 2 rotational degrees of freedom are possible. So max levels occupied will be 2, i.e the 1st and 2nd. Am I correct or these levels signify something else?

My thought is wrong. I figured it out that as temperature increases, molecules begin to occupy higher energy levels. But then tell me one thing. The total energy contribution by a single molecule for rotational is $K_bT$ at temperature $T$. So does this mean the energy level it occupies also has energy $K_bT$? But then, if we increase temperature by $t$ the new energy will be $K_b(T+t)$. But then, how do we know that it will correspond to energy of one of the band?


In diatomic gases you have 3 degrees of freedom, id est 3 contributions to total energy: translational $E_t$, rotational $E_r$ and vibrational $E_v$. Each of this energy is quantized, which means the molecule can only get a "finite" number of value (this is a result of quantum theory), more precisely: \begin{gather} E=E_t+E_r+E_v\\ E_t(n) = \frac{n^2 h^2}{2m L} \quad E_r(l) = \frac{\hbar^2 l(l+1)}{2 I} \quad E_v(m) = \hbar \omega \left(m + \frac{1}{2} \right) \quad \text{with} \quad (n,l,m) \in \mathbb{Z}^3 \end{gather}

Specific heat is a statisical mechanics concept, it is the amount of internal energy $U$ you get when giving $k_BT$ to the system. At low temperatures the typical energy scale $\Delta E$ between two levels can be rather big compare to $k_BT$. This means, you don't populate (from a statisical point of view over all the molecule of the gas) higher excited quantum states when giving $k_BT$ to the system, and $U$ doesn't change so specific heat is zero. In the other case ($\Delta E$ small in front of $k_BT$) you can show that specific heat saturates, id est $U$ grows linearly with $k_BT$, because now you don't see discretness of levels.

\begin{gather} \Delta E_t \sim \frac{h^2}{2m L} \quad \Delta E_r \sim \frac{\hbar^2}{2 I} \quad \Delta E_v \sim \hbar \omega \end{gather}

When $\Delta E \sim k_BT$ you can say that the spectrum changes from 'discrete' to 'quasi-continous', this gives you a typical temperature $\theta_r$, $\theta_v$ and $\theta_t$ for each degree of freedom. Typical value are $\theta_t \sim 10^{-10}$ K, $\theta_r \sim 10-100$ K and $\theta_v \sim 1000$ K. $\theta_t$ being really small is often considered as always 'quasi-continous'.

Then you get

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