# Different energy levels in quantum mechanics

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?

Edit:
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'.