0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.