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Topic: population of states: molecular rotational spectroscopy  (Read 7410 times)

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Offline acuben

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population of states: molecular rotational spectroscopy
« on: February 19, 2013, 08:05:19 PM »
I was trying to understand the concept of population of states so I was studying and came up to this one question.
This is population of states in terms of Molecular rotational spectroscopy, if I need to be more specific, please let me know.

"Derive an expression for the value of J corresponding to the most highly populated rotational energy level of a diatomic (linear) rigid rotor at a temperature T. Hint: the degeneracy of each level is (2J+1)."

How would I go on approaching this problem? I am very confused because I would have guessed high J, the more populated it is (because of higher degeneracy) and I don't understand how this would even relate to temperature.

I don't think I even understand the basics here. Solving the problem is no importance to me, I want to understand the concepts.

Would anyone give me an explanation? thanks.

Offline Corribus

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Re: population of states: molecular rotational spectroscopy
« Reply #1 on: February 27, 2013, 03:11:48 PM »
Dear acuben -
I hope that this post doesn’t come too late to help you.  Before explaining the specific problem, a few concepts:

As you know, quantum systems exist in specific quantized states – that is, states with discrete energies.  In a rotation system, with rigid rotor approximation, the system is allowed to exist only at certain rotational energies based on the quantum number l (often called J).  These energies are,

[tex]E_j=\frac{\hbar^2}{2I} j(j+1)[/tex], where j is an integer that ranges from 0 to infinity.  [itex]\frac{\hbar^2}{2I}[/itex]  often is represented as B, the rotational constant.

Each state has 2j + 1 degeneracy (in absence of magnetic field), because there are multiple wavefunctions for each state that have the same energy and differ only in their angular momenta (represented by quantum number m). 

Typically the spacing between these energy levels is rather small, on the order of a few wavenumbers depending on the mass of the molecule.  For this reason rotational spectra typically appear in the microwave region of the electronmagnetic spectrum.

This is the basic quantum mechanical treatment of a molecule under a rigid rotor approximation. 

The population of a state is essentially this: for a large body of molecules which have energy levels Ej, the population of the jth state is the fraction of molecules that are in that state at any given time (compared to the total number of molecules in all states).

How does this apply to your question?  Here is the answer:  In a hypothetical scenario of absolute zero temperature, all molecules would exist in the lowest energy state.  Essentially, they wouldn’t be rotating at all, unless and until they absorb a photon of the appropriate energy to promote the molecule to a higher energy level.  Molecules have to absorb energy to move to higher energy states.  In this scenario of zero temperature, we would say that the lowest energy state is completely populated and no higher energy states are populated, because the fraction of molecules in the lowest state is unity (they are all there).

However in real life the environment is not at zero temperature.  There is latent energy in the environment (heat) represented as collisions between molecules, and so forth.  What this basically means is that some molecules, just by virtue of the heat in the environment, will have enough energy to exist in a higher lying state, at any given time, even without photoexcitation.  This is just a manifestation of the classical principle of heat transfer, momentum transfer, etc.  Consider one molecule at rest and another in motion.  The one in motion bumps into the one at rest and the one at rest gets some of the kinetic energy of the one in motion.  Some of that energy makes the molecule at rest move in space, like a billiard ball being struck by another ball, but some of the energy might make it rotate or vibrate, essentially putting it into a higher vibrational or rotational state.

Therefore we conclude that the population of higher lying states is not zero when the temperature is not zero, and the population of the lowest lying state is not unity, at any given time when the temperature is not zero.  The true population of any given quantum state is a complicated function of the temperature and the energy level of the state, mediated by the Boltzmann constant.  If the state in question has an energy that is lower than the ambient heat in the environment (the external temperature), it will have a high probability of being populated.  If the state in question has an energy that is higher than the ambient heat in the environment, it will have a lower probability of being populated.  Higher temperatures give rise to more latent energy, and thus higher energy levels will have a higher probability of being populated.  In this way, excited rotational states (which are low energy states) have a higher probability of being populated at ambient temperatures than excited electronic states, which are very high energy.  You have to have a very high temperature to see population of high lying electronic states, because otherwise there simply isn’t enough latent energy to populate these states.  But, to be clear, the probably of population of any state is never strictly zero. 

This is all pretty much the basis of statistical mechanics.  Hopefully now you understand what the question is asking and why temperature is related.

To determine the answer to the specific question, you have to use what’s called a partition function.

The probability ([itex]P_j[/itex] ) of a given state ([itex]E_j[/itex] ) being populated at a given temperature (T) is 

[tex]P_j = \frac1Z e^{-\beta E_j}[/tex]

Where [itex]\beta = \frac1{k_B T}[/itex], [itex]k_B[/itex] is the Boltzmann constant, and Z is the partition function, given by

[tex]Z = \displaystyle\sum_{j=0}^{\infty} g_j e^{-\beta E_j}[/tex]

[itex]g_j[/itex] is the degeneracy of state j, and is important because it basically means that multiple states have the same energy and so the probability of population of a state of energy [itex]E_j[/itex] at that temperature is split among the number of states that have that same energy.  [itex]E_j[/itex] in this case is whatever the formula is for the quantized energy state.  Here we are talking about rotational states, but any kind of quantized state can be used – vibrational, electronic, nuclear spin, and every combination.  The complexity of your expression (and the difficulty of calculation) is the only thing that changes.

As stated above, for rigid rotor [itex]g_j = 2j + 1[/itex] and [itex]E_j = Bj(j + 1)[/itex].  So we put those in our equation and then consider the question that is being asked.

 If I asked you, for example, for the population of the j = 1 state (probability of finding a molecule in this state at temperature T), you would solve this like so:

[tex]P_{j=1} = \frac{e^{-\beta 2B}}{\displaystyle\sum_{j=0}^{\infty} (2j+1) e^{-\beta Bj(j+1)}}[/tex]

Generally speaking you would solve this for a specific system using an experimental value of B and T.
 
The question you wrote in your post wants you to find the value of J that is MOST populated at a certain temperature.  This is basically asking you to plot out [itex]P_j[/itex] as a function of j, and find the maximum value.  You can do this by taking the (partial) derivative of [itex]P_j[/itex] with respect to j, using the appropriate energy expressions for the rigid rotor.  This derivative would be dependent on the temperature you solve it at.  This would not be an easy expression to simplify or solve without a math program like mathematica but I get the impression from the question they just want you to specify the expression without solving.

(For the sake of completeness, this treatment works only for a linear diatomic molecule – larger molecules have more complicated energy levels and so the equations will change, although in principle you use the same method to write them out.  You can also incorporate vibrational , electronic and so forth states into the partition function in the same way – the energy formulae just become more complicated.)

Hope that helps.    (And hope the equations are clear and error free. :) )

EDIT: Fixed an error in the third to last paragraph, converted equations to LaTex.
« Last Edit: February 27, 2013, 05:02:28 PM by Corribus »
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline acuben

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Re: population of states: molecular rotational spectroscopy
« Reply #2 on: March 13, 2013, 06:45:03 PM »
No, it's not too late. It's not homework to turn in or anything graded in class, I was trying to understand how it's done. I'm glad you replied. I'm reading this post over and over again right now xD.
I Saw your post on February 28th, I didn't have much time to read in detail at that time. Thanks a lot for the detailed reply.


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