Take the two level system in atomic iodine where the ground state is four fold degenerate and the excited state at $7603 \,\mathrm{cm}^{-1}$ above the ground state (which is taken to have zero energy) and is two fold degenerate from the $^2P_\frac32$ and $^2P_\frac12$ terms arising from the configuration of atomic iodine.

I want to find the temperature at which the ratio of the population of the upper state to the lower state is 1:50.

Using statistical mechanics is it correct to say that this is the case when:

$$ \frac{n_u}{n_l} = \frac{g_{u}\mathrm{exp}(-E_u/k_{\mathrm{B}}T)}{g_{l}\mathrm{exp}(-E_l/k_{\mathrm{B}}T)} = \frac{2\mathrm{exp}(-7603.hc/k_{\mathrm{B}}T)}{4} = \frac{1}{50} $$

Rearranging for T and substituting in the values of the constants gives me $3401.9 \,\mathrm{K}$ which seems alarmingly high. Where have I gone wrong?

  • $\begingroup$ Doesn't seem all that high to me. $\endgroup$ – Ivan Neretin Mar 22 '17 at 15:03
  • $\begingroup$ Really? Only a 1/50 population for the lowest lying excited state at 3400K? Is my method definitely right? $\endgroup$ – RobChem Mar 22 '17 at 15:06
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    $\begingroup$ I didn't check the math, but electronic excitations typically require some thousands K. $\endgroup$ – Ivan Neretin Mar 22 '17 at 15:36
  • $\begingroup$ What units are using for everything? That's the only thing I could see causing major errors. $\endgroup$ – Tyberius Mar 22 '17 at 18:40
  • $\begingroup$ The units work alright, I'm fairly sure of that. Perhaps it's correct, I just assumed if done something wrong as it seems such a high temperature to reach such a low level of excitation $\endgroup$ – RobChem Mar 22 '17 at 18:50

I'm getting basically the same answer when I work out the math. I got $T=3398.45\ K$, but we probably just rounded the constants differently.

This is a good lesson in how much thermal energy is carried around at any given time by molecules in the air, as well as the importance of distinguishing between thermal processes and photo-processes.

That is, if we wish to sustain an ensemble of excited particles, there must be sufficient energy such that when we spread the energy out evenly, which is one of the things that this equation implies given that it claims all possible microstates are equally likely, some of the energy must be stored in the form of electronic excitations.

The reason this probably seems like a very high temperature is because we use equipment all the time that relies on electronic excitations and we never have problems with the equipment melting due to it being at thousands of degrees Celsius. That is simply because in these photo-process, meaning the energy is carried in the form of photons, the temperature never is that hot.

That is, if I shoot a photon in, or even a stream of photons at a sample with the transition you are describing, each photon carries a very small amount of energy, and even though there may be quite a lot of them, the energy is quickly dispersed because the excited state is very short-lived, so the photon gets re-emitted, and likely just makes it way out into an infinite reservoir known as the surroundings.

Thus, it is very difficult to maintain a population of excited states thermally, because the system needs to basically be a closed system or else it will eventually just radiate all the energy away.

Another way of thinking about this is that the value of $k_bT$ at room-ish temperature of $298\ K$ (in a familiar unit) is $207\ cm^{-1}$. This makes it clear then why rotationally excited states are frequently occupied at room temperature, because these transitions are on the order of tens of wavenumbers. On the other hand, we often get off free assuming everything is in the vibrational ground state because these transitions are on the order of thousands of wavenumbers.

This is also where the idea of rotational temperature and vibrational temperature come from. These can be understood physically as,

an estimate of the temperature at which thermal energy (of the order of $k_bT$) is comparable to the spacing between rotational [or vibrational] energy levels.

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  • $\begingroup$ When we make use of excited states via photons, in general how large a fraction of the population is actually excited at any one time? I know in something like a laser or an ICP-MS you're aiming to keep a majority of the relevant population excited, but for most applications the excited fraction is pretty small, isn't it? $\endgroup$ – hBy2Py Mar 23 '17 at 0:31
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    $\begingroup$ @hBy2Py Right that's what I was trying to say. It wasn't a totally pertinent point I suppose, but when exciting using photons, the fraction in an excited state should be quite small. I was just trying to make a distinction between photo excitation and thermal excitation. $\endgroup$ – jheindel Mar 23 '17 at 5:17
  • $\begingroup$ Ya and lasers are quite different because you need three states. I wonder if you can make a thermal laser? Seems unlikely and unclear what that even means. $\endgroup$ – jheindel Mar 23 '17 at 5:19
  • $\begingroup$ re. number of excited states, if the sample has an optical density of , say, 2 then 99% of light is absorbed so you can work out what fraction of molecules are excited. With a laser its not hard to completely bleach a sample and similarly it is easy to invert the population and make it into a 'gain medium' i.e. a laser amplifier. $\endgroup$ – porphyrin Mar 23 '17 at 9:55
  • $\begingroup$ <nod>, jheindel, I think the answer's really good. I was just trying to point out a minor aspect that you hadn't touched on (as best I could tell). $\endgroup$ – hBy2Py Mar 23 '17 at 11:21

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