# How to estimate the temperature required for the 2nd energy level of hydrogen to be occupied?

In my atstrophysics lecture today we were covering atomic spectra and we saw that at both the hot and cool end, the atomic spectra of stars showed a very weak balmer series. The reason being at the hot end, most of the hydrogen atoms have been ionised and so there aren't many making the specific transitions and the scattering photons. At the cooler end the electrons don't have enough energy to occupy the n=2 states of the hydrogen atom.

I wanted to then find the temperature of star at which the n=2 orbittal would be occupied because I didn't think it would be that high that even the sun didn't have many of these filled. My attempt was:

Seeing as the energy required for a transition to the n=2 orbital for hydrogen corresponds to the photons having $$E_{photon}=E_{n=2}-E_{n=1}=10.2eV$$. As $$E_{photon}=\frac{hc}{\lambda}$$ for photons and according to wiens displacement law $$\lambda_{peak}T=0.0029$$. $$E_{photon}=\frac{hcT}{0.0029}$$. So$$\frac{hcT}{0.0029}=10.2e$$, this gives $$T\approx 24000K$$.

This is much too large as apparently by this temperature the hydrogen has already been ionised. So how do I find the temperature at which this energy level is likely to be occupied?

## 2 Answers

Radiative considerations (via e.g. Wien's law) are inappropriate here ─ as a starting point, at least, the only thing you should need is Boltzmann statistics, which tell you that the probability of occupation of the $$n$$th energy level (with enrgy $$E_n$$ and degeneracy $$g_n$$) is $$P_n = \frac{1}{Z} g_n e^{-E_n/k_BT},$$ with the partition function $$Z$$ functioning as a normalization factor.

For hydrogen under these conditions, you can set $$E_1=-13.6\:\rm eV$$ and $$E_2=-3.4\:\rm eV$$, and the degeneracies are $$g_1=2$$ states for the $$n=1$$ shell and $$g_2=8$$ states for the $$n=2$$ shell. (More generally, $$E_n=E_1/n^2$$ and $$g_n=2n^2$$.) As such, the relative occupation of the $$n=2$$ shell at temperature $$T$$ is $$P_2(T) = \frac{8e^{-E_2/k_BT}}{2+8e^{-E_2/k_BT}+\cdots}.$$ If the temperature is too low, then the exponential factor kills you, but (because the $$n=3$$ and higher shells are very close in energy) there is only a very short window in temperature before the higher-lying levels start rising as well (and they have higher degeneracy).

It is reasonably easy to calculate $$P_2(T)$$ using only the bound states, and this already yields a probability that peaks at about $$T\approx 11\,000\:\rm K$$ (graphed below). The decay at temperatures higher than this peak will be made steeper once you factor in the ionized states (which makes the calculation significantly harder).

• Minor comment: you've accounted for spin degeneracy when you say $g_2=8$ but not when you say $g_n=n^2$. For consistency the latter should be $2n^2$. Which way would the peak move once adding in the continuum of bound states? My intuition is it moves to the left (low $T$). – jacob1729 Apr 16 at 14:56
• Good catch, thanks. I'm not sure the peak's maximum will shift all that much once you include continuum states - but you will definitely see a strong suppression of the tail. – Emilio Pisanty Apr 16 at 15:06
• Is it wrong to consider wiens displacement law as a starting point because the energy is not transferred via radiation? What is the primary method of energy transfer here? – Vishal Jain Apr 19 at 6:15
• @Vishal The primary mode of energy transfer here is good old mechanical collisions. Radiative interactions are presumably also important, but if you want to account for them correctly then you're likely to need to re-develop the Planck spectral law in the presence of a gas of neutral, excited, and ionised atoms, all of which interact with each mode of the EM field as a dynamical (as opposed to thermodynamical) system. – Emilio Pisanty Apr 19 at 6:26

I think you might be confusing some things about radiative energy transfer:

Keep in mind that thermal radiation always follows a distribution according to Plancks law. So at any given temperature, there will be some photons with the wavelength / energy you require for the n(1)->n(2) transition.

Wiens displacement law only gives you the wavelength at which the distribution reaches its maximum. As the energy required is $$10.2 \ eV$$, an incoming photon would need to be of wavelength ~ $$120 \ nm$$ which is at the low end of near ultra violet.

Even a 5000 K hot black body still has its maxima in the VIS-range, so the value you calculated is reasonable, but for your application the distribution matters not just the maximum of the distribution.

You have established that a photon would need to have an energy greater or equal than $$10.2 \ eV$$, which means a wavelength of $$120 \ nm$$ or shorter. Therefore integrate Plancks law from 0 to 120 nm (while treating the Temperature as constant), divide the results by the Stefan-Boltzmann Law, and you will obtain a function that will give you the fraction of photons energetic enough to faciliate the transformation for a given temperature.