# Balmer series and the Sun

1. Jun 22, 2012

### fluidistic

In a assignment problem, I had to calculate the relative number of hydrogen atoms in the fundamental state, first, second and third excited states for a temperature of 5000 K. I used Maxwell-Boltzmann's statistics and I found out that (from memory) the most populated state was the fundamental one. The first excited state was much less populated, by a factor of 10^-10 compared to the fundamental state. The 2nd excited state about 10^-12 compared to the fundamental state (thus about the order of 100 times less populated than the 1st excited state) and so on.
The next question was "explain why the Balmer series dominates over the Lymann series for the absorbtion spectrum of the Sun". I've checked out in a book and it stated that one has to assume that any transition is approximately as likely probable as any other.
Thus I do not have any idea on why if the fundamental state is the most populated, the Lymann series doesn't dominate.
As far as I remember the Lymann series is due to any transition up to the fundamental state, i.e. n--> n=1.
Why would a series with n-->n=2 be more intense than the Lymann's one?!
I'm totally stuck on this. This has nothing to do with the atmosphere (the problem specified this hold true even if we watch the Sun from space) or the Earth.
I'm totally at a loss. It's not really homework even though it was asked as an assignment more than 1 month ago. The question is haunting me.

2. Jun 23, 2012

### xlines

Hello!

Similar question arose at another forum, basically guy asked what is the color of the hydrogen at the temperature T. I provided as much help as I could and he actually found solution which I think is correct. This said, I'm not sure that the numbers you got for the population of states are good (and btw - what you call fundamental state, it is usually referred as ground state).

So, here is the link of that discussion, I'm gonna throw this into Excel and see what comes up. If this linking actually violates (?!) PF rules, here is the relevant quote:

--
Along with relative intensities, which are related to quantum amplitudes of electronic transitions, you need to take into account population of individual levels. For a given temperature T, population of level

p(i) = g(i) e^(-E(i)/kT) / sum (g(i) e^(-E(i)/kT))

sum over all relevant energies, p(i) would be percentage of population of level i and g(i) is degeneracy of level. E(i) = e(i) - e(0) where e(i) are Rydberg energies.
--

3. Jun 23, 2012

### Simon Bridge

For the Balmer series to dominate, then the chance of a transition from (say) n=4 to n=2 must be more likely than a transition from n=4 to n=1 ... which you seem to realize.

So what sort of thing could be making n=2 more favorable?
(You can check this is in fact favored by looking up the solar spectrum - sometimes that's the test.)

Since the exercise was over a month ago - shouldn't you have the model answers by now?

Anyhow ... why did you pick 5000K (that would be spectral-type K)?

4. Jun 23, 2012

### fluidistic

Hi and thanks for your replies guys.
All I found are the relative numbers, not the raw numbers.
Let n_1 be the ground state and n_2 be the first excited state. Then $\frac{n_2}{n_1}=\frac{g_2}{g_1}e^{\frac{-E_2+E_1}{k_BT}}=4e^{\frac{-10.2eV}{8.617 \times 10 ^{-5}eV\times 5000}}\approx 2.0918 \times 10^{-10}$.
Doing a similar algebra for the second excited state, I got $\frac{n_3}{n_1} \approx 5.856 \times 10^{-12}$.
Third excited state: $\frac{n_4}{n_1} \approx 2.25 \times 10^{-12}$.
I realize this but this would totally contradicts the book of Alonso and Finn (they state that the probabilities are approximately equal). Furthermore this would require a probability of transition to n=2 greater than 10^10 times the probability of transitions to n=1. That's something I don't buy at all. :)
In my opinion the point is not to verify what the problem statement says (that's something I believe), but to answer it. I mean, the point is to explain why the Balmer series dominates over the Lymann series, not to veritfy that it's true.
Yes, except that I stopped to assist to lectures around that time. I went to friend's house and study there. We called a student who study astronomy and he answered something totally wrong, namely that we see the Balmer series more than the Lymann because the spectrometers are mostly built for visible light. He missed the point of the question (unfortunately it wasn't me who was talking on the phone!); the point is, even if you build a spectrometer that has a much wider range than the visible light, the Lymann series will never be as strong as the Balmer series, even in outer space. The question is why.

Edit: Sorry it's not contradicting the Alonso-Finn's book, but the Serway/Moses/Moyer book (page 340).

Last edited: Jun 23, 2012
5. Jun 23, 2012

### Simon Bridge

However - if your theory and data agree that for the temperature specified, the Balmer series does not dominate the Lyman, then the question cannot be answered truthfully can it?

Science is not about explaining why things are the way you believe them to be but to come up with models for things the way they are. There is a famous test question in physics, where students are asked: given the principle of Archemedes, how is it that goldfish do not displace any water in their bowl?

The Balmer series dominates for the Sun ... the usual exercise is to get the student to work out the relative populations for 5000K and for 10000K. You've done the 5000K case - try the other one and see the difference.

[snip]So you failed to get the actual answers to the exercise ... i.e. the ones that the people marking the assignment used to decide if you'd got it right or not?

You usually also get a marked assignment paper back.

(My emphasis) This is not correct.

I'm curious - you did not respond to the question I asked about the spectral class of star that has a characteristic temperature of 5000K. Does this fit with the spectral class of the Sun?

Consider:
http://frigg.physastro.mnsu.edu/~eskridge/astr215/week5.html
A star with T=5000 K is a K star. K stars have very weak Balmer lines because only one atom in 10e10 is above the ground state. A factor of two increase in temperature causes a factor of 10e5 increase in excitation. As hydrogen accounts for 90% of the atoms in a star, this means that stars with T=10000 (A stars!) will have very strong Balmer lines.

6. Jun 24, 2012

### fluidistic

I don't know what you call my "theory". Here I have the problem statement right under the eyes:
1)Find the relative numbers of the hydrogen atoms that are in the ground state and in the 1st, 2nd and 3rd excited states in the solar chromosphere. Assume T=5000K and remember to include the statistical weights of the different levels.
2)Explain why the Balmer series dominates in the absorption spectrum of the Sun.
3)Calculate the partition function Z."
I typed "chromosphere" in google and apparently it should produce emission lines, not absorption lines (according to wikipedia). So the problem is maybe badly stated for this, I don't know if this change the answer to the problem.

Well the problem statement ask me a reason, not to verify that indeed the Balmer series dominates over the Lymann's one, though I'd be glad to watch both series.
It looks like the problem is trying to teach me something I don't see (yet at least). Using the model we're given, we can see that it cannot explain a phenomenon occurring in the Sun. I'm basically asked to find the reason why it fails or something like that (that's my impression).
Yes.

No mark for assignments. At my university only the final exam count for the grade (100%) of any course. The assignments are there to train us and we are free not to do them although obviously one cannot success the final exam without solving most if not all the assigned problems. We don't have to give our done assignments to the professor but we are free to ask them whether what we've done is right and ask them tips, etc.
The semester ended last Friday (1 or 2 days ago).

I meant if we look at the Sun's surface. Are you sure that in outer space (i.e. with the removal of the Earth's atmosphere which does have an effect on what a spectrometer could possibly get from the Sun) the Balmer series doesn't dominate anymore?

Now I hope it's clearer, the given temperature is given in the problem statement.

Thanks, I appreciate this interesting link. I've read the passage your quoted but I still don't understand.
For the sake of simplicity let's assume that all the star is made of hydrogen only (not only 90%). We take T=10000K. According to the quoted passage, this means that the ratio between the number of hydrogen atom in the ground state and the hydrogen atom in the 1st excited state is of the order of 10^5. In other words the ground state is still one hundred thousands times more populated than the first excited state. Wouldn't this imply that the Lymann series should dominate over Balmer's?!
As long as I don't see the 1st excited state more populated than the ground state, I don't seem to understand how Balmer series can dominate over Lymann's.
Am I missing something?

7. Jun 24, 2012

### Simon Bridge

The theory is whatever ideas are used as a basis for the calculation you did. As distinct from observation. You don't just pick equations out of the air, you have t have some reason for thinking the process is valid before you begin.
The problem statement does not always contain all the information needed to solve the problem - it does not always tel the truth about the nature of the test. You should keep this at the back of your mind: does the problem statement make sense by itself?

Well yes - you are missing the statistics for particular transitions - probably not supplied in the classes just yet so it is pretty hand-wavey for now.

I gave you that page to show you how spectral type and temperature and line-strength get related. At 5000K, specified in the problem statement, you do not expect strong Balmer lines. BTW: what is the spectral class of the Sun?

You can certainly see how the Balmer lines for 10000K would be stronger than at 5000K?

The example used the ratio n(2)/n(1) for two cases. But you need to compare for two different transitions at the same temp ... which is slightly different.