# Hydrogen spectra

1. May 21, 2004

### Hydr0matic

Let me see if I got this straight... The hydrogen spectra consists of a couple of series - Lyman, Balmer, Paschen, Brackett, etc - which all corresponds to a certain energy transition - Lyman (n>1 -> n=1), Balmer (n>2 -> n=2).. and so on... correct ?
And since n=1 is the ground state, a hydrogen atom in a low energy state will mainly emit lines in the Lyman series, right ?

Isn't it strange that the highest energy photons are emitted at the lowest energy state ?

Exactly what lines are emitted given a certain temperature (or energy state) ? I've been under the impression that all series are emitted seperately, i.e. all lines of a specific series (and only those lines) are emitted at a certain temperature. Is this true ? Or are all alfa-lines (in each series) emitted together, or perhaps just lines within a certain scope, e.g. 1000nm-2000nm ?

If the entire lyman series is emitted at a lower energy state than say the alfa-line in Balmer, does this mean that the transition from say n=100 -> n=1 is more likely to happen than the transition from n=3 -> n=2 ?
i.e. given a certain energy state, are all transitions in a lower series always more likely than any in a higher series ?

2. May 21, 2004

### mathman

You sound confused. As you had stated just before, the highest energy photons result from transitions to the ground state. Therefore the energy differences would be the largest.

3. May 21, 2004

### Hydr0matic

Oh .. by "lowest energy state" I meant the general state of the hydrogen, the temperature, if you will ...

Anything else unclear, perhaps ? let me know ....

4. May 21, 2004

### marcus

Hydr0,
AFAIK the electron's energy level in a hydrogen atom does not correspond in a simple way to temperature

a box of hydrogen gas at some temperature will have H2 molecules whizzing around at different speeds----the higher the temp the higher the average speed---and there will be some collisions too. the collisions could transfer energy to the electrons and put them in higher states.

in a simplified picture of the electron energy states, if you pick the right unit and the right energy to be the reference level (the zero on an energy scale is an arbitrary choice you make) then the energy levels are:

-1, -1/4, -1/9, -1/16

you can see the lowest is -1 and they get higher and higher as the electron's orbit absorbs more energy.

and then it can jump down
if it jumps from -1/4 down to -1 then it emits
3/4 of a unit of energy

or if it jumps down from -1/16 to -1/4 then it emits 3/16 units of energy

these are nice units of energy called Rydberg units IIRC
a rydberg unit is 13.6 eV which is about the same as 2 x 10-18 joules

that is, I guess, 2 "attojoules"

anyway you can see if it emits 3/16 of a unit by jumping down to the lower state then you can tell the photon energy in eevee if you want
by multiplying 3/16 by 13.6

5. May 22, 2004

### Hydr0matic

I understand... But there is some relationship isn't there ? I'm basing this assumption on the behaviour of most radiators. Hydrogen isn't even a good radiator, but it is, none the less, a radiator... right ? .. And I believe we can generally say that, the radiation from a certain piece (or volume) of matter reflects it's temperature, in some way. An extremely hot gas radiates more than a cold gas, and the same goes for any solid.

But the question concerning hydrogen is if there's a wavelength relationship as well - i.e. a perfect radiator will emit higher energy photons at higher temperatures, and lower respectively. Hydrogen on the other hand not only behaves unlike most other radiators, but it appears to behave the opposite !

This isn't a question concerning hydrogen actually, but rather the atomic model. It's seems rather contradictory to me that either
1. there's no correlation between the energy in a single atom and the energy in the entire system of atoms... or
2. the correlation is the opposite of the one described by Planck, i.e. the higher energy in the system, the lower energy photons emitted.

I know Nr 1 isn't true for solids like metalls and such, that follow the Planck law... which is another thing I'm not clear about by the way. If the atomic model doesn't incorporate the Planck law, why do metalls follow it ? What exactly is the source of the radiation in glowing metall ?

I'm guessing there's a big chance I've probably misunderstood something here though ..

6. May 22, 2004

### marcus

Hydr0 I have to go out so someone else will
have to respond but remember
a gas of H2 molecules is different from an isolated H atom

the molecules have a lot of other ways to absorb energy
they whiz around they collide they bounce off the walls they
tumble end over end
they can even do a springy action between the two atoms
if the temperature is high enough

at no matter what temp
the most occupied atom state is the ground state
when some accident happens and bumps the electron up in to
a higher state then it quicky radiates off the energy and falls back
to the ground state

the thermal radiation, following planck curve, must come from
many other things besides the single atom electron state hopping

so discrete spectrum of H atom is different from continuous thermal spectrum

have to go

7. May 22, 2004

### Hydr0matic

Yes, I know, but how can't there be any connection between them if the source is one and the same ? (Maybe not true for hydrogen but definately for metals). ... The radiation curve of iron for example follows Plancks law very nicely, but how is this explained if the radiation really consist of photons radiated (mainly) from the atoms ?
Like you said, at no matter what temp, electrons fall down to ground state emitting higher energy photons.

This is exactly my point. The atomic system isn't isolated from it's surroundings, and in no way is it a "stationary unit". Atoms basically bump around ALL the time, and when they do it's the electrons and protons that interact. So how could the systems not be correlated ?

8. Jan 2, 2005

### Hydr0matic

Anyone want to continue this discussion with me ?

9. Jan 2, 2005

### Kane O'Donnell

Planck's blackbody law *doesn't* apply to a single Hydrogen atom. The blackbody radiation curve arises when you do a statistical analysis of lots and lots of atoms, the most simple case is when you have rectangular prism cavity with a vacuum inside and you just consider the atoms in the walls (as classical harmonic oscillators).

Most of the derivation of the blackbody curve is actually electromagnetism, because it involves finding the energy density of the resonant cavity modes. The "Planck hypothesis", the oh-so-famous origin of h, states that the energy is absorbed or emitted in lumps of hf, and in particular the *average* energy of all the oscillators at frequency f will be:

$$E_{av} = \frac{hf}{e^{\frac{hf}{k_{B}T}}-1}$$

where k_B is the Boltzmann constant and T is the temperature of the blackbody. This average is calculated using Boltzmann statistics.

The quantum harmonic oscillator analysis shows that the hypothesis about energy coming in lumps of hf is correct. Since we used harmonic oscillators, not real atoms, there isn't really a connection to atomic energy levels that you are asking about. An analysis using a cavity made of a particular element would probably be *very* difficult.

Can someone with more experience than me make some corrections to this post? It's been a while since I've done the blackbody derivation, and I've never seen it done using atomic energy levels...

10. Jan 2, 2005

### Hydr0matic

My point is this:
.. take iron for example. When heating iron it first starts glowing red, then there's a hint of orange, then yellow, then white.. and at last a hint of blue. Analysing the intensity/wavelength relationship in the radiation as a function of temperature would reveal a planck-like curve. And YET, the same radiation would produce very discrete lines in a spectrometer.
How can photons emitted by atoms follow Plancks' law if there's no connection to atomic energy levels ?

11. Jan 2, 2005

### Kane O'Donnell

Well, in the example you give, you *wouldn't* get a very discrete spectrum, because having a gazillion atoms close together causes the energy levels to split into a gazillion finely spaced levels clustered around each of the discrete spectrum. This is a consequence of the exclusion principle. Instead of seeing energy levels, you see energy bands. In a metal such as iron, the bands overlap to a certain extent so you will in fact have a pseudocontinuous spectrum.

This is what I mean by statistical - you get a gazillion atoms together and the emission spectrum of the bulk object is *not* just an amplified version of the emission spectrum for a single atom.

So your intuition is sort of right, but you can't apply single-atom physics directly to a blackbody situation because of the exclusion principle. The energy band phenomena is studied extensively in semiconductor theory and there are plenty of websites explaining it. Try HyperPhysics for a simple explanation.

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/fermi.html

Cheerio!

Kane O'Donnell

12. Jan 3, 2005

### Hydr0matic

Thnx again, but you're sort of missing my point. Never mind the degree of discreteness in the spectrum, the radiation producing the pseudocontinuous spectrum is still mainly emitted by the atoms, right ? Even though the energy levels split within the atoms the photons are still emitted by the atoms, right ?

So why do these photons follow Planck's law ?

Here's the iron spectrum: http://www.webelements.com/webelements/elements/text/Fe/econ.html

The tiny (I guess) bulk of iron that produced this spectrum would follow Planck's law, even though the radiation obviously originates from energy transitions within the atoms.

How is this possible ? According to everybody, blackbodies have nothing to do with atomic energy levels.

13. Jan 3, 2005

### seratend

You have to understand that the iron block is a black body containing iron atoms that absorb/emit following their discrete spectrum (em field in a given volume).
When you heat your iron block, you have a black body at a given temperature with the known continuous spectrum of the black body. In superpostion, you have the iron atomes that can absorb emit/energy on the discrete spectrum.

Seratend.

Seratend.

14. Jan 3, 2005

### Hydr0matic

15. Jan 3, 2005