# Confusion about absorption spectra - cool gases absorb?

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• Cool4Kat
In summary, Kathy learned that heated low density gases produce spectral lines and cool low density gases absorb their spectral lines. Kirchhoff accidentally discovered how spectrum analysis could be used to study the sun by observing the spectral lines emitted by heated sodium.
Cool4Kat
So I was taught in school that heated low density gases produce spectral lines and cool low density gases absorb their spectral lines. I mentioned this to my husband and he asked me what the definition of hot and cool were and I had no idea. The more I thought about it, the more confused I became. Sodium that is heated in a lab emits the spectral lines and sodium on the surface of the sun (which is way hotter) absorbs the light. Also, the gasses in the atmosphere don't seem to make absorption lines in sunlight. [Edit: it turns out that the atmosphere can make absorption lines, especially at sunset or sunrise, but doesn't do it very much]

Then, I was studying how Kirchhoff accidentally discovered how spectrum analysis could be used to study the sun. In October of 1859, Kirchhoff was playing with looking at heated sodium with his spectrometer when, as a lark, he tried placing bright lamp light through it as well as the heated sodium. He was expecting a rainbow with brighter yellow doublet, instead the sodium absorbed the doublet, mimicking the sodium shadows in sunlight. So here was an example of a *heated* gas absorbing radiation!

Therefore, I have come to the conclusion that the "cool gasses absorb" statement is not correct. So what is really going on? I have a theory, and I wonder if any of you wise folks can tell me if I am on the right track or totally confused.

Theory: After a certain temperature threshold, low density gasses all emit their spectrum of light. They also, at this temperature, can absorb their spectrum of light. If they are irradiated with external radiation at these frequencies, the transmitted light has that spectrum absorbed because the emitted radiation from the gas is scattered and reflects towards the external radiation source. In Kirchhoff's experiment, the sodium gas acts like a mirror and reflects the yellow doublet back towards the lamp. If he had studied the sodium gas from an other angle the yellow doublet would be brighter as it would contain the light of the sodium and the reflected yellow light from the lamp.

Am I close? Am I full of it? Thank you a million times for your wise counsel.

Kathy

Last edited:
Heated gasses emit radiation. Radiation passing through a gaseous medium will scatter radiation as long as there are atoms in the lower level. The surface of the sun is relatively cool compared to the radiation passing through it.

I'm betting that even in a heated gas most of the atoms/molecules are in low-energy states, allowing them to absorb incoming radiation of the appropriate wavelengths. The hot gas has some small proportion of its atoms/molecules being excited by thermal motion, where they then quickly relax back down to their ground states, emitting radiation in the process. The cold gas has almost none of its atoms/molecules being excited since there isn't enough thermal energy.

In other words, a hot gas may have 1% of its atoms emitting light every second (a totally made up number) whereas in the cold gas it's more like 0.00000001% (another made up number). In either case there is still 99% or more of the gas in an unexcited state and ready to absorb incoming light of the appropriate wavelength. Hence both hot and cold gases can absorb light passing through.

Cool4Kat and lomidrevo
Cool4Kat said:
the emitted radiation from the gas is scattered and reflects towards the external radiation source.
mathman said:
Radiation passing through a gaseous medium will scatter radiation as long as there are atoms in the lower level.

As long as the discussion is about spectral lines (emission or absorption), the mechanism which is producing these lines is bound-bound transition. By absorbing incident photon, electron bound in an atom makes transition from lower energy level to higher energy level. When the electron makes the opposite transition, an photon is emitted by the atom (in random direction). In each case, the energy of the photon is exactly the same as the difference between the energy levels. Therefore we can observe the discrete spectral lines.

This mechanism is completely different from scattering, where the electrons are free - not bound in a atom. Scattering is one of the sources of the continuum opacity.

stefan r and Cool4Kat
Cool4Kat said:
So I was taught in school that heated low density gases produce spectral lines and cool low density gases absorb their spectral lines.
Issue is that there is not an exact definition of a hot gas and a cold gas, and the terms are used a bit loosely. But usually it is clear form context of particular situation what is meant by hot gas and what is meant by cold gas. The temperature of the gas itself cannot be used to make the distinction, because any gas with temperature above 0 K is indeed emitting thermal (continuous) radiation (even it can be far far below visible portion of spectrum) and also might have molecules/atoms in an excited state (to produce the emission lines).
Usually when we talk about cold gas (with absorption lines) we think of a case when there is strong source of a continuous spectrum and directly between the observer and source there is a gas in such state that is able to absorb significant portion of photons of particular frequencies (energies) via the mechanism I described in my previous post. Then we observe continuous spectrum with discrete absorption lines. An example: Sun -> Atmosphere -> observer.
For a situation with hot gas, we usually think about gas which is being excited by a nearby source of radiation, but usually it is not directly behind the gas as seen from the observer's point of view. So we only see the discrete spectrum coming from the gas (emission lines). An example: emission nebula.
I think that @Drakkith provided nice point of view on this.

DaveE and Drakkith
lomidrevo said:
Issue is that there is not an exact definition of a hot gas and a cold gas, and the terms are used a bit loosely. But usually it is clear form context of particular situation what is meant by hot gas and what is meant by cold gas. The temperature of the gas itself cannot be used to make the distinction, because any gas with temperature above 0 K is indeed emitting thermal (continuous) radiation (even it can be far far below visible portion of spectrum) and also might have molecules/atoms in an excited state (to produce the emission lines).
It would emit thermal radiation if it had opacity in infrared. Which is often lacking.
Simple example of "cool" or "hot" gas: does an open flame emit or absorb light?
Both.
Look at a candle in darkness - it emits light.
Place the candle in direct sunlight. It still burns. You can see the flame against background of sunlit surfaces.
Yet look at what happens to sunlight that falls on the flame. The flame should cast a shadow, as it absorbs sunlight.
If you use a spectroscope, you should be able to measure both the emission spectrum of candle flame (put candle in the dark and view it with spectroscope) and the absorption spectrum (put it in direct sunlight, put spectroscope in the shadow and point it at Sun through the flame).
Same gas, same temperature. Both effects. Which of them is more noticeable just depends on the incident light.

stefan r
snorkack said:
Simple example of "cool" or "hot" gas: does an open flame emit or absorb light?
Both.
Look at a candle in darkness - it emits light.
Candle is really bad example if we talk about spectral lines. Spectrum of candle can be approximated by a spectrum of blackbody (continuous) radiation at effective temperature of about 1500 K, see for example: http://www.giangrandi.ch/optics/blackbody/blackbody.shtml.

snorkack said:
I don't get your point, thermal doesn't mean infrared! It is because the everyday temperatures we encounter here on Earth means that the thermal emission of the everyday objects peaks at infrared part of the spectrum. For hot stars the thermal radiation may peak in visible or UV part of spectrum. Btw. thermal radiation and blackbody radiation are basically synonyms (at least in astrophysics). So that is for the continuous spectrums...
And regarding discrete absorption lines, they are not restricted to visible part of the spectrum only. You can have lines in all parts of the spectrum (UV, IR, MW, radio...). It depends on the composition of the gas: this is the heart of spectroscopy, thanks to which we are able to know the chemical composition of distant gases.

krater
mathman said:
Heated gasses emit radiation. Radiation passing through a gaseous medium will scatter radiation as long as there are atoms in the lower level. The surface of the sun is relatively cool compared to the radiation passing through it.

That was what I was taught as well, but the "relatively cool" thing doesn't work with lamp light going through sodium flames. The sodium is relatively hot, not cold (possibly hotter than the lamp but definitely hotter than the surrounding air). Also, the sodium is still yellow if viewed from the side, so it is still emitting its yellow light in most directions. But it is blocking the yellow light from transmitting through the hot sodium gas. So, the rule that "relative temperature" says what is absorbed or emitted just doesn't work. It seems like the rule should be if a gas is hit with radiation in one direction, it will reflect or scatter or absorb those frequencies. But which one and why? Does the light get absorbed and if so, why does the light that the sodium emit get emitted in every direction but one? If the light gets reflected, why does it reflect? So confused.

Which frequencies are absorbed depend on the energy levels of the atoms in the gas. To simplify for sodium, assume there is an upper level and a lower level (electron levels in the atom) and that yellow is emitted by going from the upper level to the lower, while absorption takes place by going from lower to upper. The gas is a mixture of atoms at these two levels, so white light passing through the gas loses photons of the wavelength (yellow) for transitioning from lower to upper level.

Cool4Kat said:
That was what I was taught as well, but the "relatively cool" thing doesn't work with lamp light going through sodium flames. The sodium is relatively hot, not cold (possibly hotter than the lamp but definitely hotter than the surrounding air). Also, the sodium is still yellow if viewed from the side, so it is still emitting its yellow light in most directions. But it is blocking the yellow light from transmitting through the hot sodium gas. So, the rule that "relative temperature" says what is absorbed or emitted just doesn't work. It seems like the rule should be if a gas is hit with radiation in one direction, it will reflect or scatter or absorb those frequencies. But which one and why? Does the light get absorbed and if so, why does the light that the sodium emit get emitted in every direction but one? If the light gets reflected, why does it reflect? So confused.

The key is to realize that once the photon is absorbed by the atom, usually it is re-emitted in a random direction in a short time after. So when you look at the gas from a point of view where the lamp is just behind the gas, the photons of corresponding wavelength (let's say yellow) are missing in the detected spectrum, due to constant absorption of those photons. But when you look at it from any different position (lamp not behind the gas), you will detect part of those constantly re-emitted photons, which will amplify the observed spectrum at the corresponding frequency. As per our example, (in a visible part of the spectrum) otherwise colorless gas might get yellowish color.

Cool4Kat said:
So, the rule that "relative temperature" says what is absorbed or emitted just doesn't work.
I suggest that thermal emission and absorption of appropriate lines from light passing through will both occur. I suppose the question is whether a 'very hot' gas will have a black body spectrum and will it look red hot, white hot etc.. If the gas is pure then I see no way that the atoms can emit anything but the characteristic spectral lines
(spread a bit due to the speeds involved). So a Sodium Vapour that's hot enough will emit the D lines at some level - the same as if it's electrically excited. But the atoms can also absorb incident D lines and re-emit them (an absorption line spectrum).
The absorption spectrum will dominate the effect for a cold gas but the visible dark lines will be filled in as the gas is heated up until they are undetectable. The density of the gas (or total mass of atoms passed through) will govern the depth of the lines but it will also affect the level of thermally sourced light. For a given temperature, will the two effects track each other?

I think the key missing insight is that if you consider the brightness of light along some ray that traces back to a source at some temperature, and you neglect scattering (which is another matter we will talk about in a moment), then any material you place in between your detector and that background source will tend to replace the brightness along that ray (which originally corresponded to the temperature of the background source) with a brightness that corresponds to the temperature of the material you are inserting. The thicker the material, the more complete is that replacement. So if the material is "optically thick" (i.e., opaque), the replacement is total, and you see the brightness of the sodium vapor regardless of the brightness of what is behind it. But the sodium vapor would only be optically thick at frequencies within the line-- if you look at the line "wings", you see through the vapor and the replacement is not complete, it's an average of the background temperature (which could be zero, or very high) and the sodium temperature. And if you look way outside the line, you just see the background temperature. So "relatively cold" means the sodium temperature relative to the background temperature, which can be anything.

Now, this assumes that temperature is all that is going on here, which means we are neglecting "scattering." Scattering is when you do not have the action of a blackbody (where incident light is absorbed, and emitted light is all due to temperature), but instead the light just bounces. A black pool ball is a blackbody that absorbs, and then re-emits based on its temperature, but a white cue ball takes a lot of the incident light and simply bounces it off. To the extent that sodium atoms only scatter existing light instead of absorbing and creating light (which depends on their density), then instead of trying to replace the brightness along a ray by the temperature of the sodium (which now does not matter at all), it tries to replace it with the average incident brightness over all directions to the outside sources (including directions that are dark, which move the average toward zero brightness and are responsible for absorption lines when there is scattering).

So that's how high density gas (which is blackbody-like) produces lines, and how low density gas (which is highly scattering) does it. High density gas tries to replace the background "brightness temperature" with the temperature of the gas (that is, the brightness corresponding to the gas temperature), and low density gas tries to replace the background brightness along a given ray with the average brightness along all directions of incident rays. That is the guts of Kirchhoff's laws, but as is all too often true, instead of arming students with the ability to figure out what is happening for themselves, we only give them cookie-cutter recipes they do not understand and cannot successfully generalize when they have curious minds and start to ask the questions that Cool4Kat is asking.

Ken G said:
That is the guts of Kirchhoff's laws, but as is all too often true, instead of arming students with the ability to figure out what is happening for themselves, we only give them cookie-cutter recipes they do not understand and cannot successfully generalize when they have curious minds and start to ask the questions that Cool4Kat is asking.
+1
The Science courses that are available for students are just not adequate to deal with the extra Science content that arrives on a daily basis. Consequently, the syllabus writers and examination boards include sophisticated but attractive topics before the basics. Would you believe that AS students were told about fundamental particles before they were even told about the electron Volt?
No one can be a Renaissance Person these days so we have to accept early specialisation in order to make time for basic 'arming' that you refer to.

sophiecentaur said:
.. If the gas is pure then I see no way that the atoms can emit anything but the characteristic spectral lines...

There will be a black body background in addition to spectral lines.

sophiecentaur
Ken G said:
I think the key missing insight is that if you consider the brightness of light along some ray that traces back to a source at some temperature, and you neglect scattering (which is another matter we will talk about in a moment), then any material you place in between your detector and that background source will tend to replace the brightness along that ray (which originally corresponded to the temperature of the background source) with a brightness that corresponds to the temperature of the material you are inserting. The thicker the material, the more complete is that replacement.
That´s a nice expression of Kirchoff´s Law. Replacement of incident radiation (whatever it is) with black body radiation depending on only the temperature of the body. In case of the black body, the replacement is complete.
Ken G said:
So if the material is "optically thick" (i.e., opaque), the replacement is total, and you see the brightness of the sodium vapor regardless of the brightness of what is behind it. But the sodium vapor would only be optically thick at frequencies within the line-- if you look at the line "wings", you see through the vapor and the replacement is not complete, it's an average of the background temperature (which could be zero, or very high) and the sodium temperature. And if you look way outside the line, you just see the background temperature. So "relatively cold" means the sodium temperature relative to the background temperature, which can be anything.

Now, this assumes that temperature is all that is going on here, which means we are neglecting "scattering." Scattering is when you do not have the action of a blackbody (where incident light is absorbed, and emitted light is all due to temperature), but instead the light just bounces. A black pool ball is a blackbody that absorbs, and then re-emits based on its temperature, but a white cue ball takes a lot of the incident light and simply bounces it off. To the extent that sodium atoms only scatter existing light instead of absorbing and creating light (which depends on their density), then instead of trying to replace the brightness along a ray by the temperature of the sodium (which now does not matter at all), it tries to replace it with the average incident brightness over all directions to the outside sources (including directions that are dark, which move the average toward zero brightness and are responsible for absorption lines when there is scattering).

So that's how high density gas (which is blackbody-like) produces lines, and how low density gas (which is highly scattering) does it.
I should not call low-density gas "highly" scattering. It has some scattering - Rayleigh one - and that´s NOT concentrated in narrow lines.

But note that for purposes of Kirchoff´s laws, all interactions that are not absorption/replacement are, well, not absorption/replacement. Could be transmission unaltered, could be refraction, could be mirror reflection, or could be scattering. All of them leave the body "white" and therefore unable to radiate, except insofar as partial absorption/replacement does occur.

stefan r said:
There will be a black body background in addition to spectral lines.
I can't help thinking that the probability of an atom being excited in the optical region is pretty much zero unless it's a spectral line. The high energy photons will not interact with any other atoms. There could be scattering which will raise the overall temperature of the gas but how would that correspond to a visible optical frequency unless the gas is very densely packed and the energy flux through it is high.
I guess it all depends on the actual conditions we are considering here.

sophiecentaur said:
I can't help thinking that the probability of an atom being excited in the optical region is pretty much zero unless it's a spectral line. The high energy photons will not interact with any other atoms. There could be scattering which will raise the overall temperature of the gas but how would that correspond to a visible optical frequency unless the gas is very densely packed and the energy flux through it is high.
I guess it all depends on the actual conditions we are considering here.

It goes by Plank's law.

f(x) = 1/(ex-1) goes toward 0 as x gets bigger. It does not actually equal 0 for any real number x.

sophiecentaur said:
...the probability of an atom being excited in the optical region is pretty much zero ...

It is not far from zero. Could be "zero observed in most experiments within in a reasonable fixed time frame", or "much less than background noise in the lab parking lot".

Post #11 said "no way that the atoms can emit anything but the characteristic spectral lines". Microwaves are a thing. Low pressure sodium lamps look very yellow to human retina based detectors. I just assumed we were talking about the whole spectrum.

sophiecentaur
stefan r said:
It goes by Plank's law.

f(x) = 1/(ex-1) goes toward 0 as x gets bigger. It does not actually equal 0 for any real number x.
But that´s just the blackbody radiation.
There is no actual requirement that a gas should be a blackbody. The absorptivity/emissivity in Kirchhoff Law can be arbitrarily low.

sophiecentaur

## 1. What is an absorption spectrum?

An absorption spectrum is a graphical representation of the wavelengths of light that are absorbed by a substance. It shows the specific wavelengths of light that are absorbed by the substance, and can be used to identify the substance or determine its concentration.

## 2. How do cool gases absorb light?

Cool gases absorb light through a process called electronic transitions. When atoms or molecules in the gas absorb light, their electrons are excited to a higher energy level. As the electrons return to their original energy level, they emit light at specific wavelengths, resulting in an absorption spectrum.

## 3. Why are cool gases used in absorption spectroscopy?

Cool gases are used in absorption spectroscopy because they have a simpler and more well-defined absorption spectrum compared to solids or liquids. This makes it easier to identify and analyze the specific wavelengths of light that are absorbed by the gas.

## 4. How does temperature affect the absorption spectrum of a gas?

The temperature of a gas can affect its absorption spectrum in two ways. First, at higher temperatures, the gas molecules have more thermal energy, causing them to move faster and broaden the absorption peaks in the spectrum. Second, at lower temperatures, the gas molecules may be in a more ordered state, resulting in sharper and more defined absorption peaks.

## 5. Can the absorption spectrum of a gas be used to determine its temperature?

Yes, the absorption spectrum of a gas can be used to determine its temperature. As mentioned earlier, the temperature of a gas can affect the shape and intensity of its absorption spectrum. By analyzing the absorption peaks and their widths, scientists can estimate the temperature of the gas. This technique is commonly used in astrophysics to study the temperature of stars and other celestial objects.

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