If we spectroscopically observe a cloud of hot gas, which is on the whole not very absorbent, and which is not illuminated by a source behind it, we observe emission lines. How does this type of spectrum form? I had thought that those lines are those in which there are transitions of atoms is true, but I don't think that's enough. Why should all the material as a whole emit like that. Why are there these electronic transitions?

And then: if there is a light source behind the material, one observes a spectrum that is in a way the negative of the one I put, i.e. the lines become absorption lines. I had thought that something different must be happening in the two cases, although I don't know what.

And then again: if we knew nothing about electronic transitions and only wanted to consider the macroscopic properties of the gas (which could be composed of complicated molecules, in which there are not only electronic transitions but also other phenomena), could we still justify the fact that the emission and absorption spectra are the negative of each other?

Thank you for any input; complex, articulate and in-depth answers are also welcome.


1 Answer 1


How does an emission line spectrum form?

It forms in materials that are optically thin (i.e. transparent) to their own radiation over most of the spectrum, but have optical thickness at the wavelengths corresponding to particular transitions of atoms and ions in the material.

Assuming some sort of local thermodynamic equilibrium then the intensity will be equal to the Planck function at the corresponding temperature in the spectral lines and equal to $\sim$ zero in the continuum regions where the material is transparent.

If you now put an illuminating continuum light source behind such an object then the light in the "continuum regions" just goes straight through. But, because the object is optically thick at the wavelengths corresponding to the spectral lines it will absorb light from the illuminating source and then re-radiate that in all directions.

In order to get an absorption spectrum you need to make the intervening material colder than the radiation field of the illuminating source - that will mean there are more photons coming towards the observer in the continuum than in the lines. If it were the same temperature you would just see the continuum spectrum of the illumnating source - any absorption will just be balanced by emission from the intervening medium, if the intervening material were hotter you would still see emission lines but with less contrast to a non-zero continuum provided by the cooler illuminating source.


I simply assumed in the above discussion that the emission and absorption properties of the transitions up or down between two energy levels are connected/proportional. This is always true in a system that has reached local thermodynamic equilibrium. The principle of detailed balance tells us that each microscopic process (in this case the rate of emission) is balanced by its inverse process. This leads to a mathematical connection between the Einstein coefficients governing absorption and emission - if the absorption cross-section is large then so too is the emission coefficient. It is this relationship that is responsible for the absorption spectrum being "the negative" of the emission spectrum (if the conditions are correct - see above).

  • $\begingroup$ I really appreciated this clarifying answer. But, let us focus on the fact that the emission spectrum is the negative of the absorption spectrum: could you explain better, though, why a molecule tends to emit more at the frequencies where it absorbs more? Could some non-obvious consequence of the second law of thermodynamics be involved? Thank you for any further comment. $\endgroup$
    – Bml
    Commented Nov 6, 2023 at 20:08
  • $\begingroup$ @Bml see my edit. $\endgroup$
    – ProfRob
    Commented Nov 6, 2023 at 20:28
  • $\begingroup$ This is a very clear and concise conceptual answer, but I have one follow-up: what does it mean for an emission line to be optically thick / thin, conceptually? $\endgroup$
    – kirklong
    Commented Dec 30, 2023 at 1:06
  • 1
    $\begingroup$ @kirklong these are standard terms in spectroscopy. As a photon travels it has a mean free path before it is absorbed. This mean free path is inversely proportional to the absorption coefficient. If the mean free path is larger than the path to escape the emitter, then we say it is optically thin at that wavelength. Most emitted photons will escape. If the mfp is much shorter than the escape path length, then the emitting object/gas is optically thick at that wavelength. $\endgroup$
    – ProfRob
    Commented Dec 30, 2023 at 3:31
  • 1
    $\begingroup$ @kirklong a gas is more likely to be optically thick at wavelengths corresponding to an atomic transition because the absorption coefficient will be higher and the mean free path lower. $\endgroup$
    – ProfRob
    Commented Dec 30, 2023 at 3:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.