If you look at the spectral lines of hydrogen, the emission lines seem to be in the exact same spot where the absorption spectrum is black. Together, they build a full spectrum of visible light. Does this happen with hydrogen only or also with other atoms?
Krish Schmitt
Answered question
2022-09-01
Does this happen with hydrogen only or also with other atoms?
Answer & Explanation
Peutiedw
Beginner2022-09-02Added 9 answers
- It is indeed true that for every emission line, there is a corresponding absorption transition that, under suitable conditions, can be observed in an absorption spectrum.
- However, it is not true that if you take an emission spectrum and an absorption spectrum from the same atom, the lines in the two must always match.
The reason is simply that not all possible lines must always be represented in any given spectrum.
- For an emission spectrum, the upper level must be populated, and the lower level must not be saturated.
- For an absorption spectrum, the lower level must be populated.
Thus, for example, if you have atomic hydrogen that's very cold, the Balmer series will be weak or absent in absorption, since there won't be enough energy for thermal excitations to populate the n=2 shell. Conversely, at high temperatures the Lyman series will weaken off, as there will be less and less population in the n=1 shell that can absorb that light. And so on.
abkapseln87
Beginner2022-09-03Added 1 answers
Under some conditions you can have states that are reached by a single absorption event (creating a high energy absorption line) that decay (with a non-trivial branching ratio) through several steps (creating several lower energy emission lines).
Indeed this kind of process is the usual scheme for setting a population inversion so that you can get lasing behaviour.
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