Why can one observe an electronic transition of the $3s$ state to the $3p$ state in the emission spectrum of the nitrogen atom, but not in its absorption spectrum?

I know that the selection rules technically don't forbid either one since $\Delta \ell=\pm 1$ is permitted.

  • $\begingroup$ For reference, chemical element symbols are never italicized (i.e. it's $\rm N$, not $N$), and hyphens are not necessary. And really, calling explicit symbols when the straight element name fits is unnecessary use of abbreviations. $\endgroup$ Commented Oct 2, 2018 at 17:39

1 Answer 1


I'm going to offer an answer which I think makes sense, but I'm not a real expert in spectroscopy, so it is quite possible that better answers will appear.

I think that this is just a question of populations. Both of those states are excited states of the atom, corresponding to one electron being taken from the 2p orbital (which has three electrons in it, in the ground state) to a 3s or 3p orbital. In fact, those electronic configurations both include several distinct quantum states: they broadly divide into spin doublets ($s=\frac{1}{2}$) and quartets ($s=\frac{3}{2}$) and you can see a diagram here. But these details are not crucial to the discussion.

Suppose an atom is excited to one of the 3p states. Then, in time, it is expected to decay to a lower lying state; some of those transitions will be to a 3s state, and will involve an emitted photon which you can see in the emission spectrum. Most of the excited state population in 3p will contribute to one emission line or another, neglecting any collisions with other atoms. So it is not surprising to see some 3p $\rightarrow$ 3s emission lines.

Suppose an atom is excited to one of the 3s states. It will likewise decay towards 2p (more lines in the emission spectrum, which don't concern us here). Because of this, its lifetime will not be particularly long, and to see the 3s $\rightarrow$ 3p transition in absorption, we are relying on a suitable photon coming along to be absorbed, before this happens. Hardly any of the population of the excited 3s state will actually live long enough to do this; the overwhelming fraction of them will decay towards the ground state, emitting a photon.

So there is a general problem with observing "excited state absorption". Practically, you would need to "pump" the transition from the ground state to the lower-lying of the two excited states (3s here) with such intense radiation that you establish a meaningful steady-state population of this state.

Some of the points above appear on the Excited State Wikipedia page.

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    $\begingroup$ This is pretty much it, though it's important to note that the previous excitation doesn't need to be with a tailored transition - the 3s-3p transition will be activated if the temperature is high enough to put a substantial population on that excited state. $\endgroup$ Commented Oct 2, 2018 at 17:42
  • $\begingroup$ Good point. Very high temperature would do it. Unless I'm mistaken, something like $10^5$ Kelvin would be needed for an excitation energy approaching 10 eV. $\endgroup$
    – user197851
    Commented Oct 2, 2018 at 17:44

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