1
$\begingroup$

When white light is shone on a cool gas the electrons absorb photons of a certain wave lengths and become excited.

Shouldn’t the electrons then return to ground and release photons with the same wavelength as they absorbed meaning that line absorption spectra would be continuous? If the electrons don't return to ground they won't be able to absorb the photons to raise them to $n = 2$.

Improve this question
$\endgroup$
7
  • 1
    $\begingroup$ I don’t quite follow “line absorption spectra would be continuous”? The absorption and emission lines are discrete but can be broadened by homogeneous and inhomogeneous processes. I think your summary is fine. They emit in the same spectral region they absorb $\endgroup$
    – boyfarrell
    Commented Dec 11, 2020 at 20:40
  • $\begingroup$ However, the emitted light goes in all directions, not just in the direction of the stimulating light, so the brightness of the detected light (in the direction of the stimulating light) is very greatly reduced. $\endgroup$
    – S. McGrew
    Commented Dec 11, 2020 at 20:44
  • $\begingroup$ @boyfarrell I meant there would be no dark bands $\endgroup$
    – jfearn
    Commented Dec 11, 2020 at 20:49
  • $\begingroup$ @S.McGrew Thankyou, that's exactly what I was missing. The textbook I'm working through uses the words "black lines" which implies no light was being diffracted to that spot. $\endgroup$
    – jfearn
    Commented Dec 11, 2020 at 20:52
  • $\begingroup$ Dark bands appear on the solar spectra because the approximate blackbody spectrum of the sun is being transmitted through and filtered by sun’s outer atmosphere. Like the commenter above said, we don’t see the fluorescence from those lines because it is re-radiated in all directions. $\endgroup$
    – boyfarrell
    Commented Dec 11, 2020 at 20:52

1 Answer 1

Reset to default
0
$\begingroup$

Yes, for an isothermal medium in thermal equilibrium, there will be just as many upward as downward radiative transitions. This is known as the principle of detailed balance.

However, if the downward transitions are via spontaneous emission, then the emitted photons go in random directions and only a small fraction will rejoin the original illuminating beam. Hence there would be absorption lines.

Note, that this typical explanation assumes that the mean free path of a photon in the gas is comparable or larger than the extent of the gas. i.e. Most of the original light makes it through. If instead the gas is "optically thick", then photons are absorbed and re-emitted many times before re-emerging and the absorption line would be at least partially infilled by emission.

Improve this answer
$\endgroup$

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.