2
$\begingroup$

I'm trying to understand the absorption spectrum in terms of what happens when an electron absorbs a photon. If we shine white light through a sample and use a prism to disperse the light, we would see black lines corresponding to the wavelengths absorbed by the electron. However, if the specific wavelength is absorbed, wouldn't it be released once it comes down anyways? Why then do we see dark lines?

My assumption is that when the electron falls back down to a lower energy level, the photon is scattered in all directions so the intensity of the light for that wavelength is reduced in comparison to the other wavelengths we can see (i.e the rest of the spectrum).

$\endgroup$
5
  • 2
    $\begingroup$ Hmmm ... despite occasional sloppy usage it is atoms (or "atomic electrons" if you wish) that are absorbing your photons—free electrons (a) can't simple absorb a photon it must scatter it and (b) would not show line spectra but continuous ones. That said, whether the re-emission is fully isotropic or not depends on the conditions of the experiment, but any re-emission angular distribution other than "always in the direction of the incident photon" would be sufficient. $\endgroup$ Commented Apr 24, 2018 at 20:23
  • $\begingroup$ I'm sorry, I dont think I'm able to follow.. $\endgroup$ Commented Apr 24, 2018 at 20:26
  • 1
    $\begingroup$ I think @dmckee is basically approving your assumption but prefers rephrasing "scattered in all directions" to "scattered in other directions". $\endgroup$
    – npojo
    Commented Apr 24, 2018 at 20:44
  • 3
    $\begingroup$ Relaxation of an excited electronic state does not necessitate emission of a photon, much less a photon of the same wavelength that created the excited state. A trivial example would be you feeling warm in the sun - photons have been absorbed and converted into heat. $\endgroup$
    – Jon Custer
    Commented Apr 24, 2018 at 21:05
  • $\begingroup$ @JonCuster unless you continue to feel warmer, and warmer and warmer,... then yes, the principle of detailed balance says that all processes are in balance at a microscopic level. $\endgroup$
    – ProfRob
    Commented Feb 9, 2020 at 10:06

2 Answers 2

1
$\begingroup$

First, an electron cannot absorb a photon, it can only scatter it. But your question stands when discussing absorption by atoms.

You are quite right, that if your "sample" is in equilibrium, that for every atomic transition that absorbs a photon, there must be a downward transition that produces a photon.

There are basically two photon production processes - spontaneous and stimulated emission.

The situation where you see absorption lines is caused by spontaneous emission, because the photons are emitted in all directions, not just in the direction of the original beam.

The net effect is to remove photons at the characteristic wavelength from the beam and thus produce an "absorption line".

Interestingly, if stimulated emission dominates, then the emitted photons would be in the same direction, at the same wavelength, as the absorbed photons and there would be no absorption lines. This is rarely the case at visible wavelengths, but often the case at radio wavelengths.

$\endgroup$
4
  • 1
    $\begingroup$ "that for every atomic transition that absorbs a photon, there must be a downward transition that produces a photon." There is also the possibility of nonradiative deexcitation. $\endgroup$
    – my2cts
    Commented Feb 9, 2020 at 10:40
  • $\begingroup$ @my2cts (and upvoter), the principle of detailed balance requires that every microscopic process is balanced by its reverse. Indeed, this is how the relationship between the various Einstein coefficients is derived. The statement is true (in equilibrium conditions), whether or not collisional (de)-excitation is important. $\endgroup$
    – ProfRob
    Commented Feb 9, 2020 at 11:12
  • $\begingroup$ @my2cts you are correct, in the case of nonradiative deexcitation, the energy gets transferred into the vibrational and rotational energies of the molecules, and sometimes no photon is emitted. $\endgroup$ Commented Feb 9, 2020 at 17:20
  • $\begingroup$ @ÁrpádSzendrei and in a case of two states in equilibrium for which radiative de-excitation is forbidden then the Einstein B absorption coefficient is also so small that collisional, rather than radiative excitation is also responsible for populating the upper level. This is the principle of detailed balance. Collisional processes are also in balance. Separately. $\endgroup$
    – ProfRob
    Commented Feb 9, 2020 at 17:43
1
$\begingroup$

First we need to clarify:

  1. when an photon interacts with an atom, three things can happen:

    1. elastic scattering, the photon keeps all its energy, and changes angle

    2. inelastic scattering, the photon gives part of its energy to the atom, and changes angle

    3. absorption, the photon gives all its energy to the atom, sending the valence electron to a higher energy level

  2. what you are wrong about is that you think the bound electrons are absorbing and emitting photons

  3. it is the atom that does it, and it, the nucleus and electron system has many available energy levels according to QM

  4. the energy of the photon has to be exactly the same as the difference between the energy levels of the valence electron

  5. the probability of the atom absorbing the photon is very high if the energy level is exactly the same.

  6. the probability is very small if the energy level of the photon is not the same

  7. that is why you see spectra

  8. the angle of elastic scattering (reflection) is the same as absorption for glass, and the opposite for mirrors (though in the case of glass, most of the photons get elastically scattered, and only few get absorbed, because elastic scattering is the only way to keep a mirror image, and the photons that get absorbed, will heat up the glass)

$\endgroup$
2
  • 4
    $\begingroup$ Sorry but I diagree with 8. there is no absorption and re- emission when images are retained, but elastic scattering, the photon through transparent media scatters elastically with the lattice. For a mirror with the reflecting atoms behind the glass, usually a level of meta. It has to be elastic so that the phases and colors are retained. Absorption implies energy changes, and possibly cascades of photons, depending on the material, and phases are lost, so no images reflected or transmitted. $\endgroup$
    – anna v
    Commented Apr 25, 2018 at 11:06
  • $\begingroup$ @annav you are correct, I edited. $\endgroup$ Commented Feb 9, 2020 at 17:17

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.