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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).

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    $\begingroup$ 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. $\endgroup$ – dmckee Apr 24 '18 at 20:23
  • $\begingroup$ I'm sorry, I dont think I'm able to follow.. $\endgroup$ – Dylan Patel Apr 24 '18 at 20:26
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    $\begingroup$ I think @dmckee is basically approving your assumption but prefers rephrasing "scattered in all directions" to "scattered in other directions". $\endgroup$ – npojo Apr 24 '18 at 20:44
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    $\begingroup$ 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 Apr 24 '18 at 21:05
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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 is the same as absorption for glass, and the opposite for mirrors

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    $\begingroup$ 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 Apr 25 '18 at 11:06

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