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As far as I know (I may be wrong), substances seem to have a definite color because they reflect/transmit all the light rays of the given color, and absorb all the lights rays of the remaining colors (of course, we are talking about the visible region only). That means, that if a substance is red in color, it absorbs all the visible region light except red, and reflects red light.

When we say light is absorbed, it means that the electrons in the molecules of the substance get excited, right? Then, since energy levels are quantized, only certain wavelengths get absorbed. An apple, for example; it always appears to be red, no matter much light shines on it. So, that means for how much ever light I shine on it, all wavelengths except those corresponding to red get absorbed, and the electrons are continuously excited. How is that taking place without any deexcitation?

Assuming that if any deexcitation takes place, the energy is released as photons only. If so, all other wavelengths should also be emitted, and the apple should appear white.

How can we definitively say, that the energy gaps are equally spaced? If deexcitation does not take place, all substances, after shining light on them for a sufficient amount of time, should eventually become conductors, but I don't think that happens. Also, if the light source is switched off, the electrons should deexcite, and the apple should appear cyan in color (complementary color of red). However that does not happen. Why?

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  • $\begingroup$ You might be interested in this video. $\endgroup$ – Ali Nov 29 '13 at 20:29
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Under normal circumstances, what you are seeing is the steady state condition where the rate of absorption is equal to the rate at which energy is conducted away or radiated away again, so the material doesn't heat up. As I understand, unless a material fluoresces, the de-excitation happens in the infrared.

Also, you should realize that just because a photon being absorbed means that an electron went from energy $E_1$ to energy $E_{10}$, there is no reason to believe that in de-excitation, the electron will go back from $E_{10}$ to $E_1$ again in one jump. The energy levels of a solid are very complicated (band structure) and has very many finely spaced energy levels that are almost like a continuum. Where there are gaps in the band structure, the material is transparent.

Here is an example of the band structure of a random material (density of states of gold in the middle). I believe colour has more to do with the conduction bands (the upper squiggles) and the relative number of available states.

Band structure of gold

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  • $\begingroup$ ok, so, if there is an energy band (which is like a continuum), why do only specific wavelengths get absorbed? $\endgroup$ – user1767282 Nov 29 '13 at 18:17
  • $\begingroup$ Well, there are many bands. $\endgroup$ – lionelbrits Nov 29 '13 at 19:22
  • $\begingroup$ Could you please elaborate? $\endgroup$ – user1767282 Nov 29 '13 at 19:43
  • $\begingroup$ Also, would only the unpaired valence shell electrons get excited or can any electron be excited? $\endgroup$ – user1767282 Nov 29 '13 at 19:45
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Assuming that if any deexcitation takes place, the energy is released as photons only. If so, all other wavelengths should also be emitted, and the apple should appear white.

The fact that electron got excited via photon absorption has nothing to do with the fact how it will deexcite itself. Deexcitation means that your electron will lower its energy, giving away the excess energy somewhere. It can go to a photon of the same color, but it can also partially go to a photon of lower energy color and partially elsewhere, or not into a photon at all.

It doesn't even have to be pure electronic excitation. By shining light (in general sense, not only visible) you can excite vibration or rotation modes in a molecule as shown for $CO_{2}$ on the image below.

Modes of carbon dioxide molecule

So, deexcitation of such molecule can mean changing from one type of vibrational motion to the other. There can be no photon involved in this transition.

How can we definitively say, that the energy gaps are equally spaced?

Energy gaps are equally spaced in quantum harmonic oscillator, but in real potentials of atoms and molecules, the potential is not quadratic, as such it is anharmonic and doesn't have equally spaced energy levels.

Anharmonic potential energy levels

If deexcitation does not take place, all substances, after shining light on them for a sufficient amount of time, should eventually become conductors, but I don't think that happens.

If we continuously shine light on a semiconductor material with photons of energy equal or higher than the energy band gap of the material, we will excite more and more electrons to higher energy up to the point at which there will be exactly as many electrons in higher as in lower state. At this point the material will appear transparent to the light at this specific wavelength. If you could get even more electrons to the higher energy state you will get population inversion. The medium will not absorb radiation but amplify it and this is the main mechanism how lasers and optical amplifiers work. You can get to this situation via different mechanisms and only in materials that have specific energy level structure for it to occur. I don't know what is the structure of a molecule building apple peel but probably nobody tried making a laser out of it. Although people made lasers from blood and jelly. As Arthur Leonard Schawlow said, “Anything will lase if you hit it hard enough”.

Population inversion

As from electronic transport side of view, yes, material can become conductor in that way too. Metallic conductors are conducting because they have many free electrons that can carry the current. In semiconductor you can optically excite electrons to the conduction band and make it conduct electricity. Basically through this effect solar panels work. By using very energetic photons or very intense light of lower energy photons, you can make an isolator conductive too.

Photovolatic cell

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