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I'm fully aware that glass absorbs light in the ultraviolet band. Moreover, glass does not absorb any light which has a longer wavelength than that of the UV light, making it transparent.

This makes the UV photon the lowest energy photon that can be absorbed by glass. In other words, the UV photon moves the electron's energy from ground state (K) to the lowest excited state (L). Thus, making the only way the electron can fall back to lower energy level by moving from L shell to K shell, releasing a photon with the same energy as the one absorbed, emitting ultraviolet light. Is this true ?

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  • $\begingroup$ Well, glass will absorb x-rays and such, just not as much as in some regions of the UV. And it depends on the specific glass. $\endgroup$ – Jon Custer Mar 9 at 17:16
  • $\begingroup$ Yes, I know that glass can absorb shorter wavelength than UV. But the only photon glass can emit is the photon of UV light. How does this make glass a good at blocking UV light when absorbed UV light get released as UV light. $\endgroup$ – Nguyen Xuan Minh Mar 9 at 17:22
  • $\begingroup$ Constituents of glass emit x-rays just fine. But what you are looking for is non-radiative transitions. Glass is a solid, so it has a band gap, and excited electrons can quite happily lose energy by many non-radiative paths. $\endgroup$ – Jon Custer Mar 9 at 17:51
  • $\begingroup$ "the UV photon moves the electron's energy from ground state (K) to the lowest excited state (L)" Actually it is better to say that the UV photon moves a valence electron from a bonding to an anti-bonding orbital. $\endgroup$ – my2cts Mar 13 at 18:17
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When a photon interacts with an atom three things can happen:

  1. elastic scattering, the photon keeps 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

In your case it is a common misconception to think that UV light will be all absorbed by the glass.

When glass reflects visible light, like a mirror image, that is 1., elastic scattering.

In your case, when UV light or IR light (non visible) enters the glass deeper into the material, that is 2., inelastic scattering, or 3. absoption. Inelastic scattering will happen as the photons give their energy to the molecules in the glass lattice and transform into the vibrational energy of the molecules (heat the glass).

Absorption can happen too with glass, and will too heat up the material.

The question is only the ratio of the three interactions, but all will happen. With glass,

  1. visible light will mostly be reflected (1. elastic scattering) or will be refracted (will travel through the glass)

  2. non-visible light will mostly be inelastically scattered deeper in the material, and some will be absorbed heating up the material.

In your case, UV light will mostly be inelastically scattered or absorbed by the glass, heating it up.

Now the UV light that glass emits, is depending on the lattice structure of the glass, and the vibrational motions (heat energy) of the molecules in the lattice. Glass can emit UV light spontaneously too, because of the structure of the lattice, but it can do it too because the absorbed light will be re-emitted at a different frequency, glass can emit UV even if no UV is absorbed.

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  1. Use black body radiation, it always emit $\textbf{some}$ light in ultra violate region.

  2. When you electron poped out of the allowed excited states, the usual classical model was to treat it as a free electron. The free electrons usually bounce around and interact with the surrounding electrons in other latices, which effectively distrusted the energy to other latices, thus you almost always won't be able to absorb glass emit UV light, rather heat and bunch of other photons.

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Moreover, glass does not absorb any light which has a longer wavelength than that of the UV light, making them transparent.

This statement isn't correct for a couple of reasons.

First of all, it's too absolute. Glass does absorb some light in the visible band. Its just that it is a poor absorber there and most visible light is transmitted. But some is not none.

Second, depending on the specifics of the glass, it will have some stronger absorbance bands in the IR. It is not completely transparent at all wavelengths longer than visible. These are much preferred modes for transmission at low energy.

So there's no mechanism for glass to emit UV under normal conditions.

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  • $\begingroup$ I'm not sure this completely answers the question. Why would glass not exhibit luminescence? $\endgroup$ – garyp Mar 13 at 19:37
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Absorption image of Glass. Not as pretty only because glass is an amorphous structure.

Hi! So it is pretty common to mistake what glass does. Glass actually absorbs some wavelengths in the upper end of the spectrum (is you look at a typical glass pane from the side, you would see hints of green due to defects, and thinking about energy transitions, the smallest energy transition you could make is from you 1s shell splittings, which leads to that big bump on the right side.)

As for the question, by itself, without any external source, glass will not (I day will not but technically, highly improbable) emit UV light (light of wavelength ~ 10 to 380 nm) When shined with a UV light source though, yes, glass will absorb UV light. However, since, as shown in the picture above, we have multiple other states that we can jump to (right side), then the absorbed UV light may rather decay to the other longer wavelengths like IR. So, if UV is absorbed, yes some UV leaks out, but most of them will come out as IR or longer)

Tldr: UV is absorbed. Yes some UV leaks out, but most of them will come out as IR or longer)

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Any object will emit radiation. The spectrum depends on its temperature and of its electronic properties. A ideal black body will emit the famous black body spectrum. A real material emits this spectrum multiplied with a wave length dependent emissivity. This emissivity simply equals the absorptivity. Glass has thus high emissivity in the UV but the blackbody intensity is extremely low at temperatures below melting. The net result is what should be a very low emission. If found a perfect blackbody at 1200K emits 2 nanowatt/m2 above 300 nm. http://www.spectralcalc.com/blackbody_calculator/blackbody.php

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  • $\begingroup$ My interpretation of the OP is that the interest is in absorption and re-emission, not thermally generated radiation. (Also, the emissivity of glass in the visible is low.) $\endgroup$ – garyp Mar 13 at 19:39
  • $\begingroup$ @garyp Well, there is no absorption and reemission. $\endgroup$ – my2cts Mar 13 at 20:32

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