What factor determine whether a body behaves like a transparent object for EM waves of a particular frequency?
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2 Answers
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It depends on what your "glass" is made out of. If you work in a chemistry lab, it's common to use cuvettes made of pure quartz, SiO2, which has a bandgap of 10.2 eV and will very happily pass ultraviolet light. However, most glass is only 80% SiO2 and its the impurities that lower the bandgap.
The "bandgap" is the distance between the ground state of the electrons and the first excited state. A bandgap of 10.2 eV means that jumping that electron from the ground state to the excited state takes 10.2 eV.
A photon that's 10.2 eV has a wavelength of $1.21\times10^{-7} m$, which is UV light. Less energetic light (e.g. blue-to-red) just doesn't have enough energy to kick up that electron. So the photons pass through because they can't be absorbed.
The impurities lower the band gap because they create many different local states for those ground-state electrons. Some of those electrons now have a bigger bandgap, and others have a smaller bandgap. The smaller the bandgap, the higher the chance that a photon passing through will be absorbed.
This is important in, say, water purification plants, where water is passed through quartz sleeves and exposed to UV light.
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2$\begingroup$ Actually, the tubes of the germicidal lamps themselves are also made of quartz for the same reason. $\endgroup$– RuslanCommented Feb 24, 2020 at 10:03
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$\begingroup$ Indeed. And people think that the germicidal lamps are made of glass, even though they are made of quartz... $\endgroup$– vy32Commented Feb 25, 2020 at 21:54
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$\begingroup$ Unfortunately, both answers are skipping the basic reason of "why/how" this works. I'm trying to understand (cf. my comment on Pieter's answer). What key words should I google/youtube to know more about it (about UVC specifically or about electromagnetism waves passing through or blocked by materials) ? $\endgroup$– JinSnowCommented Mar 11, 2020 at 8:53
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$\begingroup$ @Jinsnow, do you understand the concept of a "bandgap?" $\endgroup$– vy32Commented Mar 12, 2020 at 2:29
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1$\begingroup$ Bandgaps have an energy level. The bandgap for quartz is bigger than for glass. it's so big even UV light doesn't have enough energy to jump an electron from the bottom of the bandgap to the top. So the UV light can't get absorbed, because there is nothing to absorb it. $\endgroup$– vy32Commented Mar 17, 2020 at 2:19
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Glass has a band gap of about 5 eV, which corresponds to photon energies at UV frequencies. Photons with lower energy do not have enough energy to excite the valence electrons to unoccupied states.
The values depend a bit. Quartz glass has a larger gap than lime glass and borosilicate glass.
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$\begingroup$ Normal glass is not a crystal, so it doesn't have a band gap. It may only have a lower density of states at some energy range, but definitely not as low as it would be in a crystal. $\endgroup$– RuslanCommented Feb 24, 2020 at 10:01
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$\begingroup$ @Ruslan Normally, melting does not make much of a change in electronic structure. Think in terms of tight-binding theory. $\endgroup$– user137289Commented Feb 24, 2020 at 10:57
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$\begingroup$ Given that $\mathrm{SiO}_2$ has a bunch of polymorphs, I wouldn't say melting doesn't change electronic structure. After all, re-crystallization can result in any of these forms, in which electronic structure differs considerably due to differing symmetry. $\endgroup$– RuslanCommented Feb 24, 2020 at 11:16
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$\begingroup$ @Ruslan The differences in optical properties are small between quartz glass and crystalline quartz: hardly any difference in transparency and optical band gap. But of course the crystalline form is birefringent and optically active while the amorphous solid is isotropic. Also molten salt will have the same band gap as crystalline rocksalt. Long-range order does not matter that much. $\endgroup$– user137289Commented Feb 24, 2020 at 13:12
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1$\begingroup$ @Pieter Thanks! Britney's pictures didn't help, but your keywords "photon passing through material" lead me to the track I was looking for. $\endgroup$– JinSnowCommented Mar 17, 2020 at 9:29