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I have noticed that most objects such as the red dye in a plastic stool look black under blue or green light.
From what I understand about direct band gap semiconductors, the dominant recombination mechanism is radiative recombination. Therefore If the dye was a direct band gap semiconductor I should expect it to still appear red under blue or green light as the electron in the red dye should absorb the light, thermalize and radiatively recombine releasing red light that's equal to the band gap in its energy. Therefore if it was a direct band gap I would expect it to fluoresce red in blue or green light.

The fact that it appears black implies there is another recombination mechanism such as indirect recombination that would cause the blue or green light that is absorbed to be entirely lost due to heat rather than by radiative emission.

Based on this I would assume that all indirect semiconductors would appear black, but for direct semiconductors, they should fluoresce.

I also understand that materials can have both indirect and direct band gaps. But for wide direct band gap semiconductors that have an indirect band gap that far offsets the direct band gap in energy should fluoresce under blue or green light.

Is my understanding correct or is there something I'm missing? Many thanks.

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    $\begingroup$ Most things are not semiconductors. $\endgroup$
    – Jon Custer
    Jun 28, 2022 at 21:16
  • $\begingroup$ I agree, as most objects are composite solids. But to my understanding, all materials in their pure form (free of impurities) are either metals or semiconductors. No insulator can be colored if it is pure as its band gap is too high and sits in the UV range and should all appear white. $\endgroup$
    – terminate
    Jul 4, 2022 at 12:58

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Maybe it is better to think through your observations it in terms of energy levels and molecules. You want to pick the right set of tools and nomenclature to solve your problem.

One way to get to band theory is to start with an atom, then multiple atoms and chains of atoms, then atoms periodically spaced in crystals.

Along the way, if you’re an atomic physics person you might think carefully about energy splitting of a certain energy level as two atoms are bought together. if a chemist maybe more in terms of molecular orbitals and for conductive polymers consider HOMO and LUMO levels and how electrons might move along some chain of atoms. If a Condensed matter physicist or electrical engineer start thinking conduction and valence bands or even band structures and with that direct and indirect bandgaps. But in all these different approaches you have you quantum, Pauli exclusion principle, molecular vibrations, phonons, excitons etc. The way you problem solve and describe the problem depends on what community you are in.

The point is that depending you seems to be trying to force fit a some concepts suitable for one case to a different case.

Dye molecules are typically very good absorbers, incidentally this implies they tend to be good emitters. But the emission peak will tend to be lower in energy than the absorption peak.

But not just that, you might notice that there could be different lifetimes for the emitters light. Diagram

So for a dye molecule, since you don’t really have a semiconductor in the sense of Bloch waves periodicity etc. talking about being direct or indirect for the Dye molecule isn’t really a correct approach. However you do still have a separation of energy levels you still have phonons etc and you look at what transitions are allowed.

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  • $\begingroup$ Thank you for your response, but I'm not sure how this answered my question. I'm familiar with bloch wave theory and band structures, but to my understanding, a wide direct band gap semiconductor such as a red dye should behave just like figure 4a in the image you sent. That is, all direct band gap semiconductors should fluoresce in monochromatic light that is of higher energy than the band gap (rather than appear black). We shouldn't think about it in terms of a single dye molecule but rather the bands that is formed as the dye molecules solidify. $\endgroup$
    – terminate
    Jun 29, 2022 at 17:17
  • $\begingroup$ You don't really have electron transport from dye molecule to dye molecule if it is a solid. It won't have a band structure like a conventional semiconductor. In fact, I think for most dyes you can have quenching of the fluorescence if dye molecules start to aggregate. On the other hand, you can probably call some dyes organic semiconductors but then people tend to discuss them in terms of HOMO and LUMO levels. If you can crystalize the dye as an organic semiconductor you might have case, but even then the density of states won't resemble the band structure of a conventional semiconductor. $\endgroup$
    – UVphoton
    Jun 29, 2022 at 18:00
  • $\begingroup$ I don't understand why this would be the case. To my understanding, when dye molecules aggregate they should experience a splitting of energy levels due to the stark and Ziemann effect. The aggregation should give rise to a band structure. You can have electron transport from dye molecule to dye molecule in its solid phase if this was not the case the electron mobility would be zero and dye-sensitized solar cells would not work. I thought that HOMO and LUMO levels would apply to dyes that were in solution. $\endgroup$
    – terminate
    Jul 4, 2022 at 12:53
  • $\begingroup$ I don't think so, I think the key part of the electron transport is from the molecule adsorbed to the something like TiO2 and that the LUMO HOMO levels are aligned to have better electron transfer from the adsorbed dye molecule to the semiconductor . What I usually have seen are band diagrams of LUMO HUMO levels in relation to the semiconductor bands. I think the aggregation is likely disordered and unless you end up with some type of crystallization that band like transport from molecule to molecule is unlikely, Maybe some hopping transport. Most dye cells have electrolyte. $\endgroup$
    – UVphoton
    Jul 4, 2022 at 16:15
  • $\begingroup$ Yes, you are right about dyes being single molecules (I didn't previously know that). That came as a surprise to me that solid-state forms of dyes are difficult to obtain. Thanks for your help. $\endgroup$
    – terminate
    Jul 7, 2022 at 17:02

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