How Does Activating ZnS with Sub-ppm Ag Work?

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SUMMARY

The activation of ZnS (zinc sulfide) with sub-ppm levels of Ag (silver) is primarily due to the ability of Ag impurities to trap electron-hole pairs and facilitate recombination at lower energy levels, resulting in visible blue light emission. The band gap of ZnS is approximately 3.5 eV, which is not visible without the addition of Ag. This low concentration of Ag ensures minimal perturbation of the ZnS band structure, allowing for effective scintillation without introducing unexpected states in the band gap. The discussion highlights that even a small quantity of Ag can significantly enhance the luminescence efficiency of ZnS due to its high quantum efficiency.

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What is the mechanism that explains the activation of ZnS with less than ppm of Ag?

Bibliography from the 50’s mention of 10-7 parts of Ag in weight in order to activate the ZnS.
 
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It’s not clear from you question what exactly you want to know. How scintillators work? Why they contain activators? Why the activators are in such low concentration? If you clarify what you’re looking for, someone might be able to help you more easily.
 
Sorry! I don't understand how so few Ag atoms modify the whole product. Is like 1 Ag atom modifies the response of 10>7 molecules. What is the process? Is a field? Is a crystalline structure altered by the Ag?
 
It sounds like you don't really have an understanding of what's going on, so I'll try to start at the beginning. When ionizing radiation impinges on ZnS, it excites electrons from the valence band to the conduction band of the material. The scintillation is the light emitted when these electrons fall back down into the valence band (or, to put it another way, when they recombine with the hole in the valence band). The band gap of ZnS is ~3.5eV, which is in the UV. This means that the photoemission from electron-hole recombination in ZnS is not visible. To make it visible, you add Ag impurities. These trap electron-hole pairs and recombine them at a lower energy, so that the emission from these recombination events ends up being blue light. A low concentration of silver ensures that 1) there's a pathway for photoemission that's smaller than the band gap energy, but 2) the silver atoms in general are far enough apart that they only weakly perturb the ZnS band structure and don't interact to produce unexpected states in the band gap.
 
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Hi
Thank you for your detailed explanation.
The thing that bothers me is how a small quantity of an element may condition the whole mass of the product.
Atomic weight of Ag is 107.8 and ZnS molecular weight is 97.4, so, a 10-7 weight-to-weight presence of Ag in the product means that 1 atom of Ag modify the response of 10>7 molecules.
That amuses me.
 
I'm not an expert on zinc sulfide, but it might be the case that ZnS has a relatively low luminescence quantum efficiency on its own. This would mean that only a small number of electron-hole pairs recombine to give a photon (they could undergo radiationless recombination via a number of other processes). In this case, even a very small amount of dopant with a high quantum efficiency could dramatically increase the luminescence of the crystal.
 
Just out of interest I looked up doping levels in semiconductors. That seem to vary from about 1 in 10^9 to 1in 10^4 .
Surprising, as you say, how little is needed, but thinking in different terms, for impurities at 1 in 10^6 the linear spacing is only about 100 atoms. That doesn't sound quite so bad.
 
Merlin3189 said:
but thinking in different terms, for impurities at 1 in 10^6 the linear spacing is only about 100 atoms. That doesn't sound quite so bad.
Yes, in general the optimal spacing is determined to first order by the diffusion length of excitons through the bulk material, which tends to be on the order of 1's to 10's of nm.
 

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