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Thermally excited holes

  1. Oct 6, 2016 #1
    I am looking at a paper that takes about thermally excited holes in relation to photocatalytic destruction, another paper with the same substance (TiO2) talks about decomposition of polycarbonates with the same method (thermally excited holes). I'm a little rusty on this topic. Please could someone explain in layman's terms what thermally excited holes are and how they would aid in a decomposition process?

    Any suggestions would be appreciated. Extracts from abstracts below.

    Thanks for any help.
  2. jcsd
  3. Oct 6, 2016 #2


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    In the band theory of the electronic structure of solids, the electrons in a solid fill energy levels up to the Fermi level. In an insulator like TiO2, the Fermi level is above one band and below another. The band below the Fermi level is called the "valence band" and the one above is called the "conduction band." At absolute zero, the valence band is completely filled with electrons and the conduction band is completely empty. However, at finite temperatures, some electrons are excited from the valence band into the conduction band. These electrons are therefore thermally excited. The absence of electrons in the valence band leaves behind "holes," positively charged quasiparticles that behave a lot like charge-reversed electrons. Thus at finite temperatures, there will be thermal electrons as well as thermal holes. I googled the papers you quoted above, and what they're claiming is the following:
    1) Due to a phenomenon known as band bending, the thermal holes concentrate at the surface of the TiO2 (this part is well-known in the literature and is uncontroversial).
    2) These holes are at a suitable potential to oxidize polycarbonate at the surface of the titanium oxide. Basically, the positively charged holes tend to pull electrons off the polycarbonate (oxidation is loss of electrons), which then leads to further reactions with oxygen in the air, which breaks down the polymer. This is the main finding of their papers. I haven't read the papers too closely, but at first glance, it looks like they corroborate their claims by showing that polymer breakdown is inhibited by the presence of antioxidants.
  4. Oct 7, 2016 #3
    Ahhh I see now, I reread the second paper with regards to what you wrote above, since TiO2 has thermally excited holes the presence of such things as OH radicals with free electrons will cause them to exist in a "metastable {TiO2 + OH}" state.

    Just one thing I'm unsure about, if TiO2 and OH are reacting together why would "This oxidation potential of {TiO2 + OH} is sufficiently high for decomposition of 1,4-dioxane"? How would the combination of thermally excited holes and the free electron of the OH radical create a "oxidation potential" for another substance in the region?


    Ok I think I've kind of got it, I read the reference "Role of OH Radlcals and Trapped Holes In Photocatalysis. A Pulse Radlolysls Study" and it seems to be suggesting that the joined hole and electron make a TiO radical and it is this that can increase chemical reaction.

    I think. Would that account for an increase in reaction rates?

    ^^^^ Thanks TeethWhitener, extremely helpful!
  5. Oct 7, 2016 #4


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    Ok, I've read the paper a little more closely.
    The authors seem to be referring to an additional possibility. The main thrust of this paper is to answer the question: does photoexcitation of TiO2 from sonoluminescence increase the rate of dioxane decomposition? The idea is that sonication produces cavitation in a solvent, which leads to sonoluminescence (basically flashes of light that come from the superheating caused by collapse of solvent bubbles during sonication). This sonoluminescence can photoexcite charge carriers in TiO2, which then might act as oxidation catalysts for the breakdown of dioxane. The researchers find that the sonoluminescence is in fact not particularly important to the dioxane breakdown. Instead, they hypothesize that the thermal excitation imparted by thermomechanical heating from the sonication of the solvent is enough to excite charge carriers to catalyze dioxane decomposition. The quote above introduces a third explanation for the observed effects; namely, that instead of the thermally excited holes in TiO2 directly oxidizing dioxane, they might react with water to give OH radicals. These OH radicals themselves can oxidize dioxane. The authors mention that the OH radicals can adsorb onto the TiO2 surface, where they trade off a lower oxidation potential for a much longer lifetime in solution. The authors note that the oxidation potential, while lower than solvated OH radicals, is still high enough to oxidize dioxane. That's my take on the whole situation.
  6. Oct 7, 2016 #5
    Wow, many thanks for your time! That's a very clear interpretation! I'll take another look at the paper with regards to what you have written above and recomment if I have any further problems, but I think the above will be sufficient to understand the parts that I was struggling with! Thanks again, very much appreciated!
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