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Ionization & photoelectrons in x-ray solid state detectors

  1. Nov 11, 2015 #1
    As far as i understand, this is the principle of solid state x-ray detectors:

    1. A photon of energy E enters the detector and goes through photoelectric absorption
    2. As a result a photoelectron with energy (E - Φ) is ejected, where Φ is the binding energy required to eject the photoelectron.
    3. A number of electron-hole-pairs are created, they are collected and thus the energy of the original x-ray can be determined.

    What i'm confused about is, is the amount of electron-hole-pairs proportional to E or actually (E - Φ) ? As far as i understand, the ionization cloud (ie. electron-hole-pairs) is created by the photoelectron, which has an energy of (E - Φ). Textbooks usually cite though, that the amount of electron-hole-pairs is proportional to the original x-ray energy E, but shouldn't that mean that the energy Φ would also have to be converted into electon-hole-pairs? How exactly is the energy Φ "absorbed" in this case? (Assuming that fluorescence or Auger electron emission does not happen).

    Many thanks in advance.
  2. jcsd
  3. Nov 11, 2015 #2


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    Usually, E (i.e. the photon energy for x-rays) is in the keV range. Φ in this case is the work function of the material, and it is in the low eV range. So E-Φ is not statistically different when compared to E in this process.

    Just so you know, the "electron-hole" pairs that are created is not the electron-hole pair that was created by the original incoming x-ray photon. In a photoelectric/photoemission process, one photon creates one electron-hole pair (single-photon photoemission). However, in the case of an x-ray detector, that first photoelectron was created deep in the bulk of the material (since x-ray is very penetrating). This photoelectron has a lot of energy still, and so it "rattles around" the semiconductor, knocking off other electrons. Many of these electrons will have energy higher than Φ, and thus, they too create electron-hole pairs. The higher the incoming x-ray, the higher the energy of that first photoelectron, and the more electron-hole pairs are created in the material. This is why you detect a higher current corresponding to higher energy x-ray (up to a point).

  4. Nov 11, 2015 #3
    Thanks for the reply, however I'm still confused.

    As far as i understand: in this case of photoelectric absorption, Φ is not the work function but the binding energy, eg. for Si it has values over 100 eV.
    I'm especially interested in detection of low energy photons (250-400 eV).

    Photon of 250 eV arrives at Si.
    -> Photon ejects a photoelectron from L shell, the binding energy is 100 eV.
    -> Photoelectron of just 150 eV is ejected, correct?

    For electron-hole-pair generation via ionization (ie by a free electron if I'm correct), the mean energy required to produce one pair in Si is 3,6 eV.
    So would the amount of electron-hole-pairs detected now be just 42 on average, instead of 69? (150/3,6 = 42 and 250/3,6 = 69)
  5. Nov 11, 2015 #4
    The 100eV hole will not just stay there but decay, either emitting Auger electrons or another, softer x-ray photon that is probably re-absorbed in the detector.

    It can happen that photons of a characteristic energy "escape" the detector without being detected. This can give rise to a lower energy side peak called "escape peak",
  6. Nov 11, 2015 #5


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    I checked a few sources, and I reconfirmed what I said earlier (for example, see this Princeton Instrument document, Pg. 3).

    The Φ is the work function IF the photoelectron is to produce subsequent cascade of electron-hole pairs. It is the only range of photoelectron that has enough "leftover energy" to cause a substantial series of multiple scattering to create all these electron-hole pairs.

    For the photon that excites the core level electrons (i.e. if it has to overcome a larger binding energy beyond the valence band), then these will not result in a considerable electron-hole pairs. Rather, they will be the ones that will likely result in the Auger or photoluminescence processes.

  7. Nov 16, 2015 #6
    Many thanks for the replis, i think i understand this now better.
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