- #1
XRF_Rebel
- 3
- 0
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.
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.