Probability of Photoelectric Effect Occuring in the Context of Medical Imaging

In summary, the probability of the photoelectric effect occurring is proportional to Z^3 and E^-3, where Z is the atomic number and E is the energy of the photon. This means that the photon's energy must be close to the binding energy of the inner electron to be absorbed. However, while increasing the energy decreases the probability of interaction, a higher atomic number can actually increase the probability. This may seem contradictory since higher Z values correspond to higher binding energies, but this can be explained by finding a material with a high binding energy that matches the energy of the photon beam.
  • #1
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I was recently taught that the probability of the photoelectric effect occurring was proportional to Z^3 and E^-3 (where Z is the atomic number, and E is the energy of the photon). My understanding is that the photon's energy must be close to the binding energy of the inner electron to be absorbed, increasing the energy decreases probability because you're getting further from the binding energy value. However, I don't understand how a higher atomic number would increase the probability; I've been told that more tightly bound electrons are more likely to be ejected. Why would that be? Wouldn't tightly bound electrons be difficult to remove?
 
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Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
Thanks for the reply! It might be a weird question, I assume that most people learned the photoelectric effect in terms of KE(of the photoelectron) = hf (incident photon) - work function, but I was told in my medical imaging class that the probability of a photon being absorbed is proportional to Z^3 and E^-3 (atomic number and energy respectively). I'm sure my class talks about the photoelectric effect in terms of non-metal matter (bones have a average Z greater than soft tissue, so bones absorb more photons), while the KE=hf-(work function) equation applies to metals with "electron seas". Also, my class talks about the photoelectric effect in terms of ejecting the inner electron.

According to the wiki book below, the energy of a photon must be equal or greater than the binding energy to eject the inner electron. Higher energies decrease the probability of interaction. The book never explained it but I assume that a beam of higher energy photons will have less photons that can be fully absorbed by the electron, as most photons will be too high in energy.

http://en.wikibooks.org/wiki/Basic_Physics_of_Digital_Radiography/The_Patient

What I don't understand is that the wiki book says that the more tightly bound an electron is (high Z), the higher the probability of a photon interacting with it. This contradicts the first point (that the energy of a photon must be equal or greater than the binding energy of the electron) - a higher Z means higher binding energy so less photons are able to eject an electron.

It does make a little sense if you have a very high energy beam and you find a material with a high binding energy to match the beam, but if you keep the energy of the beam constant, I would assume that increasing the Z number would decrease probability of interaction.
 
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1. What is the probability of photoelectric effect occurring in medical imaging?

The probability of photoelectric effect occurring in medical imaging is dependent on various factors such as the energy of the incident photons, the atomic number of the material being imaged, and the thickness of the material. Generally, for medical imaging purposes, the probability is relatively low, ranging from 0.1% to 1%.

2. How does the energy of the incident photons affect the probability of photoelectric effect?

The probability of photoelectric effect increases with an increase in the energy of the incident photons. This is because higher energy photons have a greater chance of interacting with the electrons in the material being imaged, leading to the release of photoelectrons.

3. Does the atomic number of the material being imaged affect the probability of photoelectric effect?

Yes, the atomic number of the material being imaged directly affects the probability of photoelectric effect. Materials with higher atomic numbers, such as bone or metal, have a higher probability of photoelectric effect as they have more tightly bound electrons, making them more susceptible to interactions with photons.

4. How does the thickness of the material being imaged impact the probability of photoelectric effect?

The probability of photoelectric effect increases with an increase in the thickness of the material being imaged. This is because a thicker material provides more opportunities for photon-electron interactions to occur, increasing the overall probability of photoelectric effect.

5. Are there any factors that can decrease the probability of photoelectric effect in medical imaging?

Yes, there are certain factors that can decrease the probability of photoelectric effect in medical imaging. These include using lower energy photons, using materials with lower atomic numbers, and reducing the thickness of the material being imaged. Additionally, using collimation and filters can also decrease the probability of photoelectric effect by limiting the number of photons that reach the imaging detector.

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