cemtu said:
K inner shell requires more photon energy to break free, so should require higher energetic photons to break free
That's true. I'd say it slightly differently: The
minimum photon energy needed to ionise an inner shell electron is greater than the minimum photon energy needed to ionise an 'more outer' shell electron (for a given element).
cemtu said:
but photoelectric effect happens when photon energy is low
Or to be a little more precise:
The PEE is most likely to occur (in a particular element) when the incident photon energy is sufficiently large and of comparable size to the binding energies of the element's electrons. This corresponds to the low energy end of the X- ray spectrum.
cemtu said:
amd K orbital here has the greatest possibility to break free by photoelectric effect
I don't believe that is correct. Can you give a link?
cemtu said:
as it is stated as "usually"!
@mfb @Steve4Physics
I don't believe I ever said that. And I can't see any reference to the word "usually" anywhere!
cemtu said:
It seems like all things above contradict the graph below, which shows that the mass attenuation coefficient for inner shells, from outer to inner shells, decreases by increase in photon energy, which means that probability of seeing a photoelectric effect in inner shells decrease by increase in energy:
View attachment 335180
In your graph (above attachmnent) for simplicity let's take the K-edge as 100keV and the L-edge as 10keV.
If you fire 10keV photons at the material:
Probability of L-electron ionisation by PEE = x
Probability of K-electron ionisation by PEE = 0
If you fire 100keV photons at the material:
Probability of L-electron ionisation by PEE = 0 (or very low)
Probability of K-electron ionisation by PEE = y
x is about 10 times bigger than y which is not a problem as far as I know.
In Post #5 you wrote:
"and yet it is said that inner shells are the most probable for the photoelectric effect to be observed!"
This is misleading/confusing. Where is it 'said'?
For example, if 10keV photons are used, you will
never see K (innermost) electrons emitted. So the innermost shell is not the 'most probable' in those circumstances.