K-Shell, Low-Energetic Photons & Photoelectric effect

[cemtu]

TL;DR Summary

the probability of the photoelectric effect

At low photon energies, the probability of the photoelectric effect to occur increases, but the probability of the photoelectric effect to happen also increases when going towards most inner shells like K shell but inner shells require much more photon energies to be broken, so isn't there a contrast here? How is this EVEN possible?

 

[mfb]

For every given energy level where an electron can be ionized, the cross section decreases with increasing energy. If you reach enough energy to ionize additional energy levels (that are bound tighter) then you get an upward jump in the total cross section.

 

[cemtu]

For every given energy level where an electron can be ionized, the cross section decreases with increasing energy. If you reach enough energy to ionize additional energy levels (that are bound tighter) then you get an upward jump in the total cross section.

okay but it nevertheless happens in the area of lower ionization possibility.

 

[Steve4Physics]

TL;DR Summary: the probability of the photoelectric effect

At low photon energies, the probability of the photoelectric effect to occur increases,

Yes. At low photon energies the photoelectric effect (PEE) is the dominant absorption process. The overall trend is for it to decrease with increasing energy. As shown on the graph here

but the probability of the photoelectric effect to happen also increases when going towards most inner shells like K shell but inner shells require much more photon energies to be broken, so isn't there a contrast here? How is this EVEN possible?

There is a sort of 'resonance' effect. If the photon energy happens to match an inner shell's ionisation energy, there is an increased probaility of ionising an electron from that shell. As shown here. But as the energy continues to increase (passing the 'resonance') the PEE absortion continue to decrease.

I guess the 'resonance' affect is explained by quantum mechanics.

 

[cemtu]

Yes. At low photon energies the photoelectric effect (PEE) is the dominant absorption process. The overall trend is for it to decrease with increasing energy. As shown on the graph hereThere is a sort of 'resonance' effect. If the photon energy happens to match an inner shell's ionisation energy, there is an increased probaility of ionising an electron from that shell. As shown here. But as the energy continues to increase (passing the 'resonance') the PEE absortion continue to decrease.

I guess the 'resonance' affect is explained by quantum mechanics.

okay, but the spike of resonance effect in the K-shell has still a lower probability for the photoelectric effect to occur than the L-Shell, and yet it is said that inner shells are the most probable for the photoelectric effect to be observed! I do not get it.

 

[Steve4Physics]

okay, but the spike of resonance effect in the K-shell has still a lower probability for the photoelectric effect to occur than the L-Shell, and yet it is said that inner shells are the most probable for the photoelectric effect to be observed! I do not get it.

I wonder it this is simply an issue of (mis)interpretation of something you have been told or read?

For photons of energy E, the general trend is that the probability of absorption by the PEE decreases (roughly) with E³ (i.e. ∝1/E³).

But if E happens to match a shell’s ionisation energy, there is a big (e.g. factor of 10) increase in the probability of absorption due to 'resonance'.

The net result is a combination of both of these processes (∝1/E³ and resonance).

The statement “inner shells are the most probable for the photoelectric effect to be observed” is a rather general/non-specific statement.

Let’s make-up some figures. Suppose an element’s K-shell electron has an ionisation energy of 50keV; and for an L-shell electron it is 30keV. I don’t think there are any claims that the net probability of absorption of a 50keV photon by K-ionisation should greater than the probability of absorption of a 30keV photon by L-ionisation.

Edit -typo's.

 

[cemtu]

I wonder it this is simply an issue of (mis)interpretation of something you have been told or read?

For photons of energy E, the general trend is that the probability of absorption by the PEE decreases (roughly) with E³ (i.e. ∝1/E³).

But if E happens to match a shell’s ionisation energy, there is a big (e.g. factor of 10) increase in the probability of absorption due to 'resonance'.

The net result is a combination of both of these processes (∝1/E³ and resonance).

The statement “inner shells are the most probable for the photoelectric effect to be observed” is a rather general/non-specific statement.

Let’s make-up some figures. Suppose an element’s K-shell electron has an ionisation energy of 50keV; and for an L-shell electron it is 30keV. I don’t think there are any claims that the net probability of absorption of a 50keV photon by K-ionisation should greater than the probability of absorption of a 30keV photon by L-ionisation.

Edit -typo's.

HELP! Do you see the parodox here? K inner shell requires more photon energy to break free, so should require higher energetic photons to break free, but photoelectric effect happens when photon energy is low amd K orbital here has the greatest possibility to break free by photoelectric effect as it is stated as "usually"! @mfb @Steve4Physics

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:

 

[Steve4Physics]

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).

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.

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?

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!

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.

 

[cemtu]

I don't believe that is correct. Can you give a link?

Not you, look at my sources, the pictures, (I tagged you to summon you here thats all):

"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'?

Source for first picture: https://www.nrc.gov/docs/ML1121/ML11210B520.pdf
(slide 29)

Source for the second picture: https://radiopaedia.org/articles/photoelectric-effect

People have argued this it seems. And I am seeing it just now: https://www.physicsforums.com/threads/why-is-it-usually-the-k-shell-electron-that-is-ejected.706719/

• In this argument in the above link, they seem to agree on basically this:

"As the orbital number of an atom increases, its radius also increases. This is because the outer electrons are further away from the nucleus and are therefore less attracted to it. The effective nuclear charge experienced by the outer electrons is also less, as it is shielded by the inner electrons. This allows the outer electrons to occupy larger orbitals.
"
So electron density matters a lot, inner shells have larger electron density than outer shells and this makes them easier to get shot statistically -classically speaking, analogy-

 

[mfb]

K inner shell requires more photon energy to break free, so should require higher energetic photons to break free,

Correct

but photoelectric effect happens when photon energy is low

It's not limited to low energies.

amd K orbital here has the greatest possibility to break free by photoelectric effect as it is stated as "usually"!

It has the greatest probability at a given photon energy if the photon has enough energy for it.

When comparing different sources, keep in mind that the meaning of "low energy" can vary a lot. For radio astronomy, low energy might be 10^-9 eV, for particle physicists a 10¹⁰ eV photon can be "low energy". It's so low that ATLAS and CMS generally don't trigger on them.

 

[Steve4Physics]

Not you, look at my sources, the pictures, (I tagged you to summon you here thats all):

Sorry - misinterpreted what you said.

Not sure I can add much. But will note this...

“... inner shells are the most probable for the photoelectric effect to be observed"​

This is an incomplete (and potentially misleading) statement. It contains no context.

For example, if the incident photon energy is less than the innermost (K) binding-energy, the probability of a K-electron being knocked out is zero, not ‘most probable’.

On the other hand if the incident photon energy is (a little) bigger than the innermost K-energy, the probability of a K-electron being knocked becomes very significant and most of the emitted electrons would indeed be from the K-shell (with few from other shells). Maybe that's the intended message.

(Also, it's worth noting that in practice we are not dealing with monochromatic X-rays and the spectral composition will affect what happens.)