Interaction mechanisms of photons and electrons with matter

In summary: This is because incident photons carry more energy than the electrons that are used in the experiment. It's not like they just ignore secondary electrons. If you use incident photons and are interested in making a graph of secondary electron yield VS primary electron energy (ev), what would be the difference if we used incident photons instead of electrons, of the same energy?In summary, an experiment using incident photons will not produce secondary electrons.
  • #1
superduke1200
57
1
Hello everyone,

I would like to ask a couple of questions related with the interaction mechanisms of photons and electrons with matter. Through searching about this subject, I have concluded that they both have different penetration depth and different interaction mechanisms. But apart from this general notation, I have unfortunately failded to find something more specific. The obvious things that anyone could say, is that among other qualities, electrons do not have charge or mass like electrons do and thus they should undergo different interactions.

But the question is what are those differences? For example I learned thanks to Zz in the thread about Auger Vs Secondary Electrons, that in an experiment where the incident beam uses photons, we will never get Secondary electrons. Why? Because we call secondary, the electrons that come out after primary electrons have interacted with matter. But in a very specific experiment, where we use incident electrons and we are interested in making a graph of Secondary electron yield VS Primary electron energy (ev), what would be the difference if we used incident photons instead of electrons, of the same energy?

Finally, what would the result be, if we did a XRD experiment and we used electrons instead of X rays, with the same energy? Of course, I know that it is not ideal to suppose that there is a magical way to do something like that, but what I wanted to do through this example was to further clarify my question's point.

Thanks a lot for your time. I hope my question is clear enough.
 
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  • #2
superduke1200 said:
in an experiment where the incident beam uses photons, we will never get Secondary electrons. Why? Because we call secondary, the electrons that come out after primary electrons have interacted with matter.
That looks like semantics. You will get electrons if a photon beam hits a target, no matter how you call them.

The general idea: electrons and positrons produce photons, photons produce electrons and (above 1 MeV) positrons. If the target is thick enough and if the energy is sufficient, you can get multiple interactions, making an electromagnetic shower in the material.

superduke1200 said:
Finally, what would the result be, if we did a XRD experiment and we used electrons instead of X rays, with the same energy?
Then you get an electron microscope.
 
  • #3
mfb said:
That looks like semantics. You will get electrons if a photon beam hits a target, no matter how you call them.

As I said the words about secondary electrons originating only when we have primary electrons are not mine. You can check number 20 here https://www.physicsforums.com/threads/auger-electrons-vs-secondary-electrons.809979/#post-5086828 if you wish. Nevertheless Zz's notation was logical enough I think. If he sees this thread, perhaps he could help.

mfb said:
If the target is thick enough and if the energy is sufficient, you can get multiple interactions, making an electromagnetic shower in the material.

Could you be a bit more specific? Which are these multiple interactions making an electromagnetic shower in the material?

mfb said:
Then you get an electron microscope.

This is logical and profound. I know that techniques like SEM and TEM are electron microscopies. My question is: given that XRD gives someone information about a material being a crystal or not, will a technique that uses electrons instead of photons give us the same diffraction pattern for a certain material?
 
  • #4
superduke1200 said:
Nevertheless Zz's notation was logical enough I think. If he sees this thread, perhaps he could help.

After my last encounter, I did not want to touch this even with a 10-ft pole!

This is logical and profound. I know that techniques like SEM and TEM are electron microscopies. My question is: given that XRD gives someone information about a material being a crystal or not, will a technique that uses electrons instead of photons give us the same diffraction pattern for a certain material?

Look at RHEED and LEED.

Zz.
 
  • #5
ZapperZ said:
After my last encounter, I did not want to touch this even with a 10-ft pole!

Excuse me if your last encounter was not so pleasant. Your opinion is respected. Someone helps when he wants to.

ZapperZ said:
Look at RHEED and LEED.

I will definitely look at both of them. Thanks for helping
 
  • #6
superduke1200 said:
Could you be a bit more specific? Which are these multiple interactions making an electromagnetic shower in the material?
If the energy is high enough, a photon will make pair production, both electron and positron produced there will produce multiple photons via bremsstrahlung, those photons can make pair production again, ...
At lower energies, compton effect becomes relevant as well, and at very low energies you get the "normal" ionization.
 
  • #7
mfb said:
The general idea: electrons and positrons produce photons, photons produce electrons and (above 1 MeV) positrons. If the target is thick enough and if the energy is sufficient, you can get multiple interactions, making an electromagnetic shower in the material.

About your general notation.. I would not say that I agree with you so much since experiments which involve incident photons tend to give both electons( XPS) and photons ( XRD, Compton experiment). The experiment’s set up is the one that will collect either photons or electrons in order to give us information about our specimen. On the contrary, the multiple interactions that you described are clear and to some extent prove that someone gets both electrons and photons ( electromagnetic shower ).

mfb said:
a photon will make pair production, both electron and positron produced there will produce multiple photons via bremsstrahlung, those photons can make pair production again, ...

So, in order to get the electromagnetic shower, someone has to have high energy icident photons( 1mev=1,2 pm) as you said. Does this mean that we will not get these multiple interactions that cause the electromagnetic shower with an incident beam of incident electrons with the same wavelength 1,2 pm?

Furhermore, I do not know if that is true, but judging from my limited knowledge, an incident beam of photons with the same wavelength with a beam of electrons will penetrate in a target more than an incident beam of electrons with the same wavelength. And actually that was part of my question. Which of them ( electrons and photons ) tend to penetrate more and why?

Now about RHEED. Zz's answer, did perfectly answer my question. I can understand that the reason why electrons do not interact with the material's bulk but only with their surface, is the extremely small angle of interference. What I fail to understand though, is how we get a diffraction pattern of electrons that happen to interact constructively. What I can imagine though, is that incident electrons with wavelenghts similar to interatomic distances interact constructively according to Bragg's law. I have heard only about photons diffracted according to Bragg's law and I did not encounter someone talking about Bragg's law at RHEED when I searched for it, but that's just what I suppose. Anyway, this thread's purpose is not RHEED analysis.

Finally, I have to admit that even now I would not be able to answer if someone asked me what I asked: how do electrons interact with matter, how do photons interact with matter and why are their interactions different?

I know in general that when an incident beam of electrons interacts with a target, we may get: backscattered electrons, secondary electrons, absorbed electrons, elastically scattered electrons, Bremsstrahlung Xrays and inelastically scattered electrons. But I do not know when we get each of those cases. Plus I do not know what are all the possible effects when photons interact with a target. That is why I am still not even close to saying, that I know the ways that electrons and photons interact with matter.
 
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  • #8
superduke1200 said:
About your general notation.. I would not say that I agree with you so much since experiments which involve incident photons tend to give both electons( XPS) and photons ( XRD, Compton experiment). The experiment’s set up is the one that will collect either photons or electrons in order to give us information about our specimen. On the contrary, the multiple interactions that you described are clear and to some extent prove that someone gets both electrons and photons ( electromagnetic shower ).
As mentioned, it depends on the energy and the target. A very thin target will never have showers because more than one interaction is extremely unlikely. And at low energy you don't get showers either.

superduke1200 said:
So, in order to get the electromagnetic shower, someone has to have high energy icident photons( 1mev=1,2 pm) as you said. Does this mean that we will not get these multiple interactions that cause the electromagnetic shower with an incident beam of incident electrons with the same wavelength 1,2 pm?
Not with pair production for sure. You still have ionization and bremsstrahlung as processes.

superduke1200 said:
Furhermore, I do not know if that is true, but judging from my limited knowledge, an incident beam of photons with the same wavelength with a beam of electrons will penetrate in a target more than an incident beam of electrons with the same wavelength. And actually that was part of my question. Which of them ( electrons and photons ) tend to penetrate more and why?
At low energy: photons, as they are uncharged.

superduke1200 said:
why are their interactions different?
They are completely different particles, there is no reason to assume that their interactions would be the same.
superduke1200 said:
I know in general that when an incident beam of electrons interacts with a target, we may get: backscattered electrons, secondary electrons, absorbed electrons, elastically scattered electrons, Bremsstrahlung Xrays and inelastically scattered electrons. But I do not know when we get each of those cases.
In general, you get all of them, the relative fraction depends on the setup.
 
  • #9
mfb said:
At low energy: photons, as they are uncharged.

One last question, since your last answer really helped me to make some clear conclusions. In high energies which are the differences? Do we suppose that in this case electron's charge is not so important in order to give them a different approach from photons?
 
  • #10
At really high energy you need a much thicker wall of lead to stop photons than to stop electrons
 
  • #11
my2cts said:
At really high energy you need a much thicker wall of lead to stop photons than to stop electrons

Why do you say that? At multi-GeV energies and above, both electrons and photons shower, and the showers behave the same. (And since PF likes to quibble, yes, I know there is a factor of 7/9 for the very first interaction of the shower)
 
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  • #12
Vanadium 50 said:
Why do you say that? At multi-GeV energies and above, both electrons and photons shower, and the showers behave the same. (And since PF likes to quibble, yes, I know there is a factor of 7/9 for the very first interaction of the shower)
This may be so, but it does not tell me that the probability to cause a shower is equal for electrons and photons at equal energy. Is it ?
 
  • #13
my2cts said:
At really high energy you need a much thicker wall of lead to stop photons than to stop electrons

my2cts said:
This may be so, but it does not tell me that the probability to cause a shower is equal for electrons and photons at equal energy. Is it ?

These two statements are not equivalent.

Nonetheless, if you want to stop a multi-GeV photon or electron, you need enough material so that the shower probability is essentially 100%.
 
  • #14
Vanadium 50 said:
These two statements are not equivalent.

Nonetheless, if you want to stop a multi-GeV photon or electron, you need enough material so that the shower probability is essentially 100%.
Would significantly more material be required to achieve this for a photon than for an electron ?
 
  • #15
No. Something like 1/9 of a radiation length difference, where you need many radiation lengths (10+) to contain the shower.
 

What is the interaction mechanism between photons and electrons?

The interaction between photons and electrons is known as the photoelectric effect. This occurs when photons, which are packets of light energy, strike a material and transfer their energy to electrons, causing them to be emitted from the material.

How do photons and electrons interact with matter?

Photons and electrons can interact with matter in several ways, including the photoelectric effect, Compton scattering, and pair production. These interactions are dependent on the energy of the photons and the properties of the material.

What is the role of electrons in the interaction with photons?

Electrons play a crucial role in the interaction with photons as they are the particles that absorb and emit the photon energy. The energy transferred from photons to electrons can result in the emission of electrons from a material or the creation of an electron-positron pair.

How does the interaction between photons and electrons affect materials?

The interaction between photons and electrons can have various effects on materials, depending on the intensity and energy of the photons. These effects can include changes in the material's electrical and optical properties, as well as alterations in its atomic and molecular structure.

What are some real-world applications of understanding the interaction between photons and electrons with matter?

Understanding the interaction between photons and electrons with matter has many practical applications, such as in solar cells, photodetectors, and medical imaging technologies. It also plays a crucial role in fields such as quantum physics and materials science.

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