Photoelectric effect vs Compton scattering

hms.tech

What is the difference between the two ?

Maybe that in the photoelectric effect all of the energy of a photon is absorbed by the electrons ? Is that it ?

In Compton scattering, only part of the energy is absorbed hence there is a photon of lower frequency emitted !

OR

Maybe in the two processes we use e.m waves of different frequencies ?

Here is another thought : Which one of them is the closest of being the exact opposite of the production of x-rays ?

To answer the above query, I need to know :

Are the electrons (from the bremsstrahlung radiation) decelerated to exactly zero velocity (after striking the metal surface) OR are they reflected (eleastically/inelastically) from the metal surface ?

ZapperZ

You missed one very important characteristics of both of them - momentum conservation.

In Compton scattering, there is a direct relation between the momentum of the photon, and the momentum of the new photo+electron after scattering.

In the photoelectric effect, you cannot use momentum conservation between the incoming photon and the outgoing photoelectron due to the presence of the lattice ions.

Zz.

hms.tech

After reading you post, it occurred to me that momentum is not conserved during production of x-rays (via "break" radiation/bremsstrahlung) since the fast moving cathode rays strike a thick metal surface which is essentially immovable .

So does that imply the photoelectric effect is the exact opposite of x-ray production via bremsstrahlung

ZapperZ

Are you now changing the topic?

I am simply and narrowly addressing the question you posted in the topic. I hate a thread going in 20 different directions and can't come to a closure.

Zz.

hms.tech

No ! I posted this exact same query in my 1st Post:

Here is another thought : Which one of them is the closest of being the exact opposite of the production of x-rays ?




I just made a link between your reply and a concept which I couldn't understand and then asked you whether you think its true .

ZapperZ

And I ignored it because (i) it has really nothing to do with the comparison between photoelectric effect and Compton scattering and (ii) comparison between them and x-ray production is rather puzzling, because bremsstrahlung should really be a different mechanism than the two.

But this is really besides the point. Do you still need to know the differences between Photoelectric effect and Compton scattering, i.e. the thing you posted in the Title of this thread?

Zz.

hms.tech

Yes.

According to Wikipedia, both of these processes are called " light-matter interaction" and the article says that :


Low-energy phenomena:

Photoelectric effect

Mid-energy phenomena:

Thomson scattering
Compton scattering

High-energy phenomena:

Pair production


What makes Photoelectric effect a low energy phenomenon ?
Does it consist of only long wavelength e.m radiation ?

ZapperZ

I hate Wikipedia. Let's get that out of the way.

While the "standard" photoelectric effect uses photons in the visible, maybe even uv range, the general photoemission phenomenon is NOT restricted to just that range. Look up x-ray photoemission. Would you, or that Wikipedia author, still want to call this "low energy".

Zz.

Graniar

Photon wavelength thousands of times bigger than distance between atoms, photon interact not with the single atom, but with the "giant molecule" consist of bunch of electrons trapped by ion lattice.
Theoretically, there is still a probability for photon to loose only part of it's energy and pass the thin film. But, unlike Compton case, this "molecule" consist of too many parts, and such probability is much under observation.
Otherwise, for example, exposing this metal film by blue light, we would see the red light from the other side.

PS. I suppose more correct to assume system size from electron wavelength, not electron as particle (plasmon may be?)