# Electron ejected in Photoelectric effect

1. Jan 5, 2016

### Neutroniclad

Since the binding energy of an electron in outer shell is smaller than the binding energy of the electron in inner shell, why the photon kick out the electron orbiting in inner shell? It is always easier to kick out the electron with less binding energy, is it not?

2. Jan 5, 2016

### Orodruin

Staff Emeritus
Why do you think it is a photon in a lower shell which is being ejected?

3. Jan 5, 2016

### Neutroniclad

I am thinking why it is an electron in lower shell being ejected by an incident photon not an electron in higher shell being ejected, since the electron in higher shell has smaller binding energy than electron in lower shell.
There is no photon orbiting around the atom.

4. Jan 5, 2016

### Orodruin

Staff Emeritus
Yes, you said so in your OP. My question to you was why you think so.

5. Jan 5, 2016

### Neutroniclad

Because the photoelectron has kinetc energy EK=E(gamma)-E(binding), and the electron in outer shell has less E(binding). I think it supposed to be easier for the incident photon to remove an electron in outer shell.

6. Jan 5, 2016

### Orodruin

Staff Emeritus
But this is not what you said, you started with the assumption that it is the inner shell electrons which are being ejected. I asked why you start with this assumption.

7. Jan 5, 2016

8. Jan 5, 2016

### Neutroniclad

There are 3 ways that gamma-ray photons interact with matter. One of them is photoelectric absorption, during which, the incident photon conduct all of it's energy to an electron in inner shell of the atom and disappears. The electron absorbed the energy is then able to get rid of the atom become a free electron...
Normally that is what happened. I'm just thinking why it is not an electron in outer shell become the free electron.

9. Jan 5, 2016

### vanhees71

As I said, it's a question of the em. waves frequency. If it's in the range of the binding energy (divided by $\hbar$) of the inner electrons, it's likely that these are ejected. If it's at the lower frequency corresponding to the binding energy of the outer electrons, these will get more probably ejected (see the Insights article).

10. Jan 5, 2016

### Orodruin

Staff Emeritus
This still does not explain why you think the electron has to be from an inner shell in the photoelectric effect. (It doesn't! In fact the binding energy of the photoelectric effect is a property of the metal - not an atomic property. The inner shell electrons are too tightly bound to be affected.)

11. Jan 5, 2016

### vanhees71

For the inner-shell electrons it should be mostly the atomic states which are relevant (tight-binding approximation).

12. Jan 5, 2016

### Orodruin

Staff Emeritus
Agreed, but it is not what the OP is talking about. The highest electron energy according to Ek = E(gamma) - W (as given by OP in post #5) is given by the valence states which are a metal property. I agree that energy levels of the inner states will be dominated by the atomic states.

What I am trying to make the OP realise is that the work function W of a metal is not the inner state binding energy.

13. Jan 5, 2016

### vanhees71

Sure, these are the conduction electrons of a metal, but those usually should more be in the range of visible light rather than in X- or higher $\gamma$-ray range.

14. Jan 5, 2016

### Orodruin

Staff Emeritus
Yes, but the OP did not specify the energy range. I think post #5 makes it clear that we are dealing with the photoelectric effect involving the conduction states.

15. Jan 5, 2016

### Neutroniclad

Actually, I read this description about photoelectric effect from my text book then I start to think "why not an electron in outer shell? For it's less well bound"
Just curious

16. Jan 5, 2016

### Neutroniclad

Anyway, thank you guys for answering my question. :)

17. Jan 5, 2016

### Staff: Mentor

Are you (and your textbook) discussing the photoelectric effect for visible photons (a few eV energy), or gamma-ray photons (hundreds of eV)? I suspect Orodruin is thinking of the first, while vanhees71 is thinking of the second.

18. Jan 5, 2016

### ZapperZ

Staff Emeritus
There are tons of crossed signals here, and it started waaaaay back with the title of the thread itself.

"Photoelectric effect", at least the standard one that was modeled by Einstein and tested by Millikan, is done on METALS using visible-to-UV range light source. So the electrons being emitted are from the CONDUCTION BAND of the metal. This is crucial because the energy spectrum of the photoelectrons are continuous, not discrete as in an atomic orbital.

So whenever I read something like this thread, where the OP is asking about inner shell electrons, etc. (i.e. no longer about the bands), then I question on whether this is an x-ray photoemission (XPS) phenomenon where core-level states are now being probed, or if this is a photoionization phenomenon, where the material isn't a solid (metal), but rather a gas. This is NOT your typical, standard Photoelectric effect! There are many aspects of the photoelectric effect model that doesn't quite work in this regime.

In XPS, the low binding energy electrons ARE ejected as well (one can see the Fermi edge in the spectrum), but due to the favorability of the absorption cross-section, the core-level states often dominates in the spectrum. This, in itself, should answer the OPs question, I would think.

I just wish people pay a bit of attention in their use of the terminology. It might eliminate many of the confusing signals and discussion.

Zz.

19. Jan 5, 2016

### Neutroniclad

#8 has narrow down the topic to gamma-ray interact with matter, I think

20. Jan 5, 2016

### Neutroniclad

It's my fault, I did't mention it's gamma-ray photon until #8

21. Jan 5, 2016

### ZapperZ

Staff Emeritus
You also neglected to elaborate on the "medium" that the gamma-ray photons impinges on. Is it a gas? Is it a solid?

Gamma ray impacting on a solid (such as those in detectors) is not easy and can't just be called "photoelectric effect". In fact, depending on the material, the more dominant effect might be from Compton scattering, and the electrons being kicked out might be from secondary interactions.

It is obvious that you have something in mind. Rather than hide it from us, when you ask a question such as this, be as thorough and as complete as possible, especially if this is related to a particular phenomenon or application. If you had asked this and connected it to, say, the detector used in PET scans, then it would have been clear to many of us what you are talking about because we can see the CONTEXT. Or if this is something you read in an article or paper, then cite the paper.

Zz.

22. Jan 5, 2016

### vanhees71

Well, for me the photoelectric effect is defined as the process of absorbing at least a photon and exciting a bound electron into the continuum state. If you talk about kicking out an inner electron from an atom and absorbing a photon in the $\gamma$-ray range, that's as well a photoeffect.

23. Jan 5, 2016

### ZapperZ

Staff Emeritus
It is fine if you are talking about a generic "photon in, electron out" phenomenon, AND, everyone understood that this is what you meant. However, it is not OK if this is being referred to as being the same as the Photoelectric effect that we give in intro Physics labs, and the ones that Millikan tested. Core level photoemission has differences than the Einstein photoelectric effect model. This is not what Millikan tested and what history called the photoelectric effect.

The entire phenomenon is generically called "photoemission". This includes the classic photoelectric effect, angled-resolved photoemission, resonant photoemission, XPS, multiphoton photoemission... etc... etc. If you pick such a book on this topic, you'll see various TYPES of this phenomenon. If someone picks up an undergrad intro physics text, and look at what is being called the photoelectric effect, it looks different than what XPS, for example, will produce.

There is no harm is being clear, and using the more accurate name for something, especially when there is already a name for it.

Zz.

24. Jan 5, 2016

### vanhees71

Well, most introductory treatments of the photoeffect a la Einstein do more harm than good => See my insights article.

25. Jan 5, 2016

### ZapperZ

Staff Emeritus
That's a matter of taste. But at least those texts do confine themselves to the actual, relevant phenomenon, rather than try to describe something it isn't.

Zz.

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