# Photoelectric Effect - Electrons

1. Jan 26, 2009

### Chewy0087

Hey there, me again =o

I just want to discuss the photoelectric effect however more than that just to clear up my understanding of electrons really.

I understand that electrons are 'wave packets', or a quantized wave if you like, and i'd like to know if i'm right or wrong in that? Can you be concise as it really confuses me! =o

Stemming from that (if i'm right), assuming that electrons are a wave, in the photoelectric effect, one electron absorbs one photon, I know this is proof of particle like behaviour however I want to look at it from the point of view of waves if that's possible.

I might just stop here as i suspect I already have many flaws in my logic? :P

I just really like thinking about these things when i'm bored, but i think i'm getting a bit carried away, and actually making it up in my head

Thanks in advance for any help

2. Jan 26, 2009

### ZapperZ

Staff Emeritus
Electrons in a conductor are not described by "wave packets". They are, however, described via planes waves in the simplest model, or, if you want to be a bit more sophisticated than that, they are described via Bloch plane waves.

Zz.

3. Jan 26, 2009

### DeepSeeded

The wave packet describes the effects of a particle. It doesn't change the fact that the electron gains momentum when it absorbs a photon.

A wave packet is the integral of a plane wave equation over a very small range of wavelengths. It's amplitude reaches a maximum at a single point and dies out very rapidly in all other directions.

4. Jan 26, 2009

### Chewy0087

Hmmm i don't really understand what you guys are saying hmm

Am I right or not thinking of it as I am?

And if so! The question I was going to ask is that if it is wave-like, it will absorb the photon, making it a bigger wave (bigger wave magnitude) but how does that convert into kinetic energy? Or is the magnitude itself kinetic energy? And if so, how is that kinetic energy given off?

Thanks again, i just need real consiceness iguess

5. Jan 26, 2009

### ZapperZ

Staff Emeritus
Er... it appears that you don't quite understand basic QM yet and are trying to apply it to the photoemission process.

Maybe you need to start with the basic Schrodinger Equation first and solve simple situations (such as for a free particle) before you try out your ideas with the photoemission process, which is a lot more complicated than you think (i.e. there's the Spicer's Three-Step Model for photoemission that adds to the complexity of this process).

Zz.

6. Jan 26, 2009

### Chewy0087

7. Jan 26, 2009

### DeepSeeded

I don't think it is related to the amplitude (biggness) of the wave. The way I think of it, the electron wave packet is traveling in an arc around the nucleus similar to the way light travels in an arc when in a medium. Different wavelengths travel at different speeds which causes the entire wave (group velocity) to change direction constantly.

When a photon is absorbed, new wavelengths are introduced into the wave packet changing the radius of the arc of the electron wave packet as it travels around the nucleus.

In this way as the wave packet absorbs more and more photons that cause more and more interference effects and a lower total number of different frequencies in the wave packet you get less and less difference in speeds between frequencies. This makes the electron go around the nucleus in a much larger orbit because it doesn't change direction as quickly.

8. Jan 26, 2009

### ZapperZ

Staff Emeritus
Er.. you need to be careful here a bit. The photoelectric effect, which is what the OP is asking, is typically done on conductors, in which the conduction electrons are the ones involved. The atomic model no longer works here once we are dealing with a solid. The proof of this is that you get conduction BANDS, which are missing in an atomic spectra. That is why one has to deal with the Bloch wavefunction, and NOT the atomic wavefunction.

Photoemission is not the same as photoionization, which is the atomic/molecular equivalent of the photoelectric effect.

Zz.

9. Jan 26, 2009

### DeepSeeded

Yea I deviated a bit off the posting title. I am not familiar with the Bloch wf yet. Will be getting to it soon I am sure.

10. Jan 27, 2009

### Sedunov

Hi, Chewy and all participants of the discussion!
When electron absorbs the photon, it changes own frequency, but not amplitude. Old, low, frequency did not permit electron to cross the conductor's boundary. But new, higher, frequency corresponds to propagating waves through the boundary. Therefore, electron can now cross the boundary and get free.
Just compare: optical waves oriented alond the axis of the optical waveguide cannot cross the guide's surface and travel on hundreds kilometers along its axis. But waves with large angle cross the surface and introduce losses in signal.

11. Jan 27, 2009

### ZapperZ

Staff Emeritus
Er... it is not that trivial.

For example, look at the setup for the Hamiltonian for the photoemission process. See this paper, for instance, and look at Eq. 3 onwards. You will see that it requires quite a bit more than just a "changes own frequency". In fact, trying to solve this via setting up a "wavefunction" is going to be futile because it is THAT tedious when one has to sum over all cystal wave vector. That's why we instead deal with the single-particle spectral function, which is the Green's function.

Furthermore, per the Spicer's Three Step Model, just because an electron has already been promoted to an energy above the vacuum state, it doesn't mean that it will escape the bulk material. There is a statistical component to the diffusion of such electrons out of the surface. That's why we don't have a material with 100% quantum efficiency.

We no longer deal with "waves" or wavefuntion at this point to describe this phenomenon. It will be too tedious, and no one does it.

Is this a valid analogy? For example, even if I have an optical wave "oriented along the axis", if I am not at one of the mode frequency of the cavity or waveguide, there's going to be plenty of that wave leaking out. It doesn't have to have "large angle cross the surface" for that to occur.

Zz.

12. Jan 27, 2009

### Staff: Mentor

As Zz has already pointed out, with the photoelectric effect you can't think of the electron wavefunction in terms of individual atoms, but rather in terms of the solid as a whole.

I'd like to add that even for situations where you can consider atoms in isolation, you can't think of the electron as traveling as a wave packet in a circular (or other) orbit. The energy states of atomic electrons are like standing waves, not traveling waves. The probability distribution $\Psi^* \Psi$for a particular level does not change with time, and we call them "stationary states" accordingly.

13. Jan 27, 2009

### Chewy0087

Ah! That sort of makes sense, the change in frequency that is!

I know it's much more difficult than that, but the change in frequency relates to the property of electrons orbiting a nucleus that the 'wave' can only be in certain 'orbits' right? However if the frequency is increased to a great enough extent, it must move up, otherwise it risk's cancelling itself out (which can't happen)?

I understand that this is alot more complicated a process, but I just really want something to picture i suppose, am I possibly right at all =o? I'm definateley going to look into this when i get the time

14. Jan 28, 2009

### Sedunov

Yes, Chewy!
In quantum world any process of energy consumption leads to growth of the particle's frequency. And the particle with enlarged frequency has quite new propagation possibilities, including crossing the surface of metallic body.

15. Jan 28, 2009

### Chewy0087

=D

Thanks, I know it's much more intricate than that in actual fact, but bieng able to picture something like that close to the truth is nice!