Photoelectric effect vs Compton scattering

ZapperZ

I am puzzled on why this is now my claim.

You note that I was responding to sophiecentaur's claim that there's a small gap for the free electron. So I asked what is this gap! Free electrons has a continuous energy state. In a metal, there is no gap that separates the "conduction band" to the "valence band". That's why I asked for the nature of this gap.

Somehow, you now think that I'm the one claiming that a free electron has some sort of a energy gap.

Zz.

sophiecentaur

I only said small gap because someone would have pointed out that the band does not actually have zero gaps - there are just a huge number of quantum numbers involved - effectively a continuum.

ZapperZ

What bands?

If you are talking about the "band structure", then you need to tell me which material you are referring to. If you are talking about the energy band structure of a typical, standard metal that we all deal with in the first 2 chapters of Ashcroft&Mermin, then I will ask again, "What gap"?

Zz.

sophiecentaur

This is getting philosophical. My point is that a finite number of particles will have a finite set of interactions. If somebody's model introduces an integral rather than a sum then that' s fair enough and it's clearly the sensible approach. But if two copper atoms have discrete levels then so do 2e23 atoms. If no, when do you make the switch.
Remember it's all models.

ZapperZ

So instead you decided to contradict the prevailing solid state models and introduce your own model that has not been verified?

Zz.

sophiecentaur

Not at all. There is only a contradiction in your mind. I don't have access to that reference but I wouldn't mind betting that, somewhere along the line of the argument there.will be some integration. That makes an assumption of continuous variables. (I did my Maths Analysis course many years ago.)
There has to be an assumption that you can jump.from discrete to continuous at some stage i.e. when the numbers are big enough so that you can come up with an answer.
With whom could I be arguing about that- apart from you?

ZapperZ

But this is many-body physics! If you think you can, for example, derive the Fermi Liquid Theory via adding up such single interactions, then I want to see it.

The definition of a conductor here is that there is no gap at the Fermi level! Period!

Now, if you wish to argue that you can show that there is a "just finite" gap here, then I want an exact reference. It is that simple.

Zz.

Dickfore

ZapperZ, have you ever heard of a work function of a metal?

sophiecentaur

Would you also say that "main body Physics" insists that a gas is a continuum because it uses the gas laws? Those gas laws came from the statistics of large numbers of discrete particles and we all live with that.
They are only models, remember.

truesearch

I would say that it is misleading to introduce 'work function' as part of the explanation of electrical conduction in metals.

ZapperZ

Then derive using your model the DOS for free electron gas and show that it has a "just finite gap" at the Fermi energy. So far, you have made claims with ZERO materials to support them.

Zz.

sophiecentaur

Why are you being so precious about this? The support for my statement is that calculus is used for that model of 'yours'.
Btw, do you know what the word 'calculus' means?

Dickfore

I would say it isn't, since we're discussing photoelectric effect. Do you know Einstein's formula for the photoeffect?

truesearch

Yes I do and that equation describes the process of ejecting electrons from a metal, not conduction of electricity through a metal.
I wonder if SC is confusing the small ( but not zero) energy gap between donor or acceptor impurities and the conduction or valence band in semiconductors with the continuous conduction band in metals ?

sophiecentaur

There is no confusion. I should be grateful if you would point out why, for simple cases of just a few charges interacting, you get integer values of quantum numbers and, hence discrete 'levels'. And yet, increasing the number involved, changes that 'in principle'. Of course, you can't do the same sums for large quantities and naturally go to statistics and calculus to get an answer.
Tell me where the 'gear change' happens. Is it for 10 atoms, 1000, 100,000?
We are surely just looking at two ends of a range of situations. It would be daft to consider the actual value of energy gaps when a band model is more suitable just as it would be daft to assume a continuum for just a few atoms. we are not dealing with reality. We are applying the most convenient model to describe things. How is it more than that?

Dickfore

Now I see that you had made a conceptual mistake. You claim that even a macroscopic body has a discrete spectrum, but the levels are very finely spaced.

While technically this is true, it is as useless as the idealization of an infinite crystal, which gives continuous bands exactly.

The problem comes when you try to define your system! Namely, you need to make the system isolated, so that it does not exchange particles nor energy with the environment, to really calculate energy eigenvalues. But, every system is essentially open. Because of the fine spacing between the levels, you can never claim that you isolated your system so well to be able to distinguish individual discrete eigenvalues.

Let us estimate the difference between subsequent energy levels. For this, assume a crysta lsystem with a cubic unit cell with side a. The total number of atoms in the system is N, so that [itex]N = (L/a)^3[/itex]. The discrete values for the wave vector are:
[tex]
k_{i} = \frac{2 \pi \, n_i}{L}, \ (i = x, y, z)
[/tex]
Assume the free electron model so that the energy eigenvalues are:
[tex]
E = \frac{\hbar^2 \, k^2}{2 m}
[/tex]

Then, the uncertainty in energy due to a neighboring level is:
[tex]
\Delta E = \frac{\hbar^2 \, \vert k_i \vert}{2 m} \, \Delta k_i
[/tex]
The order of magnitude for the wave vector corresponds to the Fermi wavevector, found from the total electron density n:
[tex]
\frac{N_e}{a^3} = 2 \frac{1}{(2\pi)^3} \, \frac{4 \pi k^{3}_{F}}{3} = \frac{k^{3}_{F}}{3 \pi^2} \Rightarrow k_F = \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}}
[/tex]
where [itex]N_e[/itex] is the number of electrons per atom.

The uncertainty in the wave vector is given by:
[tex]
\Delta k_i = \frac{2 \pi}{L} = \frac{2 \pi}{a} \, N^{-\frac{1}{3}}
[/tex]

Then, we have:
[tex]
\Delta E = \frac{\hbar^2}{2 m} \, \frac{1}{a} \, \left( 3 \pi^2 \, N_e \right)^{\frac{1}{3}} \, \frac{2 \pi}{a} \, N^{-\frac{1}{3}}
[/tex]
[tex]
\Delta E = \frac{1}{4} \, (\frac{3}{\pi})^{1/3} \, \frac{h^2}{m \, a^2} \, \left( \frac{N_e}{N} \right)^{\frac{1}{3}}
[/tex]

For typical crystals [itex]a \sim 5 \stackrel{o}{A}[/itex]. Take an Avogadro number of atoms, and one electron per atom ([itex]N_e = 1[/itex]). We have:
[tex]
\Delta E = 0.246 \times \frac{(6.626 \times 10^{-34})^{2}}{9.11 \times 10^{-31} \times (5 \times 10^{-10})^2} \times (6.022 \times 10^{23})^{\frac{-1}{3}} \, \mathrm{J} \times \frac{1 \, \mathrm{eV}}{1.602 \times 10^{-19} \, \mathrm{J}} = 2.9 \times 10^{-5} \, \mathrm{eV}
[/tex]

sophiecentaur

Thanks for that. I take it that this bit is the 'nub' of what you're saying. That makes excellent sense, the fact being that you only need to have your system in an environment to be 'stirring things up' enough to blur out any discreteness of levels that a simple model would suggest.
I was thinking that the 'line spreading' effect as you increase the pressure in a gas (due to the additional interactions) was all that counts but the system wouldn't, even then, be isolated.

Well, at least that's something!. In a real situation, there would be quite a small upper limit to the number of atoms for what I suggested to be actually true. You have accounted for the 'gear change'.