Exclusion principle in quantum gasses

[Quant ummm?]

Hi there,

I understand (to a degree) Pauli's exclusion principle in terms of electrons in an atom but I'm a little confused about the scales involved with free electrons, say electron gasses in metals...

My textbook gives an example:

"consider a 1cm cube of copper at room temperature. The number of free electrons N can be found from Table 1.3 to be N = nV = 8.45 × 10²⁸ m⁻³ × 10⁻⁶m = 8.45 × 10²². The total number of quantum states up to energy kT, (found by using the density of states De(E) in a definite integral) has the value 2 × 10¹⁹. You can see that this number of states can accommodate only about 0.02% of the free electrons. The rest have to pile up into states of higher energy, a long way above kT. If we ask how far up the energy scale we have to go to accommodate all the free electrons, we obtain the amazing answer of about 7 eV. This is about 300kT at room temperature."

1cm is quite large compared to the de Broglie wavelength of an electron so L doesn't appear to make much difference here. To take it to an extreme, if I regard the hull of a battleship as a block of iron, the surfaces of which contain an electron gas, does the quantum state of an electron in the bow forbid an electron in the stern from having that state? Also, it would seem that with a big enough sample, some electrons would have to occupy unimaginably high states.

I suspect that separation is relevant to exclusion but I can't seem to find a rule, or I have fundamentally misunderstood the whole thing...

 

[Marty]

Hi there,

I understand (to a degree) Pauli's exclusion principle in terms of electrons in an atom but I'm a little confused about the scales involved with free electrons, say electron gasses in metals...

My textbook gives an example:

"consider a 1cm cube of copper at room temperature. The number of free electrons N can be found from Table 1.3 to be N = nV = 8.45 × 10²⁸ m⁻³ × 10⁻⁶m = 8.45 × 10²². The total number of quantum states up to energy kT, (found by using the density of states De(E) in a definite integral) has the value 2 × 10¹⁹. You can see that this number of states can accommodate only about 0.02% of the free electrons. The rest have to pile up into states of higher energy, a long way above kT. If we ask how far up the energy scale we have to go to accommodate all the free electrons, we obtain the amazing answer of about 7 eV. This is about 300kT at room temperature."

1cm is quite large compared to the de Broglie wavelength of an electron so L doesn't appear to make much difference here. To take it to an extreme, if I regard the hull of a battleship as a block of iron, the surfaces of which contain an electron gas, does the quantum state of an electron in the bow forbid an electron in the stern from having that state? Also, it would seem that with a big enough sample, some electrons would have to occupy unimaginably high states.

I suspect that separation is relevant to exclusion but I can't seem to find a rule, or I have fundamentally misunderstood the whole thing...

I'm going to take a stab at this:

The number of free electrons is approximately the number of copper atoms if you assume that all but one of the electrons in the atom are bound to their own local atom. So you have a block of positive (+1) copper ions in a sea of free electrons.

This is approximately a cubical potential well, which we solve just like the potential wells of first-year physics. The lowest energy state is a standing wave with nodes along the walls of the cube; the subsequent states divide the cube into two, three or more nodal regions.

All these regions are filled by electrons one at a time. Subsequent levels require more and more nodal surfaces, until at the limit, the nodal surfaces virtually coincide with the number of atoms. You can see this must be so by a counting argument where you compare the number of atoms in a cubical lattice to the number of standing wave modes filling the same cubical volume. So the highest energy electrons have a wavelength approximately equal to the distance between two atoms.

And that is where your 7 electron volts comes from. If you just think of the hydrogen atom (-13.6 eV) you will know that a wavelength of atomic dimensions has more or less that much energy.

The initial volume (1 cc) is irrelevant to the shape of this argument, and that is why it applies to a battleship the same way it applies to a penny.