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How large is the penetration length for static electric field into good conductors?

I have two versions: (1) few atomic spacings

$$a\sim n_{e}^{-1/3},$$

and (2) Debye length computed by Fermi energy $\varepsilon_{F}$ (not the temperature)

$$\lambda_{D}=\sqrt{\varepsilon_{F}/4\pi e^2 n_{e}}.$$

First variant is preferable since $$n_{e}\lambda_{D}^3<1.$$

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1 Answer 1

The penetration depth model does not work for a good metal, but is ok for a moderately doped semiconductor. The pentration depth you get is smaller than an Angstrom for a good conductor. This means that the surface electrons are essentially confined to the first atomic layer, with some field entering into the next few layers, and the details of the atomic orbitals and the electron electron interactions are necessary for working out exactly how the electric field goes away.

Penetration depth model

Here, you assume that the potential enters a self-consistent jellium, which is a uniform positive charge density plus almost free Fermi gas. The potential shifts the energy levels of the electrons, and you take this into account by filling the levels in the potential. But you take the electron electron interactions into account by modifying the potential according to the induced charge density.

If a jellium metal has a potential V imposed on it, the number density of the electrons at any point can be calculated semiclassically with no significant error, as follows. The occupied phase space volume for electrons in a box of side length $\Delta$ at an electrostatic potential $\phi$ is:

$${4\pi\over 3} (2m(E_f+e\phi))^{3\over 2} \Delta^3$$

where m is the mass of the electron, e is the magnitude of its charge, and $E_f$ is the Fermi energy. Dividing this by $h^3 = (2\pi \hbar)^3$ gives the number of occupied states. This immediately gives the electron density in a slowly varying potential

$$n(x) = {1 \over 6\pi^2} (\sqrt{ k_f^2 + {2em \phi\over \hbar^2}})^3$$

While the jellium model is no good for the low lying states, which see a nonuniform electric potential from the localized nuclei, it is still a good model material with a spherical Fermi surface, and is ok for order of magnitude estimates even away from a spherical Fermi surface. The point is that the actual number density is not what is important, it is only the variation in the number density given a certain V(x) which matters, and this can be accurate even when the number density itself is very wrong, because the states deep below the Fermi surface are not right in the model.

Anyway, expanding this density to lowest order in V and multiplying by -e gives the electronic charge density given an imposed external potential:

$$\rho(x) = - {1\over 6\pi^2} e k_f^3 + {2\over \pi} k_f {e^2\over 4\pi\hbar c} {mc\over \hbar} \phi(x)$$

The first term can be identified as the constant charge density of the electrons in the absence of a potential, and this is cancelled by the positive charge on the jellium. The remaining term gives a charge density proportional to the potential, which leads to screening of static fields.

When $\rho$ is proportional to V, $\rho= k V$ the Laplace equation becomes the Poisson equation

$$ \nabla^2 \phi = 4\pi \rho = 4\pi k \phi$$

The inverse screening length is found by solving the 1d version, and it is $\sqrt{4\pi k}$. For the particular problem above, the screening length $l_s$ is given by:

$$ l_s^{-1} = \sqrt{8 k_f \alpha \over \lambda}$$

Where $\alpha = {1\over 137}$ is the fine structure constant, $\lambda$ is the Compton wavelength of the electron times $2\pi$. The combination ${\alpha \over \lambda}$ is the Bohr radius, so this is really

$$ l_s^{-1} = \sqrt{8 k_f \over a_0}$$

The screening length is the geometric mean of the inverse Fermi momentum and the Bohr radius divided by the square root of eight.

For a good metal, the inverse Fermi momentum is about an Angstrom, although it can be much longer in a semiconductor. The result is less than an Angstrom, and is therefore unphysical. In this regime, the assumptions used to derive the self-consistent jellium break down.

So the penetration depth model is not correct for a good metal, and it requires a Fermi momentum which is about 10-100 times smaller than a lattice spacing to become good. This is the situation in a semiconductor, but not in good metals.

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So what is your conclusion? Penetration length is few atomic layers? –  Igor Sep 23 '11 at 8:54
    
@Igor-- yes--- and not exponential in form. The jellium screening length is called the "Fermi Thomas screening wave vector" according to Wikipedia, and although it is less than an Angstrom, maybe the decay is approximately exponential with this tiny length, I don't know, because the approximation is not really valid at lengths so small. –  Ron Maimon Sep 23 '11 at 22:54
    
one Angstrom is the magnitude of an atom layer, I think that this is against any ideas of exponential or not. –  Georg Nov 22 '11 at 12:23
    
@Georg: Well, the jellium calculation is out of its domain of validity for sure, but the actual charge distribution might still be exponential with a screening length 1/4 of an angstrom, when examined through 1.6 angstroms layers (I am just choosing approximate round numbers here), or 4 e-folds (a factor of 60 drop) every atomic layer. That seems like ridiculously strong screening, but that's what the jellium calculation gives, and I don't know a better calculation. –  Ron Maimon Nov 23 '11 at 6:55

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