# Can surface dipoles/charges change the work function of a metal?

As typically drawn in simplified band diagrams (see picture below), the metal Fermi Level is shown as the top of the conduction band, with the entire band filled.

In many situations, including joining a metal and semiconductor together to form a contact, charge is transferred between the two materials. For a metal/n-SC contact this forms a very thin layer of negative charge on the surface of the metal. However, the final band diagram is still drawn with the metal band unchanged.

Is there simply a negligibly small change of the work function at the surface in this case that is not drawn?

If so, what is the statistical meaning of the Fermi Level being above the top of the conduction band at the surface/interface (in the case of a thin sheet of negative charge)?

-
Could you show one of those simplified band diagrams? If entire band is filled it is not a conduction band, it is a valence band. And in this case it is a dielectric not metal. – Maksim Zholudev Feb 23 '12 at 8:36
Sure, here you go: iue.tuwien.ac.at/phd/ayalew/img308.png – Caedar Feb 24 '12 at 6:03
I see, thanks. There is no marks for conduction band in the metal at all. The blue line is the Fermi level. So the band is not filled completely. – Maksim Zholudev Feb 24 '12 at 7:19

I will ignore the confused statement "the metal Fermi Level is shown as the top of the conduction band, with the entire band filled" and focus on the main part of the question.

First, I'll answer within the framework of Schottky-Mott theory, which is where this diagram comes from. In any material, the difference between vacuum level and a particular electron wavefunction state is constant (a function of the electronic state and of the material, but independent of everything else). For example, in a semiconductor, the difference between vacuum level and the bottom of the conduction band is constant, and the difference between vacuum level and the top of the valence band is constant. In a metal (unlike a semiconductor), the fermi level is fixed relative to the band structure (i.e., an electron with a certain wavefunction may always be at the metal's fermi level), because the metal is always charge-neutral (except within a Debye length (few angstroms) of the surface). So if there are X protons per cm^3, then there are X electrons per cm^3, and the metal Fermi level always sits close to the X'th lowest-energy electron state.

So again, the metal and semiconductor are treated the same: The states of both stay unchanged relative to the local vacuum level. The difference is that in a semiconductor, but not a metal, the Fermi level can shift relative to the electronic states.

Sometimes people ask "Does the vacuum level of the metal shift or does the vacuum level of the bulk semiconductor shift?" This question has no answer. Only the difference in vacuum level between somewhere and somewhere else is meaningful.

A whole separate subject is "Is the framework of Schottky-Mott theory actually correct for understanding these things?" In fact, its predictions for Schottky barrier heights are pretty bad. It by-and-large captures the trend--there are systematic differences between low and high workfunction metals--but the quantitative Schottky barrier heights can be pretty far off.

On a related note, the question "What is the workfunction of Metal XYZ?" doesn't have a single answer. It depends on the detailed atomic structure of the surface. For example the (111) surface of tungsten will have a different workfunction than the (100) surface of tungsten.

-
Thanks for the explanation. I was wondering, why is the top of the conduction band of the metal never ever shown? I have not seen one diagram where it is drawn. Is there a physical reason for this? – Caedar Feb 24 '12 at 18:43
The Fermi level of a metal is somewhere in the middle of the conduction band, i.e. well above the bottom and well below the top. Exactly how far above the bottom? It doesn't matter. Exactly how far below the top? It doesn't matter. (At least, it doesn't matter if you're interested in electrical conduction and electronic devices.) Therefore the top and bottom of the band are not drawn. – Steve B Feb 29 '12 at 14:52