As in this post How would I calculate the work function of a metal, the definition is given by "the minimum thermodynamic work (i.e. energy) needed to remove an electron from a solid to a point in the vacuum immediately outside the solid surface"(from Wikipedia). But when we do photoemission spectroscopy, the work function is with respect to vacuum so that we have $\phi = h\nu -\mid E_b\mid$. Are they consistent?

Edit: I guess the reconciliation may be sought in the following way:

Without the surface double layer, the Fermi energy $E_F$ is with respect to the vacuum. Once the double layer is taken into account, the chemical potential will effectively be lowered by $W_s$, i.e., the electrochemical potential becomes $E_F - W_s$ (with respect to the vacuum). Therefore, the total energy is $E_K$ in the end and $\phi = -E_F + W_s$. Then, probably, the macroscopic field due to inequivalent surfaces (or "image interaction" if they are equivalent) is neglected in some cases.


When an electron is ejected from a metal surface it experiences an attraction due to the image force. However the attraction falls off rapidly with distance. A back of the envelope calculation tells me that (relative to infinity) the potential energy due to the image force is only around -0.001eV at a distance of a micron.

My understanding is that the image potential is normally considered part of the electron binding energy so it's included in the work function. So we're taking the expression:

to a point in the vacuum immediately outside the solid surface

to mean removing the electron to a distance outside the solid where the image force is negligable. In practice this would be a few microns, so it really isn't very far from the surface.

  • $\begingroup$ Maybe I misunderstood something. The definition of work function is $\phi = - E_F + W_s$, $W_s$ is due to double layer. And the final kinetic energy in the vacuum is $E_K = W_s + E_0$ where $E_0$ is the kinetic energy at immediate surface if no force like "image force" is involved. Then we should have the total energy $\phi + E_0 = -E_F +E_K$. Isn't it? $\endgroup$ – DarKnightS Jan 29 '15 at 9:54
  • $\begingroup$ I'm not sure about exactly how the terms are defined, but isn't the double layer normally considered to be a surface effect so it operates on length scales of a few nm? Are you taking the double layer to extend far enough that the image potential is negligable outside it? $\endgroup$ – John Rennie Jan 29 '15 at 10:06
  • $\begingroup$ Yes. This seems to be the picture indicated by Ashcroft and Mermin in their definition of work function. You can read the answer in the linked post. $\endgroup$ – DarKnightS Jan 29 '15 at 10:47

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