# Work function definition

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.

• 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? – DarKnightS Jan 29 '15 at 9:54