Electrons escaping a metal surface

[ergospherical]

Homework Statement

Electrons in a semi-infinite slab (z < 0) of metal behave as an ideal non-relativistic Fermi gas. They escape the surface if ##p_z^2/(2m) > E_F + V##, where ##E_F## is the Fermi energy and ##V## is a potential barrier - what is the current density of escaping electrons? Assume ##E_F \gg k_B T## and ##V \gg k_B T##.

Relevant Equations

N/A

In the low temperature limit ##\mu \approx E_F## and the Fermi-Dirac distribution is ##n(E) \approx g(E)/(e^{\beta(E-E_F)}+1)##. An escaping electron contributes ##\Delta j_z = -ev_z = -ep_z/m## to the current density. How can I calculate the rate that electrons escape at? I can't see how to relate ##p_z## to the Fermi-Dirac distribution (apart from ##E = p^2/(2m) = (p_x^2 + p_y^2 + p_z^2)/(2m)##, in which case I don't know what to say about the transverse component of the momentum).

 

[TSny]

How can I calculate the rate that electrons escape at?

##g(E) \large \frac{dE}{e^{\beta(E-E_F)}+1}## gives the number density of electrons with energy between ##E## and ##E + dE##.

In terms of momentum, verify that the number density of electrons with momentum in the range ##(p_x, p_y, p_z)## to ##(p_x+dp_x, p_y+dp_y, p_z + dp_z)## is $$\frac{2}{h^3} \frac{dp_x dp_y dp_z}{e^{\beta[(p_x^2+p_y^2+p_z^2)/(2m)-E_F]}+1}$$ Use this to set up an integral that gives the rate ##R## at which electrons will escape from a unit area of the surface of the metal. $$R = \int_{??}^\infty dp_z \int_{-\infty}^\infty dp_y\int_{-\infty}^\infty dp_x \rm {\,[\, integrand \,\, left \,\, for \,\, you \, :) \,]}$$

 

[ergospherical]

Cheers. Where does the factor 2 come from? I figured that if the (momentum) phase space volume of a state is ##h^3##, then ##g(E) dE \sim d^3 p / h^3##.

Then I thought about a section of the metal with small surface area ##dA##. In time ##dt##, electrons with velocities between ##v_z## and ##v_z + dv_z## reach the surface if they are within a depth ##v_z dt##, i.e. within a volume ##(p_z/m) dt dA##. There are ##dn (p_z/m) dt dA## such electrons, where ##dn = dn(p_x,p_y,p_z)## is the number density of electrons given above. So the rate of escape per unit area, given the restriction that only ##p_z > \sqrt{2m(E_F + V)}## can escape, is \begin{align*}
R = \frac{2}{h^3 m} \int_{\sqrt{2m(E_F + V)}}^{\infty} dp_z \int_{-\infty}^{\infty} dp_y \int_{-\infty}^{\infty} dp_x \ \frac{p_z}{e^{\beta(p^2/(2m) - E_F)} + 1}
\end{align*}Does that look right to you?

 

[TSny]

Cheers. Where does the factor 2 come from? I figured that if the (momentum) phase space volume of a state is ##h^3##, then ##g(E) dE \sim d^3 p / h^3##.

Electron spin allows two electrons to be in each momentum state. This is easy to forget.

Then I thought about a section of the metal with small surface area ##dA##. In time ##dt##, electrons with velocities between ##v_z## and ##v_z + dv_z## reach the surface if they are within a depth ##v_z dt##, i.e. within a volume ##(p_z/m) dt dA##. There are ##dn (p_z/m) dt dA## such electrons, where ##dn = dn(p_x,p_y,p_z)## is the number density of electrons given above. So the rate of escape per unit area, given the restriction that only ##p_z > \sqrt{2m(E_F + V)}## can escape, is \begin{align*}
R = \frac{2}{h^3 m} \int_{\sqrt{2m(E_F + V)}}^{\infty} dp_z \int_{-\infty}^{\infty} dp_y \int_{-\infty}^{\infty} dp_x \ \frac{p_z}{e^{\beta(p^2/(2m) - E_F)} + 1}
\end{align*}Does that look right to you?

Yes. Very nice.