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.