Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S16 / 2

Treat the ions as a stationary neutralizing background and displace the cold electrons by a small longitudinal distance $\xi$. Their electric polarization is $P=-n_{e}e\xi$, so the restoring electric field is $E=-P/{ϵ}_0=n_{e}e\xi /{ϵ}_0$. The electron equation of motion isFor $E=E_0\cos {\omega }_{\mathrm{pe}}t$, the oscillatory velocity is $v=-e{E}_0\sin {\omega }_{\mathrm{pe}}t/(m_e{\omega }_{\mathrm{pe}})$. HenceThe longitudinal mode exchanges electron kinetic energy and electrostatic field energy like a harmonic oscillator, whose mean kinetic and potential energies are equal over a cycle. This is a time-average statement for the coherent oscillation, not thermal equipartition with $k_{\mathrm{B}}T$.

The Debye length is the distance over which a small, slowly varying electrostatic potential is screened by mobile thermal charges. For classical electrons in a weak potential $\varphi$, their Boltzmann distribution givesTake fixed singly charged ions of density $n_0$. Then $\rho =e(n_0-n_e)\simeq -n_0{e}^2\varphi /(k_{\mathrm{B}}T)$, and Poisson's equation outside a test charge becomes ${\nabla }^2\varphi =\varphi /{\lambda }_D^2$, withThe screened point-charge potential is proportional to $e^{-r/{\lambda }_D}/r$. The derivation assumes nondegenerate statistics and $|e\varphi |\ll k_{\mathrm{B}}T$. Mobile ions would add their own contribution to ${\lambda }_D^{-2}$; the expression here is the electron screening length.

For a nonrelativistic, three-dimensional ideal electron gas with spin degeneracy two,Dividing givesWrite the density-independent coefficient as $K$. From $\hslash {\omega }_{\mathrm{pe}}/E_F=1.384$, $n_e=(K/1.384{)}^6$. In this free-electron model,The result uses the stipulated energy ratio, rather than substituting a separately tabulated aluminium density or plasmon energy.

Using the replacement prescribed in the question, $k_{\mathrm{B}}T\to E_F$, givesThe number of electrons in a Debye sphere is consequentlyA useful check eliminates $n_e$: with $r=\hslash {\omega }_{\mathrm{pe}}/E_F$ and $k_F=(3{\pi }^2{n}_e{)}^{1/3}$, $k_F{\lambda }_D=\sqrt{2}/r$ and $N_D=8\sqrt{2}/(9\pi r^3)$.

Having substantially fewer than one electron in this sphere shows that the classical many-particle Debye-screening picture is not applicable. It does not exclude collective plasmons in a degenerate electron gas. A proper zero-temperature Thomas-Fermi screening calculation uses $\partial n/\partial \mu =3n/(2E_F)$ and gives ${\lambda }_{\mathrm{TF}}^2=2{ϵ}_0{E}_F/(3n_e{e}^2)$, rather than literally replacing $T$ by $T_F$. The boxed count follows the source's stated prescription and exposes its physical limitation.