Past exam of the chemistry A course of the University of Cambridge / 2021 / Paper 2 / A9

For the free electron model, the Fermi energy is the highest occupied one-electron energy at zero temperature, measured relative to the bottom of the free-electron band. The occupied wavevectors fill a disk in two dimensions or a sphere in three dimensions. The radius is the Fermi wavevector $k_F$, withEach allowed wavevector has two spin states. At nonzero temperature, the Fermi-Dirac distribution rounds the occupation boundary; the zero-temperature Fermi energy remains a useful density-defined scale and should not be defined as the thermal energy $k_{\mathrm{B}}T$.

Metallic atomic densities place very many electrons in a small volume. The Pauli exclusion principle prevents all these fermions occupying the lowest spatial state, so filling the available states produces a Fermi energy of order electronvolts. At room temperature, $k_{\mathrm{B}}T\simeq 0.026\mathrm{eV}$, much smaller than that scale. Equivalently, ordinary metals are at $T\ll T_F=E_F/k_{\mathrm{B}}$, where $T_F$ is the Fermi temperature. Their electrons are quantum degenerate even at ordinary laboratory temperatures.

Let $\mathcal{A}$ be the total two-dimensional area. There are $\mathcal{A}/(2\pi {)}^2$ wavevector states per reciprocal-space area per spin. Including spin degeneracy two and filling a disk givesFor areal number density $n_e=N_e/\mathcal{A}$ this yieldsThe disk is centred at zero wavevector, and its boundary is a fermi circle. Omitting the two spin states would incorrectly double the energy inferred for a given number density.

Using $M_{\mathrm{Au}}\simeq 196.97\mathrm{g}\mathrm{mo}{\mathrm{l}}^{-1}$ and the Avogadro constant, the areal atom count isOne atom occupies each square cell of area $a^2$, so the inferred square lattice spacing isThis uses the given deposited mass and the assumed single square layer; it does not substitute a bulk gold lattice parameter for the measured areal count.

One mobile electron per gold atom givesThis is an areal number density; no layer thickness is required and it should not be converted into a volume density for the two-dimensional model.

Substitution in part (a) givesandThe conversion uses $1\mathrm{eV}=1.60218\times 10^{-19}\mathrm{J}$. These are the predictions of the stipulated two-dimensional free electron model, with the vacuum electron mass and one carrier per inferred square cell.

The reciprocal lattice is generated byIts points are spaced by $2\pi /a$ along both reciprocal axes. Perpendicular bisectors between the origin and its four nearest reciprocal neighbours enclose the first Brillouin zone,The original drawing labels both the reciprocal-lattice spacing and the nearer zone boundaries. Its shaded disk is the occupied region for one electron per real-space cell.

For one electron per square cell, $n_e=1/a^2$, soNumerically, $\pi /a\simeq 2.17\times 10^{10}{\mathrm{m}}^{-1}$ exceeds $k_F\simeq 1.73\times 10^{10}{\mathrm{m}}^{-1}$, so the fermi circle lies entirely inside the first Brillouin zone.

The state-counting explanation is that a spin-degenerate band has two electron states per primitive cell. One electron per cell therefore occupies half of that band. In the circular free-electron approximation the occupied reciprocal-space area isA disk of that area fits inside this square: its radius is less than the distance to an edge. Half filling alone is not a shape-independent theorem that every possible Fermi surface must avoid every zone boundary; the square geometry and circular free-electron dispersion make the statement true here.

An isolated spin-degenerate lowest energy band accommodates two electrons per atom when there is one atom per primitive cell. Starting from one per atom, the estimate for filling the band is thereforeThis corresponds to an additional areal carrier count of about $4.77\times 10^{19}{\mathrm{m}}^{-2}$. If a genuine band gap separates the filled first band from the next band, there are no partially occupied band states at low temperature, so ordinary metallic conduction is lost.

Simply touching an edge is insufficient. In the circular approximation first contact would occur at $k_F=\pi /a$, corresponding to $n_e{a}^2=\pi /2\simeq 1.57$ electrons per cell; that is a partly filled band and need not be insulating. Near zone boundaries, the periodic potential reshapes the free-electron contours. The two-electron filling estimate assumes a global gap without overlapping bands, rather than inferring an insulator merely from one local avoided crossing.