Past exam of the physics A course of the University of Cambridge / 2014 / Paper 2 / Section C

The Pauli exclusion principle confines thermally excited conduction electrons to an energy interval of order $k_{B}T$ around the Fermi surface. Their excess energy is therefore proportional to $T^2$, giving an electronic heat capacity $\gamma T$. More precisely, $\gamma =({\pi }^2/3)k_B^2{g}_{\mathrm{mol}}({ϵ}_F)$. In a three-dimensional Debye model, acoustic phonons have density of states proportional to ${\omega }^2$; their thermal energy scales as $T^4$, givingper mole of atoms. This assumes a normal metal without a lower-temperature superconducting or magnetic anomaly.

For ${\theta }_D\ll T\ll T_F$, the lattice approaches the Dulong-Petit value, so $C_V\simeq 3R+\gamma T$, usually dominated by $3R$. Each classical vibrational mode contributes $k_B$ through kinetic and potential equipartition. This is a high-temperature solid-state regime, not a claim that the solid persists to arbitrarily high temperature; anharmonicity and $C_P-C_V$ can matter.

The body-centred cubic cell has two atoms, so $n_{\mathrm{at}}=2/a^3$. Counting three acoustic branches gives $k_D^3=6{\pi }^2{n}_{\mathrm{at}}$. By the Debye-temperature definition,This is an effective Debye sound speed; separate longitudinal and transverse speeds are not determined. For isotropic branches, $3/v_D^3=1/v_L^3+2/v_T^3$.

The first Brillouin zone is $|k_x|,|k_y|\le \pi /a$. For one atom per primitive cell and two spin states, a monovalent free-electron gas hasIts circular Fermi surface lies inside the zone and encloses half its area: a partially filled band is conducting. Here odd filling obstructs a spin-degenerate band insulator: a monovalent band insulator is impossible in this spin-degenerate one-atom independent-electron model. It would require changed premises, such as unit-cell enlargement or physics beyond the free-electron picture; no insulating free-electron contour should be invented.

For divalent atoms, $k_{F}a=\sqrt{4\pi }$. The circle has the same area as the first Brillouin zone but crosses its faces and leaves its corners unfilled. Thus the free-electron case remains conducting: filling capacity alone does not remove band overlap. A nearly free-electron conductor has reconstructed electron and hole pockets; a divalent band insulator instead has one completely filled band separated from all higher bands, with no Fermi surface in the gap.

For two electrons per primitive cell,Set $A={\hslash }^2{\pi }^2/(2m_e{a}^2)=2.350\mathrm{eV}$. In the indirect gap in a uniformly split zone-boundary model approximation stipulated here, the free boundary energies are $A$ along $[10]$ at $X=(\pi /a,0)$ and $2A$ along $[11]$ at $M=(\pi /a,\pi /a)$. Opening a symmetric gap givesThe branches bend away from the free-electron crossings and flatten at their boundary extrema. Importantly, the upper band near $X$ lies below the lower band's corner energy:Therefore the crystal is metallic despite its positive boundary gaps; electron and hole occupations redistribute across overlapping bands. positive direct gaps can coexist with band overlap.

A microscopic square-lattice corner has additional plane-wave degeneracy. The quoted symmetric boundary numbers use the question's directional gap model, not an unspecified four-state corner Hamiltonian. For the indirect gap of a separable square cosine potential, a separable weak potential with the same face gap gives the lowest corner energy $2A-E_g=3.70\mathrm{eV}$, still above the $2.85\mathrm{eV}$ upper-band face energy. This refinement preserves the metallic conclusion.