Past exam of the physics A course of the University of Cambridge / 2018 / Paper 2 / Section C

Use the quantum free electron model: conduction Electrons move as independent particles in a three-dimensional box of volume $V$, with a uniform positive background replacing the ionic lattice potential. Neglect the detailed periodic potential and explicit electron-electron interactions; the interior one-electron dispersion relation is $ϵ={\hslash }^2{k}^2/(2m_e)$. The boundary confines the particles, and its detailed shape does not affect the bulk state count in a sufficiently large sample.

The Electrons are spin-$1/2$ fermions: each spatial quantum state accommodates two opposite spin states, with occupations given by the Fermi-Dirac distribution and the Pauli exclusion principle. A fixed number $N$ of conduction Electrons and thermal equilibrium are assumed. Collisions can be represented by a phenomenological momentum relaxation time for transport, but are unnecessary for the equilibrium state counting here. These assumptions distinguish the quantum model from a classical Maxwell-Boltzmann Electron gas.

Impose periodic boundary conditions on a large cubic box. The allowed wavevectors occupy a lattice of cell volume $(2\pi {)}^3/V$ in wavevector space. Including the two spin states, the number of states with magnitude at most $k$ isUsing $k=\sqrt{2m_eϵ}/\hslash$ and differentiating gives the total, spin-summed three-dimensional free-electron density of states,It counts states per unit energy; divide by $V$ for states per unit volume per unit energy. At zero temperature, the Fermi-Dirac distribution fills all states up to the Fermi energy, givingWriting $n=N/V$ givesThe energy zero is the bottom of the free-electron band. The electron Fermi temperature is $T_F=E_F/k_B$.

The occupation probability of an individual one-electron state is the Fermi-Dirac distribution,At $T=0$, it is a step: one below $E_F$ and zero above it. Heating rounds the step over an energy range of order a few $k_{B}T$, creating empty states below and occupied states above the chemical potential. At every positive temperature, $f(\mu ,T)=1/2$.

For fixed $N,V$ in the three-dimensional free electron model, the chemical potential is determined by $N=\int g(ϵ)f(ϵ,T)dϵ$; it is not exactly constant. In the degenerate regime $T\ll T_F$, the Sommerfeld expansion givesIt is therefore reasonable in a simple low-temperature sketch to set $\mu \simeq E_F$, provided this approximation is stated. The figure includes the leading small shift. At much higher temperature the fixed-$E_F$ approximation fails and the distribution approaches the classical dilute limit. The plotted occupation is $f$, rather than the energy distribution $gf$.

Only a shell of energy width of order $k_{B}T$ near the Fermi energy can change occupation: much lower states are blocked by the Pauli exclusion principle. The thermal-shell estimate of electronic heat capacity countsTo recover the rough coefficient requested in the paper, assign each of these active particles an excitation energy of order $3k_{B}T/2$. ThenThe differentiation matters: the active shell itself widens as temperature rises. This argument explains the small linear-in-$T$ electronic heat capacity; its shell width and average excitation energy are heuristic, so it does not establish a precise numerical prefactor.

A controlled calculation from the Fermi-Dirac distribution gives a nearby but different result. At fixed particle number, the Sommerfeld expansion yieldsThus $9/2$ is a rough estimate of the exact free-electron coefficient ${\pi }^2/2\simeq 4.935$, not an exact consequence of the stated density of states. Both require $T\ll T_F$.

Assume sodium supplies one conduction Electron per atom. The supplied Fermi energy givesThe gas is strongly degenerate at room temperature. For one mole, $N=N_A$ and $N_A{k}_B=R=8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. Using the requested estimate,The supplied mass density determines the answer per unit volume: $N/V=(970/0.023)N_A=2.54\times 10^{28}{\mathrm{m}}^{-3}$. ThereforeThe density is not needed for the molar heat capacity once $E_F$ is supplied. Using the precise Sommerfeld expansion coefficient instead gives $0.327\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$.

Either electronic estimate is only about $1\%$ of the stated measured total. Most of the total heat capacity comes from lattice phonons, whose high-temperature Dulong-Petit law limit is $3R=24.94\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. The remaining difference from $28\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ need not be electronic: anharmonic lattice effects and the difference between heat capacity at constant pressure and heat capacity at constant volume also contribute. The paper does not specify the measurement constraint, so the two totals should not be equated exactly.

A Bloch state in a crystal experiences a periodic lattice potential. Its band dispersion relation $ϵ(k)$ encodes the lattice response, so an effective mass allows a wavepacket's acceleration under a weak applied force to be described without explicitly following the microscopic periodic forces.

In one dimension the wavepacket group velocity and semiclassical equation for crystal momentum areDifferentiating the first equation with respect to time givesDefining $F=m^{\ast }a$ yieldsThis band effective mass is controlled by band curvature, not the slope. A band minimum has positive $m^{\ast }$; a band maximum has negative Electron $m^{\ast }$ and is often more conveniently described with positive-mass electron holes. In three dimensions the inverse effective mass is generally a tensor, $(m^{\ast -1}{)}_{ij}={\hslash }^{-2}{\partial }_{k_i}{\partial }_{k_j}ϵ$. The scalar formula assumes motion along a specified direction in an isolated band, weak fields and negligible interband transitions; the bare Electron mass has not physically changed.

Let $e>0$ denote the elementary charge, so an Electron's charge is $-e$. In the relaxation-time Drude model with positive transport effective mass,The steady drift velocity is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m^{\ast }=-\mu \mathbf{E}$, defining positive electron mobility $\mu =e\tau /m^{\ast }$. The current density is opposite to the Electron drift:By the definition of electrical conductivity, $\mathbf{J}=\sigma \mathbf{E}$, soIts units are $({\mathrm{m}}^{-3})(\mathrm{C})({\mathrm{m}}^2{\mathrm{V}}^{-1}{\mathrm{s}}^{-1})=\mathrm{S}{\mathrm{m}}^{-1}$. The mobility definition alone establishes the result in the scalar linear-response regime; the relaxation-time equation supplies a microscopic approximation for $\mu$.

Interpret “3 mm wire” as a circular wire of diameter $3\mathrm{mm}$, and take the current density as uniform. One conduction Electron per atom givesUsing $I=neA|v_d|$, the drift velocity corresponding to a current magnitude of $13\mathrm{A}$ isIt points opposite to conventional electric current. A simple distance scale over the $1/50=0.020\mathrm{s}$ cycle is $\boxed{|v_d|/f=2.71\mu \mathrm{m}}$, but this treats the magnitude as constant. For sinusoidal mains, alternating-current drift excursion gives zero net displacement after a full cycle.

If $13\mathrm{A}$ is the usual root mean square current, $v_{\mathrm{pk}}=\sqrt{2}|v_d|$ and the displacement amplitude and total absolute drift distance per cycle areIf $13\mathrm{A}$ instead denotes peak current, these become $0.431\mu \mathrm{m}$ and $1.72\mu \mathrm{m}$. The paper does not state peak versus RMS; the characteristic drift is the boxed $1.35\times 10^{-4}\mathrm{m}{\mathrm{s}}^{-1}$ on whichever current convention is chosen. It also does not explicitly say diameter; taking $3\mathrm{mm}$ as the radius would reduce all these velocities and drift distances by four. None of these distances is the microscopic path travelled at roughly the Fermi velocity between scattering events; that path cannot be obtained from the supplied current alone.

In a normal metal, the electrical resistivity can often be approximated using Matthiessen's rule,Static impurities, defects and grain boundaries scatter conduction Electrons even as $T\to 0$, giving the residual resistivity. Thermal lattice phonons add a temperature-dependent contribution through electron-phonon scattering. In a clean three-dimensional acoustic model, small-angle electron-phonon transport at low temperature gives ${\rho }_{\mathrm{phonon}}\propto T^5$ at low $T$; once thermal phonon populations are approximately classical, the resistivity is approximately linear in $T$. An electron-electron contribution proportional to $T^2$ can be relevant in some low-temperature ranges.

Carbon in mild steel, together with associated disorder, strain and microstructural changes, adds scattering and generally raises the residual resistivity relative to purer iron. Thus a dilute alloy may have a similar broad phonon-dependent rise but a larger baseline. The sketch is a generic phonon-plus-disorder model, not measured iron data. Iron's magnetic scattering and phase changes, and real steel's phases and processing history, can add structure; the simple sum is approximate rather than an exact material law. The expected explanation is greater Electron scattering from the added carbon and resulting disorder.