Free-Electron Model of Metals

Introduction

Metal atoms often contribute electrons that move through an extended crystal rather than remaining on one bond. The free-electron model makes a deliberately severe approximation: replace the complicated atomic potential by a constant interior potential and treat the conduction electrons as nearly independent particles confined to a large box. Despite ignoring the lattice's detailed periodic potential, this model introduces the central scales of metal physics: k-space filling, Fermi energy and the fact that only electrons close to the Fermi surface respond thermally to ordinary temperatures.

Core explanation

For an electron with wavevector k in the interior of the idealised box, the energy measured from the chosen potential zero is E(k) = ħ²k²/(2m e) . The dependence is isotropic: all k vectors of the same length have the same energy. Boundary conditions quantise allowed k values. In a very large specimen, points in k-space are extremely dense, so one can count them using volumes. Each spatial k state holds at most two electrons, one of each spin projection. At T = 0 K, electrons fill the lowest-energy available states first because of the Pauli exclusion principle.

Since E rises with k², occupied points form a sphere in three-dimensional k-space. If n = N/V is the conduction-electron number density and spin degeneracy is two, the counting gives k F = (3π²n)^(1/3) . The highest occupied energy is E F = ħ²k F²/(2m e) . Thus E F scales as n^(2/3), and the Fermi speed scale is v F = ħk F/m e. These equations assume a parabolic free-electron dispersion with bare electron mass, one isotropic population, and no important interactions. They describe a reference model rather than a direct universal measured property of every metal.

The Fermi energy is usually several electronvolts, corresponding to a Fermi temperature T F = E F/k B of tens of thousands of kelvin. Room temperature is small compared with this scale. It does not mean the electrons are cold in the classical sense: filled states span a wide energy range even at T = 0 because Pauli exclusion forces many electrons into higher momentum states. Heating to room temperature changes occupations only within an energy window of order k BT around E F. This explains why a metal's electronic heat capacity is much smaller than a classical calculation that assigns k BT of thermal energy to every conduction electron would predict.

The model also helps explain conduction qualitatively. Applying an electric field slightly shifts the occupation of states near the Fermi surface, producing a net current. A completely filled symmetric set of states has cancelling velocities and cannot simply shift into occupied neighbours. Electrical resistance still requires scattering, for example by phonons, impurities and defects. A uniform interior potential alone cannot predict a finite resistivity, and the free-electron picture misses the detailed bands, gaps and effective masses produced by the lattice.

For copper, a widely used free-electron density is approximately 8.47 × 10²⁸ m⁻³, treating roughly one conduction electron per atom. The model gives E F near 7.0 eV, consistent with the OpenStax free-electron treatment and its worked metal table. The MIT physical chemistry lecture on the free-electron theory of metals develops the same baseline model. Its value is not perfect realism but a clear reference against which band-structure and correlation effects can be recognised.

Step-by-step reasoning

1. Decide how many electrons per atom are treated as conduction electrons. 2. Convert atomic number density to electron number density n in m⁻³. 3. Compute k F from (3π²n)^(1/3). 4. Compute E F from ħ²k F²/(2m e), then convert joules to electronvolts if needed. 5. Check whether predictions depend sensitively on ignored periodic potentials or interactions.

Visual explanation

Draw a sphere centred at k = 0. The interior contains filled k states at zero temperature; the surface is labelled k F. Next to it draw an upward parabola E versus k, reaching E F at ±k F. A thin shell around the sphere surface is shaded to show the states whose occupation changes appreciably at ordinary temperatures or under a weak applied field.

Real-world analogy

Imagine assigning people to seats in a stadium from the centre outward, with only two people allowed at each location. Even at zero temperature, late arrivals must occupy seats far from the centre because inner seats are full. Gentle heating shuffles mostly those near the outer occupied edge; it does not displace everyone equally.

Real-world example

The estimate E F ≈ 7 eV for copper shows why k BT ≈ 0.026 eV at room temperature is a small perturbation to the electron distribution. Copper's excellent conductivity, however, is not obtained from E F alone. The frequency of scattering from vibrating atoms and defects, plus its actual band structure, determines measured resistivity.

Why?

Why is the Fermi sphere a sphere? Under the free-electron assumption, energy depends only on k , not its direction. Filling all states below the same maximum energy therefore fills all k points within one radius. An anisotropic crystal band can instead have a distorted Fermi surface.

Common misconception

"Every electron in a metal has energy E F." E F is the upper occupied energy at zero temperature. Occupied states range from the band bottom up to that level, and at finite temperature the occupancy edge is smoothed over an energy range of order k BT.

Worked example

Question: Estimate E F for a free-electron density n = 8.5 × 10²⁸ m⁻³. Use ħ = 1.055 × 10⁻³⁴ J s, m e = 9.11 × 10⁻³¹ kg and 1 eV = 1.602 × 10⁻¹⁹ J.

Reasoning: k F = (3π²n)^(1/3) ≈ (2.52 × 10³⁰ m⁻³)^(1/3) ≈ 1.36 × 10¹⁰ m⁻¹. Then E F = (1.055 × 10⁻³⁴)²(1.36 × 10¹⁰)²/[2(9.11 × 10⁻³¹)] ≈ 1.13 × 10⁻¹⁸ J. Divide by 1.602 × 10⁻¹⁹ J/eV to obtain about 7.1 eV. Rounding density and constants accounts for a small difference from a tabulated 7.0 eV.

Answer: Approximately 7 eV for this simple model.

Quick check

1. If electron density doubles, does the free-electron Fermi energy double? Answer: No. Since E F ∝ n^(2/3), it grows by 2^(2/3), approximately 1.59, under the stated assumptions.

Exam focus

State the spin factor of two in k-state counting and distinguish k F, E F and Fermi temperature. Explain why electrons far below E F contribute little to small thermal changes, and identify the constant-potential assumption as the model's major limitation.

Advanced insight

The free-electron gas predicts a three-dimensional density of states proportional to √E, measured from its band bottom. Real metals have crystal-dependent dispersions and can exhibit sharp density-of-states features near saddle points, often called van Hove singularities. Replacing m e by an effective mass is useful near a parabolic band extremum, but a single effective mass cannot repair all departures from an isotropic free-electron picture.

Summary

The free-electron model fills a spin-degenerate sphere of k states at zero temperature. Its radius and upper energy follow k F = (3π²n)^(1/3) and E F = ħ²k F²/(2m e). It captures important scales and qualitative metallic response while omitting periodic band gaps, anisotropy and detailed scattering.

Practice questions

1. Why do occupied k states form a sphere in this model? Answer: Energy depends only on the magnitude of k, so equal-energy surfaces and the filled region are spherical. 2. What sets the factor of two in counting electrons per k state? Answer: The two electron spin projections allowed for one spatial wavevector state. 3. How does E F vary with n in three dimensions? Answer: It is proportional to n^(2/3) for noninteracting free electrons. 4. Why does the model alone not predict copper's measured resistivity? Answer: It omits the actual periodic band structure and scattering by phonons, defects and impurities.