University Physics IV · Molecular and Solid-State Physics · 12.7

Band Filling, Metals & Insulators

Count the valence electrons in one primitive cell and you can usually say whether a solid conducts before computing a single energy. This lesson is about where that count is decisive, where band overlap defeats it, and why conduction still needs a scattering time and an effective mass on top of it.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Once Bloch's theorem has handed you bands, "is this a metal?" becomes a counting question before it is an energy question. Born-von Karman boundary conditions on N primitive cells allow exactly N distinct k in the first Brillouin zone, and spin doubles that, so every band holds 2N electrons: z valence electrons per primitive cell fill z/2 bands. An odd z leaves a band half full, with empty states an arbitrarily small energy above the highest occupied one — a Fermi surface, and a metal.

An even z can fill bands exactly, and a filled band carries no current at any field, because the field displaces every k alike and a full band displaced is the same full band. That classification is astonishingly cheap; it costs three things. It decides the answer only if a gap survives in every direction of k, and in three dimensions bands overlap, which is why the divalent elements are metals.

Free states alone are not conductivity: σ = n e squared τ over m* still needs a scattering time from whatever breaks the periodicity, and an effective mass from the curvature of E(k). And the whole scheme assumes independent electrons, so a solid whose on-site repulsion beats its bandwidth — NiO and every other Mott insulator — is misclassified by construction.

Simple definition: A solid is a metal when its highest occupied band is only partly filled, so empty states lie immediately above the last occupied one; it is an insulator or a semiconductor when every occupied band is exactly full and an energy gap separates it from the next empty band.
Example: Sodium gives one electron per primitive cell to a band holding two, so it is half full and sodium conducts at 2.1 x 10⁷ S m⁻¹; silicon gives eight to four bands holding eight, every band fills, a 1.12 eV gap opens, and its conductivity is 10¹¹ times smaller.

How many states a band holds2N states per band, so z electrons per cell fill z/2 bands

An odd z leaves a half-filled band and a Fermi surface — a metal — with no energy calculation performed at all.

N primitive cells give N allowed k in the zone; spin doubles it. z is counted per primitive cell, never per atom.

Current carried by one bandj = -(e/V) Σ_occ v(k), with v = (1/ℏ) dE/dk

E(k) = E(−k) forces v(−k) = −v(k), so a filled band's sum cancels state by state and j vanishes identically.

v in m s⁻¹, j in A m⁻², and the sum runs over the occupied states of that one band.

Semiclassical response to a fieldℏ dk/dt = −eE, so the sea shifts by dk = eEt/ℏ

Copper at 10 mV m⁻¹: dk = 0.38 m⁻¹ against k_F = 1.36 x 10¹⁰ m⁻¹, a tilt of one part in 3.6 x 10¹⁰.

dk in m⁻¹, E in V m⁻¹, t the scattering time in s. Every k in the band shifts by the same dk.

Effective mass from band curvature1/m* = (1/ℏ²) d²E/dk²

Tight binding E = E₀ - 2t cos ka with t = 1.0 eV and a = 0.30 nm gives m* = 0.42 mₑ. Wide band, light electron.

m* in kg, positive near a band minimum and negative near a band maximum.

Conductivity of a partly filled bandσ = n e² t / m* = n e u

Copper's 5.96 x 10⁷ S m⁻¹ at n = 8.47 x 10²⁸ m⁻³ implies t = 25 fs and a mean free path of 39 nm.

n in m⁻³, t in s, m* in kg, sigma in S m⁻¹, mobility u = et/m* in m² V⁻¹ s^-1.

Holes in a nearly filled bandj = +(e/V) Σ_empty v(k), with m*ₕ = −m*ₑ

It trades 10²⁸ electrons for 10²⁰ carriers of the opposite sign, and it is why aluminium and zinc show a positive Hall coefficient.

The sum now runs over the few empty states, each acting as a carrier of charge +e and positive mass.

A band holds 2N states, and N counts cells, not atoms

Impose Born-von Karman boundary conditions on a crystal of N primitive cells and exactly N distinct values of k fit into the first Brillouin zone — one per cell, whatever shape the dispersion takes. Spin doubles the capacity, so one band holds 2N electrons and z valence electrons per primitive cell fill z/2 bands. Sodium is body-centred cubic with one atom per primitive cell and one 3s electron, so z = 1: the band is half full and sodium is a metal before any energy is evaluated. Silicon has two atoms per primitive cell and four valence electrons each, so z = 8, four bands fill exactly, and a gap must open. The commonest slip is counting per atom or per conventional cell: silicon's conventional cubic cell contains eight atoms and its primitive cell two, and only the primitive count decides anything.

A filled band carries no current, in any field

The current from a band is j = -(e/V) times the sum of v(k) over its occupied states, with v = (1/ℏ) dE/dk. Time-reversal symmetry gives E(k) = E(−k), so v(−k) = −v(k): in a full band every state is paired with its opposite and the sum vanishes exactly. A field cannot repair that. The semiclassical equation ℏ dk/dt = −eE displaces every k by the same amount, and in a full band the displaced set of occupied states is the same set — a state pushed past the zone boundary re-enters at the opposite edge by Bragg reflection — so the occupation, and the current, are unchanged. A partly filled band behaves entirely differently: the whole Fermi sea slides by dk = eEt/ℏ, which for copper at 10 mV m⁻¹ is 0.38 m⁻¹ against a Fermi wavevector of 1.36 x 10¹⁰ m^-1. Conduction is that one-part-in-10¹⁰ asymmetry.

Counting is not destiny: bands overlap

An even z fills bands only if a gap survives in every direction of k. The energy at the zone boundary depends on direction, so in three dimensions the maximum of band n along one direction can lie above the minimum of band n+1 along another, and the two bands overlap in energy without ever touching at the same k. Calcium is face-centred cubic with one atom per primitive cell and z = 2, so the count alone predicts an insulator; its resistivity is 34 nOhm m, a better conductor than iron. The 4s band overlaps the 3d, electrons spill from one to the other, and the metal ends up with a hole Fermi surface in the lower band and an electron Fermi surface in the upper, carrying equal numbers of each. Squeeze the overlap towards zero and you reach a semimetal: bismuth carries about 3 x 10²³ carriers per cubic metre, a few millionths of copper's.

Holes are book-keeping, not new particles

Summing 10²⁸ occupied velocities to get a current controlled by a handful of empty states is absurd, so subtract the identically zero full-band sum from it. What remains is j = +(e/V) times the sum of v(k) over the empty states: the same current, now written as a sum over the few vacancies, each behaving as a carrier of charge +e. Near a band maximum the curvature d²E/dk² is negative, so the electron effective mass there is negative and the hole mass m*ₕ = −m*ₑ comes out positive; a hole therefore accelerates along E like an ordinary positive particle. Nothing physical has been added — this is one algebraic identity applied to one sum — but it is what explains the positive Hall coefficient measured in aluminium and zinc, which no free-electron model can produce.

Free states are necessary; tau and m* set the size

Empty states nearby make conduction possible, and σ = n e² t / m* = n e u says how much of it you get. The point worth pausing on is that a perfect static periodic lattice contributes no resistance at all: Bloch states are stationary states of that lattice, an electron placed in one propagates forever, and sigma would be infinite. Resistivity measures departures from periodicity — phonons, impurities, vacancies, surfaces — which combine by Matthiessen's rule as 1/t = 1/tₚₕ + 1/tᵢₘₚ, giving ρ = ρ₀ + ρₚₕ(T). Copper's measured conductivity implies t = 25 fs and a mean free path of 39 nm, about 110 lattice spacings between events. The mass is not the free-electron mass but the curvature of the band: 1/m* = (1/ℏ²) d²E/dk², so a tight-binding band E = E₀ - 2t cos ka with t = 1.0 eV and a = 0.30 nm gives m* = 0.42 mₑ.

Where the classification runs out

Insulator and semiconductor are not different structures; they are the same structure differing in one number. Diamond, silicon and germanium share the diamond lattice and z = 8, and their gaps are 5.47, 1.12 and 0.66 eV. At 300 K the factor exp(−Eg/2kT) that populates the conduction band is 1.1 x 10⁻⁴⁶, 3.9 x 10⁻¹⁰ and 2.9 x 10⁻⁶ respectively: forty orders of magnitude of behaviour from a factor of eight in the gap. Draw the line wherever you like, but know it is a line drawn on a continuum. The deeper limit is the independent-electron assumption. NiO leaves a partly filled 3d band, so band filling predicts a metal; NiO is an insulator with a gap near 4 eV, because the on-site Coulomb repulsion U exceeds the bandwidth W and the electrons localise. Mott insulators are where band filling stops being the right question to ask.

Change one variable at a time

Make the relationship visible.

Interactive model

Valence electrons per primitive cell2 e/cell

Gap at the zone edge1.1 eV

Width of each band2.4 eV

Hold 2 electrons per cell and drag the gap down through zero. Above zero the lower band is exactly full, the upper exactly empty, and nothing conducts; below zero the same two electrons spill across, the Fermi level cuts both bands, and the insulator has become a divalent metal.

LOWER BAND FILLED100 %

UPPER BAND FILLED0 %

FERMI LEVEL2.95 eV

GAP / kT AT 300 K42.5

Live interpretationLOWER BAND FILLED: 100 %. UPPER BAND FILLED: 0 %. FERMI LEVEL: 2.95 eV. GAP / kT AT 300 K: 42.5

Catch the common trap

Explain before calculating.

Misconception check

The error is to hear "small gap" and picture kT lifting electrons across it. Silicon's gap is 1.12 eV and kT at 300 K is only 0.0259 eV, so the gap is 43 kT: thermal energy is nowhere near enough to carry a typical electron over it. What populates the conduction band is the exponential tail of the occupancy, not its width — the factor is exp(−Eg/2kT) = exp(−21.7) = 3.9 x 10⁻¹⁰, which is why intrinsic silicon musters about 10¹⁶ carriers per cubic metre against copper's 8.5 x 10²⁸. Read the same point backwards and the gap looks unimportant, when in fact it sits in an exponent: at fixed temperature, raising Eg from silicon's 1.12 eV to diamond's 5.47 eV cuts the carrier density by more than 10³⁶. A semiconductor is not a solid whose gap is of order kT. It is a solid whose gap is large, but not hopelessly large, compared with kT.

Calcium is face-centred cubic with one atom per primitive cell and two valence electrons per atom, so the lowest band holds exactly the electrons available. Its measured resistivity is 34 nOhm m and it rises when the sample is heated. Which statement accounts for both facts?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium is body-centred cubic with one atom per primitive cell and one 3s valence electron per atom. A crystal contains N = 2.0 x 10²² primitive cells. How many states does the 3s band hold, how many electrons occupy it, and what does that make sodium?

Born-von Karman boundary conditions on N cells allow exactly N distinct values of k inside the first Brillouin zone — one per primitive cell, however the band disperses.
Each of those k values holds two electrons, spin up and spin down, so the band's capacity is 2N = 4.0 x 10²² states.
Electrons supplied: one per atom, one atom per primitive cell, N cells, so 2.0 x 10²² electrons.
Filling fraction = 2.0 x 10²² / 4.0 x 10²² = 0.50. The band is exactly half full, so empty states lie an arbitrarily small energy above the highest occupied one.

Answer4.0 x 10²² states, 2.0 x 10²² electrons, half filled — a partly filled band with a Fermi surface, so sodium is a metal. No energy was computed anywhere: the count alone settled it.

MediumCopper has n = 8.47 x 10²⁸ conduction electrons per cubic metre and σ = 5.96 x 10⁷ S m⁻¹ at 300 K. Take m* = mₑ = 9.11 x 10⁻³¹ kg, v_F = 1.57 x 10⁶ m s⁻¹ and k_F = 1.36 x 10¹⁰ m^-1. A field of 10 mV m⁻¹ is applied. Find the scattering time, the mean free path, the drift speed, and the fractional displacement of the Fermi sea.

Invert σ = n e² t / m*: t = σ m* / (n e²) = (5.96 x 10⁷ x 9.11 x 10⁻³¹) / (8.47 x 10²⁸ x (1.602 x 10⁻¹⁹)²) = 5.43 x 10⁻²³ / 2.17 x 10⁻⁹ = 2.50 x 10⁻¹⁴ s.
Mean free path: l = v_F t = 1.57 x 10⁶ x 2.50 x 10⁻¹⁴ = 3.9 x 10⁻⁸ m = 39 nm, about 110 lattice spacings at a = 0.361 nm. Use v_F, not the drift speed — it is the Fermi electrons that do the travelling.
Current density J = σ E = 5.96 x 10⁷ x 1.0 x 10⁻² = 5.96 x 10⁵ A m⁻², about 0.60 A per square millimetre, so v_d = J/(ne) = 5.96 x 10⁵ / (8.47 x 10²⁸ x 1.602 x 10⁻¹⁹) = 4.4 x 10⁻⁵ m s^-1.
Displacement of the sea: dk = eEt/ℏ = (1.602 x 10⁻¹⁹ x 1.0 x 10⁻² x 2.50 x 10⁻¹⁴) / (1.055 x 10⁻³⁴) = 0.38 m^-1.
Fraction: dk/k_F = 0.38 / 1.36 x 10¹⁰ = 2.8 x 10⁻¹¹, which equals v_d/v_F = 4.4 x 10⁻⁵ / 1.57 x 10⁶ — the same number reached two ways.

Answert = 2.5 x 10⁻¹⁴ s, l = 39 nm, v_d = 4.4 x 10⁻⁵ m s⁻¹ (about 16 cm per hour), and the Fermi sea shifts by 2.8 x 10⁻¹¹ of its radius. Conduction is a barely perceptible lopsidedness in a very fast, otherwise balanced, sea.

HardSilicon has Eg = 1.12 eV and, at 300 K, √(N_c Nᵥ) = 1.71 x 10²⁵ m⁻³, uₙ = 0.135 and uₚ = 0.048 m² V⁻¹ s^-1. Express the gap in units of kT, find nᵢ and σ, compare with copper's 5.96 x 10⁷ S m⁻¹, then predict σ(400 K)/σ(300 K) given N_c Nᵥ proportional to T³ and mobility to T⁻³/2.

kT at 300 K = 1.381 x 10⁻²³ x 300 / 1.602 x 10⁻¹⁹ = 0.0259 eV, so Eg/kT = 1.12/0.0259 = 43.3. The gap is 43 kT, not of order kT, and the population factor is exp(−Eg/2kT) = exp(−21.66) = 3.9 x 10^-10.
nᵢ = √(N_c Nᵥ) exp(−Eg/2kT) = 1.71 x 10²⁵ x 3.9 x 10⁻¹⁰ = 6.7 x 10¹⁵ m⁻³, within a factor of 1.5 of the measured 1.0 x 10¹⁶ m^-3.
σ = nᵢ e (uₙ + uₚ) = 6.7 x 10¹⁵ x 1.602 x 10⁻¹⁹ x 0.183 = 2.0 x 10⁻⁴ S m⁻¹, a resistivity of about 5 kOhm m.
Ratio to copper: 5.96 x 10⁷ / 2.0 x 10⁻⁴ = 3.0 x 10¹¹. Eleven orders of magnitude, and essentially all of it sits in n rather than in tau or m*.
At 400 K, kT = 0.0345 eV and exp(−Eg/2kT) = exp(−16.25) = 8.8 x 10⁻⁸, a factor 225 larger. The T³/2 rise in √(N_c Nᵥ) and the T⁻³/2 fall in mobility cancel exactly, so sigma rises by that same factor 225.

AnswerEg = 43 kT; nᵢ = 6.7 x 10¹⁵ m⁻³ and σ = 2.0 x 10⁻⁴ S m⁻¹, so copper conducts 3 x 10¹¹ times better. Heating to 400 K multiplies silicon's conductivity by 225 while copper's falls by about 30%: the sign of d(σ)/dT is the sharpest test there is.