CHEM 548: Materials Chemistry

Lecture 19: Electrical Conductivity: the Drude and Free Electron Models

Mon 11/09/2026 · Meeting 21

Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 10 §10.2

Learning Objectives: The Drude Model

Electrical Conductivity and the Drude Model

Distinguish between extrinsic (I, V, R) and intrinsic (J, E, σ, ρ) electrical properties
Apply the Drude model to describe electron behavior as an ideal gas in metals
Calculate key parameters: drift velocity, electron mobility, mean free path, and relaxation time
Derive the relationship between conductivity, carrier concentration, and mobility (σ = neμ)
Estimate conductivity parameters for simple metals using the Drude model
Recognize the limitations and failures of the Drude model in predicting temperature dependence and trends across different metals

Electrical Properties of Materials | Materials Science & Chemistry

Learning Objectives Electrical Conductivity and the Drude Model Distinguish between extrinsic (I, V, R) and intrinsic (J, E, σ, ρ) electrical properties Apply the Drude model to describe electron behavior as an ideal gas in metals Calculate key parameters: drift velocity, electron mobility, mean free path, and relaxation time Derive the relationship between conductivity, carrier concentration, and mobility (σ = neμ) Estimate conductivity parameters for simple metals using the Drude model Recognize the limitations and failures of the Drude model in predicting temperature dependence and trends across different metals Electrical Properties of Materials | Materials Science & Chemistry

To talk about conductivity as a material property, we first have to separate it from the geometry of the sample, and that starts with Ohm's law.

Ohm's Law

From Ohm's Law to Conductivity

Recasting Ohm's law into intrinsic form: J A equals E L over L rho over A, which reduces to J equals E over rho, and J equals sigma E with conductivity sigma equal to one over rho.

Conductivity of Materials

Logarithmic scales of resistivity and conductivity spanning 28 orders of magnitude, with metals at high conductivity, semiconductors in the middle, and insulators at low conductivity.

Conductivity of Select Materials

Table of conductivities in siemens per meter for selected materials, from silver, copper, and aluminum near ten to the seventh down through oxides and semiconductors to quartz and Teflon near ten to the minus sixteen and below.

In the upper left, we have elemental metals with the highest conductivity, often used as conductors.

Silver and copper: around 6×10⁷ S/m — silver is actually the most conductive element at room temperature; aluminum isn't far behind; also rhenium trioxide with quite high conductivity, other oxides, and a nitride like niobium nitride
What makes silver so much more conductive than metals like titanium or lanthanum — over an order of magnitude more conducting
What makes some oxides like rhenium trioxide and strontium molybdates so highly conducting, whereas SiO₂, aluminum oxide, and nickel oxide are highly insulating — you might say for SiO₂ and aluminum oxide all electrons are tied up in covalent bonds with no partially filled bands, and that's true, unlike in titanium oxide, niobium nitride, or rhenium trioxide
But nickel oxide: rock salt structure, Ni²⁺ is d⁸ with partially filled d orbitals; titanium oxide also rock salt with Ti²⁺ (d²), yet 13 orders of magnitude difference in conductivity — what explains that
Semiconductors like silicon and germanium: silicon's intrinsic conductivity isn't very high, yet there's a hugely important industry based on it — how do we understand semiconductor conductivity
Organics: Teflon unsurprisingly a terrible conductor, but doped polyacetylene has appreciable conductivity, not far behind some metals — how do we get polymers and organic molecules to conduct electricity
These are questions we want to answer by the end of this chapter

Our first model for answering them is the Drude model.

The Drude Model: a Free Electron Gas

A grid of metal cations with an electron label; a click traces the electron's zigzag path as it collides with ions in the lattice and redirects at each collision.

The Drude model (Paul Drude, 1900) predates quantum mechanics.

Treats metals as arrays of metal cations surrounded by a sea of valence electrons — for example, sodium as Na⁺ ions in a sea of electrons
While simple and not accurate, it provides a quantitative starting point: valence electrons as an ideal gas, kinetic molecular theory

[advance: trace the electron's path through the lattice]

The electrons are moving very fast, but in some ways they're not really going anywhere — the electron keeps colliding with the ions of the lattice
Every collision with a lattice ion redirects its path
Over time: no net flow of current in one direction or the other in the absence of an external driving force

If the electrons behave as an ideal gas, their speeds follow the Boltzmann distribution.

The Boltzmann Distribution

Boltzmann probability distributions of electron speed for a cold gas peaked at low speed and a hot gas spread to higher speeds; a click adds the average energy three halves k T and the root-mean-square velocity formula giving about ten to the fifth meters per second at 300 kelvin.

Treating electrons as an ideal gas means using the Maxwell-Boltzmann distribution to describe electron speeds versus the number of electrons at each speed.

Cold gas: distribution peaks at lower speeds, concentrated around the average
Hot gas: spreads out and shifts to higher speeds (similar effects occur for heavy versus light atoms, but here we're considering electrons)

[advance: add the average energy and the rms velocity]

Average energy E = (3/2)kT; from kinetic energy E = (1/2)mv², we can solve for velocity: v = √(2E/m)
Substituting the Boltzmann energy expression gives v = √(3kT/m_e)
At room temperature (300 K): approximately 100,000 m/s — extremely fast because electrons are so light

Between collisions, how far does an electron actually get at that speed?

Mean Free Path and Relaxation Time

The relaxation time tau equals mean free path over velocity, annotated with mean free path approximately one nanometer, velocity ten to the fifth meters per second, and relaxation time approximately ten to the minus fourteen seconds.

Drift Velocity

The electrons are moving very fast, but randomly.

To get a net flow of current in one direction we must apply an electric field. The net speed of the electrons being pushed by the electric field is called drift velocity.

[advance: add Newton's law and the force on the electron]

The force acting on the electron is its charge (e) times the electric field (E).

[advance: solve for the acceleration and the drift velocity]

Rearranging gives the acceleration of the electron, and multiplying the acceleration by the relaxation time gives the drift velocity.

[advance: add the size of the drift velocity]

From Newton's law, the force on an electron is F = ma = -eE. Rearranging: a = -eE/m_e. The drift velocity is v_d = aτ = -eEτ/m_e.

Example: with E = 100 V/m and τ ≈ 10^-14 s, v_d ≈ 0.2 m/s — much smaller than the thermal velocity (~100,000 m/s) because electrons keep colliding with ions. Increasing the relaxation time (longer travel between collisions) increases drift velocity and current.

How easily a material's electrons pick up that drift velocity is its mobility.

Electron Mobility

Conductivity: $\sigma = ne\mu$

Current density J equals n e v d with n the concentration of electrons; a click substitutes the mobility to give J equals n e mu E, compares with J equals sigma E, and identifies sigma equals n e mu.

Adding Up the Numbers: Sodium

Table of Drude-model quantities for sodium: measured conductivity 2.1 times ten to the seventh, electron concentration 2.5 times ten to the twenty-eighth per cubic meter, mobility 5.2 times ten to the minus third, relaxation time 3 times ten to the minus fourteenth seconds, rms velocity ten to the fifth meters per second, and mean free path 3 nanometers.

For sodium (one valence electron per atom): measured conductivity is 2.1×10^7 S/m.

Assuming each sodium contributes one electron: n = 2.5×10²⁸ m^-3 (from crystal structure density)
From σ = neμ, we calculate μ = 5×10^-3 m²/(V·s); from mobility, τ = 3×10^-14 s (close to our earlier estimate)
Velocity from the Boltzmann distribution and τ: mean free path = 3 nm — this seems reasonable (our initial guess was 1 nm; electrons might travel 5-6 unit cells before collision)
Limitations: the Drude model doesn't correctly predict that metals become more conductive at low temperatures (counterintuitive since electrons slow down)
Nor does it explain why alloys have reduced conductivity compared to pure metals

There is another problem, and this table exposes it.

Valence Electron Concentration and Conductivity

Table 10.2 Conductivity σ does not scale with the valence-electron concentration n.

Metal	n (m⁻³)	σ (S/m)	Metal	n (m⁻³)	σ (S/m)
Rb	1.1×10²⁸	0.8×10⁷	Ag^a	5.9×10²⁸	6.2×10⁷
K	1.3×10²⁸	1.4×10⁷	Au^a	5.9×10²⁸	4.5×10⁷
Na	2.5×10²⁸	2.1×10⁷	Cu^a	8.4×10²⁸	5.9×10⁷
Ca	4.6×10²⁸	2.9×10⁷	Zn^a	13.1×10²⁸	1.7×10⁷
Mg	8.6×10²⁸	2.3×10⁷	Sc	12.0×10²⁸	0.18×10⁷
Al	18.0×10²⁸	3.8×10⁷	Ti	20.5×10²⁸	0.25×10⁷

^a The d subshells of Cu, Ag, Au, and Zn are assumed to be full and are not counted in the valence-electron count.

Why doesn’t σ scale with valence electron concentration, n, once we move beyond the alkali metals?

Check Your Understanding: Which Velocity Carries the Current?

Question 3 asking whether the high random thermal velocity or the much slower drift velocity is directly responsible for the net flow of current, then its answer: the drift velocity, the net motion in response to the electric field.

3. The lecture describes electrons having a very high random thermal velocity (~100,000 m/s) and a much slower drift velocity (~0.2 m/s).

Which of these is directly responsible for the net flow of current (J)?
A: Neither; current is caused by the electric field itself, not electron motion
B: The drift velocity, which is the net motion in response to the electric field
C: Both contribute equally to the current

Take a moment and think about it.

[advance: reveal the answer]

B: the drift velocity, which is the net motion in response to the electric field — that's right
This slow, net movement in one direction, superimposed on the random thermal motion, is what constitutes the electrical current

That brings us to your homework on the Drude model.

Learning Objectives: Quantum Mechanics and Metallic Conductivity

Quantum Mechanics and Metallic Conductivity

Explain the transition from the Drude model to the free electron model and identify the key quantum mechanical principles incorporated in the free electron model
Describe the parabolic E(k) relationship in the free electron model and relate it to the particle-in-a-box treatment
Connect effective mass (m*) to band curvature and predict how band structure affects carrier mobility and conductivity
Apply the Fermi-Dirac distribution to explain which electrons contribute to electrical conductivity and why only states near the Fermi level matter
Explain the temperature dependence of metallic conductivity, including the role of phonon scattering and residual resistivity
Predict relative conductivities across the periodic table by analyzing band structure, particularly distinguishing between d-band and s-band contributions to transport
Justify why coinage metals (Cu, Ag, Au) exhibit exceptionally high conductivities compared to transition metals based on their electronic band structures

Learning Objectives Quantum Mechanics and Metallic Conductivity Explain the transition from the Drude model to the free electron model and identify the key quantum mechanical principles incorporated in the free electron model Describe the parabolic E(k) relationship in the free electron model and relate it to the particle-in-a-box treatment Connect effective mass (m*) to band curvature and predict how band structure affects carrier mobility and conductivity Apply the Fermi-Dirac distribution to explain which electrons contribute to electrical conductivity and why only states near the Fermi level matter Explain the temperature dependence of metallic conductivity, including the role of phonon scattering and residual resistivity Predict relative conductivities across the periodic table by analyzing band structure, particularly distinguishing between d-band and s-band contributions to transport Justify why coinage metals (Cu, Ag, Au) exhibit exceptionally high conductivities compared to transition metals based on their electronic band structures

The free electron model starts from a very simple picture of the potential an electron feels.

The Potential in the Free Electron Model

The Free Electron Model

The Bloch function psi of x equals e to the i k x times u of x; clicks set u of x to one, apply the time-independent Schroedinger equation to give E equals V zero plus h-bar squared over two m e times k squared, and draw the particle-in-a-box wave functions oscillating faster as k increases.

What is the crystal orbital wave function?

Recall from Chapter 6: the Bloch function states the wave function of a crystal orbital equals e^(ikx) multiplied by a basis set function of x. We'll work in one dimension for simplicity.

[advance: set the basis function to one]

In the free electron model, we set the basis set to unity. By assuming constant potential throughout the crystal, we effectively ignore the nuclei and their attractive forces — meaning we disregard atomic orbitals and bonding entirely. We simply confine electrons within the crystal boundaries. With this simplification, the wave function reduces to e^(ikx).

[advance: apply the Schrödinger equation]

Applying this wave function to the time-independent Schrödinger equation yields E = V₀ + (ℏ²/2mₑ)k², where k is the crystal momentum — the quantum number discussed extensively in Chapter 6.

[advance: draw the wave functions in the box]

Plotting the wave functions that solve the Schrödinger equation yields results analogous to the particle-in-a-box problem. At k = 0, the wave function equals 1 — a straight line across the crystal. As k increases, the wave function oscillates more frequently: wavelength λ = 2π/k, so larger k means shorter wavelengths and more oscillations.

Now let's examine how the energies of these wave functions vary with k.

The Free Electron Band Structure

Band Structure Comparison

Side-by-side band structures: the purely parabolic free-electron model and the tight binding model for a hydrogen atom chain, which flattens near the zone edges.

Effective Mass

The energy expression E equals V zero plus h-bar squared over two m-star times k squared; clicks add the second derivative giving m-star equals h-bar squared over the second derivative of E with respect to k, a red circle highlighting the energy expression with the takeaway that wide bands mean large curvature and small effective mass, remove the circle, and add the mobility relation mu equals e tau over m-star.

Now let's dig a little bit deeper behind this m star, the effective mass.

This is our expression for the energy of any crystal orbital in the band structure.

[advance: take the second derivative]

If I were to take the second derivative of this function with respect to k, then the term on the right is just going to be reduced to h bar squared over m star. And then the term on the left is going to be the second derivative of the e versus k function with respect to k. What does this do for us?

[advance: circle the energy expression and add the wide-bands takeaway]

So this is the expression we want to focus on.

[advance: clear the red circle]

This relationship connects effective mass to band curvature. Wide bands have large curvature: the second derivative of E versus k is large, which gives a small effective mass. Conversely, flat bands have small curvature, so the second derivative is small. This small curvature in the denominator yields a large effective mass.

[advance: add the mobility relation]

We can also relate mobility to effective mass. The mobility expression is μ = eτ/m, where e is electron charge, τ is relaxation time, and m is mass. Substituting effective mass m* shows that small effective mass yields high mobility. Since mobility is a key factor determining conductivity: wide bands with large curvature produce small effective mass, leading to high mobility and therefore high conductivity — at least that's half of the conductivity equation.

The other half of the equation is how many electrons can actually respond, and that is a question about occupation.

The Fermi-Dirac Distribution

The other half of the conductivity equation is the concentration of electrons that can move in response to an applied electric field.

The Drude model assumed all valence electrons could respond — that's not accurate
We need to consider how much energy it takes to excite an electron from a filled crystal orbital to an empty one — the Fermi-Dirac distribution function
At zero Kelvin: all states below the Fermi level fully occupied, all states above empty
At finite temperatures: the sharp boundary smears out — some crystal orbitals above the Fermi level become occupied, some below become empty
An electron very close to the Fermi level can be thermally excited into an empty state; an electron half an eV below is too far away
As temperature increases, the smearing becomes more pronounced

Multiplying that occupation function by the density of states shows which electrons actually matter.

Density of States in the Free Electron Model

$The free-electron density of states rising as the square root of energy, with the occupied states f of E times N of E shaded up to the Fermi energy; only the fraction of valence electrons near the Fermi level carries the current.$

Multiplying the Fermi–Dirac distribution by the density of states in the free-electron model produces this curve.

Density of states increases with the square root of energy — two key points emerge
First: not all valence electrons can respond to an applied field — only those near the Fermi level, within a few percent of the total
This might suggest similar carrier concentrations in different metals; but as more electrons are added, the Fermi level moves higher into the band, where the density of states is larger — the absolute number of mobile carriers rises
Explains the alkali metal trend: sodium conducts better than potassium, which conducts better than rubidium
As we move to smaller, more densely packed atoms, the same number of valence electrons occupy a smaller volume, so the electron density increases and the Fermi level is pushed up to higher energies, where the density of states is larger
Only electrons near the Fermi level can move, but a higher density of states there means more mobile electrons in total, giving better conductivity

That leaves the speed of the electrons that do respond.

Fermi Velocity

The filled free-electron parabola up to the Fermi energy; clicks add the momentum relations p equals h-bar k, m-star v equals h-bar k, and v equals h-bar k over m-star, then the Fermi velocity v F equals the square root of two E F over m-star.

Now what about the speed of the electron?

That's another parameter that's going to be important in figuring out the conductivity. In the derivative model, it was a Maxwell-Boltzmann distribution of speeds; in the free electron model, it's actually dependent upon K.

[advance: relate momentum to the wave vector]

We know the electron's momentum equals ℏ times k, its reciprocal space wave vector. Since momentum also equals mass times velocity, we can write mv = ℏk, where m is the effective mass. Solving for velocity gives v = ℏk/m*.

The key takeaway: velocity depends on k. Electrons in states near k = 0 have almost no velocity — at k = 0, exactly zero. As k increases toward the band edge where filling stops, the velocity becomes much higher. So what is the velocity of electrons at the Fermi level?

[advance: solve for the Fermi velocity]

Well, we can make an estimate of that. Take the kinetic energy expression, energy equals one half m v squared, with the energy equal to the Fermi energy; then we call the velocity of those electrons the Fermi velocity. Solving gives the square root of two times the Fermi energy divided by the effective mass.

For a real metal like copper, that velocity is startlingly fast, and here is why it matters.

Uncompensated Electrons Carry the Current

Filled free-electron parabolas with no applied field and with an applied field, where the electron distribution shifts in k and a bracket marks the uncompensated electrons near the Fermi energy; for copper the Fermi energy is 7.0 electron volts and the Fermi velocity 1.6 times ten to the sixth meters per second.

In the free electron model, we can calculate the Fermi energy from atomic density and use it to estimate the velocity of electrons at the Fermi level.

Copper: 1.6 × 10⁶ m/s — more than an order of magnitude faster than the 1 × 10⁵ m/s we estimated in the Drude model
Crucially, these fast-moving electrons near the Fermi level are the ones that matter for conductivity — here's why
Apply a field: electrons with negative k traveling leftward are attracted to the positive charge; electrons with positive k traveling rightward are repelled by the negative charge
The electron distribution shifts: more electrons moving in one direction than the other
Uncompensated electrons — the ones without a corresponding electron traveling in the opposite direction to cancel them out — carry the electrical current, and they are located near the Fermi energy
Far fewer electrons can move than in the Drude model, but those that do move travel much faster — the two effects largely compensate, giving similar conductivity predictions from both models

Where the two models really part ways is temperature.

Temperature Dependence of Metals

Resistivity of aluminum and copper versus temperature: linear decrease on cooling from 800 kelvin, then a plateau below about 20 kelvin marking the residual resistivity where defect scattering dominates; the mean free path in high-purity aluminum is 29 nanometers at 300 kelvin and 0.7 millimeters at 1 kelvin.

However, there are important differences between the models.

Resistivity of aluminum and copper versus temperature: resistivity decreases as temperature drops — metals become more conductive at lower temperatures, and for simple metals this relationship is actually linear
The Drude model predicts the opposite: it suggests electrons slow down at low temperature, which would decrease conductivity, not increase it; this contradicts experimental observations
What actually happens is not that electron speed changes significantly, or that the number of electrons responding to an external field differs much — instead, at low temperatures the atoms vibrate much more slowly, so electrons scatter less frequently from ion cores, and reduced scattering increases conductivity
Back-of-the-envelope: high-purity aluminum at room temperature has a mean free path of about 29 nm — ten times longer than the Drude estimate; at very cold temperatures, electrons can travel almost a millimeter before colliding with an ion core, which is an incredible distance
At very low temperatures conductivity levels off, nearly constant — the plateau is the residual resistivity, and it comes from impurities in the metal

Those two scattering mechanisms set up the last puzzle: the transition metals.

Conductivities of Transition Metals

Table of transition-metal conductivities in units of ten to the seventh siemens per meter across groups 3 through 11, with the coinage metals copper, silver, and gold jumping to 5.9, 6.2, and 4.5 while most transition metals sit far lower.

Electrons are scattered by two mechanisms as they move through the lattice in response to an external field: vibrating atoms displaced from their proper positions, and defects or impurities in the crystal.

At very cold temperatures, vibration scattering becomes negligible and defect scattering dominates; the transition temperature reflects crystal quality — this is the residual resistivity
Room temperature: sodium about 2 × 10⁷ S/m, magnesium similar, aluminum about 4 × 10⁷
Most transition metals: much lower — the early ones fall well below this range, except perhaps molybdenum and tungsten, which approach magnesium and sodium
Groups eight and nine show respectable values; group ten drops again; group eleven — the coinage metals — suddenly jumps up by a factor of four to six
Why are copper and silver the kings of conductivity while manganese, titanium, and yttrium are poor conductors? Let's examine the band structure

Palladium, in group ten, is the place to start.

Band Structure of Palladium

The band structure of face-centered cubic palladium and its partial density of states: the Fermi level cuts through narrow 4d bands, which translates to higher effective mass and lower mobility.

We're now coming full circle.

The Drude model was too simplistic; the free electron model is better — it incorporates quantum mechanics but ignores bonding entirely, so it can't distinguish one material from another
Palladium: face-centered cubic; a wide band with significant S character, similar to our free electron band
Below it: five narrower, less dispersed bands — the partial density of states plots show these are the palladium 4d bands
At very high energy, almost irrelevantly high: bands with P character
Key point: the Fermi level cuts through several relatively flat bands with d orbital character
What happens when we move to silver — one element to the right, one additional valence electron?

Here is silver's band structure.

Band Structure of Silver

The band structure of face-centered cubic silver and its partial density of states: the filled 4d bands sit below the Fermi level, which now cuts through the wide 5s band, translating to lower effective mass and higher mobility.

Gradescope Quiz

Practice questions

Work the Lecture 19 practice questions before the next class. They cover the intrinsic electrical properties and Ohm's law, the Drude model from drift velocity and mobility through the relaxation time and mean free path, the free electron model and its parabolic band, effective mass and band curvature, the Fermi-Dirac distribution and the Fermi velocity, the temperature dependence of metallic conductivity, and the d-band versus s-band explanation of the coinage metals.

Open the Lecture 19 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.