CHEM 548 — Lecture 19: Electrical Conductivity: the Drude and Free Electron Models

Ohm's Law

Ohm’s Law (Extrinsic)$I = \dfrac{V}{R}$

Current density$J = I/A$

Electric field intensity$E = V/L$

Resistivity$\rho = RA/L$

These intrinsic properties are independent of sample geometry.

The logical starting point for electrical properties is Ohm's law. In this form it says the current through a conductor equals the voltage drop divided by its resistance. But current, voltage and resistance are all extrinsic quantities. The resistance depends on the size of the sample, so as a copper wire gets longer its resistance goes up, and as its cross section gets smaller its resistance also goes up.

[advance: the three intrinsic quantities]

To understand and design materials we need intrinsic properties. Instead of the current we use the current density, the current divided by the cross-sectional area. Instead of the voltage drop we use the electric field intensity, the voltage drop divided by the length over which it occurs. And instead of the resistance we use the resistivity, rho, the resistance times the area divided by the length.

[advance: why these are the ones we use]

These intrinsic properties do not depend on the shape or the size of the sample.

From Ohm's Law to Conductivity

Ohm’s Law (Extrinsic)$I = \dfrac{V}{R}$

Ohm’s Law (Intrinsic)$I = \dfrac{V}{R} \;\longrightarrow\; JA = \dfrac{EL}{L\rho/A} = \dfrac{EA}{\rho}$

The lengths and the areas cancel: what is left is the material’s own property.

Resistivity, $\rho$$J = \dfrac{E}{\rho}$Ω·m

Conductivity, $\sigma = 1/\rho$$J = \sigma E$Ω⁻¹m⁻¹ = S/m

Now let's take the extrinsic form of Ohm's law, I equals V over R, and recast it in intrinsic form.

[advance: make the substitution]

We plug in the intrinsic properties we just defined. Instead of the current we write the current density times the area, instead of the voltage drop the electric field times the length, and instead of the resistance the resistivity times the length divided by the area. The lengths cancel, and there is a cross-sectional area on each side of the equation, so the areas cancel too.

[advance: J = E/ρ]

We are left with J equals E over rho. The current density equals the electric field divided by the resistivity, and resistivity has units of ohm meters.

[advance: J = σE]

Another quantity we will talk about at length is the conductivity, sigma, which is one over the resistivity, so J equals sigma E. Its units are one over ohm meters. One over ohm has its own name, the siemens, symbol S, so we write conductivity in siemens per meter.

Conductivity of Materials

Conductivity spans 28 orders of magnitude, from the best metals to the best insulators; almost no other property of a material ranges so far.

Conductivity of Select Materials

Substance	σ (S/m)	Substance	σ (S/m)
Ag	6.2×10⁷	Bi₂Ru₂O₇	2×10⁵
Cu	5.9×10⁷	LaNiO₃	1×10⁵
Al	3.8×10⁷	doped polyacetylene	8×10⁴
Na	2.1×10⁷	Fe₃O₄	2×10⁴
ReO₃	1.1×10⁷	YBa₂Cu₃O₇^*	1×10²
Ti	2.5×10⁶	Ge	2×10⁰
La	1.6×10⁶	Si	10⁻³
SrMoO₃	1.0×10⁶	NiO	10⁻⁸
Bi	7.7×10⁵	Al₂O₃	10⁻¹²
Mn	6.2×10⁵	S	10⁻¹⁵
NbN	4×10⁵	SiO₂ (Quartz)	10⁻¹⁶
TiO	3×10⁵	Teflon	10⁻²²

Why is Ag more than an order of magnitude better than Ti or La? Why do ReO₃ and SrMoO₃ conduct while SiO₂, Al₂O₃ and NiO insulate? Why are NiO and TiO 13 orders of magnitude apart? How do Si and doped polyacetylene conduct at all?

^*Conductivity in the ab plane of the crystal.

Per Solid State Materials Chemistry, section 10.2, Table 10.1

This table from our textbook lists some well-known conductors and insulators. In the upper left are the elemental metals with the highest conductivities. Silver and copper are about 6 times 10 to the 7 siemens per meter, and silver is the most conducting element at room temperature. Below them are rhenium trioxide from last lecture, other oxides, and a nitride, niobium nitride. By the end of this chapter we want to explain why silver conducts more than ten times better than titanium or lanthanum, and why rhenium trioxide and strontium molybdate conduct while silica, alumina and nickel oxide insulate. Silica and alumina have every electron tied up in covalent bonds. But nickel oxide and titanium monoxide both take the rock salt structure with partially filled d orbitals, and their conductivities differ by 13 orders of magnitude. We also want to explain how silicon carries a whole industry with a low intrinsic conductivity, and how doped polyacetylene comes close to some metals.

The Drude Model: a Free Electron Gas

We start with the Drude model, named after Paul Drude. It dates back to 1900, so it predates quantum mechanics. In general chemistry you were probably told to think of a metal as an array of cations that have lost their valence electrons, with those electrons as a sea around the cations. So sodium is a lattice of sodium plus ions in a sea of valence electrons. That picture is simple and it is not accurate, but it is a starting point. To do anything quantitative with it, the Drude model treats the valence electrons as an ideal gas and applies kinetic molecular theory to them.

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

The electrons move very fast, but in some ways they are not going anywhere. An electron moves fast, collides with one of the ions of the lattice, and every collision redirects its path. So with no external driving force there is no net flow of current in one direction or the other.

The Boltzmann Distribution

Two Boltzmann probability distributions of electron speed: a cold curve, tall and peaked at low speed, and a hot curve, lower, broader and shifted to higher speed.

Probability against speed. This is the distribution of electron energies and velocities used in the Drude model: the cold gas peaks at low speed, the hot gas is broader and shifted to higher speed.

If we treat the electrons as an ideal gas, the Maxwell-Boltzmann distribution describes their speeds. When the gas is cold the distribution peaks at a lower speed and stays close to the average value. When the gas gets hot the distribution spreads out and its maximum shifts to a higher speed. Heavy and light atoms would give the same contrast, but here we are thinking about electrons.

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

In a Boltzmann distribution the average energy of the particles is three halves k T, Boltzmann's constant times the absolute temperature. The kinetic energy is one half m v squared, so the velocity is the square root of two E over the electron mass. Putting in the average energy gives the root mean square velocity, the square root of three k T over the electron mass. Apart from the temperature these are fundamental constants, so at room temperature, 300 kelvin, we estimate about 100,000 meters per second. The electrons move that fast because they are extremely light.

Mean Free Path and Relaxation Time

$\tau = \dfrac{\ell}{v}$

Relaxation time
$\tau = $ 1×10⁻¹⁴ s

Mean free path
$\ell \approx$ 1 nm = 1×10⁻⁹ m

Velocity
$v \approx$ 1×10⁵ m/s

Mean free paththe distance an electron travels before it collides with a stationary cation

Relaxation timethe time that flight takes, so the two estimates fix $\tau$

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.

$F = ma$

$eE = m_e a$

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

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.

$F = ma$

$eE = m_e a$

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

$a = \dfrac{eE}{m_e}$

$v_d = a\tau = \left(\dfrac{eE}{m_e}\right)\tau$

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.

$F = ma$

$eE = m_e a$

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

$a = \dfrac{eE}{m_e}$

$v_d = a\tau = \left(\dfrac{eE}{m_e}\right)\tau$

The drift velocity increases as the electric field gets larger or the relaxation time gets larger ($v_d \cong 0.2$ m/s)

Now let's apply an electric field and see if we get a flow of current. What we care about is not how fast the electrons are going but their net speed in the direction the field pushes them, and we call that the drift velocity.

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

We start with force equals mass times acceleration. The force on the electron is its charge times the electric field intensity, small e times capital E, so e E equals the electron mass times the acceleration.

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

Rearranging, the acceleration of the electron is its charge times the field divided by its mass. Velocity is acceleration times time, and the time is the relaxation time, so the drift velocity is e E over the electron mass, times tau.

[advance: add the size of the drift velocity]

With a field of 100 volts per meter we get a drift velocity of about 0.2 meters per second. The drift velocity is much smaller than the thermal velocity, because the electron keeps running into the ions. Anything that increases the relaxation time, so the electron goes farther before it hits an ion, increases the drift velocity, and that should increase the current.

Electron Mobility

Electron mobility, $\mu$ is a measure of how easily the electrons move in response to an applied field.

$\mu = \dfrac{v_d}{E}$

$\mu = \left(\dfrac{eE\tau}{m_e}\right)\dfrac{1}{E}$

$\mu = \dfrac{e\tau}{m_e}$

Mobility increases as the relaxation time increases.

Mobility is intrinsicthe field cancels, so $\mu$ is a property of the material and not of the field applied to it

Conductivity: $\sigma = ne\mu$

$J = nev_d$

$n$ = concentration of electrons (m⁻³)

Now we are ready for the conductivity. In the Drude model the current density is n, the concentration of valence electrons, times the charge of the electron times the drift velocity. Check the units. The concentration is per cubic meter, the charge is in coulombs, and the drift velocity is in meters per second, so we get coulombs per second per square meter, which is a current density.

[advance: substitute the mobility and identify the conductivity]

We can write the drift velocity as the mobility times the field, so J equals n e mu E. Compare that with the intrinsic form of Ohm's law, J equals sigma E, and the conductivity is sigma equals n e mu. The conductivity is the concentration of carriers, the mobile electrons that respond to the field, times their charge times their mobility. So to make a material conduct better we need more carriers or a higher mobility.

Adding Up the Numbers: Sodium

Quantity	Method of determination	Value
Conductivity, σ	measured	2.1×10⁷ Ω⁻¹ m⁻¹
e⁻ concentration, n	from crystal structure	2.5×10²⁸ m⁻³
Mobility, μ	$\mu = \sigma/ne$	5.2×10⁻³ m²/V·s
Relaxation time, τ	$\tau = \mu m_e/e$	3×10⁻¹⁴ s
Rms velocity, v_rms	$v_{rms} = \sqrt{3kT/m_e}$	1×10⁵ m/s
Mean free path, ℓ	$\ell = v_{rms}\tau$	3×10⁻⁹ m

The numbers hang together, but the model still fails twice: it predicts the wrong temperature dependence, since metals conduct better when they are colder, and it cannot explain why an alloy conducts worse than either pure metal.

Now let's put numbers into the Drude model for sodium, a simple metal with one valence electron per atom. We measure the conductivity at room temperature, 2.1 times 10 to the 7 inverse ohm meters. Each sodium atom gives up one electron, so the crystal structure gives an electron concentration of 2.5 times 10 to the 28 per cubic meter. From the conductivity and the concentration we get the mobility, and the mobility gives the relaxation time, 3 times 10 to the minus 14 seconds, not far from our estimate. The velocity from the Boltzmann distribution times the relaxation time gives a mean free path of 3 nanometers. We guessed 1 nanometer, and five or six unit cells between collisions seems reasonable. That does not mean the Drude model is correct. Metals conduct better at low temperature, as we will see in the second half, but in the Drude model the electrons slow down as the metal cools. And alloys conduct worse than the pure metals, which the Drude model does not explain.

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?

Per Solid State Materials Chemistry, section 10.2, Table 10.2

The most damning problem shows up when we predict the conductivities of the elemental metals. In the alkali metals, rubidium, potassium and sodium, the atoms get smaller, n goes up, and the conductivity goes up. Calcium is divalent, so we might expect its conductivity to double, but it goes up only a little. Magnesium is smaller still, with more valence electrons per volume, and its conductivity goes down. Expecting sigma to scale with n assumes the mobility is the same in every metal, and nothing in the Drude model predicts the mobility. Aluminum has the highest n in its column and the highest conductivity, but not in proportion. Silver, gold and copper conduct well with modest n, while titanium, with four valence electrons per atom, has a very high n and a low conductivity for a metal. So there is more going on than the Drude model can explain.

Which velocity carries the current?

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)?

Neither; current is caused by the electric field itself, not electron motion.
The drift velocity, which is the net motion in response to the electric field.
Both contribute equally to the current.
The thermal velocity, because it’s so fast.

Learning objectives: metallic conductivity

Drude to free electronsExplain the transition from the Drude model to the free electron model and identify the key quantum mechanical principles incorporated in the free electron model

The parabolic E(k)Describe the parabolic E(k) relationship in the free electron model and relate it to the particle-in-a-box treatment

Effective massConnect effective mass (m*) to band curvature and predict how band structure affects carrier mobility and conductivity

Fermi-Dirac occupationApply the Fermi-Dirac distribution to explain which electrons contribute to electrical conductivity and why only states near the Fermi level matter

Temperature dependenceExplain the temperature dependence of metallic conductivity, including the role of phonon scattering and residual resistivity

Trends across the tablePredict relative conductivities across the periodic table by analyzing band structure, particularly distinguishing between d-band and s-band contributions to transport

The coinage metalsJustify why coinage metals (Cu, Ag, Au) exhibit exceptionally high conductivities compared to transition metals based on their electronic band structures

The Drude model predates quantum mechanics, so in the second half we bring quantum mechanics in with the free electron model, which imposes its rules but treats bonding in the most naive way possible. By the end you should be able to explain how we get from the Drude model to the free electron model, and to name the quantum mechanics it puts in. You should be able to relate its parabolic E of k to the particle in a box. You should be able to connect the effective mass to the curvature of a band, and to predict how a band structure affects mobility and conductivity. You should be able to use the Fermi-Dirac distribution to say why only the electrons near the Fermi level carry the current. You should be able to explain why a metal conducts better when it is cold. And you should be able to read a band structure and say why copper, silver and gold conduct better than the transition metals beside them.

The Potential in the Free Electron Model

Inside the crystalThe potential used for the free electron model is flat at $V = V_o$, so an electron moves through the metal freely

Leaving the crystalThe potential is effectively infinite, so the electrons are trapped inside — the particle in a box of physical chemistry, with the crystal as the box

The Free Electron Model

Crystal orbital wavefunction (Bloch function)

$\psi(x) = e^{ikx}u(x)$

Crystal orbital wavefunction (Bloch function)

$\psi(x) = e^{ikx}u(x)$

Free electron model: set $u(x) = 1$

$\psi(x) = e^{ikx}$

Crystal orbital wavefunction (Bloch function)

$\psi(x) = e^{ikx}u(x)$

Free electron model: set $u(x) = 1$

$\psi(x) = e^{ikx}$

Time independent Schrödinger eq’n

$\mathcal{H}\psi = E\psi$

$E = V_o + \dfrac{\hbar^2}{2m_e}k^2$

Crystal orbital wavefunction (Bloch function)

$\psi(x) = e^{ikx}u(x)$

Free electron model: set $u(x) = 1$

$\psi(x) = e^{ikx}$

Time independent Schrödinger eq’n

$\mathcal{H}\psi = E\psi$

$E = V_o + \dfrac{\hbar^2}{2m_e}k^2$

$\lambda = 2\pi/k$

So what is the wave function of a crystal orbital? Remember from chapter 6 that the Bloch function gives the wave function of a crystal orbital as e to the i k x times the basis function u of x. Let's work in one dimension because it is simpler.

[advance: set the basis function to one]

In the free electron model we set the basis function to one. Saying the potential is the same everywhere in the crystal means we ignore the nuclei that attract the electron. So we are not thinking about atomic orbitals or any bonding at all. We only confine the electrons within the crystal, and the wave function reduces to e to the i k x.

[advance: apply the Schrödinger equation]

Put e to the i k x into the time independent Schrödinger equation and the energy is V naught plus h bar squared over two times the electron mass, times k squared. Here k is the crystal momentum, the quantum number we discussed in detail in chapter 6.

[advance: draw the wave functions in the box]

The wave functions that solve the Schrödinger equation are the same as the particle in a box solutions. At k equals zero the wave function is one, a straight line from one end of the crystal to the other. As k gets larger the wave function oscillates more. As with the hydrogen atom chain, the wavelength is two pi over k, so as k gets larger the wavelength gets shorter.

The Free Electron Band Structure

$E = V_o + {}$$\hbar^2$$2m^*$$k^2$

↑ effective mass, $m^*$

A single upward-opening parabola of energy against k: the energy axis runs from 20 electronvolts at the top to minus 20 at the bottom, the k axis from minus pi over a through zero to pi over a, the minimum sits at k equals zero near minus 20 electronvolts, and the curve reaches about plus 20 at both zone edges.

Energy from 20 to −20 eV; $k$ from $-\pi/a$ through 0 to $\pi/a$. The parabola bottoms out at $V_o$, about −20 eV, and reaches about +20 eV at both zone edges.

Band Structure Comparison

An upward-opening parabola of energy against k, minimum at k equals zero near minus 20 electronvolts, rising to about plus 20 at both zone edges without ever levelling off. — **Free-electron model** — energy 20 to −20 eV, $k$ from $-\pi/a$ to $\pi/a$; a pure parabola that keeps climbing to each zone edge.

A cosine-shaped band of energy against k: it rises from a minimum at k equals zero and then flattens as it approaches each zone edge instead of continuing to climb. — **Tight binding model, H atom chain** — the same axes; the band flattens as it approaches each zone edge.

The tight binding band flattens at the zone edge; the free electron band never does.

Where it stopsNo p or d orbitals, no multi-atom unit cells.