Metals: Drude Model and Conductivity
(Covering Pages 2-11)
Objectives
By the end of this section you should be able to:
 Understand
the basics of the Drude model
 Determine the valence of an atom
 Determine the density of conduction electrons
 Calculate the electrical conductivity of a metal
 Estimate and interpret the relaxation time
Models of Atoms
Za = atomic number
= # of electrons
Drude Model
isolated atom
in a metal
Z valence electrons are weakly bound to the nucleus (participate in reactions)
Za – Z core electrons are tightly bound to the nucleus (less of a role)
In a metal, the core electrons remain bound to the nucleus to form the metallic ion
Valence electrons wander away from their parent atoms, called conduction electrons
Why mobile electrons appear in
some solids and not others?


According to the free electron model, the valance
electrons are responsible for the conduction of
electricity, thus termed conduction electrons.
Na11 ￫ 1s2 2s2 2p6 3s1
Core electrons


Valance electron (loosely bound)
Metallic 11Na, 12Mg and 13Al are
assumed to have 1, 2 and 3 mobile
electrons per atom respectively.
This valance electron, which occupies the third
atomic shell, is the electron which is responsible
chemical properties of Na.
Valence = Maximum number of bonds formed by
atom
Group: Find the Valence
What is the valency of iron,
atomic Number 26?
1s2 2s2 2p6 3s2 3p6 4s2 3d6
Number of outer electrons = 6
Iron would like to suck up other electrons to fill
the shell, so it has a valence of 4 when
elemental.
Valence vs. Oxidization
Fe: 1s2 2s2 2p6 3s2 3p6 4s2 3d6
Number of outer electrons = 6
The removal of a valance electron leaves a positively
charged ion.
-s -> O2O:
1s2
2s2
2p4
often
steals
two
e
8
Note: oxidization state (e.g. 2- in O) is not the
valence, but can be used in combination with
the original valence to find the new valence
How the mobile electrons become mobile

When we bring Na atoms together to form a Na
metal, the orbitals overlap slightly and the
valance electrons become no longer
attached to a particular ion, but
belong to both.
Na metal
A valance electron really belongs to the whole
crystal, since it can move readily from one ion to its
neighbor and so on.
Drude Model: A gas of electrons
X

+
+
+
+
+
+

Drude (~1900) assumed that the
charge density associated with the
positive ion cores is spread uniformly
throughout the metal so that the
electrons move in a constant
electrostatic potential.

All the details of the crystal structure
is lost when this assumption is made.
According to both the Drude and later the free
electron model, this potential is taken as zero and
the repulsive force between conduction electrons
are also ignored.
Drude Assumptions

Electron gas free and independent, meaning no
electron-electron or electron-ion interactions

If field applied, electron moves in straight line
between collisions (no electron-electron
collisions). Collision time constant. Velocity
changes instantaneously.

Positives: Strength and density of metals
Negative: Derivations of electrical and thermal
conductivity, predicts classical specific heat
(100x less than observed at RT)

Classical (Drude)Model
Drude
developed
this theory by
considering metals
as a classical gas of
electrons, governed
by MaxwellBoltzmann statistics
f MB ~ e E / kBT

Applies kinetic theory of gas to conduction
electrons even though the electron densities are
~1000 times greater
Electron Density n = N/V
6.022 x 1023 atoms per mole
 Multiply by number of valence electrons
 Convert moles to cm3 (using mass density
m and atomic mass A)

n = N/V = 6.022 x 1023 Z m /A
Group: Find n for Copper
Density of copper = 8.96 g/cm3
n = N/V = 6.022 x 1023 Z m /A
Effective Radius
density of conduction electrons in metals ~1022 – 1023 cm-3
rs – measure of electronic density = radius of a sphere whose volume
is equal to the volume per electron
4rs
V 1
 
3
N n
3
1/ 3
 3 
rs  

 4n 
~
1
n1/ 3
mean inter-electron spacing
in metals rs ~ 1 – 3 Å (1 Å= 10-8 cm)
2
a0 
 0.529 Å – Bohr radius
2
me
rs/a0 ~ 2 – 6
A model for electrical conduction

(a)
(b)
Dr. Jie Zou
Drude model in 1900
 In the absence of an electric field,
the conduction electrons move in
random directions through the
conductor with average speeds v
~ 106 m/s. The drift velocity of
the free electrons is zero. There
is no current in the conductor
since there is no net flow of
charge.
 When an electric field is applied,
in addition to the random motion,
the free electrons drift slowly (vd
~ 10-4 m/s) in a direction
opposite that of the electric field.
“Quasi-Classical” Theory of Transport
Microscopic
Macroscopic
dq
Current: i 
(Amps)
dt
q   idt
V
i
R
Charge
Current Density: J = I/A
Ohm’s Law
J
E

  E where   resistivity
 = 1/
  conductivity
Finding the Current Density
Vd = drift velocity or average velocity of conduction electrons
Current Density: J = I/A
J
E

  E where   resistivity
  conductivity
Applying an Electric Field E to a Metal
Needed from problem 1.2
F = ma = - e E, so acceleration a = - e E / m
Integrating gives v = - e E t / m
So the average vavg = - e E  / m, where  is
“relaxation time” or time between collisions
vavg = x/, we can find the xavg or mean free path 
Calculating Conductivity
vavg = - e E  / m
J =  E = - n e vavg = - n e (- e E  / m)
Then conductivity  = n e2  / m
( = 1/)
And  = m / ( n e2) ~10-15 – 10-14 s
 = m / ( n e2)
Group Exercise
vavg = x/
Calculate the average time between collisions
 for electrons in copper at 273 K.
Assuming that the average
E speed for free
 x 10
 E6 m/s
where 
 resistivity
electrons in copper is J1.6
calculate

  conductivity
their mean free path.
Resistivity vs Temperature not accounted
for in Drude model 10 20
Resistivity is
temperature dependent
mostly because of the
temperature
dependence of the
scattering time .
 = m / ( n e2)
Resistivity (Ωm)
Insulator
Diamond
1010-
Semiconductor
Germanium
100 Metal
10-100
100
200
Temperature (K)
Copper
300
• In Metals, the resistivity increases with increasing temperature.
The scattering time  decreases with increasing temperature T (due to
more phonons and more defects), so as the temperature increases ρ
increases (for the same number of conduction electrons n)
• We will find the opposite trend occurs in semiconductors but we
will need more physics to understand why.
In Terms of Momentum p(t)
Momentum p(t) = m v(t)
 So j = n e p(t) / m
 Probability of collision by dt is dt/


p(t+dt)=prob of not colliding*[p(t)+force]
(1-dt/)
= p(t)–p(t) dt/ + force +higher order terms
p(t+dt)-p(t) =–p(t) dt/ + force +higher order
Or dp/dt = -p(t)/ + force(time)
From the opposite direction