Download The Drude Model and DC Conductivity

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 Drude 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.
Warning: Confusing Terminology
There exist many forms of valence!

By valence electrons here we mean outer
electrons...the ones that could easily be
pulled from the core to form an electron gas
(In chemistry) Valence = Maximum
number of bonds formed by atom
Warning: The "oxidation state" of an atom in a molecule gives
the number of valence electrons it has gained or lost through
bonding. In contrast to the valency number, the oxidation state
can be positive or negative. Sometimes called valence in physics.
If I give you an element on the test,
how would you know if metallic?
I would give atomic number.
Note orbital filling in
Cu does not follow
normal rule
Metallic/insulating properties can be understood by how loose
(i.e. low ionization energy) outer electrons are.
Metallic or Insulating?
Most metals are formed from atoms with partially filled atomic orbitals.
e.g. Na, and Cu which have the electronic structure
Na 1s2 2s2 2p6 3s1
Cu 1s2 2s2 2p6 3s23p63d104s1
Insulators are formed from atoms with closed
(totally filled) shells e.g. Solid inert gases
He 1s2
Ne
1s2 2s2 2p6
Or form close shells by covalent bonding
i.e. Diamond
Note orbital filling in
Cu does not follow
normal rule
Simple picture. Metal have CORE electrons that are bound to the
nuclei, and VALENCE electrons that can move through the metal.
Group: Find the # of Valence
Electrons for Si
What is the valence of silicon,
atomic Number 14?
1s2 2s2 2p6 3s2 3p2
How many electrons are in the n=3 state?
Valence=4
#Valence Easy for Outer Groups
Group: 1 2
18
13
Group
Group
Group
Group
Group
Group
Group
Group
1: 1
2: 2
13: 3
14: 4
15: 5
16: 6
17: 7
18: 8 (except for helium, which has 2)
Group: Find the Valence of Fe
What is the valence of iron,
atomic Number 26?
1s2 2s2 2p6 3s2 3p6 4s2 3d6
Number of 3d electrons = 6
What about in compounds? (SrFeO3)
Except, are these the outer most?
No, the 4s is farther! So valence =2!
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.
U(r)
U(r)
Drude makes the
free electron approximation
U(r)
U(r)=0
Neglect periodic potential & scattering between electrons
Reasonable for “simple metals” (Alkali Li,Na,K,Cs,Rb)
What does this remind you of?
Drude Assumptions

He also ignored electric fields, meaning no
electron-electron or electron-ion interactions

If field applied, electron moves in straight line
between collisions (no electron-electron
collisions). Velocity changes instantaneously,
but direction random.
<v>=0, but <v2>T (gas)
Positives: Strength and density of metals
Negative: Derivations of electrical and thermal
conductivity, predicts classical specific heat
(100x less than observed at RT)



Conduction electron Density n = N/V
6.022 x 1023 atoms per mole
 Multiply by number of valence electrons (Z)
 Convert moles to cm3 (using mass density
m and atomic mass A)

n = N/V = 6.022 x 1023 Z m /A
Group: Predict 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
~ 1/ 3
n
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
If
Current Density: J = I/A
, what is the
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
Ohm’s law
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.
Electrical conductivity of materials
How to measure the
conductivity / resistivity
A two-point probe can be used but the contact
or wire resistance can be a problem, especially
if the sample has a small resistivity.
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 dt]
(1-dt/)
= p(t)–p(t) dt/ + force dt +higher order
terms in (dt)
p(t+dt)-p(t) =–p(t) dt/ + force +higher order
Or dp/dt = -p(t)/ + force(time)

Main Results from Drude
dp/dt = -p(t)/ + force(time); compare
to Netwon’s second law F=dp/dt
 There is an effective damping term due
to collisions

VD(t) = VD(0)exp(-t/p )

Also j = n e v
p(t)/p(t=0)
1.0
0.8
0.6
0.4
0.0

t/p
0.2
-1
0
1
2
3
We are going to use these next class to
understand some specific cases
4