Download Electrical Conductivity

ELECTRONIC CONDUCTION
IN METALS
1
Free Electron Theory
• A metal is defined as a solid with
valence electrons that are not tightly
bound to the atoms but are easily able
to move through the whole crystal.
• They can be called free electrons or conduction electrons.
• So according to Free Electron Model (FEM)
the valence electrons are responsible for the conduction
of electricity, & for this reason these electrons are called
“Conduction Electrons”.
2
• As an example, consider Sodium (Na).
• The electron configuration of the Free Na Atom is:
1s2 2s2 2p6 3s1
Valence Electron
Core Electrons
(loosely bound)
• The outer electron in the third atomic shell
(n = 3, ℓ = 0)
is the electron which is responsible for the
chemical properties of Na.
3
• When we bring Na atoms together to form a Na metal, we see
the picture as
Na metal
• Outer orbitals of Na atoms overlap slightly.
From this observation it follows that a valance
electron is no longer attached to a particular
ion, but belongs to both neighbouring ions at
the same time.
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

a valance electron belongs to the whole crystal, since it can
move readily from one ion to its neighbour, and then the
neighbour’s neighbour, and so on.
This mobile electron becomes a conduction electron in a solid.
+
+
+
+
+
+
– The removal of the valance electrons
leaves a positively charged ion.
– The charge density associated with the
positive ion cores is spread uniformly
throughout the metal
– The electrons move in a constant
electrostatic potential.
– Electric field due to lattice ions is uniform
so neglected
– Repulsive force b/w conducting electron
ignored
– Attractive force b/w lattice ions and
elections ignored
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•These free electrons in a metal behaves like a gas molecule
with random motion
•Why electric current is zero in the absence of electric field???
Randomness of electrons due to thermal energy???
+
+
+
+
+
+
average kinetic energy
1
3
me vth2  kT
2
2
thermal velocity of electron
vth 
3kT
me
Applied Electric Field
Drift velocity of the electrons
Electrical Conductivity
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Electrical Conductivity:
Electric field leads to a potential difference which creates a force on electron
F  eE
This force accelerates the electrons and give velocity away from the field
F  eE  me a  m
d  
Ee
dt
m
d
dt
Ee
vd   
m
Drift velocity:

Mean collision time: average time b/w two collision
during its random motion
r
Mobility : magnitude of drift velocity achieved by electron in unit electric field
e

me
vd

E
If applied field E is removed????
The drift velocity will decays exponentially to zero
The relaxation time:
r 

1  cos
r

v0  vd e
 t / r
vd
at t   r , v0 
e
the drift velocity reduces to 1/e to its peak value after switch off the E
Mean Free Path (λ): average distance travelled by electrons
b/w two collision During random motion
Electrical Conductivity:
A conductor of Length L, cross-section area A
Current flow I, potential drop V, Electric field E
I  V / R,
J 
I
V
LV / L


A
AR
AR
I
 E
A
Current density J  nedv
 
L
AR
where dv 
 e
E
m
ne 2
J
E
m
Electrical Conductivity
ne 2
But J  E 
E
m
ne 2

me
Effect of temperature and impurity on conductivity?????
Electrical Conductivity
Electrical Conductivity
ne 2

me
Mean free path   vth , where vth is the thermal velocity
   / vth  
me
3kT
ne 2
ne 2


me
3me kT
thermal velocity of electron
vth 
3kT
me
3me kT

ne 2
Failure of Classical Free Electron Theory:
• resistivity of metal vary as T1/2
But experimentally is varies with T
•It gives almost seven times value at t=300K of exp.
•Conductivity depends on free electron density n so dia or trivalent metal
should have higher conductivity
Electrical Conductivity
The kinetic energy of the electron is
1
3
2
mvth  kT
2
2
3kT
and vth 
m
At room temperature , the drift velocity imparted to the electrons by
the applied electric field is very much smaller than the average
thermal velocity. The average distance travelled by an electron
between two successive collisions in the presence of applied field is
known as free mean path . The time taken by an electron between
two successive collisions is known as mean collision time of the
electron c.
Hence the time taken by the electrons in traversing the distance  will
be decided mainly by rms velocity.

m
Now  c 

vth
3kT
Mobility
• Mobility of the electron μ is defined as the steady state drift
velocity<vd> per unit electric field.
d
e


E
m
ne 2
e

 ne.
m
m
  ne
1
m
m
 
 2
 ne ne 
Where  (resistivit y )
ne ne 2


m
m
•The electrical conductivity σ depends on two factors ,the charge
density n and their mobility. These two quantities depend on
temperature.
•In metals n is constant and μ decreases slightly with
temperature and hence with increase of temperature ,the
conductivity decreases.
•In semiconductors the exponential increase of n with
temperature is responsible for increase of conductivity with
temperature.
• In insulator n remains constant and above certain temperature
μ increase exponentially resulting in dielectric breakdown.
For the thermal conductivity of a metal in terms of temperature.
Relaxation Time
Relaxation Time can be defined as the time taken for the drift
velocity to decay to 1/e of its initial value
Let assume that the applied field is cut off after the drift velocity of
the electron has reached its steady value. Drift velocity after this
instant is governed by
d d

 m
dt

d d
dt

m
d

 d (t )   d (0) exp( t /  )
at t  
 (0)
 d (t )  d
e
Differenti ating equation

d
 d (t )   d
dt

Vd(0)is the steady state drift velocity
Collision time
vd  0
the change in average velocity on collision is opposite. Hence the rate of change of
average velocity is given by
Q.1/Tut 9
The relaxation time and root mean square velocity of the electron
at room temperature are 2.5x10-14 s and 1x105 m/s. Calculate the
value of mean free path of the electron.
Q.2/Tut 9
The resistivity of a metal at temperature 20°C is 1.69 x 10-8 ohm m
and concentration of the free electrons in metal, ne = 8.5x1028/m3.
Calculate root mean square velocity (c), relaxation time (τ), mean
free path (λ), mobility of electrons (μ) and value of electrical
conductivity (σ) on the basis of classical free electron theory.
• Therefore, these conduction electrons can be
considered as moving independently in a square well of
finite depth & the edges of the well correspond to the
edges of the crystal.
• Consider a metal with a cubic shape with edge length L:
Ψ and E can be found by solving the Schrödinger
equation:
2
Since

V
   E
2
2m
V 0
Use periodic boundary conditions
& get the Ψ’s as travelling
plane waves.
-L/2
0
L/2
 ( x  L, y  L, z  L)   ( x, y, z )
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• The solutions to the Schrödinger equation are
plane waves,
1 ik r
1 i ( kx x  k y y  kz z )
 ( x, y , z ) 
e 
e
V
V
Normalization constant
kL  2n
2 

where
,
k


 

So the wave vector must satisfy
2
2
2
k x  nx k y  n y k z  nz
L
L
L
nx , n y , nz taking any integer values; +ve, -ve or zero.
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• The wave function Ψ(x,y,z) corresponds to an
energy
2
2
k2
E
2m
E
2m
(k x 2  k y 2  k z 2 )
• The corresponding momentum is:
p  (k x , k y , k z )
• The energy is completely kinetic:
2 2
1 2
k
mv 
2
2m
mv 
2 2
2
k
2
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2
2
2
k x  nx k y  n y k z  nz
L
L
L
One unit cell corresponding to one wave
vector has sides
kx 
2
2
2
ky 
kz 
L
L
L
So volume of unit cell will be
2 2 2 8 3 8 3
kxk ykz 
 3 
L L L
L
V
V is the volume of the crystal.
The number of k values in the range
K and k+dk is
4k 2 dk
4k 2 dk
g (k )dk 

volume of unit cell 8 3 / V
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• So the number of allowed k values inside a
spherical shell of k-space of radius k is:
Vk 2
g (k )dk  2 dk ,
2
•
Here g(k) is the density
of states per unit
magnitude of k.
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The number of allowed states
per unit energy range?
• Each k state represents two possible electron states, one for
spin up, the other is spin down.
dk
g ( E )  2 g (k )
dE
g ( E )dE  2 g (k )dk
2
2
2
dE
k

dk
m
k
E
2m
g ( E )  2 g (k )
k
2mE
2
2Vk 2 m
Vk m
V m
m


2

2
2
2 2  2 k
2 2  2
2 2  2
k
g (E) 
V
2 2
2mE
2
3/ 2 1/ 2
(2
m
)
E
3
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Ground State of the Free Electron Gas
• Electrons are Fermions (s = ± ½) and obey
the Pauli exclusion principle; each state can
accommodate only one electron.
• The lowest-energy state of N free electrons is
therefore obtained by filling the N states of
lowest energy.
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• Thus all states are filled up to an energy EF,
known as The Fermi energy, obtained by
integrating the density of states between 0 and
EF, The result should equal N. Remember that
g (E) 
N
EF

0
g ( E )dE 
EF
V
2 2
V
 2
3/ 2 1/ 2
(2
m
)
E
3
(2m)
3
2
3/ 2
E dE 
1/ 2
0
V
3 2
3/ 2
(2
mE
)
F
3
• Solving for EF (Fermi energy);
 3 N 
EF 


2m  V 
2
2
2/3
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• The occupied states are inside the Fermi
sphere in k-space as shown below; the radius
is Fermi wave number kF.
EF 
kz
 3 N 


2m  V

2
Fermi Surface
E = EF
kF
2/3
2
kF 2
EF 
2me
From these two
equations, kF can be
found as,
1/ 3
 3 N 
kF  

V


2
ky
kx
2
The surface of the Fermi sphere
represents the boundary between
occupied & unoccupied k states at
absolute zero for the free electron gas.
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• Typical values may be obtained by using
monovalent potassium metal (K) as an example;
for potassium the atomic density and hence the
valence electron density N/V is 1.402x1028 m-3
so that
EF  3.40 1019 J  2.12eV
k F  0.746 A1
EF  kBTF
• The Fermi (degeneracy) Temperature TF is given
by
EF
4
TF 
kB
 2.46 10 K
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• It is only at a temperature of this order that the
particles in a classical gas can attain (gain)
kinetic energies as high as EF .
• Only at temperatures above TF will the free
electron gas behave like a classical gas.
• Fermi momentum
PF  meVF
PF  kF
PF
VF 
 0.86 106 ms 1
me
• These are the momentum & velocity values of the electrons
at the states on the Fermi surface of the Fermi sphere.
• So, the Fermi Sphere plays an important role in
the behavior of metals.
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Typical values for monovalent potassium metal;
 3 N 
EF 


2m  V 
2
2
 2.12eV
1/ 3
 3 N 
kF  

V


2
2/3
 0.746 A1
PF
VF 
 0.86 106 ms 1
me
EF
TF 
 2.46 104 K
kB
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Free Electron Gas at Finite Temperature
• At temperature T the probability of occupation
of an electron state of energy E is given by the
Fermi distribution function
f FD 
1
1  e ( E  EF ) / k B T
• The Fermi distribution function determines the
probability of finding an electron at energy E.
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• The Density of States g(E) specifies
how many states exist at a given energy E.
• The Fermi Function f(E) specifies how many of the
existing states at energy E will be filled with electrons.
•
I
• It is a probability distribution function.
33
33
Fermi-Dirac Statistics
The Fermi Energy EF is
essentially the same as the
Chemical Potential μ.
EF is called The Fermi Energy.
Note the following:
• When E = EF, the exponential term = 1 and FFD = (½).
• In the limit as T → 0:
L
L
• At T = 0, Fermions occupy the lowest energy levels.
• Near T = 0, there is little chance that thermal agitation
will kick a Fermion to an energy greater than EF.
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In the limit as T → 0
For E < EF :
1
f ( E  EF ) 
 1
1  exp ()
For E > EF :
1
f ( E  EF ) 
 0
1  exp ()
E
A step function!
EF
0
1
f(E)
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Fermi Function at T=0
& at a Finite Temperature
f FD 
1
1 e
If E = EF then f(EF) = ½
( E  EF ) / k B T
fFD(E,T)
0.5
E
E<EF
EF
E>EF
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