Download Electronic Properties of Metals

V. Electronic Properties of Metals
A. The Free Electron Gas (FEG) Model
B. Properties of the FEG: Fermi Energy EF and Density of States N(E)
C. Plasma Oscillations and Plasmons
D. Heat Capacity of the FEG
E. Electrical Resistivity of the FEG: A Derivation of Ohm’s Law
F. The Hall Effect
G. Thermal Conductivity of the FEG
H. Limitations of the FEG Model—and Beyond
Having studied the structural arrangements of atoms in solids, and the thermal
and vibrational properties of the lattice, we now consider the electronic
properties of solids in terms of a very simple model.
A. The Free Electron Gas Model
Plot U(x) for a 1-D
crystal lattice:
Simple and
crude finitesquare-well
model:
U
U=0
Can we justify this model? How can one replace the entire lattice by a
constant (zero) potential?
Assumptions of the FEG Model
1. Metals have high electrical conductivity and no apparent activation energy,
so at least some of their electrons are “free” and not bound to atoms
2. Coulomb potential energy of positive ions U  1/r is screened by bound
electrons and is weaker at large distances from nucleus
3. Electrons would have lowest U (highest K) near nuclei, so they spend less
time near nuclei and more time far from nuclei where U is not changing
rapidly
4. Therefore model the behavior of the free
electrons with U = 0 inside the volume of the
metal and a finite potential step at the surface.
Assume each atom has n0 free electrons, where
n0 = chemical valence. Assume that resistance
comes from electrons interacting with lattice
through occasional collisions:
B. Properties of the FEG: Fermi Energy and Density of States
2 2

   U   E
2m
2mE
2 2
2






   E
2

2m
Time-independent
Schrödinger Equation:
With U = 0:
 2  k 2
Solutions have

i ( k r t )
1. Wave functions:
  Ae
2. Energies:
2k 2
E
2m
Traveling waves (plane waves)
Parabolic energy “bands”
E
kx
Properties of the FEG
By using periodic boundary conditions for a cubic solid with edge L and
volume V = L3, we define the set of allowed wave vectors:
2n y
2nx
2nz
kx 
ky 
kz 
L
L
L
nx , n y , nz  1,  2,  3, ...
This shows that the volume in k-space per solution is:
And thus the density of states in k-space is: N (k ) 
Since the FEG is isotropic, the surface of
constant E in k-space is a sphere. Thus
for a metal with N electrons we can
calculate the maximum k value (kF) and
the maximum energy (EF).
 2 


 L 
3
1
V Just as in the case

2 / L 3 8 3 of phonons!
k
kF
y
kx
kz
Fermi sphere
Fermi Wave Vector and Energy
Taking into account the spin degeneracy of the electron, N electrons will be
accommodated by N/2 energy states, so:
N
V 4
 # states 

# states  
 3  k F3 
volume 
2 8  3
 volume 

And the maximum energy is easily found:
These quantities are called the Fermi
wavevector and Fermi energy in honor of
Enrico Fermi, who (along with Arnold
Sommerfeld) did the most to apply quantum
mechanics to calculate the properties of
solids in the late 1920s.
1/ 3
N

k F   3 2 
V


 3 2 n



1/ 3
2/3
 2 k F2  2
EF 

3 2 n
2 m 2m
Density of States N(E)
We often need to know the density of electron states, which is the number of
states per unit energy, so we can quickly calculate it:
The differential number of
electron states in a range of
energy dE or wavevector dk is:
This allows:
Now using the
general relation:
dN  N ( E )dE  N (k )dk
dk
N (k ) V / 8 3
mV
N ( E )  N (k )

 2
 2 3
dE dE / dk  k / m  8 k
1/ 2
 2mE 
k  2 
  
we get:
N (E) 


1/ 2
V
3
2
m
E
 2 3
Reality of the Fermi Energy
There are several spectroscopic techniques that allow the measurement of the
distribution of valence electron states in a metal. The simplest is soft x-ray
spectroscopy, in which the highest-lying core core level in a sample is ionized.
Only higher-lying valence electrons can fall down to occupy the core level, and
the spectrum of emitted x-rays can be measured:
EF  13 eV
The valence bandwidth is in reasonable agreement
with the FEG prediction of EF = 11.7 eV
Utility of the Density of States
EF
With N(E) we can immediately
calculate the average energy per
electron in the 3-D FEG system:
E 
total energy

# electrons
 N ( E ) E dE
0
EF
 N ( E ) dE
0
We can simplify by using the relation:
EF
E 
C E
0
EF
E
3/ 2
dE
C  E1/ 2 dE
0

2 5/ 2 F
E
5
0
E
2 3/ 2 F
E
3
0
3
 EF
5
N ( E )  CE1/ 2
Why the factor 3/5? A look at the
density of states curve should give the
answer:
N(E)
E
EF
C. Plasma Oscillations and Plasmons
If we take seriously the existence of the FEG, we might expect it to exhibit
collective oscillations if it is disturbed. Consider a cylindrical metal sample
with an induced polarization in its FEG:
unpolarized
A
polarized: FEG displaced by x
n
x
L
x


Induced dipole moment p  qd   neAxL iˆ

 p
p
Induced polarization
P 
 nex iˆ
V AL
We can now find the
induced E field:

 
D  0E  P  0

  P  nex
E

iˆ
0
0
E field provides a restoring
force for displaced electrons
Plasma Oscillations and Plasmons
Now for a single free electron we can write Newton’s second law:
  ne 2 x
d 2x ˆ
m 2 i  eE 
iˆ
dt
0
d 2 x  ne 2

x
2
dt
m 0
To find the oscillation frequency,
compare to the equation of motion
of the mass-spring system:
d 2x
k
2


x



x
2
dt
m
This reveals that the plasma
frequency is given by:
ne 2
 
m 0
harmonic
oscillator!
2
p
The quantum of “plasma energy” ħp is called a plasmon. Experiments with
electron beams passing through thin metal foils shows that they lose energy in
integer multiples of this energy quantum.
Experimental Evidence for Plasmons
Metal
Expt. E (eV)
Calc. ħp
(eV)
Li
7.12
8.02
Na
5.71
5.95
K
3.72
4.29
Mg
10.6
10.9
Al
15.3
15.8
What trend do you
see? Can you think
of an explanation?
Hint: look at
equation for p !
D. Heat Capacity of the FEG
19th century puzzle: each monatomic gas molecule in sample at temperature T
has energy 32 kT , so if the N free electrons in a metal make up a classical
“gas” they should behave similarly.
Eel  N ( 32 kT )
or expressed per mole:
So the electronic contribution to the molar
heat capacity would be expected to be
Eel 3 N
 2 kT  32 N A kT  32 RT
n
n
Cel 
d  Eel  3

 2R
dT  n 
This is half of the 3R we found for the lattice heat capacity at high T. But
experiments show that the total C for metals is only slightly higher than for
insulators—which conflicts with the classical theory!
Heat Capacity of the Quantum-Mechanical FEG
Quantum mechanics showed that the occupation of electron states is governed
by the Pauli exclusion principle, and that the probability of occupation of a
state with energy E at temperature T is:
f (E) 
1
e
 E    / kT
1
where  = chemical potential  EF for kT << EF
Heat Capacity of the Quantum-Mechanical FEG


So at temperature T
the total energy is:
EN ( E )
Eel   E f ( E ) N ( E ) dE    E  EF / kT
dE
e
1
0
0
And the electronic
heat capacity is:
dE
1
Cel  el  2
dT
kT


0
The exact answer to this complicated
integral is derived in more advanced texts:
E N ( E )( E  EF )e E  EF / kT
e
 E  EF  / kT

1
2
Cel  3 k 2 N ( EF )T
2
dE
Cel  T !
A Rough and Ready Estimate
We can estimate Cel in just a few lines in order to confirm the linear
dependence on temperature:
2kT
N(E)
# electrons that  2kT  N ( E ) f ( E )
F
F
can absorb
N(E)f(E)
 kT N ( EF )
thermal energy
total thermal energy
of electrons at T
FEG heat
capacity at T
E (T )  E0   32 kT kT N ( EF )
E (T )  E0  32 k 2T 2 N ( EF )
Cel 
dEel
 3k 2 N ( EF )T
dT
Remarkably close
to the exact result!
E
EF
Cel  3 k 2 N ( EF )T  T
2
But this linear dependence is impossible to measure directly, since the heat
capacity of a metal has two contributions. Now for a metal at low
temperatures we can write the total heat capacity:
C (T )  Cel  Clattice   T  T 3
Assuming we can measure C(T) for a
metal, how can we test this relationship?
Heat Capacity of Metals: Theory vs. Expt. at low T
Very low temperature
measurements reveal:
Results for simple
metals (in units
mJ/mol K) show
that the FEG values
are in reasonable
agreement with
experiment, but are
always too high:
Metal
expt
FEG
expt/ FEG =
m*/m
Li
1.63
0.749
2.18
Na
1.38
1.094
1.26
K
2.08
1.668
1.25
Cu
0.695
0.505
1.38
Ag
0.646
0.645
1.00
Au
0.729
0.642
1.14
Al
1.35
0.912
1.48
The discrepancy is
“accounted for” by
defining an effective
electron mass m* that is
due to the neglected
electron-ion interactions
E. Electrical Resistivity of the FEG: A Derivation of Ohm’s Law
The FEG model was developed by Paul Drude
(1900) in order to describe the electrical and
thermal conductivity of metals. This work greatly
influenced the course of “solid-state physics” and it
introduces basic concepts we still use today.
Since an electron that experienced a
uniform electric field would accelerate
indefinitely and imply an increasing
current, Drude proposed a collision
mechanism by which electrons make
collisions every  seconds. In each
collision he assumed that all of the
electron’s forward velocity is reduced to
zero and it must be accelerated again.
The result is a constant average velocity:
d 2x
 Fx  m dt 2  eE  Fcoll
Fcoll 
p x  mv

t

 eE 
mv

0
relaxation time
v
 eE
m
drift velocity (opposite to E)
Electrical Resistivity of the FEG: A Derivation of Ohm’s Law
The current density in the FEG can easily be calculated assuming a simple
sample geometry:
J
A
n
I 1 Q 1  ne  AL 


 nev
A A t
A
L/v
L
Combined with
the earlier result
 eE
v
m
where we define the
electrical conductivity
we get
 ne 2
  
 m



 ne 2
J  
 m

 E  E

and the electrical
resistivity is:
Ohm’s Law!

 m 
 2 
  ne  
1
The question we now turn to is: How does the resistivity depend on
temperature? What does the FEG model predict?
Temperature Dependence of the Electrical Resistivity
Clearly the temperature dependence of  enters through the relaxation time :


v
mean free path of electrons between collisions
mean velocity of electrons between collisions (not drift velocity)
For homework you will show that using the classical kinetic theory of gases to
express v gives a result that does not agree with experiment. What’s wrong?
Consider once again the manifold of energy levels occupied by the FEG:
EF
unoccupied
levels
occupied
levels
The occupied states have energies (and thus
velocities) that are essentially independent of
T. So even if we calculate an average velocity
it will not depend on T. But we can easily
show that only electrons near EF contribute to
the electrical conductivity.
Electrical Conductivity in Reciprocal Space
The Fermi sphere contains all
occupied electron states in the FEG.
In the absence of an electric field,
there are the same number of
electrons moving in the ±x, ±y, and
±z directions, so the net current is
zero.
But when a field E is applied along the
x-direction, the Fermi sphere is shifted
by an amount related to the net change
in momentum of the FEG:
k
y
kx
k
Fermi surface
y
kx
kF
kz
k x  px  mvx  eEx
k x  p x  mvx  
eE x

The shift in Fermi sphere creates a net current flow
since more electrons move in the –x direction than the
+x direction. But the excess current carriers are only
those very near the Fermi surface. So the current
carriers have velocity vF.
Analysis of Mean Free Path
Since the velocity of current-carrying electrons is essentially independent of T,
we need to examine the behavior of the mean-free-path. Naively we might
expect it to be some multiple of the distance between atoms in the solid. Let’s
dig a bit deeper:
The probability of a collision in a distance x is:
P
evF
P
x
collision cross-section = <r2>
cross-sectional area of slab = A
atomic density = na
total cross  sectional area for collision
cross  sectional area of slab
na Ax  r 2
A
 na x  r 2
Now in a distance x = , P = 1 is true,
so we can solve for :

1
na  r 2
Summary: (T)
Now if we assume that the collision cross-section is due to vibrations of atoms
about their equilibrium positions, then we can write:
r
2
 x
2
Therefore
 y
2
r2  T
And the thermal average potential
energies can be written:
and

1
T
from which   T
1
2
Cx 2 
1
2
Cy 2  12 kT
is predicted.
Except for the very lowest of
temperatures (where the
classical treatment of the
atomic vibrations breaks
down), the linear behavior is
closely obeyed.
F. The Hall Effect
This phenomenon, discovered in 1879 by
American physics graduate student (!) Edwin
Hall, is important because it allows us to
measure the free-electron concentration n for
metals (and semiconductors!) and compare to
predictions of the FEG model.
The Hall effect is quite simple to
understand. Consider a B field applied
transverse to a thin metal sample carrying
a current:
I
Hall Effect Measurements
A hypothetical charge carrier of charge q
experiences a Lorentz force in the lateral
direction:
FB  qvB
I
t
As more and more carriers are deflected,
the accumulation of charge produces a
“Hall field” EH that imparts a force
opposite to the Lorentz force:
w
FE  qEH
Equilibrium is reached when these two
opposing forces are equal in magnitude,
which allows us to determine the drift speed:
qvB  qEH
From this we can write the current density:
J  nqv 
And it is customary to define the Hall
coefficient in terms of the measured quantities:
RH 
v
nqEH
B
EH
1

JB nq
EH
B
Hall Effect Results!
In the lab we actually measure the Hall voltage VH and the current I, which gives us a
more useful way to write RH:
EH
VH / w
VH t 1
I

JA

Jwt
R




VH  EH w
H
JB I / wt B IB nq
If we calculate RH from our measurements
and assume |q| = e (which Hall did not
know!) we can find n. Also, the sign of
VH and thus RH tells us the sign of q!
The discrepancies between the FEG
predictions and expt. nearly vanish when
liquid metals are compared. This reveals
clearly that the source of these
discrepancies lies in the electron-lattice
interaction. But the results for Be and Zn
are puzzling. How can we have q > 0 ???
Stay tuned…..
RH (10-11 m3/As)
Metal
n0
solid
liquid
FEG value
Na
1
-25
-25.5
-25.5
Cu
1
-5.5
-8.25
-8.25
Ag
1
-9.0
-12.0
-12.0
Au
1
-7.2
-11.8
-11.8
*
Be
2
+24.4
-2.6
-2.53
*
Zn
2
+3.3
-5
-5.1
Al
3
-3.5
-3.9
-3.9
G. Thermal Conductivity of Metals
In metals at all but the lowest temperatures, the electronic contribution
to  far outweighs the contribution of the lattice. So we can write:
The electron mean-free path can be
rewritten in terms of the collision time:
1
3
  Cel v 2
  v
Also, the electrons that can absorb thermal energy and
therefore contribute to the heat capacity have energies very
near EF, so they essentially all have velocity vF. This gives:
1
  Cel vF2
3
Cel  3 k 2 N ( EF )T
2
From our earlier discussion the electronic heat capacity is:
It is easy to show that
N(EF) can be expressed:
N ( EF ) 
3N
2 EF
Or per unit volume
of sample,
Which gives the heat capacity per unit volume:
1
   el  Cel v
3
Cel  2 k 2 n
2
N ( EF ) 
T
EF
3n
2 EF
Wiedemann-Franz Law and Lorenz Number
 
Now the thermal conductivity
per unit volume is:
Finally…
 
 2 k 2 n
3m
1 2 2 T 2
1 2 2
T
2

k
n
v

k
n
v

F
2
2
2 F
1
3
3
EF
2 mvF
T
Now long before Drude’s time,
Gustav Wiedemann and Rudolf Franz
published a paper in 1853 claiming
that the ratio of thermal and electrical
conductivities of all metals has nearly
the same value at a given T:
Not long after (1872) Ludwig
Lorenz (not Lorentz!) realized
that this ratio scaled linearly with
temperature, and thus a Lorenz
number L can be defined:

 constant

Gustav Wiedemann

L
T
very nearly constant for all metals
(at room T and above)
The Experimental Test!
L = /T 10-8 (J/CK)2
We can readily compare the prediction of the
FEG model to the results of experiment:
 k n
2
Metal
0°C
100 °C
Cu
2.23
2.33
Ag
2.31
2.37
Au
2.35
2.40
Zn
2.31
2.33
Cd
2.42
2.43
Mo
2.61
2.79
Pb
2.47
2.56
2

LFEG 
 3m
ne 2
T
m
T

 k
T
J

LFEG  2.45  108 CK
2
This is remarkable…it is
independent of n, m, and even  !!
2
3e
2
2
Agreement with experiment is quite
good, although the value of L is about
a factor of 10 less at temperatures near
10 K…Can you speculate about the
reason?
An Historical Footnote
Drude of course used classical values for the electron velocity v and heat capacity
Cel. By a tremendous coincidence, the error in each term was about two orders of
magnitude….in the opposite direction! So the classical Drude model gives the
prediction:
J

LDrude  1.12  108 CK
2
But in Drude’s original paper, he inserted an erroneous factor of two, due to a mistake
in the calculation of the electrical conductivity. So he originally reported:
J

L  2.24  108 CK
2
!!!
So although Drude’s predicted electronic heat capacity was far too high, this
prediction of L made the FEG model seem even more impressive than it really was,
and led to general acceptance of the model.
H. Limitations of the FEG Model—and Beyond
The FEG model of Drude, augmented by the results of quantum mechanics in the years
after 1926, was extremely successful in accounting for many of the basic properties of
metals. However, strict quantitative agreement with experiment was not achieved. We
can summarize the flawed assumptions behind the FEG model as follows:
1.
The free-electron approximation
The positive ions act only as scattering centers and is assumed to have no effect on
the motion of electrons between collisions.
2.
The independent electron approximation
Interactions between electrons are ignored.
3.
The relaxation time approximation
The outcome of a electron collision is assumed to be independent of the state of
motion of an electron before the collision.
A comprehensive theory of metals would require abandoning these rather crude
approximations. However, a remarkable amount of progress comes from abandoning
only the free-electron approximation in order to take into account the effect of the lattice
on the conduction electrons.
Unanswered Questions
In addition to quantitative discrepancies between the predictions of the quantummechanical FEG model and experiment (heat capacity, resistivity, thermal conductivity,
Hall effect, etc.), the FEG model is unable to answer two simple and very important
questions. A more comprehensive theory of the solid state should be able to come to
grips with these.
1.
What determines the number of conduction electrons in a metal?
Why should all valence electrons be “free”? What about elements with more than
one valence?
2.
Why are some elements metals and others non-metals?
One form of C (diamond) vs. another (graphite); In the same family, B vs. Al