Transcript Vibrational Properties of the Lattice
V. Electronic Properties of Metals
A. The Free Electron Gas (FEG) Model B. Properties of the FEG: Fermi Energy E F 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 finite square-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 n 0 n 0 free electrons, where = 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
Time-independent Schrödinger Equation: With U = 0: 2 2
m
2 2 2
m
2
U
E
E
2 2
mE
2 2
k
2 Solutions have 1. Wave functions:
Ae i
(
k
r
t
) Traveling waves (plane waves) 2. Energies:
E
2
k
2 2
m
Parabolic energy “bands” E k x
Properties of the FEG
By using periodic boundary conditions for a cubic solid with edge L and volume V = L 3 , we define the set of allowed wave vectors:
k x
2
n x L k y
2
n y L k z
2
n z L n x
,
n y
,
n z
1 , 2 , 3 , ...
This shows that the volume in k-space per solution is: 2
L
3 And thus the density of states in k-space is:
N
(
k
) 2 1 /
L
3
V
8 3 Just as in the case of phonons!
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 (k F ) and the maximum energy (E F ).
k z y k
k F
k x 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: #
states
#
states volume
volume
N
2
V
8 3 4 3
k F
3
k F
3 2
N V
1 / 3 3 2
n
1 / 3 And the maximum energy is easily found:
E F
2
k F
2 2
m
2 2
m
2 / 3 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.
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:
dN
N
(
E
)
dE
N
(
k
)
dk N
(
E
)
dk N
(
k
)
dE
N
(
k
)
dE
/
dk
V
/ 2
k
8 3 /
m
mV
2 8 3
k
Now using the general relation:
k
2
mE
1 / 2 2 we get:
N
(
E
)
V
2 3 2
m
3
E
1 / 2
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: E F 13 eV The valence bandwidth is in reasonable agreement with the FEG prediction of E F = 11.7 eV
Utility of the Density of States
With N(E) we can immediately calculate the average energy per electron in the 3-D FEG system:
E
total energy
#
electrons
E
0
F N
(
E
)
E E
0
F N
(
E
)
dE dE
We can simplify by using the relation:
N
(
E
)
CE
1 / 2
E
C E
0
F
E
3 / 2
C E
0
F
E
1 / 2
dE dE
E F
2 5
E
5 / 2 2 3
E
3 / 2 0
E F
0 3 5
E F
Why the factor 3/5? A look at the density of states curve should give the answer: N(E) E E F
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: A n unpolarized polarized: FEG displaced by x x L Induced dipole moment Induced polarization
p
P
q d
p
V
p AL neAx
L
nex i
ˆ
i
ˆ We can now find the induced E field:
D
0
E
P
0 x
E
0
P
nex
0
i
ˆ 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:
m d
2
x i
ˆ
dt
2
e E
ne
2
x
0
i
ˆ
d
2
x dt
2
ne
2
m
0
x
harmonic oscillator!
To find the oscillation frequency, compare to the equation of motion of the mass-spring system:
d
2
x dt
2
k m x
2
x
This reveals that the plasma frequency is given by: 2
p
ne
2
m
0 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 Li Na K Mg Al Expt. E (eV) 7.12
5.71
3.72
10.6
15.3
Calc. ħ p (eV) 8.02
5.95
4.29
10.9
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
19 th century puzzle: each monatomic gas molecule in sample at temperature T has energy , so if the N free electrons in a metal make up a classical 2 “gas” they should behave similarly.
E el
N
( 2 3
kT
) or expressed per mole:
E el n
3 2
N kT n
3 2
N A kT
3 2
RT
So the electronic contribution to the molar heat capacity would be expected to be
C el
d dT E el n
3 2
R
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
)
e
E
1 /
kT
1 where = chemical potential E F for kT << E F
Heat Capacity of the Quantum-Mechanical FEG
So at temperature T the total energy is:
E el
0
E f
(
E
)
N
(
E
)
dE
0
e
E E N
E F
( /
E kT
) 1
dE
And the electronic heat capacity is:
C el
dE el dT
1
kT
2 0
E N
(
E
e
)(
E E
E
F
/
E kT F
)
e
1 2
E
E F
/
kT dE
The exact answer to this complicated integral is derived in more advanced texts:
C el
3 2
k
2
N
(
E F
)
T
C el T !
A Rough and Ready Estimate
We can estimate C el in just a few lines in order to confirm the linear dependence on temperature: 2kT # electrons that can absorb thermal energy 2
kT
N
(
E F
)
f
(
E F
)
kT N
(
E F
) total thermal energy of electrons at T N(E) N(E)f(E)
E
(
T
)
E
(
T
)
E
0 2 3
kT
kT E
0 3 2
k
2
T
2
N
(
E F
)
N
(
E F
) E F FEG heat capacity at T
C el
dE el dT
3
k
2
N
(
E F
)
T
Remarkably close to the exact result!
E
C el
3 2
k
2
N
(
E F
)
T
T
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
)
C el
C lattice
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: Li Na K Cu Ag Au Al Metal expt 1.63
1.38
2.08
0.695
0.646
0.729
1.35
FEG 0.749
1.094
1.668
0.505
0.645
0.642
0.912
expt / FEG m*/m = 2.18
1.26
1.25
1.38
1.00
1.14
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:
F x
m d
2
x dt
2
eE
F coll F coll
p
t x
mv
relaxation time
eE
mv
0
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: 1
ne
n A
J
I A
1
A
Q
t
A L
/
v
nev
L Combined with the earlier result
v
eE
m
we get
J
ne
2
m
E
E
Ohm’s Law! where we define the electrical conductivity
ne
2
m
and the electrical resistivity is: 1
m ne
2 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?
E Consider once again the manifold of energy levels occupied by the FEG: F 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 E F the electrical conductivity.
contribute to
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 z y k
k F
Fermi surface k x
k x
p x
mv x
eE x
k x
p x
mv x
eE x
y k k 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 v F .
Analysis of Mean Free Path
e 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
total cross
sectional cross
sectional area area for of collision slab
v F
P
n a A
x
A r
2
n a
x
r
2 x collision cross-section =
n a
1
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
y
2 Therefore
r
2
T
And the thermal average potential energies can be written: and 1
T
from which
T
1 2
Cx
2 1 2
Cy
2 1 2
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:
F B
qvB
As more and more carriers are deflected, the accumulation of charge produces a “Hall field” E H that imparts a force opposite to the Lorentz force:
F E
qE H
Equilibrium is reached when these two opposing forces are equal in magnitude, which allows us to determine the drift speed:
qvB
qE
t
H
From this we can write the current density: And it is customary to define the Hall coefficient in terms of the measured quantities:
J
nqv
nqE H B R H
E H JB
1
nq v
E H B
w I
Hall Effect Results!
In the lab we actually measure the Hall voltage V H more useful way to write R H : and the current I, which gives us a
V H
E H w I
JA
Jwt R H
E H JB
I V H
/ /
wt w
B
V H t IB
1
nq
If we calculate R H from our measurements and assume |q| = e (which Hall did not know!) we can find n. Also, the sign of V H and thus R H 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…..
* * Metal Na Cu Ag Au Be Zn Al n 0 1 1 1 1 2 2 3 R H (10 -11 m 3 /As) solid -25 -5.5
-9.0
-7.2
+24.4
+3.3
-3.5
liquid -25.5
-8.25
-12.0
-11.8
-2.6
-5 -3.9
FEG value -25.5
-8.25
-12.0
-11.8
-2.53
-5.1
-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:
el
1 3
C
el
v
The electron mean-free path can be rewritten in terms of the collision time:
v
1 3
C
el
v
2 Also, the electrons that can absorb thermal energy and therefore contribute to the heat capacity have energies very near E F , so they essentially all have velocity v F . This gives: 1 3
C
el
v
F
2 From our earlier discussion the electronic heat capacity is:
C el
3 2
k
2
N
(
E F
)
T
It is easy to show that N(E F ) can be expressed:
N
(
E F
) 3
N
2
E F
Which gives the heat capacity per unit volume: Or per unit volume of sample,
N
(
E F
) 3
n
2
E F C
el
2 2
k
2
n T E F
Wiedemann-Franz Law and Lorenz Number
Now the thermal conductivity per unit volume is: 1 3 2 2
k
2
n T E
F
v
2
F
1 3 2 2
k
2
n T
1 2
mv
F
2
v
2
F
Finally… 2
k
2
n
3
m 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:
constant
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:
T
L
Gustav Wiedemann very nearly constant for all metals (at room T and above)
The Experimental Test!
We can readily compare the prediction of the FEG model to the results of experiment:
L
FEG
L
FEG
2
k
2
n
T
3
m ne
2
T
2 .
45 10 8
T m
CK
2 2
k
2 3
e
2 This is remarkable…it is independent of n, m, and even !!
Metal Cu Ag Au Zn Cd Mo Pb L = / T 10 -8 (J/CK) 2 0 ° C 2.23
2.31
2.35
2.31
2.42
2.61
2.47
100 °C 2.33
2.37
2.40
2.33
2.43
2.79
2.56
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 C el . 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:
L
Drude
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:
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 quantum mechanical 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.
2.
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?
Why are some elements metals and others non-metals?
One form of C (diamond) vs. another (graphite); In the same family, B vs. Al