Transcript The Solid State

The Solid State
Covalent Crystals
Van der Waals Forces
The Metallic Bond
“It was absolutely marvelous working for Pauli. You could ask him
anything. There was no worry that he would think a particular question
was stupid, since he thought all questions were stupid.”—V. Weisskopf
10.3 Covalent Crystals
There are relatively few 100% covalent crystals in
nature.
Examples: diamond, silicon, germanium, graphite (in
plane), silicon carbide.
These materials are all characterized by tetrahedral
bonding, which involves sp3 hybrid orbitals.
Hybrid orbitals result when two or more atomic orbitals
of an isolated atom mix in covalently-bonded atoms
(we’ll discuss this in more detail later).
This is one of the four hybrid orbitals formed by
the hybridization of one s and three p orbitals.
tetrahedral bonding
http://www.siue.edu/
CHEMISTRY/resources/
orbitals/sp3.htm
Some properties of covalent crystals:
• They are brittle due to their highly directional
bonding.
• They have high melting points due to their high bond
strengths.
• They have low impurity solubility due to their
directional sp3 hybrid bonds.
• They are insoluble in polar fluids.
Diamond structure:
(two interpenetrating fcc
lattices)
Graphite structure (like diamond but only in planes)
http://cst-www.nrl.navy.mil/lattice/struk.xmol/olda9.xyz
10.4 Van der Waals Forces
Even inert gases form crystals at low temperatures.
What force holds them together?
• Ionic bonds? No, because inert gases don't ionize.
• Covalent bonds? No, because inert gases have no
"desire" to share electrons.
• Some kind of Coulomb attraction? Inert gases are are
electrically neutral, so they don't attract charged
particles.
• “Hydrogen bonding,” where asymmetry of the
molecule gives rise to a nonuniform charge distribution
and a polarity? Not if inert gas atoms are nonpolar
The answer is that, while on the
average inert gas atoms are nonpolar,
their electrons are in constant motion
and they can have instantaneous
nonuniform charge distributions.
The + and - portions of the atom can exert attractive force,
which is enough to bond at very low temperatures.
It's a Physics 24 type problem to show that a dipole of
moment p gives rise to an electric field E given by
1
E=
4ε0
 p 3 p  r  
r.
 3 5
r
r

This electric field can induce a dipole moment in a normally
nonpolar molecule.
The induced dipole moment is
p = αE ,
where  is the polarizability of the molecule or atom.
The energy of the induced dipole in the electric field is
U = - p  E ,
and it is not too difficult to show that U is proportional to -1/r6.
Things to note: the potential is negative (attractive, stable).
The force is proportional to dU/dr which is proportional to 1/r7,
so the force drops off very rapidly with distance.
Comments about van der Waals forces
• Van der Waals forces are responsible for hydrogen
bonding, such as occurs in water.
• Van der Waals forces are always present, but at even
moderate temperatures they are so relatively weak that
they can safely be neglected.
• Our brief derivation is a classical one, and, as you
might suspect, not really correct at the atomic level.
However, a full quantum mechanical derivation gives
the same r-dependence. (Harmonic oscillators!)
Hydrogen bonding:
“The attraction of the partially positive end of one highly polar
molecule for the partially negative end of another highly polar
molecule is called a hydrogen bond.”
http://207.10.97.102/chemzone/lessons/03bonding/mleebonding/hydrogen_bonds.htm
Now we see how an ionic crystal dissolves in a polar fluid like
water:
(I probably should have shown this back in the ionic crystals
section.)
The Lennard-Jones potential, which you may have
encountered in chemistry (also sometimes known as the
Lennard-Jones 6-12 potential or the 6-12 potential) describes
van der Waals forces.
 σ 12  σ 6 
ULJ  r  = 4ε   -    .
 r   r  
overlap (pauli)
van der Waals
adjustable parameters
This potential models the interaction between pairs of atoms.
In section 10.2, we used an 1/r9 dependence for ionic crystals
instead of the 1/r12 dependence here. The difference can be
“absorbed” in our adjustable parameters.
10.5 Metallic Bond
Metallic bonding is caused by electrostatic forces acting in
combination with the Heisenberg uncertainty principle and the
Pauli exclusion principle.
Metal atoms give up electrons (usually one or two,
sometimes three). The result is a lattice of positive ion
cores sitting in a "sea" or "gas" of electrons.
+
-
+
+
+
-
+
-
+
-
-
+
+
+
-
+
+
- +
+
-
+
+
+
- +
-
+
-
+
+
The electrons in the gas repel each other. The electron gas and
ion cores attract each other. Bonding occurs because the
reduction in energy due to the attraction exceeds the increase in
energy due to repulsion.
“Hang on just a minute there, pal. Don't we have a problem
here?”
If we bring 1022 or so atoms together and try to combine that
many electrons into a single "system," don't we have a problem
with the Pauli exclusion principle?
We sure do. It would seem that the energies of most of the
electrons would need to be so high (remember, energies go up
as we put electrons in successively further out shells) that the
net electron energy would be so great that bonding could never
occur.
How do we get around this? The energy levels of the
overlapping electron shells are all slightly altered. The energy
differences are very small, but large enough so that a large
number of electrons can be in close proximity and still satisfy
the Pauli exclusion principle.
We are describing the formation of energy bands, consisting of
many states close together but slightly split in energy.
These energy states are so close together that for all practical
purposes we can consider bands as a continuum of states,
rather than discrete energy levels like we have in isolated atoms
(and in the core electrons of atoms of metals).
“This sounds seriously artificial.” True, it sounds like an ad hoc
explanation. “Did you make it up—you know, just to keep from
having to explain it to us?” “Where do these bands come
from?”
We should first ask where the electronic energy
levels in atoms “come from.”
Remember, we solved the Schrödinger equation for an electron
in the potential of the hydrogen nucleus. This gave us our
energy levels and quantum numbers.
More complex atoms require more complex mathematics, but
the idea is the same: the energy levels come from the solution
of Schrödinger's equation for electrons in the potential of the
nucleus.
The same holds for metals, except now we have a periodic array
of nuclei, and a periodic potential. We still have to solve
Schrödinger's equation for an electron moving in this periodic
potential.
The problem can't be solved exactly, of course, but it can be
solved with quite reasonable accuracy.
The energy bands result quite naturally from the solution, just
as energy levels in atoms came naturally out of the solution to
Schrödinger's equation.
Actually, “we” don’t have to solve anything (unless we are
physicists). The electrons know how to do it. They solve
Schrödinger’s equation for us. We just have to discover what
they tell us.
This is a bit unsatisfying right now.
“Look, but don’t eat.” However, we
will discuss band theory in more
detail in later sections in this chapter.
Ohm’s Law
Any reasonable theory of the solid state should be able to come
up with Ohm's law. Since we are talking solids and metals, now
is a good time to give it a try.
Ohm’s law is not really a fundamental law. It is
an empirical relationship that (usually) works for
(some) solids.
V
I=
R
You’d feel a bit better about Ohm’s law if we could derive it,
wouldn’t you? Let’s try.
What do electrons in a conductor do in the absence of an
applied electric field?
+
-
+
- +
+
-
+
+
-
-
-
+
+
+
+
-
- +
- +
+
-
-
- +
+
+
- +
-
+
-
+
+
They move, very rapidly, but in random directions. There is no
net current.
I’m not going to put velocity vectors on all the electrons. I hope
you get the idea!
Let’s just follow one of the electrons. What happens when an
electric field is applied?
E
+
+
+
+
electron drift velocity
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
Electrons accelerate and gain a net "drift" velocity in some
direction. We will see that the drift velocity is actually very small
compared to the Fermi velocity.
Since an electric field accelerates electrons, why don't they
move faster and faster? Or do they do just that?
E
+
+
+
+
electron drift velocity
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
They don’t. Some force must eventually oppose the force due
to the electric field and bring the drift velocity into a "steady
state" condition.
What is this "force"? Collisions with + ions? Attractive idea,
and used in classical theory—which failed miserably.
Collisions with other electrons? No, electrons don’t “like” each
other (Pauli) and thus don’t interact like particles.
So what is this "force“ that causes resistance? Typically it is a
result of collisions of electrons with impurities.
“Impurities” means defects like we talked about earlier (point
defects, dislocations), the edges of the metal, or atoms
temporarily out of their equilibrium position because they are
vibrating like balls connected by springs.
Let's analyze this situation mathematically. Ohm's law had
better come out of the analysis.
Only electrons near the Fermi energy can be accelerated
(remember, all energy states below εF are occupied).
Those electrons near the Fermi level (which can be accelerated)
start with some random velocity (about vF in magnitude but
random in direction).
They gain “additional” velocity (in a direction opposite the
electric field direction) until they collide with an impurity.
The collision with the impurity “randomizes” the velocity, and the
process starts over again.
If we define λ as the mean free path of electrons between
collisions, then the average time between collisions is
λ
=
.
vF
An applied electric field results in a force on the electron, and
therefore an acceleration
a =
f
eE
=
.
m
m
Since the electron is accelerated only during the time , after
which it undergoes a collision and has its velocity randomized
again, the average gain in velocity, or "drift velocity," is
eE
v d = a =
.
mv F
Current is the amount of charge passing a point in conductor
during some time. Let’s do an analysis of the units:
 electrons
 charge
charge


volume


=



  electron 
time
time


 distance 


  cross - section
×  traveled  

area
of
conductor

 by electron  


charge
 electrons   charge 
= 


time
 volume   electron 

 distance traveled   cross - section
×
  area of conductor 
time



I = n  e  v d  A 
See Beiser, page 351, for an equivalent, but slightly different
approach to calculating the current.
The current in the conductor is thus
I = n  e  v d  A 
 eEλ 
I = n  e  
A
 mvF 
n A e2 E λ
I =
mv F
The potential V across a length L of the conductor is related to
the electric field by E = V / L. Thus…
V
nAe
λ
L
I =
mvF
2
 n e2 λ   A 
I = 
  V
 mvF   L 
V
I =
,
R
where
 mv F   L 
L 
R = 
  = ρ   .
2
A
 n e λ  A 
The resistivity, , of the metal is defined by the above equation.
We need to see if this works (we’ll do that in a minute). If it
does, a simple drift velocity theory, based on the electron gas
and some quantum mechanics (noninteracting electrons) has led
us to Ohm’s law.
This may not seem like much, but our inability to get within
even a factor of 10 of Ohm’s law at the start of the 20th century
was a major unsolved problem in solid state physics. See the
note on the Wiedemann-Franz law on page 354 (not for test or
quiz).
Example 10.3 The resistivity of copper at 20 ºC is 1.72x10-8
·m. Estimate the mean free path between collisions of the
free electrons in copper at 20 ºC. (The temperature is specified here only
because resistivity is temperature-dependent, so you should give a temperature when
you give a resistivity.)
The density of free electrons in copper was calculated in chapter
9 to be n = 8.48x1028 m-3. I could give you this, or give you the
density and atomic mass of copper and expect you to calculate
this.
The Fermi energy may be given, or you may need to calculate it.
From the value of n above, you can calculate F = 7.04 eV.
The Fermi velocity may be given, or you may need to calculate it
from KE = F = (1/2) m vF2. The result (page 350; there is a
missing square root sign in the calculation) is vF = 1.57x106 m/s.
“Gee whiz, that’s pretty fast!” Yup, sure is.
“Gosh, would you make me do all those calculations on an exam
or quiz?” Yup, sure would.
Now that we have all the input parameters, we use our equation
for resistivity to calculate the mean free path.
I haven’t yet written down an expression for resistivity, so let’s
do it now, and make it “official.”
mv F
 =
.
2
ne λ
mass of what
“thing”?
mvF
λ =
n e2 ρ
λ =


9.11×10-31 1.57×106

 8.48×10  1.6×10  1.72×10 
28
-19
2
-8
keep everything SI so you don’t have
to worry about units
λ = 3.83×10-8 m = 38.3 nm .
That’s the mean free path of the electrons in copper. Not very
far, huh?
“Hold it right there! The atoms in copper are about 0.26 nm
apart. An electron travels past about 38.3/0.26 = 150 of them
without being scattered!”
“So you’re saying the negative electrons don’t even know the
positive ions are there?” Yup.
Electron waves in a perfect periodic lattice interact with the
lattice only under very special circumstances.
Resistivity (which comes from collisions) is due to imperfections
in the lattice.
Resistivity comes from impurities and vibrations of atoms about
their equilibrium positions.
You can remove impurities to increase conductivity.
You can remove lattice vibrations (called “phonons”) by cooling
the conductor. This also increases conductivity.
Cryo-cooled PC’s to get the ultimate frame rate in Doom 3!
You can cool a sample to a low enough temperature where
lattice vibrations add very little to the resistivity. The remaining
resistivity is due to impurities. The “residual resistivity ratio” is a
very important way of assessing the purity of your samples.
A brief caution about resistivity calculations:
Two previous editions back, Beiser estimated the mean free path
by a different method.
This method introduced a factor of 2 into the resistivity
equation, which is no longer there.
Ignore any extra factors of 2 which you may see in old exams
(or exam papers, if your house has them on file).
10.6 Band Theory of Solids
What happens in crystalline solids when we bring atoms so
close together that their valence electrons constitute a single
system of electrons?
“You do not really understand something unless you can explain it to your
grandmother.”—A. Einstein
The energy levels of the
overlapping electron shells
are all slightly altered.
The energy differences are
very small, but enough so
that a large number of
electrons can be in close
proximity and still satisfy the
Pauli exclusion principle.
The result is the formation
of energy bands,
consisting of many states
close together but slightly
split in energy.
The energy levels are so close together that for all practical
purposes we can consider bands as a continuum of states,
rather than discrete energy levels as we have in isolated atoms
(and in the core electrons of atoms of metals).
A detailed analysis of energy bands shows that there are
as many separate energy levels in each band as there are
atoms in a crystal.*
Suppose there are N atoms in a crystal. Two electrons
can occupy each energy level (spin), so there are 2N
possible quantum states in each band.
Let’s consider sodium as an example. Sodium has a single
outer 3s electron.
*Kind of. I need to explain.
When you bring two sodium
atoms together, the 3s
energy level splits into two
separate energy levels.
Things to note: 4 quantum
states but only 2 electrons.
You could minimize electron energy by putting both 3s electrons
in the lower energy level, one spin up and the other spin down.
There is an internuclear separation which minimizes electron
energy. If you bring the nuclei closer together, energy
increases.
When you bring five sodium
atoms together, the 3s
energy level splits into five
separate energy levels.
The three new energy
levels fall in between the
two for 2 sodiums.
There are now 5 electrons occupying these energy levels.
I’ve suggested one possible minimum-energy configuration.
Notice how the sodium-sodium internuclear distance must
increase slightly.
When you bring N (some
big number) sodium atoms
together, the 3s energy
level splits into N separate
energy levels.
The result is an energy
band, containing N very
closely-spaced energy
levels.
There are now N electrons occupying this 3s band. They go
into the lowest energy levels they can find.
The shaded area represents available states, not filled states.
At the selected separation, these are the available states.
Now let’s take a closer look at
the energy levels in solid
sodium.
Remember, the 3s is the
outermost occupied level.
When sodium atoms are
brought within about 1 nm of
each other, the 3s levels in the
individual atoms overlap
enough to begin the formation
of the 3s band.
The 3s band broadens as the
separation further decreases.
3s band
begins to
form
Because only half the states in
the 3s band are occupied, the
electron energy decreases as
the sodium-sodium separation
decreases below 1 nm.
At about 0.36 nm, two things
happen: the 3s energy levels
start to go up (remember
particle in box?) and the 2p
levels start to form a band.
Further decrease in interatomic
separation results in a net
increase of energy.
3s electron
energy is
minimized
What about the 3p and 4s
bands shown in the figure?
Don’t worry about them—there
are no electrons available to
occupy them!
Keep in mind, the bands do
exist, whether or not any
electrons are in them.
What about the 1s and 2s
energy levels, which are not
shown in the figure?
The sodium atoms do not get
close enough for them to form
bands—they remain as atomic
states.
Figure 10-20 (the one on the last three slides) shows energy
levels as a function of interatomic separation.
Energy levels for a crystal structure with fixed interatomic
distances vary with different directions in space.
http://www.chem.brown.edu/chem50/Notes/classpics/onefig.html, sorry about the picture
quality.
As an aid to visualization, we often represent energy bands like
this (using sodium as an example):
This is highly schematic. Real
bands aren't boxes or lines.
Sodium has a single 3s
electron, so the 3s energy
band contains twice as many
states as there are electrons.
The band is half full.
3s
2p
2s
1s
At T=0 the band is filled exactly halfway up, and the Fermi
level, εF, is right in the middle of the band.
Sodium is a metal because an applied field can easily give
energy to and accelerate an electron.
εF
Magnesium has two 3s electrons.
You expect the 3s band to be full, 3p
the 3p band to be empty, with a
forbidden gap in between.
Magnesium should be an insulator.
(Why?)
3s
But magnesium is a metal
(actually, a “semimetal”).
The 3p and 3s bands overlap.
3p
There are many empty states
nearby into which electrons can be
3s
accelerated.
Materials which have bands either
completely full or completely
empty are insulators (unless band
overlap occurs, as was the case for
magnesium).
3p
3s
In a carbon atom, the 2p shell contains 2 electrons. There
are 6 available states, so one would expect the 2p band to be
1/3 full* and carbon to be a conductor.
But carbon is an **insulator. Why?
*2N 2p electrons in a crystal with N atoms. 6N 2p states, when you include spin.
**If the diamond in your diamond ring conducts electricity, it’s time to take it back!
Figure 10.23 shows energy
bands in carbon (and
silicon) as a function of
interatomic separation.
At large separation, there is
a filled 2s band and a 1/3
filled 2p band.
But electron energy can be lowered if the carbon-carbon
separation is reduced.
There is a range of carbon-carbon separations for which the 2s
and 2p bands overlap and form a hybrid band containing 8N
states (Beiser calls them “levels”).
But the minimum total
electron energy occurs at
this carbon carbon
separation.
At this separation there is a
valence band containing 4N
quantum states and
occupied by 4N electrons.
The filled band is separated by a large gap from the empty
conduction band. The gap is 6 eV—remember, kT is about
0.025 eV at room temperature. The gap is too large for
ordinary electric fields to move an electron into the conduction
band. Carbon is an insulator.