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.

This is one of the four hybrid orbitals formed by the hybridization of one s and three p 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).

tetrahedral bondinghttp://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 are insoluble in polar fluids.

• They have low impurity solubility due to their directional sp3 hybrid bonds.

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?

• Covalent bonds?

• Some kind of Coulomb attraction?

• “Hydrogen bonding,”

• 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

3 5

0

3 p r1 pE = - r .

4 ε r r

This electric field can induce a dipole moment in a normally nonpolar molecule.

The induced dipole moment is

where is the polarizability of the molecule or atom.

p = αE ,

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.

12 6

LJ

σ σU 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.

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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. 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?

We are describing the formation of energy bands, consisting of many states close together but slightly split in energy.

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.

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.”“This sounds seriously artificial.” True, it sounds like an ad hoc explanation. “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?”

“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.

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

Ohm’s law is not really a fundamental law. It is an empirical relationship that (usually) works for (some) solids.

VI =

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?

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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?

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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.

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E electron drift velocity

Since an electric field accelerates electrons, why don't they move faster and faster? Or do they do just that?

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"? What is this "force"? Collisions with + ions? What is this "force"? Collisions with + ions? Attractive idea, and used in classical theory—which failed miserably.

So what is this "force“ that causes resistance? 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.

Collisions with other electrons? Collisions with other electrons? No, electrons don’t “like” each other (Pauli) and thus don’t interact like particles.

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).

The collision with the impurity “randomizes” the velocity, and the process starts over again.

They gain “additional” velocity (in a direction opposite the electric field direction) until they collide with an impurity.

If we define λ as the mean free path of electrons between collisions, then the average time between collisions is

F

λ = .

v

An applied electric field results in a force on the electron, and therefore an acceleration

f eEa = = .

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

d

F

eEv = a = .

mv

Current is the amount of charge passing a point in conductor during some time. Let’s do an analysis of the units:

electronscharge chargevolume = time time electron

distancecross -section

× traveledarea of conductor

by electron

charge electrons charge =

time volume electron

cross -sectiondistance traveled ×

area of conductortime

I d = n e v A

See Beiser, page 351, for an equivalent, but slightly different approach to calculating the current.

The current in the conductor is thus

dI = n e v A

F

eEλI = n e A

mv

2

F

n A e E λI =

mv

The potential V across a length L of the conductor is related to the electric field by E = V / L. Thus…

2

F

Vn A e λ

LI = mv

2

F

n e λ AI = V

mv L

VI = ,

R

where

F2

mv L LR = = ρ .

n e λ A 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 vF

2. 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!”“Gee whiz, that’s pretty fast!” Yup, sure is.

“Gosh, would you make me do all those calculations on an exam or quiz?”“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.”

F2

mv = .

n e λ

F2

mvλ =

n e ρ

-31 6

228 -19 -8

9.11×10 1.57×10λ =

8.48×10 1.6×10 1.72×10

mass of what “thing”?

keep everything SI so you don’t have to worry about units

-8λ = 3.83×10 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?”“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

“You do not really understand something unless you can explain it to your grandmother.”—A. Einstein

What happens in crystalline solids when we bring atoms so close together that their valence electrons constitute a single system of electrons?

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.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):

1s

2s2p

3s

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.

At T=0 the band is filled exactly halfway up, and the Fermi level, εF, is right in the middle of the band.

εF

Sodium is a metal because an applied field can easily give energy to and accelerate an electron.

3s

3p

Magnesium has two 3s electrons. You expect the 3s band to be full, the 3p band to be empty, with a forbidden gap in between. Magnesium should be an insulator. (Why?)

But magnesium is a metal (actually, a “semimetal”).

3s

3pThe 3p and 3s bands overlap. There are many empty states nearby into which electrons can be accelerated.

3s

3pMaterials which have bands either completely full or completely empty are insulators (unless band overlap occurs, as was the case for magnesium).

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?

**If the diamond in your diamond ring conducts electricity, it’s time to take it back!

*2N 2p electrons in a crystal with N atoms. 6N 2p states, when you include spin.

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.