# The Solid State Band Theory of Solids “You do not really understand something unless you can explain it to your grandmother.”—A. Einstein.

## Presentation on theme: "The Solid State Band Theory of Solids “You do not really understand something unless you can explain it to your grandmother.”—A. Einstein."— Presentation transcript:

The Solid State Band Theory of Solids “You do not really understand something unless you can explain it to your grandmother.”—A. Einstein

Typo in Syllabus—News Flash—Typo in Syllabus 428 points gets you a C 429 points gets you a B I already generously rounded the cutoff percentages down! (429 is not quite 79.5%) I have added two typo points to your grades. The grades spreadsheet shows the grade you will get if you do nothing for the rest of the semester. Please make sure all earned bonus points are recorded!

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? Band structure of diamond. http://home.att.net/~mopacmanual/node372.htmlhttp://home.att.net/~mopacmanual/node372.html

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 an actual crystal structure also vary with different directions in space. http://cmt.dur.ac.uk/sjc/thesis/thesis/node39.htmlhttp://cmt.dur.ac.uk/sjc/thesis/thesis/node39.html, band structure of silicon

As an aid to visualization, we often represent energy bands like this (using sodium as an example): 1s 2s 2p 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, ε F, is right in the middle of the band. εFεF 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 3p The 3p and 3s bands overlap. There are many empty states nearby into which electrons can be accelerated.

3s 3p Materials 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.

On page 355 Beiser says there are as many levels in a band (N) as there are atoms in a crystal. In figure 10.23 the caption implies there are 2N levels in a 2s band. This is inconsistent. Let’s call a “level” an energy level and a “state” a quantum state, including spin. Then figure 10.23 should talk about 2N states in a 2s band. Silicon has a similar band structure. The forbidden gap is about 1 eV. The probability of a single electron being excited across the gap is small, proportional to exp(-E gap /kT). However, there are enough 3s+3p electrons in silicon that some of them might make it into the conduction band. We need to consider such a special case.

A semiconductor is a material which has a filled (at T=0) valence band separated by a small gap from an empty (at T=0) conduction band. Because the gap is small, at room temperature there will be a few electrons in the conduction band. These electrons can be accelerated by an applied electric field. valence band conduction band A semiconductor at T= 0 K.

CB VB FF A semiconductor at room temperature. Although e (-Egap/kT) is small, Ne (-Egap/kT) can be significant. Thus, at very low temperatures, silicon is an insulator, but at room temperature, it is a weak conductor (intermediate between conductor and insulator, hence semiconductor). A rough rule of thumb: a band gap of less than 3 eV gives rise to a semiconductor. The smaller the gap, the more charge carriers in the conduction band.

It is possible to “engineer” properties of semiconductors to make them more suitable for use in devices. We do this by doping them with impurities. Here’s an example: arsenic impurities in silicon. As has an outer electron configuration of 4s 2 4p 3. After it shares 4 of its outer electrons with neighboring silicons, the remaining electron is very loosely bound. + - Homework problems 10.21 and 10.22 (assigned although listed under section 10.8) show that kT at room temperature is more than enough energy to ionize the fifth outer electron. The electrons from the As + ions are then free to move throughout the crystal.

The arsenic impurity creates a donor impurity level. Because it takes only a few meV to ionize the As and place the resulting electron in the conduction band, the donor impurity level sits just below the conduction band. CB VB FF A few donor impurities can produce many electrons in the conduction band and give rise to significant conductivity. The Fermi level lies somewhere between the top of the valence band and the highest-energy electron in the conduction band. These semiconductors are n-type, because conduction is by n egative electrons.

Here’s another example: gallium impurities in silicon. Ga has an outer electron configuration of 4s 2 4p 1. It “wants” to borrow 3 electrons from its 4 silicon nearest neighbors. After a Ga “borrows” 3 electrons, it is easy for the 4 th silicon to “convince” the Ga to “borrow” one more electron, creating a Ga - ion. The gallium impurity creates an acceptor impurity level. - + The lattice depicted here is highly schematic, not realistic, and should be used as a visual aid only!

Because the acceptor ionization energy is very small, the acceptor levels sit just above the valence band (a small fraction of an eV). CB VB FF Electrons from the valence band can easily get to the acceptor bound states. The holes represent states into which other electrons in the valence band can move. Thus, electrons can easily move around in the presence of an applied field. Alternatively, we can look at the holes, say they move around, and say that conduction is due to holes. Electrons from the valence band can easily get to the acceptor bound states. That leaves holes in the valence band. This is called a p-type semiconductor because “p”ositive holes are the charge carriers.

Again, it turns out that there don't have to be very many acceptor atoms around to result in a significant number of holes in the valence band. What's all this fuss about holes. It seems like a hole is really just an electron missing from the valence band and sitting in an acceptor state. Why don't we just talk about conduction by these electrons moving from state to state, instead of worrying about holes? Answer: holes really are more than just missing electrons. For one thing, electrons in this case are "stuck" on acceptor atoms, but the holes are free to move about in the valence band. For another thing, we can dope semiconductors so that there are excesses of holes and electrons. A "hole" really is more than just a missing electron which is somewhere else.

Optical properties of solids are closely related to band structures. Visible photons have energies ranging from about 1 eV to about 3 eV. Metals can absorb visible photons because there are many empty states for electrons to move to. In diamond, the valence band is full and the conduction band is empty. A 3 eV photon cannot excite an electron across the 6 eV band gap. Diamond cannot absorb visible photons. Diamonds are transparent.

“Shouldn’t all insulators be transparent?” Yes. They are opaque because they contain impurities and irregularities which scatter visible light. Insulators are transparent to high-energy UV (ultraviolet) photons. Now, skipping ahead briefly…

Section 10.8 is a lengthy section which attempts to give a better justification of energy bands. The section is not required, but does contain a few key points: The periodicity of a crystal lattice leads naturally (through the solution of Schrödinger’s equation) to allowed and forbidden energy bands for electrons. Electrons cannot exist in forbidden energy states because there are no solutions to Schrödinger’s equation there. Electrons do not (in general) scatter from a perfect lattice, but they do interact with the lattice.

Sometimes the lattice “holds” the “free” electrons in place. These electrons act like they have an unusually high mass (m effective > m electron ). Sometimes the electrons act like they have an unusually small mass (m effective < m electron ). Sometimes the electrons even have a negative effective mass. Huh? Sure! Sure! Push on an electron in one direction. Sure! Push on an electron in one direction. If the push gives it enough energy to Bragg scatter off the lattice (the exception to the “no scattering” rule), the electron goes in the opposite direction. Sure! Push on an electron in one direction. If the push gives it enough energy to Bragg scatter off the lattice (the exception to the “no scattering” rule), the electron goes in the opposite direction. Negative mass!

Ceramics majors would benefit from studying section 10.8. The only material you are required to know (for exam or quiz purposes) is that given above. One more thing… the band structure I showed earlier: Notice the gap that extends in all directions through the crystal.

Let’s move on to “lecture 34” and see how far we get…

Download ppt "The Solid State Band Theory of Solids “You do not really understand something unless you can explain it to your grandmother.”—A. Einstein."

Similar presentations