- Now that I've explained electronic band structure, what are its consequences?
- Why is it that some materials conduct electricity with electrons, and some conduct with holes?
- What is so scientifically exciting about graphene that it won a Nobel Prize?

**Effective mass**

In classical physics, we have the following equation for kinetic energy:

E = 1/2 mvEarlier, I said k was the quantum analogue of velocity. I lied. k is really the quantum analogue of^{2}

**momentum**, which is mass times velocity. For some reason (silly physicists!) momentum is represented by the letter p. So this is another equation for kinetic energy:

E = 1/2 pSo if we just replace p with k, we get the quantum analogue:*^{2}/m

E = 1/2 kThe important thing is that this relationship depends on m, the mass.^{2}/m

The sharper the curve of the parabola, the smaller the mass it represents.

That's the picture of particles in a vacuum. But in a crystal, the picture is more complicated. In my explanation of electronic band structure, I left off with this image:

The four lines represent possible states for electrons within a crystal. But as I said before, I am only showing a part of the electronic band structure, namely, the "first Brillouin Zone". That's because the rest of the band structure is just repeating. Here, three copies are shown:

This picture of electrons in a crystal is very different from the electrons in a vacuum. And yet, there are some places where they're nearly the same. Near points A, C, and E, you can see the similar parabola shape that you would see in a vacuum. However, some of these parabolas are more sharply curved than others. That tells you something about the electron's "effective mass". An electron at point A would have a smaller effective mass than an electron at point E. The effective mass tells you something about how the electrons respond to a voltage.

*I am ignoring some constants. Or, as physicists say, I'm setting h-bar equal to one.

**Holes**

Notice that there are also upside-down parabolas at points D and B. Do electrons at these points behave as if they have

*negative*mass?

The answer is yes, sort of. Except you won't really find any electrons at points D and B. An electron at B will just naturally fall down towards point C, and an electron at D will fall down towards point E. The only way to really prevent this is if there are already electrons occupying those lower states.

Imagine, if you will, that there are electrons occupying all of the states on the green line. Electrons at point B cannot fall down because no two electrons are allowed to occupy the same state. But if the electrons cannot move along the curve, they can't conduct electricity!

So now imagine that electrons occupy

*nearly*all the states on the green line. There are a few empty states at the top, near B. Electrons at B have negative effective mass. But a

*lack*of an electron at B would have a positive effective mass. We call this abstraction a

**hole**. Like electrons, holes have a positive mass (because it is a lack of a negative effective mass), and unlike electrons, holes have positive charge (because it is a lack of a negative charge).

One of the easiest ways to observe holes is by measuring the Hall Effect. Without going into detail, the Hall Effect creates a voltage which is proportional to the charge of the particles that carry the electricity. So in materials that conduct by holes, the Hall voltage is in the opposite direction you would expect.

**Graphene**

Graphene is a 2-dimensional material made of carbon atoms in a honeycomb pattern.

I can't really explain all the scientifically interesting things about graphene, but I can explain one thing that has to do with its band structure. Since graphene is 2-d, we have to draw the band structure with a 3-d graph. The horizontal directions represent the two components of k, while the vertical direction represents E, the energy.

(Image credit)

That's a bit hard to look at, so we'll just zoom into the important part, at one of the corners of the black hexagon.

This is what's called the Dirac Cone. It appears because of certain symmetries in the graphene structure. What's interesting about it is that it does

It turns out that in classical physics, there is another situation when E is proportional to momentum. It's true of massless particles, such as light. And so, electrons on the Dirac Cone behave as if they are massless. (If you like, you can think of the cone as being an infinitely sharp curve.) Electrons on the Dirac Cone are relativistic particles! If that sounds exciting, imagine how it sounds to a physicist.

That concludes today's answers to physics questions you thought you'd never understand. Until next time...

*not*look like a parabola. E is not proportional to k^{2}, but is instead proportional to k. Translating to classical physics, E is proportional to momentum.It turns out that in classical physics, there is another situation when E is proportional to momentum. It's true of massless particles, such as light. And so, electrons on the Dirac Cone behave as if they are massless. (If you like, you can think of the cone as being an infinitely sharp curve.) Electrons on the Dirac Cone are relativistic particles! If that sounds exciting, imagine how it sounds to a physicist.

That concludes today's answers to physics questions you thought you'd never understand. Until next time...

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