Why do holes not contribute to conduction in metals?

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1. May 16, 2015

Nate Wellington

It seems like when an electron moves to a higher energy level, even within the same band (conduction), the empty state should be a hole.

But, we are told that in metals, electrons are the charge carriers. Do I just misunderstand what holes are? Or what is going on here?

2. May 16, 2015

spaghetti3451

This is an excellent question, and clears up some of the misunderstandings students have on this topic.

The valence band and the conduction band of a semiconductor are separated by an energy gap that is greater than the order of $k_{B}T$. Furthermore, the Fermi level of a semiconductor typically lies within the valence band of possible energies. Now, at a non-zero temperature, some (or perhaps many) of the electrons possess sufficient energy to jump across the energy gap and dive into the conduction band. In this case, there are a few (or many) electrons in the conduction band, and it is appropriate to describe these electronic states using a single-electron wavefunction for each of these individual electrons. On the other hand, the remaining electrons still reside in the valence band. So, rather than placing our attention on the many-many-many electronic states in the valence band (which complicates the calculations), we focus on the few absences of the electrons (conventionally called holes), which can simplify the subsequent mathematics.

Thus arises the concept of a hole for semiconductors. Current is conducted in a semiconductor by only electrons, but physicists find it easier to focus on the behaviour of electrons in the conduction band, and the behaviour of holes (the absence of electrons) in the valence band.

On the other hand, the valence band and the conduction band of a metal conductor overlap. Therefore, the artificial concept of a hole is not really very useful in the description of the conduction of a metal. Therefore, the conduction properties of a metal conductor are described using the behaviour of excited electrons only.

Let me know if there are any mistakes in my explanation.

3. May 17, 2015

my2cts

4. May 17, 2015

Nate Wellington

Thank you! I think this is a good response. That being said, I still have the following confusion. In semi-conductors, the Hall coefficient can have a positive or a negative sign, depending on whether the transport is dominated by electrons or holes. I am under the impression that they way you get hole-dominated transport is to put in acceptors--positive ions to which electrons from the valence band can hop.

Now, if we are talking about an undoped semiconductor, we have no donors and no acceptors. If an electron jumps from one band to another, you end up with both a hole and an electron. Should this combination not make for zero hall voltage? (Since half the transport is from holes, half from electrons).

Lastly, if we go back to the idea of a metal--I get that the concept of a hole is no longer particularly useful, but I don't understand why the "missing electron states" which we would otherwise be calling holes, which still exist, don't behave in the opposite way in an electric field--just like holes do--and mess with the Hall coefficient.

Thanks!

5. May 18, 2015

DrDu

In intrinsic semiconductors, the mobilities of electrons and holes are not the same; this gives rise to a non-zero Hall coefficient. Also some metals, e.g. Aluminium, show a negative Hall coefficient.
The book by Ashcroft and Mermin contains a carefull discussion of the mobility of wavepackets in magnetic and electric fields. Highly recommended.

6. May 18, 2015

Stanley514

Why not? How do you explain a positive Hall coeeficient in metals and semimetals, then? Aren't they associated with holes mobility?

7. May 18, 2015

spaghetti3451

Yes, they are!

I take back my earlier comment!

8. May 18, 2015

Stanley514

There is no difference between behavior of holes within metals and semiconductors?

9. May 23, 2015

DrDu

I just found the time to look this up in Ashcroft and Mermin. My understanding is the following: There are two reasons to consider holes rather than electrons.
Near the top of a band, electronic bands have a negative curvature as a function of k, so that the effective mass $\propto \partial^2E(k)/\partial k^2$ is negative. It is then more intuitive to consider holes instead which have a positive mass and charge. With this setting, at least k and v have the same direction (or an angle less than 90 degree, at least).
The other reason, which may or may not coincide with the first one, is the explanation of the sign of the Hall coefficient.
Astonishingly, the effective mass does not show up in the Hall coefficient, so the sign of the mass isn't the reason.
Usually, we take our intuition for the Hall effect from the behaviour of free electrons. Free electrons will perform a circular orbit in a magnetic field and will accquire a drift velocity if there is also an electric field present.
But for real electrons in a band, not all energy levels will give rise to closed orbits in a magnetic field. If you are lucky, either the occupied or the unoccupied electron levels correspond to closed orbits. So if the Hall coefficient is negative, orbits for filled electronic levels are closed, while if it is positive, orbits for empty levels (aka holes) are closed.

10. May 26, 2015

rogerk8

Thank you DrDu for redirecting me.

I'm new at this so bear with me.

I learned a while ago that the Bohr atom model and its descrete shell radiuses is a result of the fact that the electron orbit is an integer number of the wavelength.

I know that this is old/obsolet(?) Classical Physics and not modern Quantum Physics but it gives me some kind of idea what you are talking about.

Because, as I understand it $$k=\frac{2\pi}{\lambda}$$ and therefore strictly related to wavelength.

I am further more aware of the Lorenz Force you are talking about which makes the electrons gyrate around the lines of force (B) and even drift if there is an E.

But I was not aware that they even may do that in a metal.

In a plasma, sure, but in a metal?

Is even the space large enough?

Putting the force formula to zero, yields

$$mv'=0=q(E+vXB)$$

$$EXB=vB^2$$

$$v=\frac{EXB}{B^2}$$

where the magnitude is

$$v=\frac{E}{B}$$

In other words, this is the drift of the gyrating electrons in any medium carrying "free" electrons (my guess).

Now, let's look att

$$mv'=qvXB$$

Putting

$$v=v_0e^{jwt}$$

yields

$$mjwv_0e^{jwt}=qv_0e^{jwt}XB$$

i.e

$$mjw=qXB$$

using the magnitude

$$w_c=\frac{qB}{m}$$

and while v=wr

$$r_L=\frac{mv}{qB}$$

Which is called the Larmor radius.

Now using

$$\frac{mv^2}{2}\propto kT$$

we get

$$v=\sqrt{\frac{2kT}{m}}$$

Putting this into the Larmor radius while using

T=293K
B=2T
m=1.67E-31
q=1.61E-19
k=6,38E-23

we first get v=473000 m/s and then we get 0,25um :D

MVH/Roger
PS
Why isn't there a blog function in PF anymore?

11. May 26, 2015

DrDu

Bohr's theory is not totally wrong, which should be clear already from the fact that it gives the precise energy levels in the hydrogen atom. It is more the classical interpretation in terms of circular orbits for classical point particles.
This being said, the equations you derived seem quite reasonable to me up to the one I am citing above.
The thermal energy does not limit the speed of the electron but rather the quantization a la Bohr. Specifically you would expect some uncertainty relation $\Delta r \Delta v\approx h/m$. The discrete energy levels for electrons orbiting around magnetic field lines are called Landau levels. Go on with your calculation!

In metals, especially in pure samples at low temperatures, electrons may orbit hundredth of times before they get scattered.
However, this is only possible as long as their wavelength (velocity) does not fulfill the Bragg condition for reflection.
The sign of the Hall coefficient depends on whether these states, which don't behave like free electron states lie at energies where the band is empty or full. In the first case, we can look at the electrons, in the second case at the holes.

12. May 26, 2015

rogerk8

I have never heard of Landau levels before.

Extremely interesting!

I am really guessing now, is it Heisenberg you are quoting?

You make me understand that I obviously was a little to fast stating the kT-equation.

The derivation of the Larmor radius demands a speed to compute.

So I took the speed that I thought I knew and used it.

But as you so nicely have pointed out, this speed does not apply (I now think).

It has been proven that

$$n\lambda=2\pi r$$

which means that the length of the electron orbit has to be an integer number of times the wavelength.

With the use of the de Broglie wavelength

$$\lambda=h/p$$

and

$$\hbar=h/2\pi$$

the above equation may be rewritten as

$$mvr=n\hbar$$

Referring to the basic force relationship where the centrifugal force is equal to the electromagnetic force we may write

$$\frac{mv^2}{r}=\frac{ke^2}{r^2}$$

Solving for v yields

$$v=\sqrt{\frac{ke^2}{mr}}$$

Integrating the electromagnetic force gives the potential energy as

$$E_p=-\frac{ke^2}{r}$$

The kinetic energy may as usual be written

$$E_k=\frac{mv^2}{2}$$

Adding Ep with Ek with the use of the expression for v above then yields

$$E_{tot}=E_p/2=-\frac{ke^2}{2r}$$

Now,

$$mrv=mr\sqrt{\frac{ke^2}{mr}}=\sqrt{ke^2mr}==n\hbar$$

Solving for r yields

$$r=\frac{n^2\hbar^2}{ke^2m}$$

For n=1 this is called the Bohr Radius and for Hydrogen it can be shown that this is some 0,5Å.

Best regards, Roger
PS
I hope this isn't too much :)

I have quoted my "own" Wikipedia article.

The important thing is that, while the radius is descrete, the velocity is strictly not thermal.

I love the Bohr model because it gives me the illusion that I understand things :)

Could you please explain de Broglie?

13. May 27, 2015

rogerk8

Just want to add (from the above equations):

$$r=\frac{n^2\hbar^2}{ke^2m}$$

$$v=\sqrt{\frac{ke^2}{mr}}$$

Which yields

$$v=\frac{ke^2}{n\hbar}$$

where

$$k=\frac{1}{4\pi \epsilon_0}$$

It is interesting to note that the velocity is independent of electron mass and inversely proportional to shell number (n).

Maybe I started out wrongly.

Maybe the Coulomb Force should read

$$F_c=\frac{k(Ae)e}{r^2}$$

Where A is the Atom Number.

But this would mean an atom where all electrons except one are absent.

Is it even possible to ionize a general atom that "hard"?

You now understand how much I understand about physics :D

Best regards, Roger

14. May 30, 2015

Stanley514

If electrons and holes have the same mobility in some material, for example in graphene, doesn't it mean it suppose to have zero thermoelectric effect?