I'm tutoring senior high school students. So far I've explained them the concepts of atomic structure (Bohr's model & Quantum mechanical model) very clearly. Now the next topic to be taught is semiconductors.

I myself am not conviced with the concept of electron holes. If there is no electron then there is no electron. How can it be a hole. We define a hole when there is some thing every where except at a place. But inside an atom how can we define a HOLE.

Kindly explain it with the help of Bohr's model. What was the need of introducing such abstract concept in semi conductors?

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    $\begingroup$ Electrical conduction is like water moving through a hose; when you put water in at the spigot water comes out the other end but it is not the same water. In order for conduction to occur, the electrons have to be able to move along the material. Electron holes are like spaces that the electrons can jump to, or move through. When we talk about electron holes moving, it's like how the available space moves in a game of Chinese checkers. $\endgroup$ – AdamRedwine Oct 12 '11 at 19:41
  • $\begingroup$ @Adam: That's a good explanation. I would only add that metals have "loose" electrons in their valence shells, and whenever an electron wanders away from its "home" atom, the place left behind is a positively charged "hole". So the places where electrons are not can carry charge just as well as the electrons that aren't there. $\endgroup$ – Mike Dunlavey Oct 12 '11 at 21:04
  • $\begingroup$ @Mike, Yes, this is also the case for n-type doped semiconductors as well, but it is less obvious what the "hole" is in that case. I agree with OP that the whole concept is very difficult to grasp, which is why I responded as a comment rather than a full answer. $\endgroup$ – AdamRedwine Oct 12 '11 at 21:17
  • $\begingroup$ @Adam: actually that's even more reason to post what you did as a full answer, IMO, because for a difficult-to-grasp concept we're correspondingly less likely to get additional good answers. $\endgroup$ – David Z Oct 12 '11 at 21:20
  • $\begingroup$ A related question has a great answer physics.stackexchange.com/q/10800 $\endgroup$ – Larry Harson Oct 13 '11 at 12:36

The notion of a particle in nonrelativistic quantum mechanics is very general: anything that can have a wavefunction, a probability amplitude for being at different locations, is a particle. In a metal, electrons and their associated elastic lattice deformation clouds travel as a particle. These effective electron-like negative carriers are electron quasiparticles, and these quasiparticles have a negative charge, which can be seen by measuring the Hall conductivity. Their velocity gives rise to a potential difference transverse to a wire in an external magnetic field which reveals the sign of the carriers.

But in a semiconductor, the objects which carry the charge can be positively charged, which is physically accurate--- a current in such a material will give an opposite sign Hall effect voltage.

To understand this, you must understand that the electron eigenstates in a periodic lattice potential are defined by bands, and these bands have gaps. When you have an insulating material, the band is fully filled, so that there is an energy gap for getting electrons to move. The energy gap generically means that an electron with wavenumber k will have energy:

$$ E= A + B k^2 $$

Where A is the band gap, and B is the (reciprocal of twice the) effective mass. This form is generic, because electrons just above the gap have a minimum energy, and the energy goes up quadratically from a minimum. This quadratic energy dependence is the same as for a free nonrelativistic particle, and so the motion of the quasiparticles is described by the same Schrodinger equation as a free nonrelativistic particle, even though they are complicated tunneling excitations of electrons bound to many atoms.

Now if you dope the material, you add a few extra electrons, which fill up these states. These electrons fill up k up to a certain amount, just like a free electron Fermi-gas and electrons with the maximum energy can be easily made to carry charge, just by jumping to a slightly higher k, and this is again just like a normal electron Fermi gas, except with a different mass, the effective mass. This is a semiconductor with a negative current carrier.

But the energy of the electrons in the previous band has a maximum, so that their energy is generically

$$ E = -Bk^2$$

Since the zero of energy is defined by the location of the band, and as you vary k, the energy goes down. These electrons have a negative nonrelativistic effective mass, and their motion is crazy--- if you apply a force to these electrons, they move in the opposite direction! But this is silly--- these electron states are fully occupied, so the electrons don't move at all in response to an external force, because all the states are filled, they have nowhere to move to.

So in order to get these electrons to move, you need to remove some of them, to allow electrons to fill these gaps. When you do, you produce a sea of holes up to some wavenumber k. The important point is that these holes, unlike the electrons, have a positive mass, and obey the usual Schroedinger equation for fermions. So you get effective positively charged positive effective mass carrier. These are the holes.

The whole situation is caused by the generic shape of the energy as a function of k in the viscinity of a maximum/minimum, as produced by a band-gap.

Bohr model holes

You can see a kind of electron hole already in the Bohr model when you consider Moseley's law, but these holes are not the physical holes of a semiconductor. If you knock out an electron from a K-shell of an atom, the object you have has a missing electron in the 1s state. This missing electron continues to orbit the nucleus, and it is pretty stable, in that the decay takes several orbits to happen.

The many-electron system with one missing electron can be thought of as a single-particle hole orbiting the nucleus. This single particle hole has a positive charge, so it is repelled by the nucleus, but it has a negative mass, because we are not near a band-gap, it's energy as a function of k is the negative of a free electron's energy.

This negative-mass hole can be thought of as orbiting the nucleus, held in place by its repulsion to the nucleus (remember that the negative mass means that the force is in the opposite direction as the acceleration). This crazy system decays as the hole moves down in energy by moving out from the nucleus to higher Bohr orbits.

This type of hole-description does not appear in the literature for Moseley's law, but it is a very simple approximation which is useful, because it gives a single particle model for the effect. The approximation is obviously wrong for small atoms, but it should be exact in the limit of large atoms. There are unexplained regularities in Moseley's law that might be explained by the single-hole picture, although again, this "hole" is a negative mass hole, unlike the holes in a positive doped semiconductor.

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The convention is that current flows in an electrical circuit from positive to negative. This was decided before electrons were discovered and before they were discovered to be negatively charged.

You can identically consider the flow of electrical charge as either the movement of a negative electron from left to right, or the movement of the empty place an electron would be from right to left. If you think in terms of the movements of the gaps (holes) then you have a positive current flowing which matches the normal definition of electrical current.

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  • $\begingroup$ This is not quite right, because it is a question of what are the actual physical carriers--- there is a difference between a P type and N type semiconductor which is not just a definition. $\endgroup$ – Ron Maimon Oct 13 '11 at 5:45
  • $\begingroup$ -1: Sorry, this is wrong, the mass of the carriers is not just a definition--- it's the mass of the carriers. $\endgroup$ – Ron Maimon Jul 19 '12 at 16:58

It is impossible to answer in terms of Bohr model. The important part of the concept which is out of the scope of the school approach is the degenerate electron gas. Below Fermi level all states at zero temperature are filled. If you take an electron out of this filled place you get a hole. Which behaves exactly as an "anti-electron". The key is to realize that a lot of electrons in the degenerate gas below Fermi level behave as there is nothing there.

If you really intend to give this idea at the school-level, you could try to bring an analogy with bubbles in the water. They behave exactly (well, more or less actually) as particles with negative mass. If you want to study the situation when you have a lot of water and relatively small amount of bubbles, it is constructive not to solve equations for the whole mass of water, but replace them with the equations for the bubbles. It is more or less why people use hole language.

And, technically, these holes are not "in atoms". Valence electrons which conduct in semiconductors and metals are collective. Thus, "missing valence electron" also does not belong to any particular atom.

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  • $\begingroup$ This is true, but you can make a hole description of Moseley's law, so long as you remember that the holes have negative mass. The reason is that the region near the nucleus is a very deep potential well, and it looks the same as a filled Fermi gas near there. The orbits of the hole don't leave this region for 1 or 2 shell holes in heavy atoms. $\endgroup$ – Ron Maimon Oct 13 '11 at 6:56
  • $\begingroup$ @Ron Maimon, with all respect you really have no idea what you are talking about. Valence electrons are different from core electrons and do not see the potential near nucleus. Hole may be more or less localized, but first of it all, by definition it originates from the delocalized states. $\endgroup$ – Misha Oct 13 '11 at 7:40
  • $\begingroup$ @Misha, the ability to re-write you theory in terms of holes arises whenever there is a gap that that limits the electron energiy spectrum from above. The you cahnge sigs and "top" becomes the "bottom". Localization does not matter. $\endgroup$ – Slaviks Oct 13 '11 at 10:37
  • $\begingroup$ @Slaviks, localization is irrelevant. However, we are talking about semiconductors. Where holes are charge carriers which move freely and formed by valence electrons. For holes which are localized in atoms the story would be a bit different: they would not "move freely", had no momentum, etc. $\endgroup$ – Misha Oct 13 '11 at 10:57
  • $\begingroup$ @Misha, ok, but... an acceptor-doped semiconductor is still better described in terms of holes, even if they become localized at low T because of the Anderson-Mott metal-to-insulator transition. Localized holes in this regime would be just ionized donors $\endgroup$ – Slaviks Oct 13 '11 at 11:15

What was the need of introducing such abstract concept in semi conductors?

Electrical physics and electrical engineering have been around much longer than knowledge of the electron has. For most things in electrical engineering, the circuits they design work exactly the same whether the charge-carrier is a negative-particle moving in one direction, or a positive-particle moving the other way!

Since they didn't know whether the charge-carrier was positive or negative at the time, and it didn't really make a difference to them, electrical engineers chose to make all their electrical diagrams and symbols under the assumption that the charge-carrier was positive. Of course, it turns out they were wrong.

However, by the time they discovered this, their notation had been pretty much set in stone - circuit diagrams even today are drawn as though the charge-carrier is positive, and even some elementary- and middle-school students are taught that electricity flows "from the positive terminal to the negative terminal" in a battery.

If electrons are negatively-charged, what is this "positively charged particle" that flows in a circuit? It is not a real particle - there is no positively charged particle flowing in the circuit (the protons stay relatively still). Rather, the "positive charge" is really a lack of negative-charge - an empty space with higher electric-potential, which the electrons are necessarily attracted towards.

(Electrons moving to the right necessarily cause holes to flow to the left)
(Electrons moving to the right necessarily cause holes to flow to the left)

Electron holes are not a real "thing" - they are just a concept used to help explain why it doesn't matter (to electrical engineers) whether it is a positive- or negative-charge flowing, and to help them cope with the fact that all their diagrams use the "wrong" charge-carrier :)

See this blog post for more info on this (and other good electricity-related questions).

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    $\begingroup$ This is incorrect. Electron holes are a "real thing" in that they give a positive carrier in a positively doped semiconductor. The Hall voltage is opposite. It is not a matter of convention whether the charge carrier is negative of positive--- there are real physical effects when the carriers switch signs. $\endgroup$ – Ron Maimon Oct 14 '11 at 21:42
  • $\begingroup$ -1: Gotta downvote, this is wrong. $\endgroup$ – Ron Maimon Jul 19 '12 at 16:58

Yes, they are positive.The hole is the absence of electron(hole is not a particle but a vacancy) According to quantum mechanics anything that can have a wave function, a probability amplitude for being at different locations is a particle, so we assume that hole is a particle because it satisfies the definition of a particle according to Quantum Mechanics. When a hole is created in a crystal the neighbouring electrons can move into that vacancy and fill it up.Similar to the analogy of an air bubble in water, the bubble doesn't move, the water surrounding it moves.So the hole current is nothing but the flow of electrons(but a lot slower)...the term hole is created to simplify some of the various concepts of semiconductor physics.It is an assumption to make things simpler.

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When the elctron is free in outermost orbit that empty place is called is holes..

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  • $\begingroup$ Technically true, but this really doesn't say anything useful... $\endgroup$ – David Z Jul 19 '12 at 4:39

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