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Throughout my highschool classes, I have been made to learn that metals have free electrons that's why they are able to conduct electricity.. But I never understood why. Is that related to metallic bonding... Correct me if I am wrong but even if that's the case.... I am just not able to understand the concept of free electrons

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    $\begingroup$ Isn't it mobile electrons (electrons bound to the metal as a whole) rather than free (unbound)? $\endgroup$ Aug 25, 2020 at 13:59
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    $\begingroup$ This question is one of the great mysteries of physics. Why do electrons in metals behave as if they are free when they are swimming a matrix of positive charges? Others have pointed out that the electrons are mobile not free. True, but many of the properties of metals can be explained by a model in which the electrons are really truly free. It's called the free electron model. Others have pointed out that in metals the outermost electrons are loosely bound. That's part of it, but the reason the free electron model is so good can only be explained by quantum mechanics.(+1 to @Lost) $\endgroup$
    – garyp
    Aug 25, 2020 at 14:30
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    $\begingroup$ I don't think this is even remotely answered until you get to electron orbitals in university. $\endgroup$
    – DKNguyen
    Aug 25, 2020 at 15:25
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    $\begingroup$ @garyp give indistinguishability, degeneracies and a whole lot of electrons crammed in a “small” space to quantum mechanics. Out comes conductivity. I think the effectiveness of free electron models comes from the periodic nature of crystals. Effectively a period boundary condition mimicking an infinite lattice. $\endgroup$ Aug 25, 2020 at 15:28
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    $\begingroup$ There's a "by definition" answer possible: Elements which have free electrons in their usual solid state are called "metals". This is not an unreasonable definition, but there are others. The other definitions typically focus on the properties of metals, which are a result of those free electrons. $\endgroup$
    – MSalters
    Aug 27, 2020 at 10:52

6 Answers 6

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Without getting into the quantum mechanical details, here’s a cartoon depiction of what’s going on. The vertical axis represents energy.

enter image description here

Like other answers have already pointed out, metals don’t have actual free electrons. In the cartoon this is given by the grey region. If electrons have enough energy to be in the grey region, they’re free.

In individual independent atoms (gaseous state), the energy levels below a certain energy are discrete. This is depicted by the lines in the cartoon. This means the energy is fixed, rigid. The electrons in this state can’t conduct electricity.

In solids however, the discrete states of multiple neighbouring atoms “merge” into a continuum and create what is called as bands. For further details you may look at my answers here.

With this, there exist a continuum of states called the conduction band where the electrons are not bound to any single atom of the solid. They are mobile. The fascinating property of these states is that it is possible for electrons to respond to an external electric field. These states are called Bloch waves.

In insulators there is a big energy gap between the filled states (valence) and the empty states (conduction). So without sufficient external field, they are unable to conduct electricity.

In metals however, the energy gap is absent and thus electrons can easily go into the conduction band and respond to external electric field.


Some details

The reason why mobile electrons seem like free electrons has to do with crystal symmetries. Specifically translational symmetry. In a crystal the atoms are arranged in a regular periodic manner. In the bulk (non boundary) of the metal if you go from one atom to another, the neighbourhood looks identical. This is known as translational symmetry. And a consequence of this is that the electrons have well defined momentum, just as a free electron does. This is encapsulated in the band structure.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – tpg2114
    Aug 28, 2020 at 10:50
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    $\begingroup$ This is a phenomenal explanation — could you perchance add a little bit about semiconductors too? $\endgroup$ Aug 28, 2020 at 13:04
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In high school classes , free electrons is used interchangeably with mobile electrons even though they are not exactly same. Free electron is an electron is essentially out of the electron -positive site bound system. While mobile electrons are those which are loosely bound to the positive site so even though they are not free from the positive site's influence they can essentially hover above other positive sites among the sea of electrons in such a way that as a whole the conductor has no charge.

For example consider the electron of a Bohr hydrogen atom. If its given energy exactly equal to its ground state energy it becomes free in the sense that it is not bound to the nucleus any longer , if given any more energy whole of it will appear as KE.

On the other hand conductors have many orbits and the electrons in the last orbits are loosely bound to the nucleus due to screening and also because they are far from the nucleus. These are mobile electrons. But your book is calling them free electrons. Since these are loosely bound they can easily constitute a current.

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    $\begingroup$ You, of course, are right to point out the ambiguity in the term “free electron”. I like “mobile electron” as a more rigorous alternative, although I’ll admit that I have written “free electron” to refer to the mobile charges in a metal in reputable physics journals. So, it is a common usage in the professional community, not merely a euphemism in high-school texts. $\endgroup$
    – Gilbert
    Aug 25, 2020 at 20:36
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I am afraid that the answer is not high-school level at all, let me try to explain anyway.

First of all, we talk about free electrons when they can move freely in space without external forces acting on them. Also we assume that they do not interact with one another (this is really tricky because they are charged particles, but I won't discuss this assumption). Clearly in a metal there is a crystal made of ions that acts with external forces on the electrons. However it turns out that the electrons with higher energy act as if they were free electrons, and since only the electrons with higher energy are responsible for conduction, we can treat metals as free electrons systems.

Technical details

Free electrons have energy levels described by the dispersion relation $\varepsilon_k = \hbar^2k^2/2m$, where $m$ is the mass and $k$ is the momentum. At zero temperature the particles occupy all the available electronic states from $k=0$ up to $k=k_F$, known as Fermi momentum, corresponding to an energy $\varepsilon_F=\hbar^2k_F^2/2m$ and when $k\sim k_F$ the dispersion relation is approximately linear in momentum $\varepsilon_k \sim \varepsilon_F + \hbar^2k_F(k-k_F)/m$.

When electrons are bounded by a periodic potential as in metals, the dispersion relation changes completely, in particular one can prove that for a one dimensional lattice with step $a$ the dispersion relation reads $\varepsilon_k = \mp 2t \cos{(ka)}$, where $k$ lies in the first (reduced) Brillouin zone $k \in ]-\pi/2a,+\pi/2a]$ and $t$ is called hopping and it's given by $t = \hbar^2 \eta /2ma^2$, $\eta$ being a dimensionless parameter which depends on the specific ions. You can plot the dispersion relation as an exercise. At zero temperature the electrons occupy all the energy states described by $-2t \cos{(ka)}$ and thus the Fermi momentum is $k_F = \pi/2a$ and the Fermi energy $\varepsilon_F=0$. Now you can expand the dispersion relation close to the Fermi momentum and you get $\varepsilon_k \sim 2t (k - k_F)a = \hbar^2 \eta (k-k_F)/m a$. You can easily rewrite the latter formula as $\varepsilon_k = \varepsilon_F + \hbar^2 k_F(k-k_F)/m^*$, where $m^* = (\pi/2\eta)m$.

As you can see, close to the Fermi level the dispersion relation of bounded electrons is formally equivalent to that of Free electrons with an effective mass $m^*$. Now since only electrons close to the Fermi level are affected by excitations, scattering process etc. due to Pauli exclusion principle, you can see that the "important" electrons can be treated as if they were free electrons with some effective mass. This is precisely the reason why the free electron model describes quite well the properties of metals. Finally notice that insulators and semiconductors have completely different dispersion relations, so this similarity is no longer true.

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  • $\begingroup$ is this the same k as in crystallography, the wave vector? $\endgroup$ Aug 26, 2020 at 14:33
  • $\begingroup$ Yes, to be very precise it is often called quasi-momentum and it represents the quantum number associated to the translational symmetry of the system $\endgroup$
    – Matteo
    Aug 26, 2020 at 17:08
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Without getting too far away from the high school level: the electrons in metals are not really free. But they are free to move in the sense that there are energy states where they can jump, if an electric field is applied.

If you are comfortable with the concept of a hole as a unfilled chemical bond between ions (as it is explained in high school physics) than a metal can be thought of as a semiconductor with lots of holes, so that thinking of them as particles doesn't make sense.

Update
To expand the second paragraph above:

  • Let us first think of a covalent bond in a hydrogen atom: two electrons (one from each atom) are shared by the two atoms.
  • Carbon can form four such bonds, which is why it can form complex chained and branched molecules. In a diamond each carbon atom is bound to four other atoms - we say that its bonds are saturated in the sense that all the electrons a participating in bonds in order to hold the crystal together, i.e. none of the electrons a "free".
  • One way to think of a metal is as of a material were some of the electrons are not participating in bonding.
  • Let us consider semiconductors, such as silicon or Germanium - they are in the same group as carbon in the periodic table and form diamond-like crystals with saturated bonds. If one bond is broken, an electron (or even two) is liberated and start wondering around crystal - this is a "free" electron. We also have left an empty space - a hole. An electron from another bond can jump into the empty space, which can be though of as the hole moving.
  • If we replace some of the Si/Ge atoms by atoms with more or less than four valence electrons, we will have a material with inherently more electrons or more holes than a pristine semiconductor. We refer to it as n-type or p-type semiconductors, depending on whether electrons or wholes are in excess.
  • A metal than can be thought of as a n-type semiconductor with lots of excess electrons. (One rarely speaks of excess holes, since the concept of a hole doesn't really make sense in this case).
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  • $\begingroup$ I went to high school a long time ago, but I'm pretty sure I didn't get taught "the concept of a hole as a unfilled chemical bond between ions". I don't think semi-conductors were explained as anything other than "conductors that can only conduct electricity in one direction" $\endgroup$
    – Flydog57
    Aug 26, 2020 at 0:15
  • $\begingroup$ @Flydog57 I'm surprised you were taught anything about semiconductors at ll in high school $\endgroup$
    – DKNguyen
    Aug 26, 2020 at 1:26
  • $\begingroup$ Ok, it may have been "junior college" (in Quebec, Canada you go through a two year college program before university) $\endgroup$
    – Flydog57
    Aug 26, 2020 at 3:33
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    $\begingroup$ High school carricula may differ, especially in different countries. However, logically I would expect pn-junction (conducting electricity in one direction) come after holes (p-type semiconductors). Anyhow, I tried to give an explanation without discussing the band structure, which is certainly beyond high school level. $\endgroup$ Aug 26, 2020 at 4:55
  • $\begingroup$ @Flydog57 I expanded the "high-school explanation" in terms of holes in my answer. $\endgroup$ Aug 26, 2020 at 7:24
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Let us put it like this: the lattice structure typical of a piece of metal causes the outer orbital electrons of the metal atoms to behave as if they were owned by all the atoms of the metallic lattice structure in common, rather than by each individual atom. The consequence is the "loosening" and de-individualization of each electron from its place in what would have been its original electron cloud.

I think this is the simplest answer which can be given.

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  • $\begingroup$ This lattice structure (aka crystal) is not proper to metals. And furthermore, is not even required to have band structures, a Fermi surface, and more importantly, free electrons. $\endgroup$ Aug 28, 2020 at 16:30
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You are asking for an explanation, and as we can see from the other answers, there are no understandable explanations.

Here is a description at a simple level.

Take a flat piece of glass and a flat piece of gold. Touch them together, and then pull them apart. The glass will have a positive charge, because the gold atoms hold onto electrons better than some kind of atoms in the glass. That's true for any two materials, though some are almost equal in their ability to hold on.

Ranking materials

Suppose you used polyurethane instead of gold. It is even better at grabbing electrons. But there is a difference. The electrons the polyurethane grabs are stuck where they are. Other atoms in the polyurethane have a hard time taking them from the ones that first got them from the glass. It takes a long time for the electrons to leak away. But the gold passes electrons from one gold atom to another very readily.

Different materials do that at different speeds, and resist electron movement to different degrees. Manganese has around 80 times the resistance of silver. Amorphous sulfur resists $10^{23}$ times as much.

Resistance of elements

It's more complicated than just elements -- carbon has different resistance in diamond form than as graphene, etc.

And alloys or chemical compounds resist in ways that can be unintuitive.

What causes the difference? I couldn't begin to say. Superfast Jellyfish provided a description with a picture. You can see the bands overlapping. That is another description of what happens. It doesn't say anything about what makes the bands overlap.

It's possible that someone with a godlike understanding of quantum mechanics could work out the equations for each element and chemical compound from first principles, and understand how the QM gives rise to those bands, and understand what parts of the math result in conductivity.

But I strongly suspect that such a person could not explain it to mortals. And it's hard to understand even what "understanding" means at that level.

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