8
\$\begingroup\$

An electric current is a flow of free electrons. Are these free electrons totally free from the orbits of metal atom or they moving by jumping from one orbit to another orbit of the atoms?

If they are totally free, what does enforce them to keep staying in (or on the sufrace of) the metal

Thanks

\$\endgroup\$
2

4 Answers 4

8
\$\begingroup\$

I'm pretty thankful for Jack's answer – because it explains that you might not want to stick to a model with "separate atoms" and "bouncing" electrons for a metal. So here goes what I'd like you to get the idea of regarding electron movement in a metal:

The moment you realize that these electrons aren't free to move anywhere, you must admit that the word "free electron" isn't 100% accurate.

So far, so good. Hold on, this will hurt just a bit.

The orbits you know are just a model. They don't exist as things with a shape where a "point-shaped" electron circles around. The moment you need to describe electron movement in a metal, that model breaks down, as you've noticed.

Instead, we have to understand that an electron bound to a nucleus only is bound because "fleeing" would require an external impulse, as well as "crashing" into the nucleus. For now, imagine the electron in circular motion (just like a satellite around a planet), and if no external force is applied, it'll stay at that path.

Now, take a step back. You might have heard of Heisenberg's Uncertainty principle – you can't know the exact location of something and its exact impulse at the same time. That's exactly what's happening here – we know the rotational impulse of the electron pretty exactly (because we can calculate how much impulse it needs not to crash nor to flee), and thus, the knowledge of its position must be uncertain to a specific degree.

Hence, an electron like that doesn't actually have a place on the orbit – it has a place probability distribution. It turns out that the probability is an effect (or, rather, an operator applied to) Schrödinger's Equation (for a non-near-speed-of-light single particle), which is

$$i \hbar\frac{\partial}{\partial t} \Psi(\mathbf{r},t) = \left [ \frac{-\hbar^2}{2\mu}\nabla^2 + V(\mathbf{r},t)\right ] \Psi(\mathbf{r},t)$$

(I swear, I'm not trying to scare you – the formula will look far less threatening when you've studied electrical engineering for one and a half years – you'd typically have a course called "solid-state physics / electronics", where this is explained in much more depth and with background, and a lot of mandatory math courses that explain how to deal with this kind of equation, especially with the differential Laplacian operator \$\nabla^2\$. I just need the formula below.)

So, now back from the single electron to the metal:

A metal is composed of an electron lattice – that is, the atoms are arranged in a repetitive pattern. Now, looking at Schrödinger's equation, you'll see a \$V\$ there – that's Potential, and potential is practically "distance to positive charges" for an electron – and since we know the positive charges are in a nice periodic pattern in the metal, \$V\$ is periodic!

Now, what's this \$\Psi\$? It's what we call the position-space wave function. It's the solution for Schrödinger's Equation – the function that makes the "\$=\$" above true!

Now, for a specific, periodic \$V\$, only a specific set of wave functions can exist; we can apply a different operator to the wave function \$\Psi\$ (the Hamiltonian) and get these states; they are the so-called Bloch states. Within these, an electron actually doesn't have a specific "identity" or "place" – it just contributes to the fact that things are periodic.

That's what you mean when you talk of "conduction bands" in metals – states that electrons are a) able to exist and b) are free to move around in.

Now, if you apply an electrical field, which is what you do to, macroscopically, make charges (electrons) flow, you change \$V\$; it's now a sum of a periodic function and a linear function. That leads to a change in the solution for \$\Psi\$ – and macroscopically, this means that electrons move to one end.

\$\endgroup\$
5
\$\begingroup\$

First, electric current is a flow of charges. Often those charges are electrons, but don't have to be.

Second, think of the conduction band electrons in a metal, for example, as being somewhat loose. They can hop from atom to atom relatively easily. However, they can't just all fall off or something because of electric charge. If a bunch of electrons clumped together away from the atoms they came from, there would be a negative charge at the clump and positive charge where the atoms with the missing electrons are. This charge would pull the electrons back.

There is some random motion of electrons, but they don't ever get too far out of balance else a electric field will bring them back. When we apply a external electric field, like connecting the ends of a wire to a battery, then the electrons will move. That's what we call "current".

\$\endgroup\$
3
  • 3
    \$\begingroup\$ I love the diversity of the answers here – your's is pretty practical with respect to "you probably have heard of conduction bands, now imagine them a bit more elastic", Jack's answer adresses the "an atom has a core and orbits" model aspect, and I try to convince OP to look behind things at a Schrödinger level :) \$\endgroup\$ Commented Dec 6, 2016 at 12:20
  • \$\begingroup\$ @Marcus: Yes, I'm trying to give a high level conceptual answer, certainly not a rigorous physics answer. In my opinion, that matches the level of the question better. \$\endgroup\$ Commented Dec 6, 2016 at 12:38
  • \$\begingroup\$ certainly not arguing with that! \$\endgroup\$ Commented Dec 6, 2016 at 12:57
3
\$\begingroup\$

It's complicated

If you look at the history of physics, you quickly see that prior to the discovery of Quantum Mechanics, the theory of conduction in solids had some pretty big holes. The truth is that a proper understanding of electrons in metals requires a good understanding of quantum mechanics. On the plus side, there are some simpler models which yield a reasonable approximation of the behaviour of the electrons, even if they don't really represent the actual behaviour.

The Fermi Gas Model

This is the simplest model of a metal which gives a reasonable approximation of the behaviour, but it's not easy to understand unless you already have a background in QM - the kind you only usually get from the first two years of a physics degree. Because of its complexity, I'm not going to try to explain it here, I'm just going to note that it exists, then move on. There's another model called the "Fermi Liquid", which is even slightly better, but also even more complex.

The Drude model

This is an older model, which predates Quantum Mechanics. It works fairly well, in terms of the predictions it makes, but it's not really representative of what's actually going on inside the material. It has these main features:

  • There is an energy barrier which prevents electrons passing the surface of the metal. This is known as the "work function" but without getting into quantum mechanics, it's hard to see why it exists. One approach would be to say that we have taken the outer shells of the atoms and smeared them into one big energy band, which is still lower energy than a truly free electron would have.
  • The atomic cores, with most of their electrons in bound states, are scattered through the material. The combination of atomic core + most of the electrons is called an ion.
  • The electrons from the outermost shell of the atom (and occasionally the next shell in too) are separated from the atom and flow through the lattice much like the metal balls in a pinball machine.
  • The electric field accelerates the electrons, and the electrons decelerate when they hit and bounce of an atom. They settle into some equilibrium velocity which depends on the electric field and the number and size of ions to scatter off.

All in all, it's not a bad model, and you can use it to make predictions if you don't want to get stuck into QM.

The model of electrons jumping from atom to atom is not a good one for metals, it leads to several wrong predictions, such as conductivity going up with temperature. It is a decent model for leakage current in some near-insulators, just not for metals.

\$\endgroup\$
7
  • \$\begingroup\$ Nice answer, could have used it to base my answer on, but of course I wasn't in your head at that time :) \$\endgroup\$ Commented Dec 6, 2016 at 12:12
  • \$\begingroup\$ hint, you don't need 2 years of physics – a little more than a year of EE does it, too :) \$\endgroup\$ Commented Dec 6, 2016 at 12:13
  • \$\begingroup\$ Probably because we spend the first year in physics working on Newtonian mechanics, and wired stuff like Gyroscopes, which I have never needed to know about since... :-) \$\endgroup\$
    – Jack B
    Commented Dec 6, 2016 at 12:18
  • \$\begingroup\$ :) don't worry, we do that kind of stuff, too, and off-topic kind of stuff, also, but physics students do get a much greater understanding of the mechanics behind. Also, I get the impression that they are generally forced to hand in more worksheets in order to train them to apply differential operators etc in their sleep – which highly likely comes in handy later on \$\endgroup\$ Commented Dec 6, 2016 at 12:20
  • \$\begingroup\$ Ah, and if you're in fact some third year physics student (and as we EEs shared the same math course as them): you need all of the math they teach, don't despair! Aside from a few solutions to types of differential equations I had to learn by heart, all the first three years of math were necessary for my degree, so rejoyce in the fact that you're not boring yourself in vain :) \$\endgroup\$ Commented Dec 6, 2016 at 12:22
0
\$\begingroup\$

From the tenor of the questions, it may be practical to give you a very simple model that does not involve QM and yet it helps you understand the net result.
First, you need to understand that the electrons in a molecule are not free. Even though they "move" around their respective nucleus, they are "trapped" by it.

In a solid (such as a metal), the molecules reach a "steady state" such that it is the equivalent of all molecules being frozen solid. Therefore, when you connect a battery to a piece of metal, one electron is removed by the positive battery terminal from the molecule "next" to it. This causes the molecule to become positive, and with help of the electric field, "steals" an electron from a neighboring molecule.
This repeats until the negative battery terminal is reached and it provides the missing electron to the molecule.

The net effect is that, since for each electron that goes in another goes out, this gives the appearance that electrons are free flowing.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.