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Can someone in a simple way explain me what the Fermi level is and what does it have to do with conductivity. My teacher said that Cu conducts electric current better than Al because of something in relation to Fermi level, I didn't get him very well, so please explain me what it is all about, but in a simple way.

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  • $\begingroup$ There is an ambiguity. When people talk about "high Fermi level" they might just be referring to a low applied voltage. In this case, however, your teacher is probably referring to band-referenced Fermi level (called ζ in the wikipedia article) which is an intrinsic property of materials. $\endgroup$ – Nanite Mar 12 '14 at 7:58
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Fermi energy

If you operate at zero temperature, $T = 0$ K and fill the energy-states of a system according to the Pauli-exclusion-principle, the Fermi energy is the boundary at which all lower states are full and all higher states are empty. At $T = 0$ this boundary is a sharp line.

For example, say you have ladder with five steps which you have to “fill” with ten electrons. Due to the Pauli-exclusion principle, each step can only take two electrons. Now you fill up the ladder: 2 electrons in the first step, the next two in the second step and so forth until you put the last two electrons in the 5th step. The energy at this step (the 5th step) is your Fermi energy.

Metals have Fermi energies of several electron-volts (eVs). (Cu: 7 eV, Al: 11 eV) For comparison, the thermal energy at room temperature is about $k_B T \sim $ 0.025 eV.

Connection to the conductivity

First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process. Why? I mentioned earlier that at $T = 0$, the Fermi energy is a sharp line. At $T > 0$, this sharp line gets “washed out” and you get something like this:

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This means that instead of only full and empty states, you now have half empty states above and below the Fermi energy. This is turn means that you are now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work on them.

But in consequence of the small contributions of the thermal and electric energies (thermal $ \sim $ 0.025 eV, electric less than that), you are only able to excite electrons VERY CLOSE to the Fermi energy ($E_F$). So only electrons close to $E_F$ will contribute to the conduction.

With this $E_F$ you can associate a velocity, the Fermi velocity: $$ v_F = \sqrt{2 E_F / m} $$

Now let's turn to the conductivity:

The conductivity $\sigma$ is defined as $$ \sigma = \frac{n e^2 \tau}{m} $$ where $n$ is the number of electrons, $e$ is the electron charge, $\tau$ is the time between two collisions and $m$ is the mass of an electron.

One can obtain $\tau$ from $\mathscr{l}$, the mean free path between two collisions, given as $\mathscr{l} = v_F \tau$ or inversely, $\tau = \mathscr{l}/v_F$. During this time, the electron is accelerated. A LARGE $v_F$ will therefore result in a SHORT acceleration time and LESS gain of speed for the electron in a given direction.

Finally you can write the conductivity as: $$ \sigma = \frac{n e^2 \mathscr{l}}{m v_F} $$

Remember that $v_F$, the Fermi velocity, is directly related to the Fermi Energy $E_F$.

As an instance, consider this table of Fermi energies: Copper has a LOWER Fermi energy (7 eV) than aluminum (11 eV), so it has a LOWER Fermi velocity; hence, the time between two collisions ($\tau$) is LONGER than that in aluminum. This in turn means that the electron has MORE time to accelerate in a given direction which finally explains why copper is a better conductor.

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  • $\begingroup$ Hi NoEigenvalue, welcome to Physics.SE! Please consider using $\TeX$ in the future. See meta.physics.stackexchange.com/questions/804/… $\endgroup$ – Brandon Enright Mar 12 '14 at 0:35
  • $\begingroup$ In your answer $m$ is not the mass of the electron but the effective mass of the charge carrier. This is a reasonable quantity to be considered if you are near a band edge because then the dispersion relation is approximately quadratic: $E(k) \approx \frac{\hbar^2 k^2}{2m}$. This situation applies if you consider conductivity in semiconductors. In a metal like Cu or Al bands cross the Fermi energy so that an approximation for the band edge does not necessarily apply. In general you have to consider the group velocity $v_g = \frac{dE}{\hbar dk}$ at $E_F$ for bands crossing the Fermi level. $\endgroup$ – Gregor Michalicek Jun 23 '18 at 9:19
  • $\begingroup$ The quantities $\tau$ and $l$ you use in your answer are material dependent. They depend on aspects like imperfections in the crystal and so on. Also n is not the number of electrons but the number of electrons contributing to the conduction. These are only electrons near the Fermi level and can be obtained from the density of states near the Fermi level. In conclusion the final equation you show is not adequate to explain the different conductivities in Cu and Al. With $n$, $l$, $m$, and $v_F$ it contains 4 quantities that depend on the materials in nontrivial ways. $\endgroup$ – Gregor Michalicek Jun 23 '18 at 10:05
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I suppose it depends upon how you define "simple," but in my opinion the best answer is no; explaining thousands of pages from books about QM, solid state physics, electronics, rheology, etc. "in a simple way" as the question requires implies these works could be reduced to a few pages, which would seem unlikely at face value.

Gregor Michalicek has argued (see comments below) that one should overlook this issue, and answer a different version of the question in which the simple clause was removed. I disagree. I think there are many examples in history of science when bad science was done because people glossed over issues or ignored implied or hidden assumptions in their research.

At any rate, my answer of "no" being the best answer is probably impossible to prove. But, to attempt to make my case, what I am going to do is see if the other answer here is both correct and simple. If I am right, it will either be wrong or not simple.

As a preliminary remark, the other answer here by NoEigenvalue currently has a green checkbox, 5 upvotes, an edit by a reviewer, and a favorable comment. So, at first glance, one would think my answer is wrong, since if I am right it suggests that not just one person here but the community as well is wrong also.

Nevertheless, let's see if there are any problems with NoEigenvalue's answer or if it is not as simple as it seems.

The answer starts out attempting to explain what the Fermi energy is using the Bohr model of an atom. However, it then skips to saying what Fermi energies are for some metals. It does this by referencing external tables. It entirely skips the derivation based on free electron densities, and makes no attempt to connect them. I will show that this poses a problem later, because an unstated dependence affects NoEigenvalue's argument.

Then NoEigenvalue's answer claims, "First you must know that only electrons with an energy close to the Fermi energy can participate to the conduction process...At $T>0$...you are [only] now able to excite electrons in higher energy states, which you need to do if you want to accelerate them in a direction; i.e., if you want to do work [i.e, produce current] on them." (emphasis supplied) This argument is not explained. It is actually backwards as well. I debunked this misconception here. It is obviously false when you consider either of these metals at low temperatures, where according to the argument there would be no electrons to carry current, and conduction would be zero. In reality this is where conduction is maximized and resistivity (the inverse) approaches a minimum:

Pure copper actually has almost no resistance at $T=0K$

Next the author claims that $\sigma$ is defined by $ne^2\tau/m$ where $n$ is the number of electrons [per unit volume]... Wrong. The definition of conductivity is the ratio of current to field, or the ease of current flow when pushed. The cited expression largely just gives an illusion of confidence, introducing a free parameter, $\tau$, with one new equation, which doesn't necessarily give any new physical information. It thus really defines $\tau(\sigma)$, which can be argued to be a momentum exchange time scale of some electron cloud with an ionic lattice.

Next we see $\sigma=ne^2l/mv_F$ and an argument about how if $v_F$ is lower in copper than in aluminum, then the current will be higher. Not stated is that this relies upon $n$ and $l$ being the same in both metals, something that is unlikely. Indeed, this was my point made above, that a derivation of $E_F(n)$ was missing. It seems rather convenient given that it changes the later argument; aluminum is triply ionic instead of singly ionic, so $n$ gets a factor of three from that, and it ends up having a free electron density that is a factor of $21.1\times 10^{22} {\rm cm}^{-3}/8.49\times 10^{22} {\rm cm}^{-3}\simeq 2.5$ that of copper. This relatively large factor overtakes the energy/velocity one created by NoEigenvalue in an apparent attempt to make the final argument to answer a question in a simple way because that factor of a mere $11/7\simeq 1.6$ is further softened to 1.3 due to the square root function in converting from energies to velocities. So, at face value, if you include the variance of $n$ in NoEigenvalue's argument it actually would say aluminum has a greater conductivity. This clearly is a problem with the answer given that the question is why does copper have a greater conductivity.

To sum up, I think I've shown that the allegedly simple answer is both both too simple and also incorrect in many aspects. This is not a proof that the best answer is no, but it is probably a strong case that "no" is a reasonable answer if not the best answer.

At the end of the day, I think nature isn’t always so trivial as the fictional version people sometimes come up with due to their own arrogance in wanting to think they personally know what's going on and, apparently, the version of reality you seek via your question. Humility is an important partner of curiosity and good research, and one should not leave it behind in a selfish desire to trick oneself into moving forward.

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  • $\begingroup$ Maybe as an explanation for the observed temperature dependence of the resistance you could point out that in such metals at room temperature the major part of the resistance is given by electron-phonon scattering. This part of the resistance is proportional to the temperature. At very low temperatures other scattering processes become more relevant. In the answer of @NoEigenvalue this translates into a temperature dependence of $\tau$ and $l$. In semiconductors you observe a different temperature dependence and growing inherent conductivity with temperature. $\endgroup$ – Gregor Michalicek Jun 23 '18 at 14:20
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    $\begingroup$ Gregor, thanks for your comment. I think you are furthering my point. But, before going there let me remind you that my answer here was to address the literal question. I focused on the verb, "can" in "Can someone explain in a simple way..." My answer was no. You, perhaps think my answer was, "Sure, but first let's first note that X was wrong." No. My answer was no. That phonons are not even discussed by NoEigenvalue (when they should be) just furthers my argument that "no" is probably a good answer to the actual question that was asked. $\endgroup$ – Jason Arthur Taylor Jun 23 '18 at 14:33
  • $\begingroup$ To be honest, I think people come to physics stack exchange because they are curious about physics and not because they want to hear people's opinions about other peoples (didactic) abilities and physics knowledge. If you answer the literal question here you are misinterpreting the questioner's intention. Also, if you focus on problem's in other people's answers your points should be stated as a comment to that answer, not as an own answer. Besides that: Pointing out problems in a single answer does not imply that there cannot be an answer that fulfills the questioner's demands. $\endgroup$ – Gregor Michalicek Jun 23 '18 at 16:58
  • $\begingroup$ Gregor, I agree to a limit. The danger in answering intended questions instead of actual questions is it can mask important issues due to errors in the implied assumptions of the questions. Here, it is suggesting to the questioner that the first simple answer was OK, or, if the content of my answer is injected, just needs some minor fixing. No. In this case, one needs to hit the brakes and address that this isn't a simple set of questions. Let's not reward people for oversimplifying. Simplification is good, but oversimplification is wrong. Here, it leads to misguided views of conduction. $\endgroup$ – Jason Arthur Taylor Jun 23 '18 at 17:30

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