electricitymetals
If a metal is a Fermi sea, what makes different metals better conductors?
Clearly, one valance electron dominates.
All things being equal, I would have assumed that the bigger atoms, with more electrons and more distance between the valence electron and the nucleus, would conduct the best. That would imply gold, then silver, then copper. But silver conducts best, followed by copper, then gold. Why?
These are all good questions! Based on your description I assume you haven't had an introduction to solid state physics yet? Let's take your image of an electron that "jumps"
from atom to atom. In my understanding I wouln't describe it that way, to me it's a wavefunction of the electron that is almost independent from the valence electrons and you
can use the free electron gas approximation. Why is this band independent? See the following picture for an intuitive understanding how the atomic potentials define the possible energy levels within a periodic arrangement of atoms:
I think most of your questions will be easier to answer if you make yourself familiar with basic concepts and approximations people use to describe electrons in a solid first. Sure, a lot of things can be understood if we consider electrons to be little spheres that scatter from bigger spheres (ions), but you said you want to understand on the atomic level -> it's good to see the electron as a wave and see how this wave behaves in a lattice with certain boundary conditions.
First, I would read about a crystal. Atoms are arranged in a periodic lattice (assume a nice crystal for a first simple picture) and you can make assumptions based on this periodicity. You can define a unit cell and the Brillouin zone. You will see that the energy levels will sometimes split up in different bands and based on the filling of these bands you end up wih a metal, insulator etc. Electrons are fermions, can two electrons be in the same state? This defines the Fermi velocity.
This filling of the available energy levels describes the Fermi surface, a very useful tool to describe other more advanced concepts. Then you will see what happens if you change the arrangements of the atoms or why in different spatial directions electrons can move due to the bonding of different atomic orbitals.
This could be a good start ; ) -> http://britneyspears.ac/lasers.htm
There are other introductions out there, most of them describe the basics really well.
The characteristic feature of a conducting medium is its presence of free electric charges on its surface, which are unbound. Unbound, free electrons do not have a restoring force and therefore, have no natural frequencies; however, they will always oscillate at the driving frequency. When an impinging EM wave oscillates these free electrons at optical frequencies, the free electrons will re-emit light with exactly the same driving frequencies. Since all of the visible frequencies are re-emitted, the metal surface looks “silvery."
The question is a lot more interesting when we address why gold or copper is colored, not silvery. Intuitively, if the free electrons are the cause of the silvery-look, then it must be the bound electrons in, say gold, which is responsible for its golden color. What else can it be? Bound electrons do have restoring forces and therefore, they have resonant frequencies. Because bound electrons in conductors behave the same as bound electrons in dielectrics, I can use this link to model the index of refraction as
$$n^2(ω) = 1 + \frac{Ne^2}{m ϵ_0}\left[\frac{-f_e}{ω^2-iβω} + \frac{f_j}{ω_{0j}^2-ω^2+iβ_jω}\right]$$
The first bracketed term is the contribution from the free electrons (note no natural frequencies $ω_0$, so no absorption) whereas the second term arises from the bound electrons (natural frequencies $ω_{0j}$ imply absorption). So for gold to appear a reddish-yellowish color (λ ≥ 500 nm), the blues and greens (λ ≤ 500 nm) must interact more with the bound electrons. I then expect gold to have larger skin depth values for λ ≤ 500 nm and smaller ones for λ ≥ 500 nm. On the other hand, I expect silver to have smaller skin depth values for all optical frequencies since there is apparently no absorption of any wavelengths. So the key is the skin depth values.
Skin Depth Values:
The skin depth $\delta$ is determined from the absorption coefficient where $\delta = 1/\alpha$. Using the website refractiveindex.info provided by Stefan Bischof’s post, I obtained $\alpha$-values and plotted $\delta$ vs. $\lambda$ for six different metals (Au, Ag, Cu, Al, Ni and Pt) using Excel. (Note that I am not showing the origin in the plot.)
Focusing on gold and silver only, the data shows exactly what I expected. Between 400 nm ≤ λ ≤ 500 nm, gold’s skin depth is increasing with longer wavelength to just about 490 nm, implying that the blues and greens are more likely to interact with the bound electrons and get absorbed. After the 500 nm peak, the skin depth steadily decreases implying that the red end of the visible spectrum is getting re-emitted by the free electrons. Silver on the other hand, has smaller δ-values (especially around 500 nm) suggesting that there is little interaction with the bound electrons and reflects all frequencies alike. Overall, the trend is that the silvery metals (Ag, Al, Ni and Pt) all have smaller skin depths while the colored metals (Au and Cu) have larger skin depths, as expected. I believe that this suggests that among other things, penetration depth will help to determine whether bound electrons are playing a role in the color of the metal.
Many thanks to Stefan Bischof for approaching the question in the manner he did.
Best Answer
Several things contribute to conductivity. I think two are particularly relevant to the question. First concerns band structure. Metals can conduct easily because there are empty states very close in energy to the Fermi surface, so that it is easy to promote an electron to a higher-energy level. In an electric field, the newly-occupied states all (loosely speaking) have momenta in the direction of the field, so there is net movement of the electrons in the direction of the field. How well this works depends on the density of states at the Fermi level. The more states available, the better the conductivity. I've looked for these data for the noble metals but haven't located them.
The second concerns the dominant loss mechanism at room temperature, the interaction of the electrons with the lattice. In a simple model of a solid the positions of the ion cores are taken to be fixed. In reality they are not, and a passing electron will excite lattice excitations (phonons) and lose energy. How well this works depends on the strength of the electron-phonon coupling in the solid, and that's complicated business.
Having said all that, I don't have data at hand to make an argument about the noble metals. But I think it should be clear that your picture of conductivity needs to be augmented, and that might be enough to answer your question.
Update I found data on density of states, and the order of densities of states is Ag, Cu, Au, from highest to lowest.