What's so special about Cu, Ag, and Au? fermi level? band gap?

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SUMMARY

Copper (Cu), Silver (Ag), and Gold (Au) are unique among metals due to their electronic structure, particularly the behavior of their d electrons. These d electrons are not tightly bound, allowing for significant hybridization between the s and d bands, which lowers the energy of the s band near the L point in the Brillouin Zone. This hybridization results in an open Fermi surface topology, contributing to their exceptional electrical conductivity at room temperature, with Silver leading, followed by Copper and Gold.

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  • Familiarity with Fermi levels and Fermi velocity
  • Knowledge of band theory and conduction bands
  • Basic grasp of the Brillouin Zone and its significance
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i know this is a simple question, but i have been looking for a while (maybe 15 min) and i cannot find what is so special about Cu, Ag, and Au? i know there is something special about these 3 metals. something that none of the other metals have. i remember it having to do with fermi velocity / femi levels and conduction bands but i have completely forgotten and cannot find it.

thanks guys.
 
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From the notes for my solid state course: The d electrons in those three metals are not tightly bound to the atom, so they can't be treated as core electrons. "Six bands are required to accommodate the eleven additional electrons. We expect to find a broad s band and a narrow d band, and these bands overlap and hybridize ... The result of the hybridization is to lower the energy of the s band near the L point in the Brillouin Zone. This has an important effect on the Fermi surface topology, which now 'necks out' to touch the edge of the Brillouin zone." This means the Fermi surface is open.
 
They have the highest electrical conductivities at room temperature of any materials, in this order: Ag, Cu, Au
 

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