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we know, copper has 1 electron on the last shell(smallest ionization energy).

we also know that in a copper wire (don't imagine copper wire with electricity, imagine no electricity exists), A cubic centimeter of copper (about the size of a thimble) contains approximately 8.4 × 1022 free electrons at room temperature.

I know that that free electron for each atom is that electron on that last shell I mentioned. What would exactly cause in a copper wire that those valence electrons to become free?

My impression is the following: Is it because at room temperature, atoms have kinetic energy, hence they move, meet each other, one atom's 29 protons attract another atom's valence electron and since that electron needs so little ionization, it jumps from one atom to another, then the 3rd atom would cause the same for 2nd atom and electron would jump from 2nd to 3rd atom and so on, hence they end up being free electrons.

Qmechanic
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Giorgi
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1 Answers1

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When we talk about electrons in the last shell, the implicit model is the application of the Schroedinger equation in a potential with spheric symmetry. That means: an isolated atom of copper in this case.

A real chunk of copper consists (in general) of several small crystals, each one with a periodic arrangement of atoms. The Schroedinger equation in this case requires a periodic potential, and the spherical symmetry no longer exists.

The result is a band structure, where the electrons of the crystal occupy a range of energies instead of a single value. That structure is present even close the absolute zero, where the movement of the nuclei tends to zero. So, it is not this movement the cause of electrons being 'shared' between atoms.

Claudio Saspinski
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