As an ion, copper can give off 1, 2, 3 or 4 electrons. But it has 1 s electron in the last shell and 10 d electrons. So as a metal, how many of those are delocalised and free to move around, and how many are staying with the atom?
3
-
$\begingroup$ All electrons that go out on one end of the wire, are replaced with the same quantity at the other end, so the net loss is 0 $\endgroup$– Raoul KesselsCommented Jun 7, 2018 at 15:41
-
$\begingroup$ @RaoulKessels Sure, but I'm interested in the amount of electrons that can freely move inside the wire. $\endgroup$– NebulontCommented Jun 7, 2018 at 17:48
-
$\begingroup$ $I=\frac{q}{t}$ and the charge of one electron is $1.6\times10^{-19}$C $\endgroup$– Raoul KesselsCommented Jun 7, 2018 at 17:57
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
This is a number which can be measured via the Hall effect. This reference give the Hall coefficient as $-5.4\times10^{-11}\,\mathrm{m^3/C}$ for a number density of charge carriers as $$n_\mathrm e=\frac1{\left(-5.4\times10^{-11}\,\mathrm{m^3/C}\right)\left(-1.602\times10^{-19}\,\mathrm C\right)}=1.16\times10^{29}/\mathrm m^3$$ The number density of copper ions is $$n_\ce{Cu}=8920\,\frac{\mathrm{kg}}{\mathrm{m^3}}\times\frac{1000\,\mathrm g}{\mathrm{kg}}\times\frac{1\,\mathrm{mol}}{63.546\,\mathrm g}\times\frac{6.022\times10^{23}}{\mathrm{mol}}=8.45\times10^{28}/\mathrm m^3$$ So that works out to about $1.37$ charge carriers per ion.
$\begingroup$
$\endgroup$
A good first guess is that there's a gap between the 3d and 4s bands in the electronic structure of solid Cu, and since the 3d band is filled, and the 4s band half-filled, that means only the 4s electron can be considered nearly free. (Recall Cu = [Ar] 3d10 4s1.)
answered Jun 7, 2018 at 17:56