Calculate the approximate number of conduction electrons So i have the following problem: A cube of gold 0.1 meters on an edge, calculate the approximate number of conduction electrons whose energies lie in the range from 4.0 ev to 4.025 ev.
But I'm not clear on how to start. could someone offer any help?
 A: I wasn't going to answer since I don't know exactly what your professor had in mind, but since I've been prodded by Sofia here is my suggestion. However don't treat this as gospel as I may have completely the wrong idea of the question.
I would guess that you're supposed to treat the cube as an infinite 3D potential well aka particle in a box. In that case the system has discrete energy levels given by:
$$ E_{ijk} = \frac{\hbar^2}{2m}k_{ijk}^2 \tag{1} $$
where:
$$ k_{ijk}^2 = \frac{\pi^2}{\ell^2}(i^2 + j^2 + k^2) $$
where in this case $m$ is the mass of an electron and $\ell$ is the size of the box.
The density of gold is 19.3 g/cc and the atomic weight is 197, so the mass of the gold cube is 19.3 kg and dividing by the atomic weight (in kg) gives us the number of moles, 97.7. Multiplying by Avagadro's number gives the number of atoms in our cube as about $5.9 \times 10^{25}$. We'll assume that each atom contributes one electron to the conduction band.
In principle you could start counting up the energy states starting at $111$ and counting upwards until you reach 4eV. However this isn't a practical way to do the problem. You need to knoiw the expression for the density of states. I probably learned how to derive this for the particle in the box, but I have long since forgotten the details so I just Googled to find:
$$ g(E) = \frac{\pi \sqrt{E}}{2E_{111}^{3/2}} \tag{2} $$
The density of states tells us the number of states between two energies $E_a$ and $E_b$ is:
$$ N_{ab} = \int_{E_a}^{E_b} g(E)dE $$
So the question requires you to do this integration with $E_a = 4$eV and $E_b = 4.025$eV. The integral is straightforward and gives:
$$ N_{ab} = \left[ \frac{\pi}{3} \left(\frac{E}{E_{111}}\right)^{3/2}\right]_{E_a}^{E_b} \tag{3} $$
As a sanity check let's put $E_a = 0$ and $E_b = 4$eV and see how this compares with the number of electrons in the cube. I get $E_{111} = 1.13 \times 10^{-16}$ eV, and putting this into equation (3) I get the number of states between zero and 4ev to be about $7 \times 10^{24}$. Reassuringly, this is comparable to our estimate of $5.9 \times 10^{25}$ electrons in the conduction band.
NB remember we can fit two electrons into each state, one spin up and one spin down.
And we're basically done. If we set $E_a = 4$eV and $E_b = 4.025$eV in equation (3) we get the answer:
$$ N \approx 6.6 \times 10^{22} $$
And the number of electrns in the energy range is just twice this.
A: mmm.  I don't think that's quite right.  I'd rather use the equation for the number of electrons in any given energy state, 
$N(E) = \int_0^{\infty} \frac{1}{e^{(E - E_F) / K_B T}} \frac{V (2 m) ^{3 / 2}}{2 \hbar ^3 \pi ^2}\sqrt{E} dE$ where $E_f$ is the fermi energy and $E$ is the energy of the electron.  To do this, you'd need to know the temperature of the gold, but without that the number of electrons in any given energy isn't really hugely meaningful anyway.
Once you have a temperature, if it's sufficiently small (and if they're asking you this question, it probably will be) $(E - E_F)  / K_b T$ will become either infinity or negative infinity (depending on whether E is greater or less than the fermi energy).   Then you can just integrate. 
