Sekonda
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Hey,
My question is displayed below
I have had a few questions like these and I just want to check my approach is correct, but I'm not sure - I know that the Fermi Energy of Cu is something like 7eV.
Now we are told that there are 4 atoms per cube side - meaning 8 atoms but cubic volume, I'm not sure if I should assume 1,4 or 8 electrons per cubic volume i.e.
[tex]N_{electrons}=\frac{8}{(0.4nm)^3}\times L^{3}[/tex]
Where N is the total number of electrons for the Cu for a volume given by L^{3}, and where 8 is the number of electrons per cubic volume. Though I think this may be 4? I'm not sure!
Anyway, we can also find the total number of electrons in the volume by using k-states and the Fermi-Sphere
[tex]N_{electrons}=2\times \frac{4}{3}\pi k_{F}^{3}\times \frac{1}{(\frac{2\pi}{L})^{3}}[/tex]
Where the 2 is due to 2 electrons per k-state, the next factor is the k-state volume of the fermi-sphere and the final factor is the reciprocal of the 'volume per k-state'. Making these equal I find the Fermi energy as 9.15eV... which isn't bad I suppose.
If I assume 4 electrons per cubic volume I get roughly 5.9eV, so I'm not sure which way is technically correct.
Basically, how many electrons do I assume per cubic volume and is my problem solving even the correct method?
Cheers,
SK
My question is displayed below
I have had a few questions like these and I just want to check my approach is correct, but I'm not sure - I know that the Fermi Energy of Cu is something like 7eV.
Now we are told that there are 4 atoms per cube side - meaning 8 atoms but cubic volume, I'm not sure if I should assume 1,4 or 8 electrons per cubic volume i.e.
[tex]N_{electrons}=\frac{8}{(0.4nm)^3}\times L^{3}[/tex]
Where N is the total number of electrons for the Cu for a volume given by L^{3}, and where 8 is the number of electrons per cubic volume. Though I think this may be 4? I'm not sure!
Anyway, we can also find the total number of electrons in the volume by using k-states and the Fermi-Sphere
[tex]N_{electrons}=2\times \frac{4}{3}\pi k_{F}^{3}\times \frac{1}{(\frac{2\pi}{L})^{3}}[/tex]
Where the 2 is due to 2 electrons per k-state, the next factor is the k-state volume of the fermi-sphere and the final factor is the reciprocal of the 'volume per k-state'. Making these equal I find the Fermi energy as 9.15eV... which isn't bad I suppose.
If I assume 4 electrons per cubic volume I get roughly 5.9eV, so I'm not sure which way is technically correct.
Basically, how many electrons do I assume per cubic volume and is my problem solving even the correct method?
Cheers,
SK