I always seem to struggle with questions like these, my understanding is we know the spacing of K-states throughout the lattice is given by
[tex]\frac{2\pi}{L}[/tex]
Such that the fermi wavevector divided by this spacing and multiplied by 2 due to the fact there are 2 electrons per k-state gives us the total number of electrons:
[tex]N=\frac{k_{F}}{\frac{2\pi}{L}}\times 2[/tex]
We also know that the total no. of electrons is given by the electron density multiplied by the the length of the lattice
[tex]N=nL=\frac{3L}{a}[/tex]
Where I have said the electron density is equal to 3 conduction electrons divided by the unit cell length.
Equating the expressions to eliminate 'N' I attain
[tex]k_{F}=\frac{3\pi}{a}[/tex]
Which to find the fermi energy I think I'd just use this equation but I'm not sure?
[tex]E_{F}=\frac{\hbar^{2}k_{F}^{2}}{2m_{al}}[/tex]
Where 'm_al' is the mass of an aluminium atom?
Though I'm not sure I've done this correctly... I think I may need to have the fermi wavevector divided by 2 for it to be correct - not sure why I would need to do this though.
Anyway thanks for any help!
SK
Source: http://www.physicsforums.com/showthread.php?t=661585&goto=newpost
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