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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