What is the Radius of Fermi Sphere for a 2D metal?

misterpickle
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Homework Statement


Problem 9.2(B) from Kittel Solid State Physics.

A two-dimensional metal has one atom of valence one in a simple rectangular primitive cell of a1 = 2Å and a2 = 4Å. Calculate the radius of the free electron Fermi sphere and draw this sphere to scale on the drawing of the Brillouin zones.

A 2D solution that seems to be correct is posted http://physics.unl.edu/~tsymbal/tsymbal_files/Teaching/SSP-927/HW/Homework%2008_solution.pdf" . Can anyone tell me what is wrong with my approach? Also, some equations weren't working for the latex. Sorry.

The Attempt at a Solution



First I find the electron concentration in terms of k_{F}[\latex]<br /> <br /> V=(4/3) \pi k^{3}<br /> <br /> N=2*(4/3)*\frac{\pi k^{3}}{V_{k}}<br /> <br /> where<br /> <br /> V_{k}=\frac{2 \pi}{a}*\frac{2 \pi}{a}*\frac{2 \pi}{b}=\frac{4 \pi^{3}}{a^{3}} (latex code didn&#039;t work for this)<br /> <br /> which is the k-space volume. The factor of 2 is the electron spin degeneracy.<br /> <br /> The electron concentration, N, is then:<br /> <br /> N=\frac{8}{3}\frac{\pi k^{3}a^{3}}{4\pi^{3}}=\frac{2k^{3}a^{3}}{3\pi^{2}}<br /> <br /> which gives<br /> <br /> k=(\frac{3\pi^{2}N}{2a^{3}})^{1/3}<br /> <br /> using N=1 and plugging in the values for a and b I get.<br /> <br /> k=(3\pi^{2})/16 (the latex code is screwing up for some reason)<br /> <br /> This gives me 1.23 A^-1.<br /> <br /> Where did I go wrong?
 
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