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Radius of Fermi Sphere

by misterpickle
Tags: fermi, radius, sphere
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misterpickle
#1
May1-10, 07:04 PM
P: 12
1. The problem statement, all variables and given/known data
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 = 2 and a2 = 4. 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 here . Can anyone tell me what is wrong with my approach? Also, some equations weren't working for the latex. Sorry.

3. The attempt at a solution

First I find the electron concentration in terms of [itex]k_{F}[\latex]

[itex]V=(4/3) \pi k^{3}[/itex]

[itex]N=2*(4/3)*\frac{\pi k^{3}}{V_{k}}[/itex]

where

V_{k}=\frac{2 \pi}{a}*\frac{2 \pi}{a}*\frac{2 \pi}{b}=\frac{4 \pi^{3}}{a^{3}} (latex code didn't work for this)

which is the k-space volume. The factor of 2 is the electron spin degeneracy.

The electron concentration, N, is then:

[itex]N=\frac{8}{3}\frac{\pi k^{3}a^{3}}{4\pi^{3}}=\frac{2k^{3}a^{3}}{3\pi^{2}}[/itex]

which gives

[itex]k=(\frac{3\pi^{2}N}{2a^{3}})^{1/3}[/itex]

using N=1 and plugging in the values for a and b I get.

k=(3\pi^{2})/16 (the latex code is screwing up for some reason)

This gives me 1.23 A^-1.

Where did I go wrong?
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nickjer
#2
May1-10, 11:21 PM
P: 675
Why is your momentum space 3d when the problem is 2d?


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