misterpickle
May1-10, 07:04 PM
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 = 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 here (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.
3. The attempt at a solution
First I find the electron concentration in terms of k_{F}[\latex]
[latex]V=(4/3) \pi k^{3}
N=2*(4/3)*\frac{\pi k^{3}}{V_{k}}
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:
N=\frac{8}{3}\frac{\pi k^{3}a^{3}}{4\pi^{3}}=\frac{2k^{3}a^{3}}{3\pi^{2}}
which gives
k=(\frac{3\pi^{2}N}{2a^{3}})^{1/3}
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?
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 here (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.
3. The attempt at a solution
First I find the electron concentration in terms of k_{F}[\latex]
[latex]V=(4/3) \pi k^{3}
N=2*(4/3)*\frac{\pi k^{3}}{V_{k}}
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:
N=\frac{8}{3}\frac{\pi k^{3}a^{3}}{4\pi^{3}}=\frac{2k^{3}a^{3}}{3\pi^{2}}
which gives
k=(\frac{3\pi^{2}N}{2a^{3}})^{1/3}
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?