(I copy-pasted the question)
Consider N non interacting electrons inside a 3D box similar to the one we saw
in class, but this time it is not cubical, i.e. Lx = Ly [tex]\neq[/tex] Lz.
1. Calculate the fermi energy as a function of N;Lx;Lz;me...
2. What happens if we take Lz --> 0, i.e. the limit of a 2D system. Is there
3. Calculate the fermi energy in a 2D system (Lx = Ly). Do you obtain a
4. It would be a great conflict in case the 2D result cannot be obtained by
taking the limit Lz-->0. Did you nd such a conflict? If so, explain how
it can be avoided.
I solved the 3 first questions. I hope I didn't make a calculation mistake, but anyhow, I don't think it'll fundimentally change the problem of question 4...
So here are the answers I've got:
For the 3D box, I marked Lx=Ly=L, and Lz stayed...
Ef = (1/2m)*((6N((hbar)^2)*pi)/(Lz*L^2))^3/2
For the 2D problem I got:
Ef = (4Npi*(hbar)^2)/(mL^2)
The Attempt at a Solution
Problem is, obviously, it seems that Ef-->infinity as Lz --->0. I have no idea how to resolve this conflict!!! :)
Thank you very very very much for reading and helping!