Fermi wave vector compared to 1st Brilloin zone.

[philip041]

For a = 5Angstrom and n = 3 calculate kf and compare its magnitude with the dimension of the Brillouin zone, (all in 2D, square lattice).

Relevant eqn:

kf = (2*pi*n)^1/2

Here is my solution:

kf = (2*pi* (3/a^2))^1/2
= ((6*pi)^1/2)/a
=8.7x10^9 m^-1

1st BZ

= (2*pi/a)*1/2

= pi/a = 6.3x10^9 m^-1

Are the wave vectors and 1st BZ dimensions meant to be massive? I guess so as they are 1/(a really small number).

Do these values sound correct, to me they seem right in proportion, but the fact they are not really small worries me. When you deal with reciprical space does that mean big numbers?

Also my other big worry is that in kf = (2*pi*n)^1/2, n=m/A, where m=number of electrons and A = unit area. Should I really have taken the question to mean m=3 not n=3, so that I can put the values of a in?

Cheers

 

[NateD]

Yes, the wave vectors and 1st BZ dimensions are meant to be massive as they are inversely proportional to the lattice constant. The values you have obtained seem correct. You are right that n=m/A, where m is the number of electrons and A is the unit area. So in this case n=3 and a=5 Angstrom. Thus, your answer is correct.