Homework6_Solutions - Homework#6 Solutions Spring 2014 1.a...

Homework #6 Solutions Spring 2014
2

1 st Zone:
2
2
2 π
a
2 nd Zone: k
f
=
2 π
a
3 rd Zone: k
f
=
5
2
2 π
a
4 th Zone: k
f
=
5 2
6
2 π
a
5 th Zone: k
f
=
10
2
2 π
a
6 th Zone: k
f
=
10
2
2 π
a
7 th Zone: k
f
=
10
2
2 π
a
Based on this, our table for the first 13 electrons looks like:
Brillouin Zone
m
k
f
/
2 π
a
1 2 3 4 5 6 7
1 .399 P E E E E E E
2 .564 P P E E E E E
3 .691 P P E E E E E
4 .798 F P P P E E E
5 .892 F
P P P
E E E
6 .977 F
P P P
E E E
7 1.055 F F
P P
P E E

8 1.128 F F F P P P E
9 1.197 F F F F P P E
10 1.262 F F F F P P P
11 1.323 F F F F P P P
12 1.382 F F F F P P P
13 1.438 F F F F P P P
At m=12 electrons, the Fermi radius in units of
2 π
a
is 1.382. The shortest distance to the
edge of the 7
th
BZ (i.e. the point at which the eighth BZ would begin to be filled) occurs
along the diagonal of the reciprocal lattice (see Figure 9.15). At that point, the radius of
the Fermi circle would be:
k
f
=
2 π
a
2 = 1.404
2 π
a
This Fermi wavevector is generated between m=12 electrons and m=13 electrons. Thus
for m=12 electrons only zones less than or equal to 7 are occupied while for m=13
electrons we begin to fill the eight and higher zones.
2. a.