d ¼
2p
a
1
2
2
þ
1
2
2
þ 0
ð Þ
2
"
# 1=2
¼
ffiffi ffi
2
p p
a
¼ 1:414
p
a
The above calculations, give us that the value of k F lies between
0\k F \d
This indicates that the Fermi sphere is entirely contained within the first B-Z.
Further, we have
k F
d
¼
1:24
p
a
À Á
1:414
p
a
À Á ¼ 0:88 ¼ 88%
This shows that the Fermi sphere covers 88% of the shortest distance. Finally,
we have
d À k F ¼ 1:414
p
a
À 1:24
p
a
¼ 0:174
p
a
This confirms that the Fermi sphere is separated by a distance of 0:174
p
a
À Á
from
the zone boundaries.
Example 4 Determine the radius of the Fermi sphere for a face-centered cubic
crystal of side a. Show that (i) the Fermi sphere is entirely contained within the first
B-Z (ii) it covers 90 % of the distance (of the hexagonal face) from the center of the
zone, and (iii) it is separated by a distance of 0:169
p
a
À Á
from the zone boundaries.
Solution Given: An fcc crystal, number of atoms per unit cell, n = 4, k F = ?
The radius of the Fermi sphere can be obtained by
k F ¼
3p
2 n
a 3
1=3
¼
3p
2 4
a 3
1=3
¼
12
p
1=3 p
a
¼ 1:563
p
a
Further, the first B-Z for an fcc crystal lattice has the truncated octahedron shape
(Fig. 2.31b). It has fourteen faces, eight hexagonal and six squares.
The shortest distance of a square face from the center of the zone is:
d S ¼
2p
a
ð1Þ
2 þ ð0Þ
2 þ ð0Þ
2
h
i 1=2 ¼
2p
a
Similarly, the shortest distance of the hexagonal face from the center of the zone
is
2.4 Fermi Surface and Brillouin Zone
85