Area of the hexagon ¼ 6 Â area of the triangle ABC
¼ 6 Â
1
2
Â
R
sin 60
 R
¼ 2
ffiffi ffi
3
p  R
2
Therefore,
Efficiency
ð
Þ h ¼
Area of the circle
Area of the hexagon
¼
pR
2
2
ffiffi ffi
3
p
R
2
¼
p
2
ffiffi ffi
3
p ¼ 90:7%
Example 2 Determine the number of atoms in the unit cells of simple cubic (sc),
body-centered cubic (bcc) and face-centered cubic (fcc) crystal structures. Calculate
the packing efficiency in each case.
Solution: In a cubic crystal system, the number of atoms per unit cell is given by
n ¼
N c
8
þ
N f
2
þ N b
where
N c ¼ Number of corner atoms
N f ¼ Number of face À centered atoms
N b ¼ Number of body À centered atoms ¼ 1
With the help of the above equations and the knowledge of corner and face
centered atoms, the number of atoms in the given unit cell can be easily obtained.
Therefore, for
sc; n ¼ 1 ðN c ¼ 8; N f ¼ 0; N b ¼ 0Þ
bcc; n ¼ 2 ðN c ¼ 8; N f ¼ 0; N b ¼ 1Þ
fcc; n ¼ 4ðN c ¼ 8; N f ¼ 6; N b ¼ 0Þ
The relationships between the lattice parameter “a” and radius of the atom “R”
for three cubic cases are shown in Fig. 3.13. Accordingly, the volumes of three
cubic unit cells are:
sc; a ¼ 2R; so that V SC ¼ ðaÞ
3 ¼ ð2RÞ
3 ¼ 8ðRÞ
3
bcc;
ffiffi ffi
3
p
a ¼ 4R; so that V BCC ¼ ðaÞ
3 ¼
4R
ffiffi ffi
3
p
3
¼
64R
3
3
ffiffi ffi
3
p
fcc;
ffiffi ffi
2
p
a ¼ 4R; so that V FCC ¼ ðaÞ
3 ¼ ð2
ffiffi ffi
2
p
RÞ
3 ¼ 16
ffiffi ffi
2
p
R
3
122
3 Unit Cell Calculations