where
N c ¼ Number of corner atoms
N f ¼ Number of face À centered ðtop and bottomÞ atoms
N b ¼ Number of body À centered atoms
With the help of the above equations and the knowledge of corner, face-centered
and body-centered atoms, the number of atoms in the given unit cell can be easily
obtained. Therefore, from Fig. 3.14, we have:
SH; n ¼ 3 N c ¼ 12; N f ¼ 2; N b ¼ 0
ð
Þ
HCP; n ¼ 3 N c ¼ 12; N f ¼ 2; N b ¼ 0
ð
Þ
The relationships between the lattice parameter “a” and radius of the atom “R”
for SH and HCP cases are shown in Fig. 3.14. Accordingly, the volumes of the two
hexagonal unit cells (where h is the separation between the consecutive layers) are
obtained as:
V SH ¼ Area of the hexagonal base  height ðhÞ
¼ 6 Â
1
2
 a
2 sin 60
 h
¼
3
ffiffi ffi
3
p
a
2 h
2
¼
3
ffiffi ffi
3
p  4R
2
 2R
2
¼ 12
ffiffi ffi
3
p
R
3
ðfor SH, a ¼ h ¼ 2RÞ
V HCP ¼ Area of the hexagonal base  height ðhÞ
¼
3
ffiffi ffi
3
p
a
2 h
2
¼
3
ffiffi ffi
3
p  4R
2 Â
ffiffi ffi
2
p  2R
2
ffiffi ffi
3
p
¼ 12
ffiffi ffi
2
p
R
3
ðfor HCP; h ¼
ffiffi ffi
2
p
ffiffi ffi
3
p a; a ¼ 2RÞ
Also, volume of the atom in both cases is:
Fig. 3.14 Unit cell of a SH b HCP
124
3 Unit Cell Calculations