The bond length for an atom placed in an octahedral void in a close packing is
thus 0.88 times the layer separation. Substituting p ¼ r þ R; h ¼
ffiffi ffi
2
p =
ffiffi ffi
3
p
À
Á
a and
a = 2R we have,
r þ R ¼
ffiffi ffi
2
p
R
therefore, r ¼ ð
ffiffi ffi
2
p À 1Þ R ¼ 0.414R
or r
R ¼ 0.414
ðiiÞ
This gives the critical radius ratio representing the octahedral void.
Example 5 Show that the critical radius ratio for a simple cubic coordination is
0.732.
Solution: Let us consider a void surrounded by eight spheres of radius R, the
centers of the spheres lie at the corners of a simple cube of side a = 2R as shown in
Fig. 1.35. Let a small sphere of radius r just fits into the void. The body diagonal
and side of the cube are related through the equation
2 r þ R
ð
Þ¼
ffiffi ffi
3
p
a ¼
ffiffi ffi
3
p  2R
or r þ R ¼
ffiffi ffi
3
p
R
or r ¼
ffiffi ffi
3
p À1
À
Á
R ¼ 0.732R
or r
R ¼ 0:732
This gives the critical radius ratio representing the simple cubic void.
Example 6 Calculate the void space in a close packing of n spheres of radius
1.000, n spheres of radius 0.414 and 2n spheres of radius 0.225.
Solution: Let us consider the close packing of identical spheres of radius R in a
face-centered cubic unit along [111] direction. In this case, side of the unit cell and
the radius of the sphere are related (Fig. 1.36) through the equation
Fig. 1.34 Projection of centers of spheres with their position coordinates
32
1 Unit Cell Composition