ffiffi ffi
2
p
a ¼ 2d þ x or d ¼
ffiffi ffi
2
p
a À x
2
Again, consider a triangle (Fig. 2.3c) whose normal lies in the hexagonal plane
and intersects the top diagonal, its base is one-fourth of the body diagonal of the
cube. From this triangle, we can write
cos h ¼
ffiffi ffi
3
p
a=4
ð
ffiffi ffi
2
p
a À xÞ=2
ðiiÞ
From Eqs. (i) and (ii), we have
ffiffi ffi
2
p
ffiffi ffi
3
p ¼
ffiffi
3
p a
4
ffiffi
2
p aÀx
2
¼
ffiffi ffi
3
p
a
4
Â
2
ffiffi ffi
2
p
a À x
À
Á
or 3a ¼ 2
ffiffi ffi
2
p
ffiffi ffi
2
p
a À x
¼ 4a À 2
ffiffi ffi
2
p
x
or x ¼
a
2
ffiffi ffi
2
p ¼
ffiffi ffi
2
p
4
Example 5 Construct a Wigner–Seitz unit cell for an fcc lattice.
Solution: Draw two fcc unit cells one above the other. The atom lying at the center
of a square plane is taken as the reference atom. Connect this atom with all first
nearest neighbor atoms. Mark the midpoints on these lines. Draw planes through
Fig. 2.4 Showing 110
½ Šand 11 1
½
Š
44
2 Unit Cell Construction