Example 2 Construct B-Z corresponding to a given plane hexagonal lattice with a
= b and c ¼ 120
and show that the shortest distance of a zone face from the center
of the zone is 1:155
p
a
À Á :
Solution: Given: Plane hexagonal lattice with a = b and c ¼ 120
: Follow the given
procedure, we can obtain:
(i) A primitive cell in the given lattice.
(ii) Draw normal to each axis and obtain a* and b* as
a
Ã
¼ b
Ã
¼
2p
asinc
¼
2p
a
ffiffi ffi
3
p =2
À
Á¼
4
ffiffi ffi
3
p
p
a
(iii) Obtain reciprocal lattice net and construct a W–S unit cell. This is the
required B-Z for the given plane hexagonal lattice.
The geometric relationship between the direct lattice, the reciprocal lattice and
the B-Z is shown in Table 2.3. From the figure, we observe that the shortest
distance of a zone face from the center of the zone is
d ¼
a
Ã
2
¼
2
ffiffi ffi
3
p
p
a
¼ 1:155
p
a
Example 3 Construct B-Z corresponding to a given plane rectangular lattice with
a 6 ¼ b and c ¼ 90
and show that the zone faces are at p/a and p/b, respectively,
from the center of the zone.
Solution: Given: Plane oblique lattice with a 6 ¼ b and c ¼ 90
: Following the given
procedure, we obtain:
(i) A primitive cell in the given lattice.
(ii) Draw a normal to each axis and obtain a* and b* as
a
Ã
¼
2p
a
and b
Ã
¼
2p
b
(iii) Obtain the reciprocal lattice net and construct a W–S unit cell. This is the
required B-Z for the given primitive oblique lattice.
The geometric relationship between the direct lattice, the reciprocal lattice and
the B-Z is shown in Table 2.3. From the figure, we observe that the two zone faces
are at p/a and p/b, respectively, from the center of the zone.
Example 4 Construct B-Z corresponding to a simple cubic lattice with a = b = c
and a ¼ b ¼ c ¼ 90
: Show that the six equivalent zone faces are at a distance p/a
from the center of the zone.
74
2 Unit Cell Construction
= b and c ¼ 120
and show that the shortest distance of a zone face from the center
of the zone is 1:155
p
a
À Á :
Solution: Given: Plane hexagonal lattice with a = b and c ¼ 120
: Follow the given
procedure, we can obtain:
(i) A primitive cell in the given lattice.
(ii) Draw normal to each axis and obtain a* and b* as
a
Ã
¼ b
Ã
¼
2p
asinc
¼
2p
a
ffiffi ffi
3
p =2
À
Á¼
4
ffiffi ffi
3
p
p
a
(iii) Obtain reciprocal lattice net and construct a W–S unit cell. This is the
required B-Z for the given plane hexagonal lattice.
The geometric relationship between the direct lattice, the reciprocal lattice and
the B-Z is shown in Table 2.3. From the figure, we observe that the shortest
distance of a zone face from the center of the zone is
d ¼
a
Ã
2
¼
2
ffiffi ffi
3
p
p
a
¼ 1:155
p
a
Example 3 Construct B-Z corresponding to a given plane rectangular lattice with
a 6 ¼ b and c ¼ 90
and show that the zone faces are at p/a and p/b, respectively,
from the center of the zone.
Solution: Given: Plane oblique lattice with a 6 ¼ b and c ¼ 90
: Following the given
procedure, we obtain:
(i) A primitive cell in the given lattice.
(ii) Draw a normal to each axis and obtain a* and b* as
a
Ã
¼
2p
a
and b
Ã
¼
2p
b
(iii) Obtain the reciprocal lattice net and construct a W–S unit cell. This is the
required B-Z for the given primitive oblique lattice.
The geometric relationship between the direct lattice, the reciprocal lattice and
the B-Z is shown in Table 2.3. From the figure, we observe that the two zone faces
are at p/a and p/b, respectively, from the center of the zone.
Example 4 Construct B-Z corresponding to a simple cubic lattice with a = b = c
and a ¼ b ¼ c ¼ 90
: Show that the six equivalent zone faces are at a distance p/a
from the center of the zone.
74
2 Unit Cell Construction