Example 1 Find the equivalent rhombohedral planes for the following fcc planes:
(111), ð11 1Þ and (200).
Solution: Given: fcc planes: (111), ð11 1Þ and (200).
Let us take them one by one.
Case I: fcc plane: (HKL) (111).
Substituting the values of H, K and L in Eq. 5.6, we obtain (hkl) (111) for
rhombohedron. Again, substituting the values of h, k and l in Eq. 5.4, we obtain
(HKL) (111) for fcc, this is the same indices with which we started. This also
shows that for (111) plane, both crystal systems have identical indices.
Case II: fcc plane: (HKL) ð11 1Þ.
A similar operation with the given H, K and L values will provide us (hkl)
(100) for rhombohedron. Again, substituting the values of h, k and l in Eq. 5.4, we
obtain (HKL) ð11 1Þ for fcc. This confirms the validity of Eqs. 5.4 and 5.6.
Case III: fcc plane: (HKL) (200).
A similar operation with the given H, K and L values will provide us (hkl)
(110) for rhombohedron. Again, substituting the values of h, k and l in Eq. 5.4, we
obtain (HKL) (200) for fcc. This confirms the validity of Eqs. 5.4 and 5.6. Using
these equations, a one to one correspondence of other Miller indices can be
obtained.
2. Simple Hexagonal and Orthorhombic
The axial relationships between the translation vectors of simple hexagonal and
orthorhombic unit cells are shown in Fig. 5.2. Let us consider a crystal plane which
is referred to as (hkil) in a simple hexagonal system of axes a, b, c and (HKL) in
orthorhombic system of axes a′, b′, c′. Then, the orthorhombic axes in terms of the
hexagonal axes are:
a
0
¼ 2a þ b; b
0
¼ b; c
0
¼ c
Now, using these equations, we can write the following transformation equations
H ¼ 2h þ 1k þ 0l
K ¼ 0h þ 1k þ 0l
L ¼ 0h þ 0k þ 1l
ð5:7Þ
The matrix form of these equations is
H
K
L
0
@
1
A ¼
2 1 0
0 1 0
0 0 1
0
@
1
A
h
k
l
0
@
1
A
ð5:8Þ
The determinant of the matrix in Eq. 5.8, D ¼ 2: Therefore using Eq. 5.3, we
can obtain the volume relationship between the two unit cells
182
5 Unit Cell Transformations
(111), ð11 1Þ and (200).
Solution: Given: fcc planes: (111), ð11 1Þ and (200).
Let us take them one by one.
Case I: fcc plane: (HKL) (111).
Substituting the values of H, K and L in Eq. 5.6, we obtain (hkl) (111) for
rhombohedron. Again, substituting the values of h, k and l in Eq. 5.4, we obtain
(HKL) (111) for fcc, this is the same indices with which we started. This also
shows that for (111) plane, both crystal systems have identical indices.
Case II: fcc plane: (HKL) ð11 1Þ.
A similar operation with the given H, K and L values will provide us (hkl)
(100) for rhombohedron. Again, substituting the values of h, k and l in Eq. 5.4, we
obtain (HKL) ð11 1Þ for fcc. This confirms the validity of Eqs. 5.4 and 5.6.
Case III: fcc plane: (HKL) (200).
A similar operation with the given H, K and L values will provide us (hkl)
(110) for rhombohedron. Again, substituting the values of h, k and l in Eq. 5.4, we
obtain (HKL) (200) for fcc. This confirms the validity of Eqs. 5.4 and 5.6. Using
these equations, a one to one correspondence of other Miller indices can be
obtained.
2. Simple Hexagonal and Orthorhombic
The axial relationships between the translation vectors of simple hexagonal and
orthorhombic unit cells are shown in Fig. 5.2. Let us consider a crystal plane which
is referred to as (hkil) in a simple hexagonal system of axes a, b, c and (HKL) in
orthorhombic system of axes a′, b′, c′. Then, the orthorhombic axes in terms of the
hexagonal axes are:
a
0
¼ 2a þ b; b
0
¼ b; c
0
¼ c
Now, using these equations, we can write the following transformation equations
H ¼ 2h þ 1k þ 0l
K ¼ 0h þ 1k þ 0l
L ¼ 0h þ 0k þ 1l
ð5:7Þ
The matrix form of these equations is
H
K
L
0
@
1
A ¼
2 1 0
0 1 0
0 0 1
0
@
1
A
h
k
l
0
@
1
A
ð5:8Þ
The determinant of the matrix in Eq. 5.8, D ¼ 2: Therefore using Eq. 5.3, we
can obtain the volume relationship between the two unit cells
182
5 Unit Cell Transformations