Question

In an orthorhombic crystal, a lattice plane cuts intercepts in the ratio 1:2:3 along a,b and c axes. Find the miller indices of the plane. Sketch the plane and calculate the interplanar spacing, given that a=1A, b=2A and c=3A.

Answer 1

Hint: We have to calculate Miller indices by taking the reciprocals of intercepts and for calculating the interplanar spacing, we have to use the formula,
${\displaystyle \frac{1}{{d_{hkl}}^2}}={\displaystyle \frac{4}{3}}\left({\displaystyle \frac{h^{+}+hk+k^2}{a^2}}\right)+\left({\displaystyle \frac{l^2}{c^2}}\right)$
Here, h, k, and l are miller indices.

Complete step by step answer:
We know that the orthorhombic crystal system is one of the 7 crystal systems.
Orthorhombic lattices comes from enlarging a cubic lattice along two of its orthogonal pairs by two factors, that leads in a rectangular prism with a rectangular base (a by b) and height (c), such that a, b, and c are different.
The intersection of all three bases at 90° angles, so the three lattice vectors remain mutually orthogonal.
We know that Miller indices of a plane are the reciprocals of the intercepts of that corresponding to unit length.
Thus, intercepts are a:b:c=1:2:3.
So let us now take the reciprocals:
${\displaystyle \frac{1}{a}}:{\displaystyle \frac{1}{b}}:{\displaystyle \frac{1}{c}}={\displaystyle \frac{1}{1}}:{\displaystyle \frac{1}{2}}:{\displaystyle \frac{1}{3}}$
(or) We can take L.C.M and by taking L.C.M, we get the value of miller indices as 6,3,2.
The value of h is 6.
The value of k is 3.
The value of l is 2.
We can represent the miller indices as $\left(hkl\right)=\left(632\right)$
Let us now calculate the interplanar spacing for orthorhombic crystals.
${\displaystyle \frac{1}{{d_{hkl}}^2}}={\displaystyle \frac{4}{3}}\left({\displaystyle \frac{h^{+}+hk+k^2}{a^2}}\right)+\left({\displaystyle \frac{l^2}{c^2}}\right)$
Let us now substitute the values of a, c, h, k, and l to calculate the interplanar spacing.
${\displaystyle \frac{1}{{d_{hkl}}^2}}={\displaystyle \frac{4}{3}}\left({\displaystyle \frac{{\left(6\right)}^2+\left(6\right)\left(2\right)+{\left(2\right)}^2}{{\left(1\right)}^2}}\right)+\left({\displaystyle \frac{{\left(2\right)}^2}{{\left(3\right)}^2}}\right)$
$\Rightarrow$${\displaystyle \frac{1}{{d_{hkl}}^2}}={\displaystyle \frac{760}{9}}$
$\Rightarrow$${d_{hkl}}^2={\displaystyle \frac{9}{760}}$
$\Rightarrow$$d_{hkl}={\displaystyle \frac{3}{\sqrt{760}}}$
The inter-planar spacing is ${\displaystyle \frac{3}{\sqrt{760}}}$.
The plane is sketched as,

Note:
We have to know that in two dimensions there are two orthorhombic Bravais lattices: primitive rectangular and centered rectangular. In three dimensions, primitive orthorhombic, base-centered orthorhombic, body-centered orthorhombic, and face-centered orthorhombic are the four orthorhombic Bravais lattices.