Question

The inter-planar spacing between the (2 2 1) planes of a cube lattice of edge length 450 pm is:
(A) 50 pm
(B) 150 pm
(C) 300 pm
(D) 450 pm

Answer 1

Hint: Inter planar spacing between two planes is given by the formula:
${ d\ =\ \frac { a }{ \sqrt { { h }^{ 2 }+{ k }^{ 2 }+{ l }^{ 2 } } } }$
Here h, k, l usually occurs as (h k l) and these are the miller indices. ‘a’ in the equation is the edge length. By substituting the values given in the question, we will get the interplanar distance. Interplanar distance is the perpendicular distance between two successive planes.

Complete step by step answer:
The inter planar spacing between two planes is given by the formula:
${ d\ =\ \frac { a }{ \sqrt { { h }^{ 2 }+{ k }^{ 2 }+{ l }^{ 2 } } } }$

Where, h, k, l are Miller indices; a is the lattice parameter and d is the interplanar distance.

Here in this question, it has been given that a = 450 pm.

The interplanar spacing or interplanar distance is the perpendicular distance between two successive planes in a family (h k l). It is commonly indicated as d_hkl and corresponds to the reciprocal of the length of the corresponding vector in reciprocal space.

Here, in the question, it has been provided that (2 2 1) plane. So, here value of h = 2, k = 2 and l = 1

Substitute these values in the formula:
${ d\ =\ \frac { a }{ \sqrt { { h }^{ 2 }+{ k }^{ 2 }+{ l }^{ 2 } } } }$
${ \ =\ \frac { 450 }{ \sqrt { { 2 }^{ 2 }+{ 2 }^{ 2 }+{ 1 }^{ 2 } } } }={ \frac { 450 }{ \sqrt { 9 } } =150\ pm }$

Hence, the answer is option (B) 150 pm.

Additional Information:
Miller indices: Miller indices form a notation system in crystallography for planes in crystal (Bravais) lattices. In particular, a family of lattice planes is determined by three integers h, k, and l, the Miller indices.

Lattice constants: The lattice constant, or lattice parameter, refers to the physical dimension of unit cells in a crystal lattice. Lattices in three dimensions generally have three lattice constants, referred to as a, b, and c. For example, The lattice constant for diamond is a = 3.57 angstrom units at 300 K.

Note: The interplanar distance will remain the same for all the combinations for particular value of h,k,l. i.e, for eg: (2 2 1), (1 2 2),(1 2 1) all have the same interplanar distances here.