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# 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

Last updated date: 20th Jun 2024
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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.

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 dhkl 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.