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  4. The inter-planar spacing between the (2 2 1) planes of a cubic lattice of length 450 pm is

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Q. The inter-planar spacing between the $(2 \,2\, 1)$ planes of a cubic lattice of length $450\, pm$ is

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Solution:

Interplanar distance,
$d=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$
Given, length of cubic lattice = $450 \,pm$
For $(221)$ plane,
$h = 2 , $
$k = 2 , $
$l= l$
$\therefore d=\frac{450}{\sqrt{\left(2\right)^{2}+\left(2\right)^{2}+\left(1\right)^{2}}}$
$=\frac{450}{\sqrt{9}}=\frac{450}{3}=150\,pm$