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The inter-planar spacing between the ( 221$)$ planes of a cubic lattice of length $450 \mathrm{pm}$ is $-$
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Verified Answer
The correct answer is:
$150 \mathrm{pm}$
$\begin{array}{l}
\mathrm{d}_{\mathrm{hkl}}=\frac{\mathrm{a}}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}+1^{2}}} \\
=\frac{450}{\sqrt{4+4+1}}=\frac{450}{\sqrt{9}}=150 \mathrm{pm}
\end{array}$
\mathrm{d}_{\mathrm{hkl}}=\frac{\mathrm{a}}{\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}+1^{2}}} \\
=\frac{450}{\sqrt{4+4+1}}=\frac{450}{\sqrt{9}}=150 \mathrm{pm}
\end{array}$
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