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Find the interplaner spacing between the family of planes (111) in a crystal of lattice constant 3A?.
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written 3.3 years ago by
teamques10
★ 66k
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Given –
$a=3A^{\circ}=3\times10^{-10}$
$h=1, k=1, l=1$
To find – $d$ Solution – $d=\dfrac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$ $=\dfrac {3\times 10^{-10}}{\sqrt {(1)^{2}+(1)^{2}+(1)^{2}}}$ $=\dfrac {3\times 10^{-10}}{\sqrt {3}}$ $d=1.73 \times 10^{-10}$ ∴ Interplanar Spacing is $1.73A^{\circ}$
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