Find the interplaner spacing between the family of planes (111) in a crystal of lattice constant 3A?.

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written 3.9 years ago by

teamques10 ★ 69k

• modified 3.9 years ago

engineering physics

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written 3.9 years ago by

teamques10 ★ 69k

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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