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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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Given –

a=3A=3×1010

h=1,k=1,l=1

To find – d Solution – d=ah2+k2+l2 =3×1010(1)2+(1)2+(1)2 =3×10103 d=1.73×1010 ∴ Interplanar Spacing is 1.73A

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