Question

Calculate the value of d spacing for (100) plane in a rock salt crystal of lattice constant a = 2.814 ˚A

Answer 1

Answer:

The value of d spacing for (100) plane in a rock salt crystal of lattice constant a = 2.814 ˚A  will be $2.814 \times 10^{-8} cm$

Explanation:

The d- spacing or the distance between the consecutive lattice planes of a cubic lattice  is given by,

$d_{h k l}=\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}$ ...(1)

Where,

h,k,l -  Miller indices

a      - lattice constant

Given,

( hkl )  = (100)

h = 1

k = 0

l  = 0

a = $2.814^{0}$ A

Equation (1) becomes,

$d_{100}=\frac{2.814 \AA}{\sqrt{1^{2}+0^{2}+0^{2}}}$

       $=2.814 \AA=2.814 \times 10^{-8} \mathrm{~cm}$