Question

Potassium chloride crystallises with a body centred cubic lattice. Calculate the distance between the 200, 110 and 222 planes. The length of the side of the unit cell is 5.34 Å.

Answer 1

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

For 200 plane;

\(d_{200} = \frac{5.34}{\sqrt{2^2 + 0^2 + 0^2} } = \frac{5.34}{\sqrt 4} = 2.67 Å\)

For 110 plane;

\(d_{110} = \frac{5.34}{\sqrt{1^2 + 1^2 + 0^2} } = \frac{5.34}{\sqrt 2} = 3.77 Å\)

For 222 plane;

\(d_{222} = \frac{5.34}{\sqrt{2^2 + 2^2 + 2^2} } = \frac{5.34}{\sqrt {12}} = 1.54 Å\)