# Question 38b86

Jan 24, 2015

The interplanar distance $d$ is given by:

d=(a)/(sqrt(h^2+k^2+l^2)#

Where $a$ is the lattice parameter.

For the (1, 0, 1) case:

${d}_{1} = \frac{a}{\sqrt{{1}^{2} + 0 + {1}^{2}}} = \frac{a}{\sqrt{2}}$

For the (1, 1, 1) case:

${d}_{2} = \frac{a}{\sqrt{{1}^{2} + {1}^{2} + {1}^{2}}} = \frac{a}{\sqrt{3}}$

So $\frac{{d}_{1}}{{d}_{2}} = \frac{a}{\sqrt{2}} \times \frac{\sqrt{3}}{a}$

$\frac{{d}_{1}}{{d}_{2}} = \frac{\sqrt{3}}{\sqrt{2}}$

So ${d}_{1} / {d}_{2} = \frac{1.732}{1.414} = 1.22$

This shows, therefore, that ${d}_{1} > {d}_{2}$