Question #38b86

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Answer 1 · 2015-01-24T20:12:21

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:

$d_1 = (a)/(sqrt(1^2+0+1^2))=(a)/(sqrt(2))$

For the (1, 1, 1) case:

$d_2=(a)/(sqrt(1^2+1^2+1^2))=(a)/(sqrt(3))$

So $(d_1)/(d_2)= (a)/(sqrt2)xx(sqrt3)/a$

$(d_1)/(d_2)=(sqrt3)/(sqrt2)$

So $d_1/d_2=(1.732)/(1.414)=1.22$

This shows, therefore, that $d_1>d_2$