Question

Consider a simple cubic lattice with lattice constant 0.5 nm. Calculate the spacing between two adjacent (211) planes. Give your answer in unit of nmnm. Values within 10% error will be considered correct

Answer 1

Given :

Lattice constant = 0.5 nm.

Indices = [211]

To find :

Inter planer spacing.

Solution :

The inter planer spacing is denoted as:

$d _{hkl}$

where h,k and l are miller indices.

in this case

[hkl] = [211]

so

h = 2,

k = 1,

l = 1.

and it is given that lattice constant a = 0.5 nm.

Inter planer spacing is the distance between parallel sets of planes represented by miller indices.

It depends on structure and shape of lattices.

The formula which relates inter planer spacing with miller indices and lattice constant is :

$\bf \: { (\frac{1}{d _{hkl} }) }^{2} = \frac{ ({h}^{2} + {k}^{2} + {l}^{2} ) }{ {a}^{2} }$

or

$\bf \: {d _{hkl} }= \frac{a }{ \sqrt{({h}^{2} + {k}^{2} + {l}^{2} )} \: }$

Putting the given values of a, h, k and l in above formula, we get

$\bf \: {d _{hkl} }= \frac{0.5 }{ \sqrt{({2}^{2} + {1}^{2} + {1}^{2} )} \: } \\ \bf \: {d _{hkl} }= \frac{0.5 }{ \sqrt{(6 )} \: } \: = 0.2041$

So the inter planer spacing in this case : 0.2041 nm