First slide

Crystal lattices and unit cells

Question

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in metallic crystal will be:

Moderate

A

$\sqrt{2 a}$
B

$\frac{a}{\sqrt{2}}$
C

2a
D

$2 \sqrt{2 a}$

Solution

for fcc
then $b = 4 r = \sqrt{2} a$
$a = \frac{4 r}{\sqrt{2}} = 2 \sqrt{2} r \Rightarrow r = \frac{a}{2 \sqrt{2}}$
$Therefore, distance of closest approach = 2 r = 2 \times \frac{a}{2 \sqrt{2}} = \frac{a}{\sqrt{2}}$

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