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Q.

An element crystallises in a face centred cubic (fee) unit cell with edge length ‘a’. The distance between the centres of the two nearest octahedral voids in the crystal lattices is.

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a

$\sqrt{2} a$

b

a

c

$\frac{a}{2}$

d

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

answer is A.

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Detailed Solution

The distance between two nearest octahedral voids in FCC lattice is the distance between edge centre and the body centre.
$\begin{array}{l} ∴ Dis \tan ce = \sqrt{{(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2}} \\ = \sqrt{\frac{a^{2}}{4} + \frac{a^{2}}{4}} = \frac{a}{\sqrt{2}} \end{array}$

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