Question

An element crystallises in a face-centred cubic ($\mathrm{fcc}$) unit cell with a cell edge $\mathrm{a}$. The distance between the centres of the two nearest octahedral voids in the crystal lattice is:

• A. $\mathrm{a}$
• B. $\frac{a}{2}$
• C. $\sqrt{2a}$
• D. $\frac{a}{\sqrt{2}}$

Answer 1

The correct option is D

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

The Explanation for the correct option:

In a face-centred cubic system, the octahedral void remains in the edge centre and body centre.

The distance between the centres of the two nearest octahedral voids

=:$\sqrt{\left[{\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2\right]}$

=$\sqrt{\frac{a^2}{4}+\frac{a^2}{4}}$

=$\sqrt{\frac{2a^2}{4}}$

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

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

Hence option (D) is correct.