An element crystallises in a face-centred cub | vedclass

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(C) In an $fcc$ unit cell,octahedral voids $(O.V.)$ are located at the centre of each edge and at the body centre.The distance between two nearest oct

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

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(C) In an $fcc$ unit cell,octahedral voids $(O.V.)$ are located at the centre of each edge and at the body centre.The distance between two nearest octahedral voids located at the centres of two adjacent edges is calculated using the Pythagorean theorem.The coordinates of the octahedral voids at the centres of two adjacent edges (e.g.,at $(a/2, 0, 0)$ and $(0, a/2, 0)$) result in a distance of:$d = \sqrt{(\frac{a}{2} - 0)^2 + (0 - \frac{a}{2})^2 + (0 - 0)^2}$$d = \sqrt{(\frac{a}{2})^2 + (\frac{a}{2})^2}$$d = \sqrt{\frac{a^2}{4} + \frac{a^2}{4}} = \sqrt{\frac{2a^2}{4}} = \sqrt{\frac{a^2}{2}} = \frac{a}{\sqrt{2}}$

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

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