An element crystallises in a face-centred cubic (fcc) unit cell with edge length $a$ . The distance between the centres of two nearest octahedral voids in the crystal lattice is:
(JEE Main 2020)

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Solution

Octahedral void in fcc is present at edge centre and body centre.

Minimum distance between two nearest octahedral void is: $= \sqrt{{(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2}} = \frac{a}{\sqrt{2}}$

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An element crystallises in a face-centred cubic (fcc) unit cell with edge length a. The distance between the centres of two nearest octahedral voids in the crystal lattice is: (JEE Main 2020)

Octahedral void in fcc is present at edge centre and body centre. Minimum distance between two nearest octahedral void is:=√(a2)2+(a2)2=a√2

An element crystallises in a face-centred cubic (fcc) unit cell with edge length $a$ . The distance between the centres of two nearest octahedral voids in the crystal lattice is:
(JEE Main 2020)

Octahedral void in fcc is present at edge centre and body centre.

Minimum distance between two nearest octahedral void is: $= \sqrt{{(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2}} = \frac{a}{\sqrt{2}}$