Question

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)

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

Answer 1

The correct option is A $\frac{a}{\sqrt{2}}$

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



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