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Chemistry Question on The solid state

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

A

a

B

2\sqrt{2} a

C

a2\frac{ a }{\sqrt{2}}

D

 a 2\frac{\text { a }}{2}

Answer

a2\frac{ a }{\sqrt{2}}

Explanation

Solution

distance between nearest octahedral voids(O.V.)

=(a2)2+(a2)2=\sqrt{\left(\frac{a}{2}\right)^{2}+\left(\frac{a}{2}\right)^{2}}
⇒=a2\Rightarrow=\frac{a}{\sqrt{2}}