The shortest distance between an octahedral and tetrahedral void in F.C.C. metallic lattice in terms of radius of F.C.C. packed atom would be:

Octahedral void present at the centre of cube and tetrahedral void is present ${\left(\frac{1}{4}\right)}^{\text{th }}$ of the distance along each body diagonal.

$\therefore \frac{\sqrt{3}a}{2}=2\times$distance between octahedral and tetrahedral void and $a=2\sqrt{2}R$.