An element crystallises in `f.c.c.` lattice having edge length `400 p m`. Calculate the maximum diameter, which can be placed in interstitial sites without disturbing the structure.

Explanation

Interstitital sites can be either tetrahedral or octahedral voids. As octahedral voids are bigger than tetrahedral voids, maximum diameter can fit into octahedral voids. If R is the size of atoms in the fcc packing and r is the size of the octahedral void, r=0.414 R
In fcc lattice, face diagonal =`sqrt2a`
As in fcc, atoms along the face diagonal are touching each other .
`4R=sqrt2a "or" R=(sqrt2a)/4`
Required diameter of the interstitial site =2 r = 2 x 0.414 R =2 x 0.414 x `sqrt2/4` x 400 pm
=117.1 pm (`because` a=400 pm )