In face - centred cubic (fcc) crystal lattice, edge length is 400 pm. Find the diameter of the greatest sphere which can be fitted into the interstital void without distortion of the lattice.

Explanation

For fcc, radius of atom (R)=`a/(2sqrt2)=400/(2sqrt2)` pm =141.4 pm
As octahedral void is bigger in size than the tetrahedral void, the greatest sphere will fit into octahedral voids
Radius of octahedral void (r)=0.414 R=0.414 x 141.4 pm = 58.54 pm
`therefore` Diameter of the greatest sphere fitting into the void =2 x 58.54 pm =117.08 pm