Question

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

Answer 1

In fcc unit cell,

$a\sqrt{2}=4r$

$r=\frac{a\sqrt{2}}{4}=\frac{400\times \sqrt{2}}{4}=141.4pm$

For octahedral void,

$2(r+R)=a$

$2R=a-2r=400-2\times 141.4=117.16pm$

Diameter of greatest sphere $=117.16pm$