Question

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

Answer 1

For an octahedral void a = 2 (r + R) In fcc lattice the largest void present is octahedral void. If the radius of void sphere is R and of lattice sphere is r. Then,

$r=\sqrt{2}\times 400/5=141.12pm$ $(a=400pm)$

applying condition for octahedral void, 2 (r + R) = a

∴ $2R=a-2\times r=400–2\times 141.12$

∴ Diameter of greatest sphere = $117.16$ pm