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.
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Solution
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=√2×400/5=141.12pm(a=400pm)
applying condition for octahedral void, 2 (r + R) = a