A metal crystallises in ccp (fcc) with edge length $410 pm$ . Taking $2 = 1.414$ , find the radius in pm of the largest sphere that fits in an octahedral void (void along the cell edge, between atoms at the two ends and the edge centre), nearest integer.

Numerical answer type, answer in pm, tolerance 1.

Answer: 60 pm

Atomic radius

r = a / (22) = 410/2.828 = 144.98 pm

. Along an edge:

a = 2 r + 2 r_{void}

, so

r_{void} = a /2 - r = 205 - 144.98 = 60.02 \approx 60 pm