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The nearest distance between two octahedral voids of a cubic close packing of atoms with radius r is 2r

The distance between two tetrahedral voids of a cubic close packing of atoms with radius r is 6×r at the body diagonal axes

The nearest distance between tetrahedral and octahedral void is 32×r , where is the radius of atom forming the close packing

The distance between the surfaces of atoms in fcc arrangement along an edge is 22−1r

answer is A.

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Detailed Solution

(1) dnearest O.V=a2=22r2=2r (correct)(2) (correct)(3) dO.V & T.V nearest =3a4=34×22r=32r (correct)(4) (4) a−2r=22r−2r=22−1r (Correct)

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