JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Solid State

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An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centres of two nearest tetrahedral voids in the lattice is:

(A) $\frac{a}{2}$ ✓ Correct
(B) $2 a$
(C) $\frac{3}{2} a$
(D)a

Answer: (A)

Solution

In an fcc lattice the tetrahedral voids sit at the centres of the eight small cubes of edge

\frac{a}{2}

into which the unit cell divides. The two nearest voids are the centres of two edge-sharing small cubes, so the line joining them is just one small-cube edge long. That distance is

\frac{a}{2}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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