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

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  1. √2a 
  2. a
  3. a/2
  4. 3a/2

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Option 3 : a/2
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12.04.2019 Shift 1 Synergy JEE Mains D74

The distance between the centers of two nearest tetrahedral voids in the lattice and also the minimum distance between two tetrahedral voids is \(\frac{a}{2}\)..

In fcc (face centered cubic), tetrahedral voids are located on the body diagonal at a distance of \(\frac{{\sqrt {3a} }}{4}\) from the corner. Together they form a smaller cube of edge length\(\frac{a}{2}\).

Therefore, distance between centres of two nearest tetrahedral voids in the lattice is also \(\frac{a}{2}\).

Face-centered cubic (fcc) refers to a crystal structure consisting of an atom at each cube corner and an atom in the center of each cube face. It is a close-packed plane in which on each face of the cube atoms are assumed to touch along face diagonals.

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