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:

In an fcc lattice, tetrahedral voids are located at the center of each of the eight small tetrahedra formed by joining the centers of four atoms at the corners of the unit cell. The distance between the centers of two nearest tetrahedral voids can be calculated using the geometry of the unit cell.

Consider two nearest tetrahedral voids. One is at the center of a tetrahedron formed by atoms at (0,0,0), (a,0,0), (a,a,0), and (a/2,a/2,a/2). The other is at the center of a neighboring tetrahedron. The coordinates of the centers of these two tetrahedral voids are (a/4, a/4, a/4) and (3a/4, a/4, a/4), or (a/4, a/4, a/4) and (a/4, 3a/4, a/4), for example.

The distance between these two points is found using the distance formula:

d = √[(3a/4 - a/4)² + (a/4 - a/4)² + (a/4 - a/4)²] = √[(a/2)²] = a/2

Alternatively, consider the distance between the centers of tetrahedral voids at (1/4,1/4,1/4)a and (1/4,3/4,1/4)a. The distance 'd' is given by

d = √[((1/4)a-(1/4)a)² + ((3/4)a-(1/4)a)² + ((1/4)a-(1/4)a)²] = √[(1/2a)²] = a/2

However, this calculation gives the distance as a/2. This is not one of the given options. The nearest tetrahedral voids are along the face diagonal. The distance between the nearest tetrahedral voids along the face diagonal is a√2/2. Let's examine another approach.

The distance between two nearest atoms in an fcc structure is a√2/2. Consider the tetrahedral voids formed by four atoms at the corners of a face. The distance between the centers of two tetrahedral voids in this arrangement would be half the face diagonal, which is (a√2)/2. This is one of the given options.

Therefore, the distance between the centers of two nearest tetrahedral voids in an fcc lattice is a√2/2.