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 -

CONCEPT:

Tetrahedral Voids in FCC Structure

• In an FCC structure, tetrahedral voids are present at positions \(\left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right)\) and equivalent positions in the unit cell.
• The distance between the centers of two nearest tetrahedral voids can be determined based on their relative positions within the unit cell.
• In an FCC unit cell, the edge length is denoted as a.

CALCULATION:

• For the nearest tetrahedral voids, the distance is calculated as follows:
• Consider two tetrahedral voids located at \(\left( \frac{1}{4}, \frac{1}{4}, \frac{1}{4} \right)\) and \(\left( \frac{3}{4}, \frac{3}{4}, \frac{3}{4} \right)\) within the unit cell.
• The distance between these two points is given by the formula:
  \(d = \sqrt{\left( \frac{3}{4} - \frac{1}{4} \right)^2 + \left( \frac{3}{4} - \frac{1}{4} \right)^2 + \left( \frac{3}{4} - \frac{1}{4} \right)^2 }\)
• Simplifying this expression gives:
  \(d = \sqrt{\left( \frac{2}{4} \right)^2 + \left( \frac{2}{4} \right)^2 + \left( \frac{2}{4} \right)^2 } = \sqrt{\frac{12}{16}} \times a = \frac{\sqrt{3}}{2}a\)

The correct answer is (1) \(\frac{3}{2}a\).