Question

An element crystallises in a face-centred cubic (fcc) unit cell with cell edge a. The distance between the centres of two nearest octahedral voids in the crystal lattice is

• A. a
• B. $\sqrt{2}$ a
• C. $\frac{ a }{\sqrt{2}}$
• D. $\frac{\text { a }}{2}$

Answer 1

The Correct Option is C

To solve this problem, we need to understand the structure and properties of a face-centred cubic (fcc) unit cell.

1. The fcc unit cell has atoms located at each of the corners and at the centers of each face of the cube. In an fcc structure, the atoms touch each other along the face diagonals.
2. Let the edge length of the cube be a. The face diagonal of the cube, which holds two atoms, is \sqrt{2}a.
3. The body diagonal of the cube, on the other hand, is \sqrt{3}a. However, in an fcc lattice, the octahedral voids are located at the edge centers and the body center.
4. For an fcc lattice, common positions of octahedral voids are at:
   • The center of the cube (body center).
   • The center of each edge.
5. The distance between two neighboring octahedral voids that are located on the edge of the cube can be determined as half the edge length. Thus, the distance is \frac{a}{2}.
6. The question asks for the distance between the centers of two nearest octahedral voids. In the fcc structure, nearest voids are located at edge centers separated by half the face diagonal distance.
7. Therefore, the distance between the nearest octahedral voids is \frac{a}{\sqrt{2}}, which is obtained by dividing the face diagonal \sqrt{2}a by 2.

Thus, the correct answer is: \frac{a}{\sqrt{2}}.