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:

In a face-centred cubic (fcc) lattice, the octahedral voids are located at two kinds of positions inside every unit cell.

• The first kind lies at the body centre with fractional coordinates $$\left(\frac12,\frac12,\frac12\right).$$

• The second kind lies at the centres of all twelve edges. For example, one such void is at $$\left(\frac12,0,0\right),$$ another at $$\left(0,\frac12,0\right),$$ and so on.

To find the shortest possible distance between any two octahedral voids, we choose the pair of voids that are geometrically closest. Inspection of the coordinates shows that the void at $$\left(\frac12,0,0\right)$$ is nearest either to the body-centre void $$\left(\frac12,\frac12,\frac12\right)$$ or to the adjacent edge-centre void $$\left(0,\frac12,0\right).$$ We shall calculate one of these distances explicitly; both give the same result.

Using the three-dimensional distance formula,

$$d \;=\; a\,\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2},$$

we substitute $$\bigl(x_1,y_1,z_1\bigr)=\left(\frac12,0,0\right)$$ and $$\bigl(x_2,y_2,z_2\bigr)=\left(0,\frac12,0\right):$$

$$\begin{aligned} d &= a\,\sqrt{\left(0-\frac12\right)^2 + \left(\frac12-0\right)^2 + (0-0)^2} \\ &= a\,\sqrt{\left(\frac12\right)^2 + \left(\frac12\right)^2 + 0} \\ &= a\,\sqrt{\frac14 + \frac14} \\ &= a\,\sqrt{\frac12} \\ &= \frac{a}{\sqrt2}. \end{aligned}$$

No other pair of octahedral voids is closer than this value, so $$\dfrac{a}{\sqrt{2}}$$ is the minimum (nearest-neighbour) separation required by the question.

Hence, the correct answer is Option A.