An element crystallizes in a face-centred cubic (FCC) lattice. The distance between the nearest neig — Solid State Chemistry Question

💡 Solution & Explanation

In an FCC lattice, atoms touch along the face diagonal. The face diagonal is $\sqrt{2}a$ and contains four radii, so the nearest-neighbour distance $2r=\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}$.