Final answer:
The distance between nearest neighbours depends on the crystal structure. In an FCC structure, the distance is √2 times the atomic radius, while in a BCC structure, it is √3/2 times the atomic radius. In a simple cubic structure, this distance is twice the atomic radius.
Step-by-step explanation:
The relationship between the radius of atoms and the distance between the nearest neighbours varies depending on the crystal structure: simple cubic (SC), face-centred cubic (FCC), and body-centred cubic (BCC). For an FCC structure, atoms touch across the face diagonal, meaning the face diagonal is 4 times the atomic radius (d = 4r). Since the face of the cube is a square, we can apply Pythagoras theorem where the diagonal (d) is √2 times the side of the cube (a), so d = a√2. Therefore, the relation between the atomic radius (r) and the side of the cube (a) is a = 2√2r, which follows that the distance between nearest neighbours in an FCC structure is √2 times the radius.
For a BCC structure, the central atom touches the corner atoms forming a body diagonal which is 4 times the atomic radius. Since the cube's body diagonal is √3 times the side of the cube, the relation between the radius (r) and the cube edge (a) is given by a = 4r/√3. Thus, in a BCC structure, the distance between nearest neighbours is √3/2 times the radius. Simple cubic structures have the most straightforward relationship; the distance between nearest neighbours is equal to the side of the cube, which is twice the atomic radius.