**Interatomic
Force Constant**

Behaviour of solids with respect to external forces is such that if their atoms are connected to springs. When an external force is applied on a solid, this distance between its atoms changes and interatomic force works to restore the original dimension.

The ratio of interatomic force to that of change in interatomic distance is defined as the interatomic force constant.

The interatomic force F’ developed between the atoms is directly proportional to the change in distance between them.

F’ = kΔr. Where, k = Interatomic Force Constant.

Let, L be the length of a wire and r₀ be the interatomic distance between the atoms of the wire. Suppose, when a force F is applied, the length of the wire increases by l, and the distance between its atoms increases from r₀ to r₀ + Δr. Then, longitudinal strain,

\(\frac{l}{L}\,\,=\,\,\frac{\Delta r}{{{r}_{0}}}\).

If A is the area of cross section of the wire, it will have A/ r² chains of atoms. Therefore, the interatomic force is:

\(F’\,\,=\,\,\frac{External\,\,Force}{Number\,\,of\,\,chains}=\frac{F}{A/{{r}^{2}}}=\frac{F{{r}_{0}}^{2}}{A}\).

Interatomic force constant is:

\(k=\frac{{{F}^{‘}}}{\Delta r}=\frac{F{{r}_{0}}^{2}/A}{\Delta r}=\frac{F}{A}\frac{{{r}_{0}}^{2}}{\Delta r}\).

\(k=\frac{F}{A}\frac{L}{l}{{r}_{0}}=\frac{F/A}{l/L}{{r}_{0}}\).

But by definition:

\(\frac{F/A}{l/L}=Y\).

Where, Y = Young’s Modulus of the material of the wire.

∴ k = Yr₀

Therefore, Interatomic force constant k is equal to the product of Young’s modulus of the wire and the normal distance r₀ between the atoms of the wire.