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The following equation gives the variation of interatomic energy with atomic distance: V(r)=4ε Where V(r) is the potential energy and r is interatomic distance. ε and σ are constant. Find the value of interatomic distance where energy is minimum?

The following equation gives the variation of interatomic energy with atomic distance: V(r)=4ε Where V(r) is the potential energy and r is interatomic distance. ε and σ are constant. Find the value of interatomic distance where energy is minimum? Correct Answer (σ/r12)−(σ/r6)

For minimum energy => d(V(r))/d(r)=0 => d]/dr=0 => 4 ε=0 => 2 σ6=r6 => r = 2(1/6) σ => r = 1.2 σ.

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