Prove the interatomic distance in lennard Jones 6-12 potential?

Answer 1

The Lennard-Jones potential is expressed as ( V(r) = 4\epsilon \left[ \left( \frac{\sigma}{r} \right)^{12} - \left( \frac{\sigma}{r} \right)^{6} \right] ), where ( r ) is the interatomic distance, ( \epsilon ) is the depth of the potential well, and ( \sigma ) is the distance at which the potential is zero. To find the equilibrium interatomic distance, we take the derivative of ( V(r) ) with respect to ( r ) and set it to zero: ( \frac{dV}{dr} = 0 ). This leads to the condition ( r = 2^{1/6}\sigma ) for the minimum potential energy, indicating the distance at which atoms experience a balance between attractive and repulsive forces.