Question

Fig shows the variation of potential energy U between two molecules as a function of intermolecular separation x between them. The values of x at points A, B and C respectively are 0.58 A; 1.3 A and 1.9 A. The intermolecular forces at these points respectively are

(1) repulsive; zero; attractive

(2) zero; repulsive; attractive

(3) zero; repulsive; attractive

(4) attractive; zero; repulsive

Answer 1

(1) repulsive; zero; attractive

The force F = -(dU/dr) therefore

At A; potential energy in positive. So the force is repulsive.

At B; PE is minimum, therefore F = -dU/dr = zero

At C; potential energy is negative. So the force is attractive.