Question

The potential energy function for either one of the two atoms in a diatomic molecule is often approximated by U(x)=a/x12−b/x6 where x is the distance between the atoms.

(a) At what distance of separation does the potential energy have a local minimum (not at x=[infinity])?
(b) What is the force on an atom at this separation?
(c) How does the force vary with the separation distance?

Answer 1

The local minimum of potential energy in a diatomic molecule occurs at the bond length, calculated by setting the derivative of the Lennard-Jones potential to zero. The force on an atom at this point is determined by taking the negative derivative of the potential energy. The force varies with the separation distance, being attractive at distances smaller than the equilibrium and repulsive at larger distances.

Step-by-step explanation:

The question involves finding the distance at which the potential energy of a diatomic molecule has a local minimum and understanding the nature of the force acting on the atoms at that distance. According to the Lennard-Jones potential, the equilibrium distance (often called the bond length) occurs where the derivative of the potential energy function with respect to the separation distance is zero. This is found by taking the derivative of U(x) = a/x12 − b/x6 with respect to x and setting it to zero.

(a) To find the point of local minimum, set the derivative dU/dx equal to zero and solve for x.
(b) The force on an atom at this separation can be calculated by taking the negative gradient of U, which is −dU/dx.
(c) The force varies with separation distance such that when x < xmin, the force is positive (attractive), and when x > xmin, the force is negative (repulsive). The force reaches a maximum when the atoms are too close and decreases as they move far apart, corresponding to x approaching infinity.