# Simple question about potential energy of two-atom system

1. Homework Statement
Two atoms approach each other. One is part of a moving tip (IE like an STM tip), one is part of a fixed surface (so only one atom is moving). The two atoms interact via the Lennard-Jones potential, IE U(r) = $$-A/(r^6) + B/(r^12)$$ where r is the distance between the two atoms. The bottom atom, as said, is fixed; while the top atom can be modeled as if suspended from the end of a spring of stiffness k (IE, this model is used in place of having to calculate the total potential of the system adding all the L-J potentials of all the atoms in the STM tip and the surface).

The question of the problem is, at what distance between the two atoms will instability and occur and the tip "jumps" into contact with the surface?

However, I am not having a problem with that part. I am not quite sure how to construct the potential energy function of this system to begin with! (Note: this problem is from Israelachvili's Intermolecular and Surface Forces, which I've just begun reading).

2. Homework Equations

My real problem here is my utter inability with springs. I know that if the two atoms are separated by a distance r, the L-J potential will be $$-Ar^-6 + Br^-12$$, where the first term is the attractive vdw interaction and the second term is the repulsive electron overlap (at smaller distances). But what is the contribution to system potential due to the spring? Is it $$(1/2)kr^2$$?

3. The Attempt at a Solution

My attempt is: the potential, U(r) = $$(1/2)kr^2 - A(r^-6) + B(r^-12)$$

This potential btw doesn't return the right numerical values (IE for the distances of instability), so I am doubting it's right. Any help would be great.