Simple question about potential energy of two-atom system

[Signifier]

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).

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 $$(1/2)kr^2$$?

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.

 

[Signifier]

Still need some help... Anyone?

 

[m2003]

I would first commend you for taking the initiative to seek help and clarification on a topic you are struggling with. It shows dedication and a desire to fully understand the material.

To address your question, the potential energy of a two-atom system can be calculated by summing the individual potentials of each atom. In this case, we have one fixed atom and one moving atom, so the potential energy would be the potential energy of the moving atom plus the potential energy of the fixed atom.

For the moving atom, we have the Lennard-Jones potential, U(r) = -A/(r^6) + B/(r^12), where r is the distance between the two atoms. This potential represents the interaction between the two atoms due to van der Waals forces and electron overlap.

For the fixed atom, we can assume that it has no potential energy since it is not moving. Therefore, the total potential energy of the system would be U(r) = -A/(r^6) + B/(r^12).

Now, to incorporate the contribution of the spring, we can add another term to the potential energy equation. The potential energy of a spring is given by (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, the displacement x would be the distance between the two atoms, r. Therefore, the potential energy of the spring would be (1/2)kr^2.

Combining these two potentials, we get the total potential energy of the system as U(r) = (1/2)kr^2 - A/(r^6) + B/(r^12).

To determine the distance at which instability occurs and the tip "jumps" into contact with the surface, we can use the concept of equilibrium. At equilibrium, the forces acting on the system are balanced, meaning that the potential energy of the system is at a minimum. So, we can find the distance at which the potential energy is at its minimum by taking the derivative of the potential energy equation and setting it equal to 0. This will give us the distance at which the tip will jump into contact with the surface.

I hope this helps clarify the potential energy function of the two-atom system. If you are still struggling, I suggest seeking help from your instructor or a tutor. Keep up the good work!