Interaction Between Two Atoms
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A commonly used potential energy function to describe the interaction between two atoms is the Lennard-Jones potential
U=e((Ro/R)^12-2(Ro/R)^6)
a) Show that the radius at the potential minimum is Ro, and that the depth of the potential well is e.
b) Find the frequency of small oscillations about equilibrium for 2 identical atoms of mass m bound to each other by the Lennard-Jones interaction.
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Solution Summary
The radius at the potential minimum is computed. The solution finds the frequency of small oscillations about equilibrium for 2 identical atoms.
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A commonly used potential energy function to describe the interaction between two atoms is the Lennard-Jones potential
U=e ((Ro/R) ^12-2(Ro/R) ^6)
a) Show that the radius at the potential minimum is Ro, and that the depth of the potential well is e.
The Potential Energy U varies with R as given by the Lennard-Jones equation
U = e [(R0/R) 12 – 2(R0/R) 6]
In order for the Potential Energy function to be a minimum, it is necessary that the derivative of U with respect to R is zero.
Differentiating both sides with respect to R, we get the following:
dU/dR = e { [ (R0) 12* -12* (1/R)13 ] – [ 2*R06*-6*(1/R)7 ] }
I am not sure how familiar you are with differential equations, so, I will, at the risk of sounding boring, repeat the formula:
e is a constant. So, e can stay outside. When you differentiate (R0/R) 12 with respect to R, it is similar to differentiating R012* 1/R12. R012 is a constant and 1/R12 is the same as
R-12. Differential of Rn is n*R (n – ...
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