The potential energy of two atoms in a diatomic molecule approximately by U(r) =

Question

The potential energy of two atoms in a diatomic molecule approximately by U(r) = a/r^12 - b/r^(Lennard-Jones potential), where r is the spacing between atoms, and a and b are positive constants. a) Find the force F(r) on one atom as a function of r. Make two graphs, one of U(r) versus r and one of F(r) versus r. b) Find the equilibrium distance between the two atoms. c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it, that is, to separate the two atoms by an infinite distance? This is called the dissociation energy of the molecule.

Explanation / Answer

Given,

U(r)= a/(r)^12 -b/(r)^6

r= Spacing Between the atoms

a, b= Constant

According to the formula,

Fdx = dWork ,

and PE = -Work ,

SO, F = - dW/dx

Therefore,

Force applied on one atom can be calculated by:-

F(r) = -dU(r)/dr

In this case,

dU(r)/dr = -12a/r^13 + 6b/r^7

F(r) = -dU(r)/dr = 12a/r^13 - 6b/r^7 (Ans)

when at equilibrium, the potential energy is minimum (because the force is zero), Therefore,

0 = 12a/r^13 - 6b/r^7

r^6 = 2*a/b

There are 2 real solutions for this equation, and only the root with a positive value of r has meaning.

Let us assume that a and b are both positive numbers:

r = (2*a/b)^(1/6) (ans)

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