Question

The Lennard-Jones potential energy, U(x), is a function of “x” the separation distance between a pair of atoms in a molecule. It is frequently called the van der Walls potential between atoms for dipole-dipole interactions. The potential is

U(x) = C12/ x¹² - C6/x⁶,                                                                    [1]

Where

C₁₂ = 1.51 x 10^-134 J∙m¹²

C₆ = 1.01 x 10^-77 J∙m⁶.

(a) Determine the shape of the potential energy curve vs. distance of separation (x) between the atoms using Equation [1] by plotting the equation to scale on graph paper for “x” ranging from 0.3 – 1.0 nm.

(b) Calculate the equilibrium distance “x_e” between the atoms at the minimum potential energy from Equation [1].

(c) Assuming the atoms are spherical balls with a spring between them, what is the spring constant (k) of the bond as atoms are pulled apart from their equilibrium position? Recall that F (force) = dU/dx, and F = kx.

Answer 1

Solution:

The Lennard-Jones model consists of two 'parts'; a steep repulsive term, and smoother attractive term, representing the London dispersion forces, for this ,model the equation is given by,

                 ----> {C₁₂ = 1.51E-134 J/m¹² , C₆ = 1.01 E-77 J/m⁶}

The values are tabulated and the plot is represented graphically,

C12	1.51E-134	J/m12
C6	1.01E-77	J/m6
x (nm)	U(x)
3.00E-10	1.46E-20
4.00E-10	-1.57E-21
5.00E-10	-5.85E-22
6.00E-10	-2.10E-22
7.00E-10	-8.48E-23
8.00E-10	-3.83E-23
9.00E-10	-1.90E-23
1.00E-09	-1.01E-23

                                                 

b) from the plot Xe = 0.406 nm

c) Force

by substiyuting th values we get,

F(x) = 2.17E-21