Lennard-Jones potential

The chemical bond between two atoms results from the interaction between them.

Let us consider two atoms at a distance $r$. An empirical potential that describes the interaction between them is the potential Lennard-Jones,

$$ V(r) = D\left(\frac{R}{r}\right)^{12} - 2 \left(\frac{R}{r}\right)^6 $$

where $D, R$ are constants.

[This form of a potential describes Van der Waals bonds between atoms, that is, dipole-dipole interaction.]

Note that

$$ \begin{aligned} & V(r) \to \infty,\qquad \text{when}\;\; r\to 0 \\ & V(r) \to 0,\qquad\,\, \text{when}\;\; r\to \infty. \end{aligned} $$

The derivative is

$$ \frac{dV}{dr} = \frac{12}{r}\,\left[ \left(\frac{R}{r}\right)^6 - D \left(\frac{R}{r}\right)^{12} \right] $$

Solid line: Lennard-Jones potential. Dashed line: Parabolic approximation. (Values of parameters $D=1, R=5$.)

We have a minimum when

$$ \frac{dV}{dr}\Big|_{r=r_0} = 0 \Rightarrow r_0 = D^{1/6} R. $$

Its value is

$$ V(r=r_0) = -\frac{1}{D}. $$

[We also find $\frac{d^2V}{dr^2}(r=r_0) = \frac{72}{R^2 D^{4/3}} > 0$.]

• The parameter $D$ sets the depth of the minimum of the potential, that is, it defines how strong is the chemical bond.
• The parameter $R$ sets the distance between atoms when they are in the equilibrium position.

Questions.

• What is the motion of the systems when its energy has the value $E_0 = -1/(2D)$?
• Can the energy have the value $E_0 = -2/D$;
• What is the motion of the systems when its energy has the value $E_0 > 0$?

Solid line: Lennard-Jones potential. Dashed line: Parabolic approximation. (Values of parameters $D=1, R=5$.)