Question

The potential energy of a diatomic molecule (a two-atom system like H₂ or O₂) is given by $\mathit{U}\mathbf{=}\frac{\mathbf{A}}{{\mathbf{r}}^{\mathbf{12}}}\mathbf{-}\frac{\mathbf{B}}{{\mathbf{r}}^{\mathbf{6}}}$Where, ris the separation of the two atoms of the molecule and Aand Bare positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation, that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

Short answer

1. The distance of equilibrium separation between the atoms is,$r_{eq}=1.12{\left(\frac{A}{B}\right)}^{\frac{1}{6}}$.
2. Theforce for$r<r_{eq}$will be positive or repulsive.
3. The force for role="math" localid="1661225764262" $r>r_{eq}$will be negative or attractive.

Step-by-step solution

Step 1: Given Data

The equation of potential energy of diatomic molecule is,

$U=\frac{A}{r^{12}}-\frac{B}{r^6}$

Determining the concept

Use the equation of force relating to the potential energy to determine${\mathit{r}}_{\mathbf{e}\mathbf{q}}$. Use the condition for minima to check whether the potential energy at${\mathit{r}}_{\mathbf{eq}}$is minimum or maximum. Then, state the nature oftheforce directly from the property of minima.

Formula is as follow:

$F=-\frac{dU}{dr}$

Step 3(a): Determining the distance of equilibrium separation between the atoms

The equation for the force relating with potential energy function is,

$F=-\frac{dU}{dr}$

So, applying this equation for$r=r_{eq}$

$$\begin{gathered} F=-\frac{12A}{r^{13}}+\frac{6B}{r^7} \\ =0 \end{gathered}$$

Solving this equation,

$r_{eq}=1.12{\left(\frac{A}{B}\right)}^{\frac{1}{6}}$

Hence,the distance of equilibrium separation between the atoms is, $r_{eq}=1.12{\left(\frac{A}{B}\right)}^{\frac{1}{6}}$.

Step 4(b): Determining the nature of force if the separation is smaller than the equilibrium

Now, check whether the value of$r_{eq}$has minimum or maximum potential energy.

So, to check it, take the double derivative ofthepotential energy function,

${\left[\frac{d^{2}U}{d{r}^2}\right]}_{r=r_{eq}}=\frac{dF}{dr}=\frac{12\times 13A}{r^{14}}-\frac{6\times 7B}{r^8}>0$

This proves that, the equilibrium separation has minimum potential energy.

So, for the$r<r_{eq}$ the force will be positive or repulsive.

Step 5(c): Determining the nature of force if the separation is larger than the equilibrium

As calculated in part b, as $r=r_{eq}$is minimum, the force for $r>r_{eq}$will be negative or attractive.

Therefore, the equation for force relating potential energy and the condition of minima can be used to solve this problem.