Question

The potential energy function for the force between two atoms in a diatomic molecule can be expressed approximately as $U\left( r \right) = \dfrac{a}{{{r^{12}}}} - \dfrac{b}{{{r^6}}}$ where $a$ and $b$ are constants and $r$ is the separation between the two atoms.
A) Determine the force function $F\left( r \right)$ .
B) Find the value of $r$ for which the molecule will be in stable equilibrium.

Answer 1

The force between the two atoms in the given diatomic molecule is due to the interaction between the electrons and the nuclei of each atom. It is a function of the distance of separation between the two atoms in the given diatomic molecule. The force function can be obtained by taking the first derivative of the potential function with respect to the distance of separation between the two atoms.

Formula used:
-The force between the two atoms in a molecule is given by, $F\left( r \right) = \dfrac{{dU}}{{dr}}$ where $U$ is the potential energy due to the force between the atoms and $r$ is the distance of separation between the atoms.

Complete step by step solution:
Step 1: Express the given potential energy function for the force between the two atoms and obtain the expression for the force function.
The potential energy function for the force between two atoms in a diatomic molecule is given to be $U\left( r \right) = \dfrac{a}{{{r^{12}}}} - \dfrac{b}{{{r^6}}}$ --------- (1) where $a$ and $b$ are constants and $r$ is the separation between the two atoms.
Take the derivative of equation (1) with respect to $r$ so as to obtain the force function $F\left( r \right)$ .
i.e., $F\left( r \right) = \dfrac{{dU\left( r \right)}}{{dr}}$ -------- (2)
Substituting equation (1) in (2) we get, $F\left( r \right) = \dfrac{d}{{dr}}\left[ {\dfrac{a}{{{r^{12}}}} - \dfrac{b}{{{r^6}}}} \right] = \dfrac{d}{{dr}}\left[ {a{r^{ - 12}} - b{r^{ - 6}}} \right]$
Taking the derivative we get, $F\left( r \right) = - 12a{r^{ - 13}} - \left( { - 6} \right)b{r^{ - 7}} = 6b{r^{ - 7}} - 12a{r^{ - 13}}$
a) The force function is obtained as $F\left( r \right) = 6b{r^{ - 7}} - 12a{r^{ - 13}}$ .
Step 2: Apply the condition for stable equilibrium to obtain the required value of $r$ .
For a stable equilibrium, $F\left( r \right) = 0$ .
$\Rightarrow F\left( r \right) = 6b{r^{ - 7}} - 12a{r^{ - 13}} = 0$
Simplifying by taking the common factors to the LHS of the above equation we get, $6{r^{ - 7}}\left( {b - \dfrac{{a{r^{ - 6}}}}{2}} \right) = b - \dfrac{{a{r^{ - 6}}}}{2} = 0$
$\Rightarrow {r^{ - 6}} = \dfrac{{2b}}{a}$ or, $r = {\left( {\dfrac{{2b}}{a}} \right)^{ - 1/6}}$
b) Thus at $r = {\left( {\dfrac{{2b}}{a}} \right)^{ - 1/6}}$ the molecule exists at a stable equilibrium.

Note: For a large distance of separation between the two atoms, the force is attractive in nature. But as the distance of separation decreases and crosses the equilibrium distance, the force becomes repulsive in nature. The mathematical expression of the obtained force function $F\left( r \right) = 6b{r^{ - 7}} - 12a{r^{ - 13}}$ , suggests this double nature of the force between the two atoms. When the molecule is in a stable equilibrium, the potential energy drops to zero. So we take the force between the two atoms to be zero.