Question

The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $\begin{array}{l}U\left(x\right)=\frac{a}{x^{12}}–\frac{b}{x^6}\end{array}$, where $a$ and $b$ are constants and $x$ is the distance between the atoms. If the dissociation energy of the molecule is $\begin{array}{l}D=\left[U\right(x=\mathrm{\infty })–U_{atequilibrium}]\end{array}$, then $D$ is equal to?

Answer 1

Step 1: Given data

$\begin{array}{l}U\left(x\right)=\frac{a}{x^{12}}–\frac{b}{x^6}\end{array}$ ; Here $U$ is potential energy, $a$ and $b$ are constant, and $x$ is distance between atom

$\begin{array}{l}U(x=\mathrm{\infty })=0\end{array}$

Step 2: Formula used

$\begin{array}{l}F=-\frac{dU}{dx}\end{array}$ ; Where $dU$ is change in potential energy, $x$ is distance between atom, and $F$ is force.

Step 3: Calculation for D

$\begin{array}{l}U\left(x\right)=\frac{a}{x^{12}}–\frac{b}{x^6}\end{array}$ ------ (i)

As $\begin{array}{l}F=-\frac{dU}{dx}\end{array}$

$\begin{array}{l}F=-\frac{dU}{dx}=-\left[\frac{d}{dx}\left(\frac{a}{x^{12}}\right)-\frac{d}{dx}\left(\frac{b}{x^6}\right)\right]\end{array}$

$\begin{array}{l}F=-\frac{dU}{dx}\end{array}=\left[-\frac{12a}{x^{13}}+\frac{6b}{x^7}\right]$

$\begin{array}{l}F\end{array}=-\frac{12a}{x^{13}}+\frac{6b}{x^7}$

$atequilibrium,F=0$

Therefore,

$$\begin{gathered} 0=-\frac{12a}{x^{13}}+\frac{6b}{x^7} \\ \Rightarrow \frac{12a}{x^{13}}=\frac{6b}{x^7} \end{gathered}$$

$\Rightarrow \frac{x^{13}}{x^7}=\frac{12a}{6b}$

$\Rightarrow \begin{array}{l}{x}^6=\frac{2a}{b}\end{array}$

Now putting value $\begin{array}{l}{x}^6=\frac{2a}{b}\end{array}$in equation (i)

$$\begin{gathered} U_{atequilibrium}=\frac{a}{{\left(\frac{2a}{b}\right)}^2}–\frac{b}{\left(\frac{2a}{b}\right)} \\ \Rightarrow U_{atequilibrium}=\frac{a-b\left(\frac{2a}{b}\right)}{{\left({\displaystyle \frac{2a}{b}}\right)}^2} \\ \Rightarrow U_{atequilibrium}=\frac{a-2a}{{\displaystyle \left(\frac{{\displaystyle 4a^2}}{{\displaystyle b^2}}\right)}} \\ \Rightarrow U_{atequilibrium}=\frac{-a}{{\displaystyle \left(\frac{{\displaystyle 4a^2}}{{\displaystyle b^2}}\right)}} \\ \Rightarrow U_{atequilibrium}=-a\left(\frac{b^2}{4a^2}\right) \\ \Rightarrow U_{atequilibrium}=-\frac{b^2}{4a} \end{gathered}$$

$\begin{array}{l}D=\left[U\right(x=\mathrm{\infty })–U_{atequilibrium}]=\frac{-b^2}{4a}\end{array}$

Therefore, the value will be $\begin{array}{l}D=\frac{-b^2}{4a}\end{array}$.