Question

The potential energy function for the force existing between two atoms in a diatomic molecule is approximately given by $U(x)=\dfrac{a}{{{x}^{12}}}-\dfrac{b}{{{x}^{6}}}$ , where “a” and “b” are constants and (x) is the distance between the two atoms. If the dissociation energy of the molecule is given by: $D=[U(x=\infty )-{{U}_{at\text{ }equilibrium}}]$ , then D is equal to:
$\begin{align}
  & (A)\dfrac{{{b}^{2}}}{2a} \\
 & (B)\dfrac{{{b}^{2}}}{12a} \\
 & (C)\dfrac{{{b}^{2}}}{4a} \\
 & (D)\dfrac{{{b}^{2}}}{6a} \\
\end{align}$

Answer 1

Hint: In order to calculate the dissociation energy, we shall first calculate both the terms in its equation. The potential energy at infinity can be calculated by substituting the distance as infinity in the given expression. Also, in order to calculate the potential energy at equilibrium, we will first find the force by differentiating the potential energy expression and equating it to zero to get the condition for equilibrium.

Complete answer:
The given expression of potential energy as a function of distance is:
$\Rightarrow U(x)=\dfrac{a}{{{x}^{12}}}-\dfrac{b}{{{x}^{6}}}$
Now, to calculate the potential energy at infinity, we put$x=\infty $. Therefore, we get the potential energy at infinity as:
$\Rightarrow U(x=\infty )=0$
Now, the force can be deduced from the potential energy expression as:
$\Rightarrow F=-\dfrac{d[U(x)]}{dx}$
Putting the expression and simplifying, we have:
$\begin{align}
  & \Rightarrow F=-\dfrac{d\left( \dfrac{a}{{{x}^{12}}}-\dfrac{b}{{{x}^{6}}} \right)}{dx} \\
 & \Rightarrow F=-\left[ -\dfrac{12a}{{{x}^{13}}}+\dfrac{6b}{{{x}^{7}}} \right] \\
 & \Rightarrow F=\dfrac{12a}{{{x}^{13}}}-\dfrac{6b}{{{x}^{7}}} \\
\end{align}$
Since, at equilibrium $F=0$, thus we have:
$\begin{align}
  & \Rightarrow \dfrac{12a}{{{x}^{13}}}-\dfrac{6b}{{{x}^{7}}}=0 \\
 & \Rightarrow {{x}^{6}}=\dfrac{2a}{b} \\
\end{align}$
Therefore potential energy at equilibrium is equal to:
$\begin{align}
  & \Rightarrow {{U}_{equilibrium}}=\dfrac{a}{{{\left( \dfrac{2a}{b} \right)}^{2}}}-\dfrac{b}{\dfrac{2a}{b}} \\
 & \Rightarrow {{U}_{equilibrium}}=\dfrac{{{b}^{2}}}{4a}-\dfrac{{{b}^{2}}}{2a} \\
 & \Rightarrow {{U}_{equilibrium}}=-\dfrac{{{b}^{2}}}{4a} \\
\end{align}$
Thus, the Dissociation energy, that is, the difference in potential energies of the diatomic molecule at infinity and equilibrium, can now be calculated as:
$\Rightarrow D=[U(x=\infty )-{{U}_{at\text{ }equilibrium}}]$
Putting the values of both the potential energy, we get:
$\begin{align}
  & \Rightarrow D=\left[ 0-\left( -\dfrac{{{b}^{2}}}{4a} \right) \right] \\
 & \therefore D=\dfrac{{{b}^{2}}}{4a} \\
\end{align}$
Hence, the Dissociation energy comes out to be $\dfrac{{{b}^{2}}}{4a}$ .

Hence, option (C) is the correct option.

Note:
In this problem, we did not see the use of any formulas or its derivation, but a simple condition that, at equilibrium the net force is equal to zero. This in turn with some data given in the problem guided us in our complete solution. So, it’s important to remember such conditions like, when will a body be in equilibrium. Is it stable or unstable, so on.