Question

The potential energy between two atoms in a molecule is given by $U\left( x \right) = \dfrac{a}{{{x^{12}}}} - \dfrac{b}{{{x^6}}}$, where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atom is in stable equilibrium when
A. $x = \root 6 \of {\dfrac{{11a}}{{5b}}} $
B. $x = \root 6 \of {\dfrac{a}{{2b}}} $
C. $x = 0$
D. $x = \root 6 \of {\dfrac{{2a}}{b}} $

Answer 1

Hint:In this question, the value of potential difference is given as $U\left( x \right) = \dfrac{a}{{{x^{12}}}} - \dfrac{b}{{{x^6}}}$. We will calculate the value of $x$ by using the condition of stable equilibrium of forces which is the first derivative of the potential energy will be zero. For this we will calculate the first derivative of the potential energy of the atoms.

Formula used:
The condition used for calculating the value of $x$ is given below
$F\left( x \right) = - \dfrac{{dU}}{{dx}} = 0$
This is the condition of stable equilibrium of forces.
Here, $F\left( x \right)$ is the force between the two atoms, $x$ is the distance of separation between the atoms, $a$ and $b$ are positive constants

Complete step by step answer:
Consider two atoms in a molecule that is separated by a certain distance $x$. The potential energy between the two atoms in a molecule is given below
$U\left( x \right) = \dfrac{a}{{{x^{12}}}} - \dfrac{b}{{{x^6}}}$
Here $a$ and $b$ are the positive constants and $x$ is the distance between the atoms.
Now, we will apply the condition of stable equilibrium to find the value of $x$, which is given below
$F\left( x \right) = - \dfrac{{dU}}{{dx}} = 0$
Now, we will calculate the first derivative of the potential energy equation as shown below
$\dfrac{{dU}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{a}{{{x^{12}}}} - \dfrac{b}{{{x^6}}}} \right)$
$ \Rightarrow \,\dfrac{{dU}}{{dx}} = \dfrac{d}{{dx}}\left( {a{x^{ - 12}} - b{x^{ - 6}}} \right)$
$ \Rightarrow \,\dfrac{{dU}}{{dx}} = \left( { - 12a{x^{ - 13}} + 6b{x^{ - 7}}} \right)$
$ \Rightarrow \,\dfrac{{dU}}{{dx}} = 6b{x^{ - 7}} - 12a{x^{ - 13}}$
Putting this value in the above condition, we get
$F\left( x \right) = 6b{x^{ - 7}} - 12a{x^{ - 13}} = 0$
$ \Rightarrow \,6b{x^{ - 7}} = 12a{x^{ - 13}}$
$ \Rightarrow \,\dfrac{{{x^{13}}}}{{{x^7}}} = \dfrac{{12a}}{{6b}}$
$ \Rightarrow \,{x^6} = \dfrac{{2a}}{b}$
$ \therefore \,x = \root 6 \of {\dfrac{{2a}}{b}} $
Therefore, the atom is in stable equilibrium when $x = \root 6 \of {\dfrac{{2a}}{b}} $.

Hence, option D is the correct option.

Note:Now, for a large distance of separation between the two atoms, the force will be attractive in nature. But when the distance of separation between the atoms will decrease and cross the distance of equilibrium, the force will become repulsive in nature. Now, as we know that when the molecule will be in stable equilibrium, the potential of the molecule will drop to zero. That is why, we have taken the force between the two atoms of molecules to be zero.