Question

If the potential energy of two molecules is given by, $ U = \dfrac{A}{{{r^{12}}}} - \dfrac{B}{{{r^6}}} $ then at the equilibrium position, its potential energy is equal to:
A. $ \dfrac{{{A^2}}}{{4B}} $
B. $ - \dfrac{{{B^2}}}{{4A}} $
C. $ \dfrac{{2B}}{A} $
D. $ 3A $

Answer 1

Hint
We know that the derivative of the potential energy is the force and the force zero at equilibrium condition. So, utilizing these, we can obtain the expression for the potential energy at equilibrium condition.
In this solution we will be using the following formula,
 $ \Rightarrow F = - \dfrac{{dU}}{{dr}} $
where $ F $ is the force and $ U $ is the potential energy.

Complete step by step answer
The formula of force in terms of the potential energy is given as,
 $ \Rightarrow F = - \dfrac{{dU}}{{dr}} $
According to the question, the potential energy of two molecules is given by, $ U = \dfrac{A}{{{r^{12}}}} - \dfrac{B}{{{r^6}}} $
So now substituting this value of potential energy in the formula for the force we get,.
 $ \Rightarrow F = - \dfrac{d}{{dr}}\left( {\dfrac{A}{{{r^{12}}}} - \dfrac{B}{{{r^6}}}} \right) $
Upon performing the differentiation operation on both the terms individually, the above expression reduces to the below forma as,
 $ \Rightarrow F = \dfrac{{12A}}{{{r^{13}}}} - \dfrac{{6B}}{{{r^7}}} $
The net force, that is, the sum of the forces acting on a body equals to zero, in an equilibrium condition. So, equate the above equation to zero.
Thus, we get,
 $ \Rightarrow 0 = \dfrac{{12A}}{{{r^{13}}}} - \dfrac{{6B}}{{{r^7}}} $
Hence we can write this as,
 $ \Rightarrow \dfrac{{12A}}{{{r^{13}}}} = \dfrac{{6B}}{{{r^7}}} $
Now cancelling $ {r^7} $ from both the sides and then taking reciprocal we get,
 $ \Rightarrow \dfrac{{{r^6}}}{{12A}} = \dfrac{1}{{6B}} $
Therefore on rearranging we get,
 $ \Rightarrow {r^6} = \dfrac{{2A}}{B} $
Substitute the above expression in the given expression of the potential energy of the two molecules.
So, the expression further reduces as follows.
 $ \Rightarrow U = \dfrac{A}{{{{\left( {{\raise0.7ex\hbox{ $ {2A} $ } \!\mathord{\left/
 {\vphantom {{2A} B}}\right.}
\!\lower0.7ex\hbox{ $ B $ }}} \right)}^2}}} - \dfrac{B}{{\left( {{\raise0.7ex\hbox{ $ {2A} $ } \!\mathord{\left/
 {\vphantom {{2A} B}}\right.}
\!\lower0.7ex\hbox{ $ B $ }}} \right)}} $
we can then remove the brackets to get,
 $ \Rightarrow U = \dfrac{A}{{{\raise0.7ex\hbox{ $ {4{A^2}} $ } \!\mathord{\left/
 {\vphantom {{4{A^2}} {{B^2}}}}\right.}
\!\lower0.7ex\hbox{ $ {{B^2}} $ }}}} - \dfrac{{{B^2}}}{{2A}} $
On cancelling the similar terms,
 $ \Rightarrow U = \dfrac{{{B^2}}}{{4A}} - \dfrac{{{B^2}}}{{2A}} $
On calculating we get,
 $ \Rightarrow U = \dfrac{{{B^2} - 2{B^2}}}{{4A}} = \dfrac{{ - {B^2}}}{{4A}} $
Thus, the value of the potential energy equals $ - \dfrac{{{B^2}}}{{4A}} $
 $ \therefore $ If the potential energy of two molecules is given by, $ U = \dfrac{A}{{{r^{12}}}} - \dfrac{B}{{{r^6}}} $ then at the equilibrium position, its potential energy is equal to $ - \dfrac{{{B^2}}}{{4A}} $ .
Thus, the option (B) is correct.

Note
The potential energy of a body is the energy it possesses with respect to its position, state or arrangement. In this case the potential energy is negative because energy needs to be provided to it in order to break the bond between them.