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Question: In a Van Der Waals interaction: \( U = {U_0}\left[ {{{\left( {\dfrac{{{R_0}}}{R}} \right)}^{12}} - 2...

In a Van Der Waals interaction: U=U0[(R0R)12−2(R0R)6]U = {U_0}\left[ {{{\left( {\dfrac{{{R_0}}}{R}} \right)}^{12}} - 2{{\left( {\dfrac{{{R_0}}}{R}} \right)}^6}} \right]
A small displacement xx is given from equilibrium position r=R0r = {R_0} . Find the approximate PE function.
\left( A \right)\dfrac{{36{U_0}}}{{{R^2}}}{x^2} - {U_0} \\\ \left( B \right)\dfrac{{24{U_0}}}{{{R_0}}}x - {U_0} \\\ \left( C \right)\dfrac{{96{U_0}}}{{{R_0}^2}} - {U_0} \\\ \left( D \right)None\,of\,these \\\

Explanation

Solution

In order to solve this question, we are going to take the potential energy function for the particle at a small displacement xx from the equilibrium position r=R0r = {R_0} , then, we will simplify the equation obtained and use the binomial expansion to get the required approximate PE function.
As it is given in the question, we know that the Van Der Waals interaction is given by:
U=U0[(R0R)12−2(R0R)6]U = {U_0}\left[ {{{\left( {\dfrac{{{R_0}}}{R}} \right)}^{12}} - 2{{\left( {\dfrac{{{R_0}}}{R}} \right)}^6}} \right]
The binomial expansion equation for (1+x)n{\left( {1 + x} \right)^n} is given by the equation,
(1+x)n=1+nx+n(n−1)x22!+.......{\left( {1 + x} \right)^n} = 1 + nx + \dfrac{{n\left( {n - 1} \right){x^2}}}{{2!}} + .......

Complete step by step solution:
As it is given in the question, that the Van Der Waals interaction is given by:
U=U0[(R0R)12−2(R0R)6]U = {U_0}\left[ {{{\left( {\dfrac{{{R_0}}}{R}} \right)}^{12}} - 2{{\left( {\dfrac{{{R_0}}}{R}} \right)}^6}} \right]
Where, UU is the potential energy function for a distance RR and R0{R_0} is the equilibrium position with the equilibrium potential function U0{U_0}
Now, for the small displacement xx , the potential energy function is given by:
U=U0[(R0R0+x)12−2(R0R0+x)6]U = {U_0}\left[ {{{\left( {\dfrac{{{R_0}}}{{{R_0} + x}}} \right)}^{12}} - 2{{\left( {\dfrac{{{R_0}}}{{{R_0} + x}}} \right)}^6}} \right]
Now we are going to simplify this function in order to get an approximate PE function.
U = {U_0}\left[ {\dfrac{{{R_0}^{12}}}{{{R_0}^{12}}}{{\left( {\dfrac{1}{{1 + \dfrac{x}{{{R_0}}}}}} \right)}^{12}} - 2\dfrac{{{R_0}^6}}{{{R_0}^6}}{{\left( {\dfrac{1}{{1 + \dfrac{x}{{{R_0}}}}}} \right)}^6}} \right] \\\ \Rightarrow U = {U_0}\left[ {{{\left( {\dfrac{1}{{1 + \dfrac{x}{{{R_0}}}}}} \right)}^{12}} - 2{{\left( {\dfrac{1}{{1 + \dfrac{x}{{{R_0}}}}}} \right)}^6}} \right] \\\ \Rightarrow U = {U_0}\left[ {{{\left( {1 + \dfrac{x}{{{R_0}}}} \right)}^{ - 12}} - 2{{\left( {1 + \dfrac{x}{{{R_0}}}} \right)}^{ - 6}}} \right] \\\
Now, using the Binomial expression for the two terms (1+xR0)−12{\left( {1 + \dfrac{x}{{{R_0}}}} \right)^{ - 12}} and (1+xR0)−6{\left( {1 + \dfrac{x}{{{R_0}}}} \right)^{ - 6}} , we get
\Rightarrow U = {U_0}\left[ {\left( {1 - \dfrac{{12x}}{{{R_0}}} + \dfrac{{66{x^2}}}{{{R_0}^2}}} \right) - 2\left( {1 - \dfrac{{6x}}{{{R_0}}} + \dfrac{{15{x^2}}}{{{R_0}^2}}} \right)} \right] \\\ \Rightarrow U = {U_0}\left[ {\dfrac{{36{x^2}}}{{{R_0}^2}} - 1} \right] \\\ \Rightarrow U = \dfrac{{36{x^2}{U_0}}}{{{R_0}^2}} - {U_0} \\\
Hence, option (A)36U0R2x2−U0\left( A \right)\dfrac{{36{U_0}}}{{{R^2}}}{x^2} - {U_0} is the correct answer.

Note:
The Van Der Waals interactions occur when adjacent atoms come close enough that their outer electron clouds just barely touch. These actions induce charge fluctuations that result in non-directional, non-specific attractions. But after a specific distance, if the atoms come too close to each other, then the atoms start to repel each other which makes the Van Der Waals interactions complex.