Question

If the potential energy between two molecules is given by $U=-\frac{A}{r^6}+\frac{B}{r^{12}}$, then at equilibrium, the separation between molecules, and the potential energy are:

• A. [2B / A]^⅙, [-A² / 4B]
• B. [2B / A]^⅙, [-A² / 2B]
• C. (B / A)^⅙, 0
• D. (B / 2A)^⅙, [-A² / 4B]

Answer 1

The correct option is A

[2B / A]^⅙, [-A² / 4B]

Step 1. Given Data,

$U=-\frac{A}{r^6}+\frac{B}{r^{12}}$

Step 2. Formula Used,
We can write force as,
$\Rightarrow F=-{\displaystyle \frac{dU}{dr}}$
where $F$ is the force,

$U$ is the potential energy and $r$is the separation between molecules

Step 3. Evaluating the separation between molecules,

Potential energy,$U=-\frac{A}{r^6}+\frac{B}{r^{12}}$
$\Rightarrow F=-\frac{d}{dr}\left(\frac{B}{r^{12}}-\frac{A}{r^6}\right)$
Differentiating we get,
$\Rightarrow F=\frac{12B}{r^{13}}-\frac{6A}{r^7}$
Net force = $0$,

From the equilibrium condition, we get,
$\Rightarrow 0=\frac{12B}{r^{13}}-\frac{6A}{r^7}$
$\Rightarrow \frac{12B}{r^{13}}=\frac{6A}{r^7}$
$\Rightarrow \frac{r^6}{12B}=\frac{1}{6A}$
$\Rightarrow r^6=\frac{2B}{A}$

$\Rightarrow r={\left(\frac{2B}{A}\right)}^{\frac{1}{6}}$

Step 4. Evaluation of the Potential Energy,
$U=-\frac{A}{\left({\displaystyle \frac{2B}{A}}\right)}+\frac{B}{{\left({\displaystyle \frac{2B}{A}}\right)}^2}$

$U=-\frac{A^2}{2B}+\frac{B}{{\displaystyle \frac{4B^2}{A^2}}}$
$U=-\frac{A^2}{2B}+\frac{A^2}{{\displaystyle 4B}}$
On calculating we get,
$\Rightarrow U=\frac{A^2-2A^2}{4B}=\frac{-A^2}{4B}$

The potential energy is $\frac{-A^2}{4B}$.and the separation between molecules is ${\left(\frac{2B}{A}\right)}^{\frac{1}{6}}$ at the equilibrium position
Thus, option (A) is correct.