Question

The potential energy variation between two atoms as a function of the distance between them is given by $U\left(x\right)=U_0[{\left(\frac{a}{x}\right)}^{12}-2{\left(\frac{a}{x}\right)}^6],U_0,a>0$.
Find the equilibrium position and state the nature of equilibrium (stable or unstable).

• A. $x_0=2a$ and stable
• B. $x_0=a$ and stable
• C. $x_0=2a$ and unstable
• D. $x_0=a$ and unstable

Answer 1

The correct option is B $x_0=a$ and stable

For equilibrium, $\frac{dU}{dx}=0$
$\frac{dU}{dx}=U_0[-12\frac{a^{12}}{x^{13}}+12\frac{a^6}{x^7}]$
$\frac{dU}{dx}=0$ at $x=x_0$
$\Rightarrow 12a^{12}{x}_0^7=12a^6{x}_0^{13}$
$\Rightarrow a^6=x_0^6$
$\Rightarrow x_0=a$

For checking the nature of equilibrium,
$\because$ For stable equilibrium, the $U-x$ curve should be concave upwards, so that $U_{\text{min}}$ exists at equilibrium position.


$\therefore \frac{d^{2}U}{d{x}^2}>0$ (for stable equilibrium)
Simillarly, $\frac{d^{2}U}{d{x}^2}<0$ (for unstable equilibrium)

Now, differntiating $\frac{dU}{dx}$ w.r.t $x$
$\Rightarrow \frac{d^{2}U}{d{x}^2}{|}_{x=x_0}=U_0[12\times 13\frac{a^{12}}{x_0^{14}}-12\times 7\frac{a^6}{x_0^8}]$
Now, put $x_0=a$
$\frac{d^{2}U}{d{x}^2}{|}_{x=a}=\frac{U_0}{a^2}[12\times 13-12\times 7]$
$\frac{d^{2}U}{d{x}^2}{|}_{x=a}=\frac{U_0}{a^2}[156-84]=\frac{72U_0}{a^2}>0$
(As $U_0$ & a both are positive values)
$\Rightarrow \frac{d^{2}U}{d{x}^2}{|}_{x=a}$ is positive.
$\therefore$It is a stable equilibrium