Question

The potential energy between two atoms in a molecule is given by $U=a{x}^3-b{x}^2$ where $a$ and $b$ are positive constants and $x$ is the distance between the atoms. The atom is in stable equilibrium when $x$ is equal to

• A. $0$
• B. $\frac{2b}{3a}$
• C. $\frac{3a}{2b}$
• D. $\frac{3b}{2a}$

Answer 1

The correct option is B $\frac{2b}{3a}$

Given: $U=a{x}^3-b{x}^2$
$\therefore$ Force, $F=-\frac{dU}{dx}=-(3a{x}^2-2bx)$
$=2bx-3a{x}^2$

Now, for equilibrium,
$F=0$
$\Rightarrow 2bx-3a{x}^2=0$
$\Rightarrow x=0,\frac{2b}{3a}$

For stable equilibrium, $U$ should be minimum
i.e $\frac{d^{2}U}{d{x}^2}=6ax-2b$
For $x=\frac{2b}{3a}$,
$\frac{d^{2}U}{d{x}^2}=6a\times \frac{2b}{3a}-2b=2b>0$
$\therefore$ at $x=\frac{2b}{3a}$, P.E is minimum
i.e a condition of stable equilibrium.