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

Question

Two wires $A$ and $B$ made of different materials have lengths $6.0$ cm and $5.4$ cm, and areas of cross-sections $3.0\times10^{-5}\,\text{m}^2$ and $4.5\times10^{-5}\,\text{m}^2$, respectively. They are stretched by the same magnitude under the same load. If the ratio of Young’s modulus of $A$ to that of $B$ is $x:3$, find the value of $x$.

Answer 1

To find the equilibrium separation between two molecules and their potential energy, we start by analyzing the given potential energy formula:

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

At equilibrium, the force between the molecules should be zero. The force F can be found by differentiating the potential energy with respect to the separation r:

$$F = -\frac{dU}{dr}$$

Performing the differentiation:

$$\frac{dU}{dr} = \frac{d}{dr}\left(- \frac{A}{r^{6}} + \frac{B}{r^{12}}\right)$$

This gives:

$$\frac{dU}{dr} = 6\frac{A}{r^{7}} - 12\frac{B}{r^{13}}$$

At equilibrium, set this derivative to zero:

$$6\frac{A}{r^{7}} = 12\frac{B}{r^{13}}$$

Simplifying this equation:

$$6Ar^{6} = 12B$$ $$r^{6} = \frac{2B}{A}$$

Therefore, the equilibrium separation is:

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

Now, we substitute this equilibrium separation back into the potential energy equation to find the potential energy at equilibrium:

$$U = - \frac{A}{\left(\left(\frac{2B}{A}\right)^{1/6}\right)^{6}} + \frac{B}{\left(\left(\frac{2B}{A}\right)^{1/6}\right)^{12}}$$

Simplifying each term:

$$U = - \frac{A}{\frac{2B}{A}} + \frac{B}{\frac{4B^2}{A^2}}$$ $$U = - \frac{A^2}{2B} + \frac{A^2}{4B}$$

This simplifies to:

$$U = -\frac{A^2}{4B}$$

Therefore, the correct answer is:

$$\left(\frac{2B}{A}\right)^{1/6}, -\frac{A^2}{4B}$$

Answer 2