Question

If the potential energy of a gas molecule is $U = \dfrac{M}{{{r^6}}} - \dfrac{N}{{{r^{12}}}}$ , M and N being positive constants, then the potential energy and equilibrium is:
(A) 0
(B) $\dfrac{{N{M^2}}}{4}$
(C) $\dfrac{{M{N^2}}}{4}$
(D) $\dfrac{{{M^2}}}{{4N}}$

Answer 1

In this question, we are given the function of the potential field as a function of radius of the gas molecule. First, we need to find the forces acting on the gas molecule. Then we need to use the fact that at equilibrium, the sum of net forces acting on the particle is 0. This will give us the distance at which the force is 0. Substituting this distance in the function of potential energy we will get our solution.

Complete step by step solution
We know that when a body is placed in a field of some potential energy, it experiences force. This force is equal to the negative gradient of the potential field at that point.
$F = - \dfrac{{dU}}{{dr}}$

$$F = - \dfrac{{d(\dfrac{M}{{{r^6}}} - \dfrac{N}{{{r^{12}}}})}}{{dr}} \\\ F = - (\dfrac{{ - 6M}}{{{r^7}}} - \dfrac{{ - 12N}}{{{r^{13}}}}) \\\ F = \dfrac{{6M}}{{{r^7}}} - \dfrac{{12N}}{{{r^{13}}}} \\\$$

We also know that at linear equilibrium, the net forces acting on the body is equal to 0. Equating this force to 0 we get:

$$0 = \dfrac{{6M}}{{{r^7}}} - \dfrac{{12N}}{{{r^{13}}}} \\\ \dfrac{{6M}}{{{r^7}}} = \dfrac{{12N}}{{{r^{13}}}} \\\ {r^6} = \dfrac{{2N}}{M} \\\$$

Substituting the value of r in the equation for potential energy we get:

$$U = \dfrac{{{M^2}}}{{2N}} - \dfrac{{N{M^2}}}{{{{(2N)}^2}}} \\\ U = \dfrac{{{M^2}}}{{2N}} - \dfrac{{{M^2}}}{{4N}} \\\ U = \dfrac{{{M^2}}}{{4N}} \\\$$

Therefore the option with the correct answer is option D

Note

Potential energy of an electric or magnetic field will always have a negative sign with them. This negative potential energy means that work must be done to move the charge or the mass of a body against this electric or gravitational field.