Problem 46 - Master Resource Book in JEE Main Physics

Question

The potential energy as a function of the force between two atoms in a diatomic molecules is given by \(U(x)=\frac{a}{x^{12}}-\frac{b}{x^{6}}\), where \(a\) and \(b\) are positive constants and \(x\) is the distance between the atoms. The position of stable equilibrium for the system of the two atoms is given (a) \(x=\frac{a}{b}\) (b) \(x=\sqrt{\frac{a}{b}}\) (c) \(x=\frac{\sqrt{3 a}}{b}\) (d) \(x=\sqrt[6]{\left(\frac{2 a}{b}\right)}\)

Step-by-Step Solution

Answer

The position of stable equilibrium is \(x = \sqrt[6]{\left(\frac{2 a}{b}\right)}\).

Step 1: Understand the Concept of Stable Equilibrium

For the position of stable equilibrium, the potential energy function must satisfy two conditions: the derivative of the potential energy with respect to distance \(x\) must be zero, \(\frac{dU}{dx} = 0\), indicating a minimum point; and the second derivative must be positive, \(\frac{d^2U}{dx^2} > 0\), to ensure it is a point of minimum energy.

Step 2: Compute the First Derivative of the Potential Energy

Calculate the first derivative of the potential energy function \(U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}\). \[ \frac{dU}{dx} = -\frac{12a}{x^{13}} + \frac{6b}{x^7} \]

Step 3: Set the First Derivative to Zero

To find the points of equilibrium, set the first derivative \(\frac{dU}{dx} = 0\): \[ -\frac{12a}{x^{13}} + \frac{6b}{x^7} = 0 \]

Step 4: Solve for x

Rearrange the equation from Step 3: \[ \frac{6b}{x^7} = \frac{12a}{x^{13}} \]Simplify and solve for \(x\): \[ 6b x^{13} = 12a x^7 \]\[ x^6 = \frac{2a}{b} \]\[ x = \left(\frac{2a}{b}\right)^{\frac{1}{6}} \]

Step 5: Verify the Second Derivative is Positive

Compute the second derivative \(\frac{d^2U}{dx^2}\) to ensure it is a point of stable equilibrium:\[ \frac{d^2U}{dx^2} = \frac{156a}{x^{14}} - \frac{42b}{x^8} \]Substitute \(x = \left(\frac{2a}{b}\right)^{\frac{1}{6}}\) into \(\frac{d^2U}{dx^2}\) to verify it is positive.

Key Concepts

Stable Equilibrium

In physics, stable equilibrium refers to a state where a system naturally tends to return to after a small displacement. Imagine placing a ball at the bottom of a bowl. If you gently nudge it, the ball rolls back to the bottom. This is stable equilibrium.
In the context of diatomic molecules, stable equilibrium is determined by the potential energy function. The position at which two atoms settle is when the potential energy is at a minimum.
For this position, two mathematical conditions must be met:

• The first derivative of the potential energy function with respect to distance must be zero. This condition implies that any small displacement from this point will not create a force to move away more from this position.
• The second derivative must be positive. This ensures that any displacement either to the left or right results in a force bringing the atom back to this point, akin to our ball rolling back to the center of the bowl.

Potential Energy Function

The potential energy function is crucial for understanding interactions between atoms in diatomic molecules. It describes how the energy of the system changes with the distance between the two atoms.
In our exercise, this function is represented as:\[U(x) = \frac{a}{x^{12}} - \frac{b}{x^6}\]where:

• \(a\) and \(b\) are positive constants,
• \(x\) is the distance between the two atoms.

This function combines attractive and repulsive forces. The first term, \(\frac{a}{x^{12}}\), generally represents a repulsive force, making it harder to compress atoms too closely. The second term, \(\frac{b}{x^6}\), describes an attractive force, pulling atoms closer together.
At equilibrium, these forces balance, leading to stable conditions for the molecule.

Derivative of Potential Energy

To determine the equilibrium points, we need to calculate the derivative of the potential energy function. The derivative tells us the rate at which energy changes with distance, which is crucial for finding stable positions.
For our potential energy function:\[\frac{dU}{dx} = -\frac{12a}{x^{13}} + \frac{6b}{x^7}\]Setting this derivative to zero helps locate where the net force is zero, indicating potential equilibrium points. This calculation essentially pinpoints where energy is not increasing or decreasing, highlighting possible locations for stable equilibrium.
Equilibrium occurs when the force between the atoms is balanced, causing them not to push or pull away from each other.

Minimum Energy Point

At the minimum energy point, the system is most stable, with atoms experiencing the least potential energy. It's where potential energy is at its lowest, indicating an equilibrium where attractive and repulsive forces perfectly balance.
To ensure this point is indeed a minimum, the second derivative of the potential energy function must be positive. For our exercise, this is expressed as:\[\frac{d^2U}{dx^2} = \frac{156a}{x^{14}} - \frac{42b}{x^8}\]When we substitute the equilibrium point value \(x = \left(\frac{2a}{b}\right)^{\frac{1}{6}}\) into this expression and confirm it is positive, we validate that it is indeed a point of minimum energy.
At this point, the system returns to equilibrium if disturbed slightly, ensuring stability of the molecule.