Problem 12 - An Introduction to Mechanics

Question

A commonly used potential energy function to describe the interaction between two atoms is the Lennard-Jones \(6-12\) potential given by $$ U=\epsilon\left[\left(\frac{r_{0}}{r}\right)^{12}-2\left(\frac{r_{0}}{r}\right)^{6}\right] $$ (a) Find the position of the potential minimum and its value. (b) Near the minimum the atoms execute simple harmonic motion. Find the frequency of oscillation.

Step-by-Step Solution

Answer

The potential minimum is at \( r = r_0 \) with value \( -\epsilon \). The frequency is \( \nu = \frac{1}{2\pi} \sqrt{\frac{114 \epsilon}{r_{0}^2 m}} \).

Step 1: Differentiate the Potential Function

To find the position of the potential minimum, set the derivative of the potential function, \( U(r) \), to zero. Differentiate with respect to \( r \):\[\frac{dU}{dr} = \epsilon \left[ -12 \left(\frac{r_{0}}{r}\right)^{13} \frac{1}{r} + 12 \left(\frac{r_{0}}{r}\right)^{7} \frac{1}{r} \right] = 0 \]

Step 2: Simplify and Solve for the Minimum Position

Simplify and solve for \( r \) at which \( \frac{dU}{dr} = 0 \):\[12 \left(\frac{r_{0}}{r}\right)^{13} = 12 \left(\frac{r_{0}}{r}\right)^{7}\]Divide by \( 12 \) and simplify:\[\left(\frac{r_{0}}{r}\right)^{6} = 1 \]Solving for \( r \), we get:\[ \frac{r_{0}}{r} = 1 \rightarrow r = r_{0} \]

Step 3: Evaluate the Potential at the Minimum

Substitute \( r = r_{0} \) back into \( U(r) \) to find the potential's minimum value:\[U(r_{0}) = \epsilon \left[ 1 - 2 \cdot 1 \right] = -\epsilon\]Thus, the potential minimum value is \( U_{ ext{min}} = -\epsilon \).

Step 4: Approximate Potential Near the Minimum

Around the minimum, approximate \( U(r) \) to a quadratic form for simple harmonic motion. The expansion is:\[U(r) \approx U(r_{0}) + \frac{1}{2} U''(r_{0}) (r - r_{0})^2 \]

Step 5: Find the Second Derivative and Evaluate it

Compute the second derivative of \( U(r) \):\[\frac{d^2U}{dr^2} = \epsilon \left[ 156 \left(\frac{r_{0}}{r}\right)^{14} \frac{1}{r^2} - 42 \left(\frac{r_{0}}{r}\right)^{8} \frac{1}{r^2} \right] \]Evaluate at \( r = r_{0} \):\[U''(r_{0}) = \epsilon \left[ 156 - 42 \right] = 114 \left(\frac{\epsilon}{r_{0}^2}\right) \]Therefore, \( U''(r_{0}) = 114 \frac{\epsilon}{r_{0}^2} \).

Step 6: Determine the Frequency of Oscillation

From the quadratic approximation, use the formula for the angular frequency \( \omega \) of simple harmonic motion:\[\omega = \sqrt{\frac{U''(r_{0})}{m}} = \sqrt{\frac{114 \epsilon}{r_{0}^2 m}} \]Thus, the frequency \( u \) is given by:\[u = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{114 \epsilon}{r_{0}^2 m}} \]

Key Concepts

Understanding Potential Energy

The Lennard-Jones potential is a model that describes the interaction between two atoms. This model captures how potential energy, which is the energy stored within a system, varies with distance between atoms. The Lennard-Jones potential is represented by the equation: \[U=\epsilon\left[\left(\frac{r_{0}}{r}\right)^{12}-2\left(\frac{r_{0}}{r}\right)^{6}\right]\]In this equation:- \( U \) is the potential energy.- \( \epsilon \) is the depth of the potential well, indicating the strength of interaction.- \( r_0 \) is the equilibrium distance where the potential energy is at its minimum.- \( r \) is the current distance between the two atoms.When the distance \( r \) between two atoms equals \( r_0 \), the potential energy reaches a minimum value of \(-\epsilon\). This means that the system is in a stable state, as the attractive and repulsive forces between the atoms are balanced.
As the atoms move closer than \( r_0 \), the repulsive forces dominate, increasing potential energy. Conversely, if the atoms are further apart, the attraction diminishes quickly, also affecting the potential energy and drawing atoms back toward \( r_0 \).

The Concept of Simple Harmonic Motion

The term "simple harmonic motion" (SHM) describes a type of periodic oscillation where the restoring force is directly proportional to the displacement. When atoms are at the potential energy minimum, small displacements lead to SHM.In the context of Lennard-Jones potential, around the equilibrium distance \( r_{0} \), the potential function can be approximated to a quadratic form. This approximation simplifies the behavior of atoms to resemble a harmonic oscillator, just like a mass on a spring. The relevant equation is:\[U(r) \approx U(r_{0}) + \frac{1}{2} U''(r_{0}) (r - r_{0})^2\]In this context:- \( U(r_{0}) \) is the potential energy minimum.- \( U''(r_{0}) \) corresponds to the stiffness of the "spring" formed by atomic bonds.- The second term, \((r - r_{0})^2\), indicates small deviations from the equilibrium position.These factors together mimic a spring's characteristics, enforcing the atoms to undergo SHM. This approximation is valid only when oscillations remain small near the equilibrium position \( r_{0} \).

Understanding Oscillation Frequency

The frequency of oscillation is an important property in simple harmonic motion. It indicates how fast an object returns to its equilibrium position during oscillation cycles. In this exercise, the frequency is derived from the properties of the Lennard-Jones potential at the potential energy minimum.The angular frequency \( \omega \) can be calculated using:\[\omega = \sqrt{\frac{U''(r_{0})}{m}}\]Where:- \( U''(r_{0}) \) is the second derivative of the potential energy at equilibrium, reflecting the system's "stiffness."- \( m \) is the mass of one of the atoms.The actual frequency \( u \) is then:\[u = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{114 \epsilon}{r_{0}^2 m}}\]The above formula shows that the frequency depends on:- The depth of the potential well \( \epsilon \).- The characteristic distance \( r_{0} \).- The mass of the atoms involved.Thus, the oscillation frequency provides insight into how rapidly a system returns to its equilibrium after being disturbed, illustrating the sensitivity of molecule interactions governed by Lennard-Jones potential.