Problem 49 - Calculus Single Variable

Question

The potential energy of a diatomic molecule is \(U=\) \(A\left(b / r^{12}-1 / r^{6}\right),\) where \(A\) and \(b\) are positive constants, and \(r\) is the inter- atomic distance. What value of \(r\) minimizes \(U ?\)

Step-by-Step Solution

Answer

The value of \( r \) that minimizes \( U \) is \( r = (2b)^{1/6} \).

Step 1: Understand the Problem

We need to find the inter-atomic distance \( r \) that minimizes the potential energy function given by \( U=A\left(\frac{b}{r^{12}}-\frac{1}{r^{6}}\right) \). This involves finding the value of \( r \) for which \( U \) is at its minimum.

Step 2: Differentiate the Potential Energy

To find the minimum, we differentiate \( U \) with respect to \( r \). The derivative is \( \frac{dU}{dr} = A\left(-12\frac{b}{r^{13}} + 6\frac{1}{r^{7}}\right) \).

Step 3: Set the Derivative to Zero

To find the critical points, set \( \frac{dU}{dr} = 0 \). This gives the equation \( -12\frac{b}{r^{13}} + 6\frac{1}{r^{7}} = 0 \).

Step 4: Solve the Equation

Simplify and solve the equation from Step 3: \( -12\frac{b}{r^{13}} = -6\frac{1}{r^{7}} \). By canceling out the negatives and simplifying, we get \( 12b = 6r^6 \), which further simplifies to \( r^6 = 2b \). Therefore, \( r = (2b)^{1/6} \).

Key Concepts

Diatomic Molecule

A diatomic molecule is composed of two atoms bonded together. These molecules, such as oxygen (O2) and nitrogen (N2), are the simplest molecules and have unique physical and chemical properties.

The arrangement and interaction between the two atoms often determine the properties of the molecule. In the context of the potential energy equation, these molecules' energy levels vary based on the distance between the two atoms.

• Examples include H2, O2, and N2.
• The most stable configuration often minimizes energy.
• Understanding these interactions is crucial for fields like molecular physics and chemistry.

Potential Energy

The potential energy of a diatomic molecule describes the stored energy related to the position of the atoms. The function provided in the exercise, \[U = A\left(\frac{b}{r^{12}} - \frac{1}{r^{6}}\right), \]relates the potential energy U to the inter-atomic distance r. Here, A and b are constants. These constants are intrinsic to the molecular properties and experiment results.

In a physical sense, potential energy is the energy due to position or configuration. For diatomic molecules, we are concerned with how closely or far apart the atoms are and how this spacing affects interaction forces.

• Higher potential energy suggests less stability.
• The goal is to find a point where potential energy is minimized, indicating stable molecular configuration.

Inter-atomic Distance

Inter-atomic distance refers to the space between the nuclei of two bonded atoms within a molecule. In diatomic molecules, it is a crucial factor in determining molecular properties and energy configurations.

The equation \[U = A\left(\frac{b}{r^{12}} - \frac{1}{r^{6}}\right) \]assesses how changes in r (distance) affect potential energy, implying interactions like repulsion and attraction.

• Close proximity can result in repulsion due to overlapping electron clouds.
• Greater distances might weaken the attractive forces holding the molecule together.
• Finding optimal distance is key to stability.

Derivative

In calculus, a derivative represents the rate at which a function changes as its input changes. For this exercise, the derivative of potential energy U with respect to inter-atomic distance r is crucial to find the minimum energy configuration.

The derivative given is \[\frac{dU}{dr} = A\left(-12\frac{b}{r^{13}} + 6\frac{1}{r^{7}}\right). \]It helps identify where changes in r stop affecting potential energy, which signals a stable point.

• A zero derivative indicates a "flat spot" on the curve, where potential changes stop.
• This can indicate a maximum, minimum, or saddle point.

Critical Points

Critical points occur where the derivative of a function equals zero or becomes undefined. For our potential energy function, these points can help locate the minimum energy configuration of a diatomic molecule.

Setting the derivative to zero, we simplify the equation:\[-12\frac{b}{r^{13}} + 6\frac{1}{r^{7}} = 0. \]Solving this results in\[r^6 = 2b\]This calculation reveals the inter-atomic distance that minimizes potential energy.

Critical points are essential in finding stable configurations in many physical systems.

• Zero derivative means potential energy has no change at that point.
• Identifying whether it's a maximum or minimum requires further analysis (e.g., second derivative test).