Problem 32

Question

The potential energy of a pair of hydrogen atoms separated by a large distance \(x\) is given by \(U(x) = -C_6/x^6\), where \(C_6\) is a positive constant. What is the force that one atom exerts on the other? Is this force attractive or repulsive?

Step-by-Step Solution

Verified
Answer
The force is \(-6C_6/x^7\), and it is attractive.
1Step 1: Understanding the Formula for Potential Energy
The potential energy for a pair of hydrogen atoms is given as a function of distance:\[ U(x) = -\frac{C_6}{x^6} \] where \( C_6 \) is a positive constant and \( x \) is the distance between the atoms.
2Step 2: Applying the Relation Between Force and Potential Energy
The force \( F \) exerted by one atom on the other can be found by taking the negative derivative of the potential energy \( U(x) \) with respect to \( x \): \[ F(x) = -\frac{dU}{dx} \] This relation follows from basic principles of mechanics, where force is the negative gradient of potential energy.
3Step 3: Calculating the Derivative of the Potential Energy
We need to calculate the derivative \( \frac{dU}{dx} \) of the function \( U(x) = -\frac{C_6}{x^6} \). Differentiating with respect to \( x \) gives: \[ \frac{dU}{dx} = 6\frac{C_6}{x^7} \] This result uses the power rule for differentiation, where differentiating \( x^{-n} \) yields \( -n x^{-(n+1)} \).
4Step 4: Determining the Force from the Derivative
Substitute the derivative result back into the equation for force, this gives: \[ F(x) = - \left( 6\frac{C_6}{x^7} \right) = -6\frac{C_6}{x^7} \] Hence, the force between the atoms is \( -6\frac{C_6}{x^7} \).
5Step 5: Analyzing the Nature of the Force
The negative sign of the force \( -6\frac{C_6}{x^7} \) indicates that the force is attractive. This means the force vectors point towards each other (reduction in distance), consistent with a negative derivative of potential energy.

Key Concepts

Force Between AtomsDerivative in PhysicsAttractive Force
Force Between Atoms
Understanding the force between atoms is pivotal in physics, particularly in molecular interactions. When two hydrogen atoms are separated by a distance, the force between them is derived from their potential energy.
In the provided exercise, potential energy is represented by the equation:
  • \(U(x) = -\frac{C_6}{x^6}\)
Where \(C_6\) is a positive constant and \(x\) represents the distance between the two atoms.
The key to understanding the force is knowing that it results from a change in potential energy as the distance between atoms changes. Therefore, to find this force, we utilize the relationship between force and potential energy, where force is the derivative of potential energy with respect to distance.
Derivative in Physics
The derivative is a fundamental concept in calculus and physics, often used to determine the rate of change of quantities. In the context of atomic forces, the derivative helps us find how potential energy changes with distance. The negative derivative of potential energy with respect to distance tells us the force exerted.
The exercise requires calculating the derivative of the potential energy function \(U(x) = -\frac{C_6}{x^6}\) with respect to \(x\). Applying the power rule of calculus, this gives:
  • \(\frac{dU}{dx} = 6\frac{C_6}{x^7}\)
Once differentiated, this result is substituted back into the force formula:
  • \(F(x) = -\frac{dU}{dx}\)
This substitution effectively provides the measure of force between the atoms.
Attractive Force
The nature of the force between atoms—whether attractive or repulsive—depends on the sign of the force value derived from potential energy differentiation. When the calculated force has a negative value, as seen here:
  • \(F(x) = -6\frac{C_6}{x^7}\)
This force is considered attractive.
An attractive force indicates that the atoms tend to pull towards each other. This inward pull is a result of the potential energy minimizing when the atoms are closer together. Such forces are important for understanding molecular structures and chemical bonding, as they fundamentally govern how atoms and molecules maintain bond formations.