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Chapter 7: Problem 32

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?

Short Answer

Expert verified
The force is \(-6C_6/x^7\), and it is attractive.

Step by step solution

01

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.
02

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.
03

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 \( -n x^{-(n+1)} \).
04

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 \( -6\frac{C_6}{x^7} \).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攚hether attractive or repulsive鈥攄epends 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.

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Most popular questions from this chapter

A 62.0-kg skier is moving at 6.50 m/s on a frictionless, horizontal, snow- covered plateau when she encounters a rough patch 4.20 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high. (a) How fast is the skier moving when she gets to the bottom of the hill? (b) How much internal energy was generated in crossing the rough patch?

An ideal spring of negligible mass is 12.00 cm long when nothing is attached to it. When you hang a 3.15-kg weight from it, you measure its length to be 13.40 cm. If you wanted to store 10.0 J of potential energy in this spring, what would be its \(total\) length? Assume that it continues to obey Hooke's law.

A force parallel to the \(x\)-axis acts on a particle moving along the x-axis. This force produces potential energy \(U(x)\) given by \(U(x) = \alpha x^4\), where \(\alpha =\) 0.630 J/m\(^4\). What is the force (magnitude and direction) when the particle is at \(x = -0.800\) m?

A ball is thrown upward with an initial velocity of 15 m/s at an angle of 60.0\(^\circ\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

A 10.0-kg microwave oven is pushed 6.00 m up the sloping surface of a loading ramp inclined at an angle of 36.9\(^\circ\) above the horizontal, by a constant force \(\overrightarrow{F}\) with a magnitude 110 N and acting parallel to the ramp. The coefficient of kinetic friction between the oven and the ramp is 0.250. (a) What is the work done on the oven by the force \(\overrightarrow{F}\)? (b) What is the work done on the oven by the friction force? (c) Compute the increase in potential energy for the oven. (d) Use your answers to parts (a), (b), and (c) to calculate the increase in the oven's kinetic energy. (e) Use \(\sum \overrightarrow{F} = m\overrightarrow{a}\) to calculate the oven's acceleration. Assuming that the oven is initially at rest, use the acceleration to calculate the oven's speed after the oven has traveled 6.00 m. From this, compute the increase in the oven's kinetic energy, and compare it to your answerfor part (d).
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