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The potential energy of a diatomic molecule (a two-atom system like \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\) ) is given by $$ U=\frac{A}{r^{12}}-\frac{B}{r^{6}} $$ where \(r\) is the separation of the two atoms of the molecule and \(A\) and \(B\) are positive constants. This potential energy is associated with the force that binds the two atoms together. (a) Find the equilibrium separation-that is, the distance between the atoms at which the force on each atom is zero. Is the force repulsive (the atoms are pushed apart) or attractive (they are pulled together) if their separation is (b) smaller and (c) larger than the equilibrium separation?

Short Answer

Expert verified
The equilibrium separation is \( r_e = (2A/B)^{1/6} \). The force is repulsive for \( r < r_e \) and attractive for \( r > r_e \).

Step by step solution

01

Understand the Problem

We are given the potential energy function for a diatomic molecule and need to find the equilibrium separation, where the force on the atoms is zero. Then we determine if the force is repulsive or attractive for separations smaller and larger than this equilibrium distance.
02

Find the Force Expression

The force on the atoms can be determined from the potential energy by finding the derivative of the potential energy with respect to the separation distance, i.e., \[ F(r) = -\frac{dU}{dr} = -\frac{d}{dr}\left(\frac{A}{r^{12}} - \frac{B}{r^6}\right). \]
03

Calculate the Derivative

Compute the derivative:\[ \frac{dU}{dr} = \frac{12A}{r^{13}} - \frac{6B}{r^7}. \]Thus, the force on the atoms becomes:\[ F(r) = -\left( \frac{12A}{r^{13}} - \frac{6B}{r^7} \right). \]
04

Find Equilibrium Separation

Set the force to zero to find equilibrium separation, i.e.,\[ \frac{12A}{r^{13}} - \frac{6B}{r^7} = 0. \]Rearranging gives:\[ \frac{12A}{r^{13}} = \frac{6B}{r^7} \]Upon simplification, we find:\[ r_e = \left( \frac{2A}{B} \right)^{1/6}. \]
05

Determine Force for r < r_e

For separations smaller than the equilibrium distance \(r < r_e\), the potential energy function term with \(r^{12}\) (which is repulsive and dominant at short distances) becomes larger than the \(r^6\) term. Thus, the force is repulsive.
06

Determine Force for r > r_e

For separations larger than the equilibrium distance \(r > r_e\), the attractive \(r^6\) term outweighs the repulsive \(r^{12}\) term, resulting in a net attractive force.

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Key Concepts

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

Equilibrium Separation
In molecular systems, the equilibrium separation refers to the optimal distance between two atoms within a diatomic molecule, such as \(\mathrm{H}_{2}\) or \(\mathrm{O}_{2}\). At this distance, the force acting on each atom is zero, meaning they are neither pushed apart nor pulled together. This balance arises due to the interactions between repulsive and attractive forces within the molecule.

To find this equilibrium in a potential energy function represented by \( U = \frac{A}{r^{12}} - \frac{B}{r^{6}} \), where \( r \) is the separation distance, we calculate where the net force \( F(r) = -\frac{dU}{dr} \) equals zero. Solving the equation \( \frac{12A}{r^{13}} = \frac{6B}{r^{7}} \) yields the equilibrium separation \( r_e = \left( \frac{2A}{B} \right)^{1/6} \). At \( r_e \), the repulsive and attractive forces cancel out, creating a stable molecular configuration.
Repulsive and Attractive Forces
In diatomic molecules, both repulsive and attractive forces play crucial roles in determining potential energy and atomic behavior. Repulsive forces emerge when atomic nuclei are extremely close, due to electron cloud interactions, and they act to push the atoms apart. Mathematically, these forces are represented by terms like \( \frac{A}{r^{12}} \), where a smaller distance \( r \) produces a significantly higher repulsive force.

On the other hand, attractive forces stem from forces like chemical bonds or van der Waals forces, which draw atoms together even when they are spaced apart. The potential energy component \( -\frac{B}{r^6} \) represents this relationship, with forces increasing as the separation decreases. The equilibrium separation is the point where these two competing forces balance out, stabilizing the molecule.
Diatomic Molecules
Diatomic molecules are composed of two atoms that are covalently bonded, forming species like \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\).Such molecules can be analyzed through their potential energy curves, which show how energy changes as the distance between the atoms varies. These curves are highly indicative of how molecules behave under different conditions.

In the potential energy function given, the combination of repulsive \( \frac{A}{r^{12}} \) and attractive \( -\frac{B}{r^{6}} \) terms describes the behavior of diatomic molecules. Understanding the balance and interplay of these forces allows us to appreciate how diatomic molecules maintain their integrity and stability. This is particularly important in chemistry and physics, where predicting how these molecules react under various conditions can lead to breakthroughs in technology and material science.
Force and Potential Energy Relationship
The relationship between force and potential energy is pivotal in understanding molecular behavior. Simply put, force is the agent that acts upon particles, while potential energy represents the stored energy within a system. The ability to derive force from potential energy through differentiation allows us to predict how a system will react to changes.

In diatomic systems, like those expressed through the potential function \( U = \frac{A}{r^{12}} - \frac{B}{r^{6}} \), the force is calculated as \( F(r) = -\frac{dU}{dr} \). By deriving the potential energy, we find the expression for force: \( F(r) = -\left( \frac{12A}{r^{13}} - \frac{6B}{r^{7}} \right) \). This equation shows how both attractive and repulsive forces vary with atom separation. For molecular equilibrium and stability, understanding this force-potential energy relationship is essential, as it highlights how and when interactions change from attractive to repulsive.

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Most popular questions from this chapter

You drop a \(2.00 \mathrm{~kg}\) book to a friend who stands on the ground at distance \(D=10.0 \mathrm{~m}\) below. Your friend's outstretchedhands are at distance \(d=1.50 \mathrm{~m}\) above the ground (Fig. 8-19). (a) How much work \(W_{g}\) does the gravitational force do on the book as it drops to her hands? (b) What is the change \(\Delta U\) in the gravitational potential energy of the book-Earth system during the drop? If the gravitational potential energy \(U\) of that system is taken to be zero at ground level, what is \(U\) (c) when the book is released, and (d) when it reaches her hands? Now take \(U\) to be \(100 \mathrm{~J}\) at ground level and again find (e) \(W_{g}\), (f) \(\Delta U\), (g) \(U\) at the release point, and (h) \(U\) at her hands.

At a lunar base, a uniform chain hangs over the edge of a horizontal platform. A machine does \(1.0 \mathrm{~J}\) of work in pulling the rest of the chain onto the platform. The chain has a mass of \(2.0\) \(\mathrm{kg}\) and a length of \(3.0 \mathrm{~m}\). What length was initially hanging over the edge? On the Moon, the gravitational acceleration is \(1 / 6\) of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

A stone with a weight of \(5.29 \mathrm{~N}\) is launched vertically from ground level with an initial speed of \(20.0 \mathrm{~m} / \mathrm{s}\), and the air drag on it is \(0.265 \mathrm{~N}\) throughout the flight. What are (a) the maximum height reached by the stone and (b) its speed just before it hits the ground?

A worker pushed a \(23 \mathrm{~kg}\) block \(8.4 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block - floor system?

A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=-4 x e^{-x / 4} \mathrm{~J} $$ where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J}\). (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) the particle can reach and (d) the greatest value of \(x\) the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine an expression in newtons and meters for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0\) ?

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