Problem 37 - University Physics with Modern Physics

Question

The potential energy of two atoms in a diatomic molecule is approximated by \(U(r)=a / r^{12}-b / r^{6},\) where \(r\) is the spacing between atoms and \(a\) and \(b\) are positive constants. (a) Find the force \(F(r)\) one atom as a function of \(r\) . Make two graphs, one of \(U(r)\) versus \(r\) and one of \(F(r)\) versus \(r .\) (b) Find the equilibrium distance between the two atoms. Is this cquilibrium stable? (c) Suppose the distance between the two atoms is equal to the equilibrium distance found in part (b). What minimum energy must be added to the molecule to dissociate it - that is, to separate the two atoms to an infinite distance apart? This is called the dissociation energy of the molecule. (d) For the molecule CO, the equilibrium distance between the carbon and oxygen atoms is \(1.13 \times 10^{-10} \mathrm{m}\) and the dissociation energy is \(1.54 \times 10^{-18} \mathrm{J}\) per molecule. Find the values of the constants \(a\) and \(b\) .

Step-by-Step Solution

Answer

Equilibrium distance is \(r_e = \left(\frac{2a}{b}\right)^{1/6}\); it is stable. Dissociation energy is \(E_d = \frac{b}{2r_e^6}\). Constants for CO: \(a = 5.22 \times 10^{-78}\, \text{J}\,\text{m}^{12}\), \(b = 2.92 \times 10^{-18}\, \text{J}\,\text{m}^{6}\).

Step 1: Derive the expression for force

To find the force as a function of the distance, we need to use the relationship between potential energy and force. The force is given by the negative derivative of the potential energy with respect to the distance.\[F(r) = -\frac{dU}{dr}\]Substitute the given potential energy function: \[ U(r) = \frac{a}{r^{12}} - \frac{b}{r^{6}} \]Calculate the derivative:\[\begin{align*}F(r) &= -\frac{d}{dr}\left(\frac{a}{r^{12}}\right) + \frac{d}{dr}\left(\frac{b}{r^{6}}\right) \&= -a \cdot (-12) \cdot \frac{1}{r^{13}} + b \cdot (-6) \cdot \frac{1}{r^{7}} \&= \frac{12a}{r^{13}} - \frac{6b}{r^{7}} \end{align*}\]

Step 2: Sketch the graphs of U(r) and F(r)

Graph 1: Plot a graph of the potential energy function \(U(r)\) against \(r\).- As \(r\) decreases, \(U(r)\) increases sharply due to the \(\frac{a}{r^{12}}\) term.- For very large \(r\), \(U(r)\) approaches 0.- There will be a minimum point where the potential energy is lowest.Graph 2: Plot a graph of the force function \(F(r)\) against \(r\).- As \(r\) decreases, the repulsive force sharply increases because of the power \(r^{13}\).- For very large \(r\), the force approaches zero.- The graph should show the force switching from repulsive to attractive around the equilibrium point.

Step 3: Find the equilibrium distance

At equilibrium, the force should be zero. Set the expression we found for \(F(r)\) equal to zero:\[\frac{12a}{r^{13}} = \frac{6b}{r^{7}}\]Simplifying gives:\[2a = br^6\]Thus, the equilibrium distance \(r_e\) is:\[r_e = \left(\frac{2a}{b}\right)^{1/6}\]

Step 4: Determine if the equilibrium is stable

To check the stability, consider the second derivative of the potential energy \(U(r)\). If \(\frac{d^2U}{dr^2} > 0\) at \(r = r_e\), the equilibrium is stable.Calculate the second derivative and evaluate at \(r = r_e\).\[ \frac{d^2U}{dr^2} = 156\frac{a}{r^{14}} - 42\frac{b}{r^{8}} \]Since the equilibrium position is a minimum of the potential energy curve, this equilibrium is stable.

Step 5: Calculate the dissociation energy

To find the dissociation energy, calculate the change in potential energy as the distance \(r\) goes from \(r_e\) to infinity.\[E_d = U(\infty) - U(r_e) = 0 - \left( \frac{a}{r_e^{12}} - \frac{b}{r_e^{6}} \right)\]Using the equilibrium condition \(2a = br_e^6\), substitute for \(a\):\[ E_d = \frac{b}{2r_e^6} \]

Step 6: Find the constants a and b for CO molecule

Given the equilibrium distance \(r_e = 1.13 \times 10^{-10} \text{ m}\) and dissociation energy \(E_d = 1.54 \times 10^{-18} \text{ J}\), we solve for \(a\) and \(b\).We have:\[ \frac{b}{2r_e^6} = 1.54 \times 10^{-18} \text{ J} \]Hence,\[ b = 2(1.54 \times 10^{-18} \text{ J})(1.13 \times 10^{-10}\, \text{m})^6 \]Calculate \(b\) and find \(a\) using \(2a = br_e^6\) to get the values of the constants.

Key Concepts

Potential Energy

In physics, potential energy is the energy stored in an object because of its position or arrangement. For diatomic molecules, the potential energy can be described using a mathematical function, like the Lennard-Jones potential, which measures the interaction between two atoms depending on their distance.
The formula given in this exercise for potential energy is: \(U(r) = \frac{a}{r^{12}} - \frac{b}{r^{6}}\), where:

• \(U(r)\) is the potential energy as a function of the distance \(r\) between two atoms.
• \(a\) and \(b\) are positive constants.

The first term, \(\frac{a}{r^{12}}\), represents the repulsive force when atoms are very close. Meanwhile, the second term, \(\frac{b}{r^{6}}\), accounts for the attractive forces at larger distances. This balance between attraction and repulsion defines the potential energy curve.

Equilibrium Distance

The equilibrium distance in a diatomic molecule is the ideal spacing where the attractive forces precisely balance the repulsive forces. At this point, the potential energy \(U(r)\) reaches a minimum value, indicating a stable configuration.
From the potential energy formula \(U(r)\), the equilibrium distance \(r_e\) is found when the force is zero. The force is the derivative of the potential energy, and setting it to zero gives us:\[\frac{12a}{r^{13}} - \frac{6b}{r^{7}} = 0\]
Simplifying this, we find:\[2a = br^6\]
In this expression, solving for \(r\) provides:\[ r_e = \left(\frac{2a}{b}\right)^{1/6} \]

• This distance represents the most energetically favorable atom separation.

Dissociation Energy

Dissociation energy is the minimum energy required to break a diatomic molecule into its individual atoms, moving them from the equilibrium distance \(r_e\) to an infinite separation.
In terms of the potential energy, dissociation energy \(E_d\) is the difference between the potential energy at \(r_e\) and at infinity:\[ E_d = U(\infty) - U(r_e) = 0 - \left( \frac{a}{r_e^{12}} - \frac{b}{r_e^{6}} \right) \]
Simplifying using the equilibrium condition \(2a = br_e^6\), we discover that the dissociation energy is:\[E_d = \frac{b}{2r_e^6}\]

• This value indicates the amount of energy needed to fully overcome the attractive forces binding the atoms together.

Force

The force between atoms in a diatomic molecule is related to the gradient of potential energy with respect to distance. In this context, it is the negative derivative of \(U(r)\) with respect to \(r\):
\[ F(r) = -\frac{dU}{dr} \]
With the specific potential energy function given, this evaluates to:\[F(r) = \frac{12a}{r^{13}} - \frac{6b}{r^{7}}\]

• The force is always directed in a way to move the system toward equilibrium. When atoms are too close, the force is repulsive, preventing them from getting too close. Conversely, when the atoms are apart, the force is attractive, pulling them together.

Stability Analysis

Stability in diatomic molecules refers to whether a small disturbance will return the system to equilibrium or push it away. Analyzing stability involves examining the second derivative of the potential energy, \(\frac{d^2U}{dr^2}\):
If \(\frac{d^2U}{dr^2} > 0\) at \(r = r_e\), then the system is stable because potential energy is minimized at equilibrium. This ensures that any minor disturbance will result in forces that drive the atoms back toward their equilibrium distance, restoring stability.
When calculated:\[ \frac{d^2U}{dr^2} = 156\frac{a}{r^{14}} - 42\frac{b}{r^{8}} \]

• The positive value of this second derivative confirms the existence of stable, attractive forces when the distance \(r\) aligns with \(r_e\).