Problem 35 - 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 \(F(r)\) on one atom as a function of \(r\). Draw 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 equilibrium 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\(^{-10}\) m and the dissociation energy is 1.54 \(\times\) 10\(^{-18}\) J per molecule. Find the values of the constants \(a\) and \(b\).

Step-by-Step Solution

Answer

The force is given by \( F(r) = \frac{12a}{r^{13}} - \frac{6b}{r^7} \). Equilibrium distance \( r = (2a/b)^{1/6} \) is stable. The dissociation energy is \( |U(r_0)| \). For CO, solve equations using provided data for \( a \) and \( b \).

Step 1: Find Force from Potential Energy

The force on one atom as a function of spacing is given by the negative gradient of potential energy. We can express this as \( F(r) = -\frac{dU}{dr} \). Calculate the derivative of \( U(r) = \frac{a}{r^{12}} - \frac{b}{r^6} \).\\[ F(r) = -\frac{d}{dr} \left( \frac{a}{r^{12}} - \frac{b}{r^6} \right) = \frac{12a}{r^{13}} - \frac{6b}{r^7} \]

Step 2: Equilibrium Distance

At equilibrium, the force \( F(r) \) is zero, meaning there is no net force acting on the atoms. Set \( \frac{12a}{r^{13}} - \frac{6b}{r^7} = 0 \) and solve for \( r \).\\[ \frac{12a}{r^{13}} = \frac{6b}{r^7} \] Divide both sides by \( \frac{6}{r^7} \) to simplify: \[ 2a = br^6 \quad \Rightarrow \quad r^6 = \frac{2a}{b} \quad \Rightarrow \quad r = \left( \frac{2a}{b} \right)^{1/6} \]

Step 3: Stability of Equilibrium

To determine stability, analyze the second derivative of \( U(r) \). If \( \frac{d^2U}{dr^2} \) is positive at equilibrium, the equilibrium is stable. Calculate \( \frac{d^2U}{dr^2} = \frac{156a}{r^{14}} - \frac{42b}{r^8} \). Substitute the equilibrium \( r = \left( \frac{2a}{b} \right)^{1/6} \) into \( \frac{d^2U}{dr^2} \) to confirm that it is positive.

Step 4: Dissociation Energy

The dissociation energy is the energy required to change the potential energy from the equilibrium value to zero (at infinite separation). At equilibrium \( r = r_0 = \left( \frac{2a}{b} \right)^{1/6} \), \( U(r_0) = \frac{a}{r_0^{12}} - \frac{b}{r_0^6} \). Since at infinity \( U(\infty) = 0 \), the dissociation energy is: \[ \Delta E = \left| U(r_0) \right| = \left| \frac{a}{r_0^{12}} - \frac{b}{r_0^6} \right| \]

Step 5: Determine Constants for CO

Use given data for CO (\( r_0 = 1.13 \times 10^{-10} \) m and dissociation energy \( 1.54 \times 10^{-18} \) J). Substitute \( r_0 \) in the potential equation to find \( a \) and \( b \). Solve: \[ U(1.13 \times 10^{-10}) = \frac{a}{(1.13\times 10^{-10})^{12}} - \frac{b}{(1.13\times 10^{-10})^6} = -1.54 \times 10^{-18} \] The values of \( a \) and \( b \) can be determined from these conditions.

Key Concepts

Diatomic Molecules

Diatomic molecules are composed of two atoms that are bonded together. These atoms can be the same, like in oxygen (O extsubscript{2}), or different, as in carbon monoxide (CO). The bond between these atoms involves complex interactions governed by molecular potential energy. This describes how energy changes with the distance between the atoms. In a diatomic molecule, understanding how potential energy affects the forces between atoms is crucial in exploring what makes bonds stable or how they break.

Equilibrium Distance

The equilibrium distance between two atoms in a diatomic molecule is a specific spacing where the atoms experience no net force. It can be seen as the sweet spot where forces of attraction and repulsion balance out.

To find this distance, scientists calculate where the force derived from the potential energy formula becomes zero. In our exercise, this involved solving the equation for when the derivative of potential energy with respect to distance, also known as the force, is zero. The resulting formula for the equilibrium distance is: \[ r = \left( \frac{2a}{b} \right)^{1/6} \]

At this distance, minor movements do not result in substantial changes in energy, providing a stable configuration for the diatomic molecule.

Dissociation Energy

Dissociation energy is the amount of energy needed to separate two bonded atoms to an infinite distance, effectively breaking the bond. This is an essential concept for understanding the strength of a molecular bond in a diatomic molecule.

Calculated at the equilibrium distance, the dissociation energy tells us how much energy must be added to overcome the attractive forces holding the atoms together. For the molecule CO, this energy amounted to 1.54 \( \times \ 10^{-18} \ J \) per molecule when calculated using the provided equilibrium distance and the potential energy equation.

The energy can be conceptualized as the minimum energy required to overcome the potential energy barrier holding the two atoms in a stable molecular bond.

Force Calculation

Force calculation in diatomic molecules involves determining how much force acts between two atoms at any given distance. One can achieve this by taking the mathematical derivative of the molecular potential energy with respect to distance.

In the given exercise, the potential energy function is expressed as:\[ U(r) = \frac{a}{r^{12}} - \frac{b}{r^6} \]

The force, therefore, is derived as:\[ F(r) = \frac{12a}{r^{13}} - \frac{6b}{r^7} \]

This negative gradient indicates how the force changes with the atomic spacing, showing that as atoms move closer or further apart, the forces of attraction and repulsion adjust accordingly. Positive force pushes the atoms apart, while negative force pulls them together.

Stability of Equilibrium

Stability of equilibrium in a diatomic molecule means that the atoms tend to return to their equilibrium spacing if slightly disturbed. It defines whether the equilibrium is resistive to changes due to small perturbations.

For an equilibrium to be stable, the second derivative of potential energy with respect to distance must be positive at the equilibrium point. This was verified by calculating:\[ \frac{d^2U}{dr^2} = \frac{156a}{r^{14}} - \frac{42b}{r^8} \] and confirming it yields a positive result at the given equilibrium distance.

Thus, this ensures that any displacement from equilibrium results in a restoring force, indicating a stable balance is reached and maintained unless considerable energy is added, like in dissociation.