Problem 77 - Physics for Scientists and Engineers with Modern Physics

Question

(III) The potential energy of the two atoms in a diatomic (two-atom) molecule can be written $$ U(r)=-\frac{a}{r^{6}}+\frac{b}{r^{12}} $$ where \(r\) is the distance between the two atoms and \(a\) and \(b\) are positive constants. (a) At what values of \(r\) is \(U(r)\) a minimum? A maximum? (b) At what values of \(r\) is \(U(r)=0 ?\) (c) Plot \(U(r)\) as a function of \(r\) from \(r=0\) to \(r\) at a value large enough for all the features in \((a)\) and \((b)\) to show. ( \(d\) ) Describe the motion of one atom with respect to the second atom when \(E<0,\) and when \(E>0 .(e)\) Let \(F\) be the force one atom exerts on the other. For what values of \(r\) is \(F>0, F<0, F=0 ?\) (f) Determine \(F\) as a function of \(r\).

Step-by-Step Solution

Answer

(a) Minimum at \( r = \left(\frac{2b}{a}\right)^{1/6} \). (b) \( r = \left(\frac{b}{a}\right)^{1/6} \). (e) \( F=0 \) at \( r = \left(\frac{2b}{a}\right)^{1/6} \).

Step 1: Differentiate U(r)

To find the values of \( r \) where \( U(r) \) is at a minimum or maximum, we need to differentiate \( U(r) \) with respect to \( r \) and set it to zero. This will give us the critical points.\[ U'(r) = \frac{d}{dr}\left(-\frac{a}{r^{6}} + \frac{b}{r^{12}}\right) = \frac{6a}{r^7} - \frac{12b}{r^{13}} \] Set \( U'(r) = 0 \):\[ \frac{6a}{r^7} = \frac{12b}{r^{13}} \] Simplifying:\[ 6ar^{6} = 12b \]\[ ar^{6} = 2b \] \[ r^{6} = \frac{2b}{a} \]\[ r = \left(\frac{2b}{a}\right)^{\frac{1}{6}} \]

Step 2: Analyze U(r) to Identify Maximum/Minimum

We already found the critical point \( r = \left(\frac{2b}{a}\right)^{\frac{1}{6}} \). To determine if this is a minimum or maximum, consider the second derivative:\[ U''(r) = \frac{d^2}{dr^2}\left(-\frac{a}{r^{6}} + \frac{b}{r^{12}}\right) = -\frac{42a}{r^8} + \frac{156b}{r^{14}} \] Evaluate \( U''(r) \) at \( r = \left(\frac{2b}{a}\right)^{\frac{1}{6}} \).\[ U''\left(\left(\frac{2b}{a}\right)^{\frac{1}{6}}\right) eq 0 \] If \( U''(r) > 0 \) then the function is concave up (minimum). Since the value is positive, \( r = \left(\frac{2b}{a}\right)^{\frac{1}{6}} \) is a minimum.

Step 3: Set U(r) = 0

To find the \( r \) values where \( U(r) = 0 \):Set \( -\frac{a}{r^{6}} + \frac{b}{r^{12}} = 0 \).\[ \frac{a}{r^{6}} = \frac{b}{r^{12}} \] Simplifying:\[ ar^{6} = b \]\[ r^{6} = \frac{b}{a} \]\[ r = \left(\frac{b}{a}\right)^{\frac{1}{6}} \]

Step 4: Plot U(r)

Visualize \( U(r) = -\frac{a}{r^6} + \frac{b}{r^{12}} \). The plot should show features such as the minimum point at \( r = \left(\frac{2b}{a}\right)^{\frac{1}{6}} \) and the roots at \( r = \left(\frac{b}{a}\right)^{\frac{1}{6}} \). When \( r\) is just above 0 and approaching 0, \( U(r) \) approaches infinity. As \( r \) increases, \( U(r) \rightarrow 0 \).

Step 5: Describe Motion for E 0

If \( E < 0 \), the molecule is in a bound state where the atom oscillates around the equilibrium distance \( r = \left(\frac{2b}{a}\right)^{\frac{1}{6}} \).If \( E > 0 \), the atom can escape the potential well, leading to unbound motion.

Step 6: Determine Force F as function of r

The force \( F \) is the negative gradient of \( U(r) \). \[ F = -\frac{dU}{dr} = -\left(\frac{6a}{r^7} - \frac{12b}{r^{13}}\right) \]\[ F = -\frac{6a}{r^7} + \frac{12b}{r^{13}} \]

Step 7: Determine when F>0, F=0, F<0

Force \( F \) is zero when \( \frac{6a}{r^7} = \frac{12b}{r^{13}} \), which already simplifies to \( r = \left(\frac{2b}{a}\right)^{1/6} \).For \( F > 0 \), \( \frac{6a}{r^7} < \frac{12b}{r^{13}} \), or \( r > \left(\frac{2b}{a}\right)^{1/6} \).For \( F < 0 \), \( \frac{6a}{r^7} > \frac{12b}{r^{13}} \), or \( r < \left(\frac{2b}{a}\right)^{1/6} \).

Key Concepts

Potential Energy Function

In the context of a diatomic molecule, the potential energy function characterizes the interaction between two atoms. The function for potential energy, often denoted as \( U(r) \), is dependent on the distance \( r \) between the atoms. It is given by the equation:\[ U(r) = -\frac{a}{r^6} + \frac{b}{r^{12}} \]This equation includes two terms:

• The first term \(-\frac{a}{r^6}\) represents an attractive force.
• The second term \(\frac{b}{r^{12}}\) reflects a repulsive force.

These two opposing forces combine to form the potential energy function in a diatomic molecule. The balancing of these forces determines the molecule's equilibrium position, which is the result of minimizing overall potential energy.

Equilibrium Distance

The equilibrium distance in a diatomic molecule is the distance where the system's potential energy is at a minimum. To find this distance, differentiate the potential energy function \( U(r) \), and solve for critical points where the derivative equals zero. This leads us to: \[ r = \left(\frac{2b}{a}\right)^{\frac{1}{6}} \]At this particular distance:

• The attractive and repulsive forces are balanced.
• The molecule is most stable, meaning it is at rest when undisturbed.
• Potential energy reaches its lowest point, indicating maximum stability.

The equilibrium distance is highly important in chemistry and physics as it plays a crucial role in a molecule's behavior and its reactive properties.

Force Between Atoms

The force between atoms in a diatomic molecule can be understood by looking at the negative gradient of the potential energy. This gives us the expression for force \( F \):\[ F = -\frac{dU}{dr} = -\left(\frac{6a}{r^7} - \frac{12b}{r^{13}}\right) \]When analyzed, we can determine:

• \( F = 0 \) occurs at \( r = \left(\frac{2b}{a}\right)^{1/6} \), meaning the forces are balanced at this equilibrium distance.
• \( F > 0 \) when the atoms are farther apart than the equilibrium, indicating a net attractive force drawing them together.
• \( F < 0 \) when the atoms are closer than the equilibrium, indicating a net repulsive force pushing them apart.

These relationships help understand how atoms in a molecule interact, and how they return to equilibrium after any disruption.

Bound and Unbound States

The terms "bound" and "unbound" refer to whether the atoms in a diatomic molecule can escape each other's influence. This depends on the total energy \( E \) of the system relative to the potential energy:

• If \( E < 0 \), the molecule is in a bound state. Here, the atom's energy is insufficient to overcome the potential energy, resulting in oscillations around the equilibrium distance \( r = \left(\frac{2b}{a}\right)^{1/6} \).
• If \( E > 0 \), the molecule is in an unbound state. The atom has enough energy to escape the potential energy "trap" and move freely, meaning it is not restricted to its equilibrium position.

These concepts are key in understanding phenomena such as chemical bonding, molecular stability, and reactivity, as well as the transition between different states of a molecular system.