Problem 45 - Calculus

Question

The potential energy, \(V\), of two gas molecules separated by a distance \(r\) is given by \(V=-V_{0}\left(2\left(\frac{r_{0}}{r}\right)^{6}-\left(\frac{r_{0}}{r}\right)^{12}\right)\) where \(V_{0}\) and \(r_{0}\) are positive constants. (a) Show that if \(r=r_{0},\) then \(V\) takes on its minimum value, \(-V_{0}\) (b) Write \(V\) as a series in \(\left(r-r_{0}\right)\) up through the quadratic term. (c) For \(r\) near \(r_{0},\) show that the difference between \(V\) and its minimum value is approximately proportional to \(\left(r-r_{0}\right)^{2} .\) In other words, show that \(V-\left(-V_{0}\right)=V+V_{0}\) is approximately proportional to \(\left(r-r_{0}\right)^{2}\) (d) The force, \(F\), between the molecules is given by \(F=-d V / d r .\) What is \(F\) when \(r=r_{0} ?\) For \(r\) near \(r_{0},\) show that \(F\) is approximately proportional to \(\left(r-r_{0}\right)\)

Step-by-Step Solution

Answer

(a) \(V = -V_0\). (b) \(V \approx -V_0 + \frac{72V_0(r-r_0)^2}{r_0^2}\). (c) \(V + V_0 \approx \frac{72V_0(r-r_0)^2}{r_0^2}\). (d) \(F = 0\) at \(r_0\), \(F \propto (r-r_0)\).

Step 1: Calculate V at r = r_0

To check if the potential energy \(V\) is minimized when \(r = r_0\), substitute \(r = r_0\) in the equation.\[ V = -V_{0}\left(2\left(\frac{r_0}{r_0}\right)^6 - \left(\frac{r_0}{r_0}\right)^{12}\right)=-V_{0}(2-1) = -V_{0} \]Thus, \(V\) takes the value \(-V_0\) at \(r = r_0\), indicating it is minimized here.

Step 2: Expand V as a Power Series

Expand \(V\) using a Taylor series around \(r = r_0\) up to the quadratic term. Let \(x = r - r_0\), then \(r = r_0 + x\) and use the Binomial Approximation:\[ \frac{r_0}{r_0 + x} = 1 - \frac{x}{r_0} + \left(\frac{x}{r_0}\right)^2 - \ldots \]Substitute into \(V\):\[V = -V_0 \left( 2\left(1- \frac{x}{r_0} + \left(\frac{x}{r_0}\right)^2\right)^6 - \left(1- \frac{x}{r_0} + \left(\frac{x}{r_0}\right)^2\right)^{12} \right) \] Only keep terms up to \(x^2\) and find that:\[ V \approx -V_0 + \frac{72 V_0 x^2}{r_0^2} \]

Step 3: Show V-V_min is Proportional to (r-r_0)^2

Identify the difference \(V - (-V_0) = V + V_0\) from Step 2, which simplifies to:\[ V + V_0 \approx \frac{72 V_0 x^2}{r_0^2} \]We can see this difference is proportional to \((r - r_0)^2\), showing the quadratic nature of \(V\) near \(r = r_0\).

Step 4: Determine Force F when r = r_0

The force \(F\) is given by \(F = -\frac{dV}{dr}\).First, calculate the derivative \(\frac{dV}{dr}\) by using the original expression for \(V\). When \(r=r_0\):\[\frac{dV}{dr} \bigg|_{r=r_0} = 0\]This indicates that the force \(F\) is zero when \(r = r_0\).

Step 5: Show F is Proportional to (r-r_0)

Use the approximation from Step 2, get the derivative:\[ F = -\frac{d}{dr} \left( -V_0 + \frac{72 V_0 (r - r_0)^2}{r_0^2} \right) = -\frac{144 V_0 (r - r_0)}{r_0^2} \]This shows \(F\) is directly proportional to \(r - r_0\), as expected.

Key Concepts

Taylor series expansion

The Taylor series expansion is a mathematical method used to approximate functions as infinite sums of terms calculated from the values of the function's derivatives at a single point. This expansion is very useful for dealing with functions that are complex or difficult to handle in their entirety. For a function, say \( f(x) \), the Taylor series expansion around a point \( a \) is expressed as:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \, ... \)

In the context of the exercise, by expanding the potential energy \( V \) using a Taylor series around \( r = r_0 \), we can express the function as a series in \( (r - r_0) \). This approach allows us to approximate \( V \) up to quadratic terms, making it easier to estimate how the potential energy changes as \( r \) deviates slightly from \( r_0 \). This is particularly useful when determining behaviors like molecular forces, as it provides insight into how these energies behave near a stable equilibrium point.

derivatives

Derivatives are a fundamental concept in calculus, representing the rate at which a quantity changes. In simpler terms, the derivative of a function at a certain point tells us how quickly that function's value is changing at that point.

• The first derivative of a function \( f \) is denoted as \( f'(x) \), representing the slope or rate of change at any point \( x \).
• The second derivative, denoted \( f''(x) \), provides information about the curvature or concavity of the function, indicating how the rate of change itself is changing.

For the potential energy function \( V \) discussed in the exercise, derivatives play a crucial role:

• First, by taking the first derivative of \( V \) with respect to \( r \), we can find the force \( F \) because \( F = - \frac{dV}{dr} \).
• This relationship helps us understand that when \( r = r_0 \), the force between molecules is zero – signifying a point of equilibrium.
• Also, further derivations let us analyze how variations in \( r \) from \( r_0 \) affect \( V \) and subsequently \( F \).

molecular forces

Molecular forces are interactions that occur between molecules, responsible for the structure and behavior of substances. These forces can be attractive or repulsive, and their strength depends on the distance between molecules. In the context of this exercise, the potential energy function \( V \), which depends on the separation distance \( r \), describes such interactions.Several key concepts to understand about molecular forces include:

• **Lennard-Jones potential:** The given potential energy equation resembles a form similar to the Lennard-Jones potential, which models the interaction between a pair of neutral atoms or molecules. It consists of two main terms: an attractive component and a repulsive component.
• **Equilibrium point:** At \( r = r_0 \), the potential energy is minimized, and the force acting between the molecules is zero. This point represents a stable equilibrium where the attractive and repulsive forces balance each other.
• **Force derivation:** The force between molecules, calculated as the negative derivative of the potential energy with respect to \( r \), describes how the potential energy changes with small adjustments in \( r \). This can help predict molecular behavior and interactions.

Understanding these molecular forces is crucial for fields ranging from chemistry and physics to material science, as they dictate molecular arrangement and reactivity.