(b) Explain why you expect there to be a value of r that minimizes V(r). Choose the correct answer below. O A. Find the first derivative of the given equation. Find r by equating the first derivative to 0. Now, find the second derivative and substitute the value of r. As the second derivative is negative, you expect there to be a value of r that maximize V(r). O B. Find the first derivative of the given equation. Find r by equating the first derivative to 0. Now, find the second derivative and substitute the value of r. As the second derivative is negative, you expect there to be a value of r that minimizes V(r). O C. Find the first derivative of the given equation. Find r by equating the first derivative to 0. Now, find the second derivative and substitute the value of r. As the second derivative is positive, you expect there to be a value of r that minimizes V(r). O D. Find the first derivative of the given equation. Find r by equating the first derivative to 0. Now, find the second derivative and substitute the value of r. As the second derivative is positive, you expect there to be a value of r that maximize V(r).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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16-2:

(b) Explain why you expect there to be a value of \( r \) that minimizes \( V(r) \). Choose the correct answer below.

- ∘ A. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is negative, you expect there to be a value of \( r \) that maximizes \( V(r) \).

- ∘ B. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is negative, you expect there to be a value of \( r \) that minimizes \( V(r) \).

- ∘ C. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is positive, you expect there to be a value of \( r \) that minimizes \( V(r) \).

- ∘ D. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is positive, you expect there to be a value of \( r \) that maximizes \( V(r) \).
Transcribed Image Text:(b) Explain why you expect there to be a value of \( r \) that minimizes \( V(r) \). Choose the correct answer below. - ∘ A. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is negative, you expect there to be a value of \( r \) that maximizes \( V(r) \). - ∘ B. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is negative, you expect there to be a value of \( r \) that minimizes \( V(r) \). - ∘ C. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is positive, you expect there to be a value of \( r \) that minimizes \( V(r) \). - ∘ D. Find the first derivative of the given equation. Find \( r \) by equating the first derivative to 0. Now, find the second derivative and substitute the value of \( r \). As the second derivative is positive, you expect there to be a value of \( r \) that maximizes \( V(r) \).
**Lennard-Jones 6-3 Potential Model**

One popular model for the interactions between two molecules is the Lennard-Jones 6-3 potential. According to this model, the energy of interaction between two molecules that are distance \( r \) apart is given by the following function. Molecules will attract or repel each other until they reach a distance that minimizes the function \( V(r) \). The coefficient \( A \) is a positive constant. Complete parts (a) through (c).

\[ V(r) = \frac{1}{r^6} - \frac{A}{r^3}, \; r > 0 \]

**Explanation:**

The function \( V(r) \) models how the potential energy changes with the distance \( r \) between the molecules. It has two main terms:

1. **Repulsive Term**: \( \frac{1}{r^6} \)
   - This term becomes significant at short distances and leads to positive potential energy, indicating repulsion between the molecules as they get very close to each other.

2. **Attractive Term**: \( -\frac{A}{r^3} \)
   - This term dominates at moderate distances and results in negative potential energy, reflecting an attractive force pulling the molecules together.

The balance between these two terms results in a minimum potential energy at a certain distance, which is the equilibrium distance where the molecules are most stable.
Transcribed Image Text:**Lennard-Jones 6-3 Potential Model** One popular model for the interactions between two molecules is the Lennard-Jones 6-3 potential. According to this model, the energy of interaction between two molecules that are distance \( r \) apart is given by the following function. Molecules will attract or repel each other until they reach a distance that minimizes the function \( V(r) \). The coefficient \( A \) is a positive constant. Complete parts (a) through (c). \[ V(r) = \frac{1}{r^6} - \frac{A}{r^3}, \; r > 0 \] **Explanation:** The function \( V(r) \) models how the potential energy changes with the distance \( r \) between the molecules. It has two main terms: 1. **Repulsive Term**: \( \frac{1}{r^6} \) - This term becomes significant at short distances and leads to positive potential energy, indicating repulsion between the molecules as they get very close to each other. 2. **Attractive Term**: \( -\frac{A}{r^3} \) - This term dominates at moderate distances and results in negative potential energy, reflecting an attractive force pulling the molecules together. The balance between these two terms results in a minimum potential energy at a certain distance, which is the equilibrium distance where the molecules are most stable.
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