Question

The potential energy of a system of two atoms is given by the expression \( U = -A/r^2 + B/r^{10} \). A stable molecule is formed with the release of 8.0 eV of energy, when the interatomic distance is 2.8 Å. The values of A and B are:

• A. A = \(1.22 \times 10^{-37}\) J m\(^2\) and B = \(9.52 \times 10^{-115}\) J m\(^{10}\)
• B. A = \(1.22 \times 10^{-23}\) J m\(^2\) and B = \(1.52 \times 10^{-115}\) J m\(^{10}\)
• C. A = \(1.22 \times 10^{-23}\) J m\(^2\) and B = \(2.52 \times 10^{-115}\) J m\(^{10}\)
• D. A = \(1.22 \times 10^{-27}\) J m\(^2\) and B = \(3.53 \times 10^{-115}\) J m\(^{10}\)

Hint:

For this type of problem, always set up the two main equations first: \( dU/dr = 0 \) at \(r=r_0\) and \( U(r_0) = -E_c \). Solving this system algebraically before plugging in numbers can simplify the calculations, as shown by finding expressions like \( B = E_c r_0^{10}/4 \) and \( A = 5E_c r_0^2/4 \).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A stable molecule is formed at the equilibrium separation \(r_0\), where the potential energy \(U\) is at a minimum. This implies two conditions:
1. The net force is zero: \( F = -\frac{dU}{dr} = 0 \) at \( r = r_0 \).
2. The potential energy at this point is equal to the negative of the binding energy (energy released): \( U(r_0) = -E_c \).
Step 2: Key Formula or Approach:
Given \( U(r) = -\frac{A}{r^2} + \frac{B}{r^{10}} \).
First, find the derivative and set it to zero:
\[ \frac{dU}{dr} = \frac{2A}{r^3} - \frac{10B}{r^{11}} \]
At \( r = r_0 \), \( \frac{dU}{dr} = 0 \):
\[ \frac{2A}{r_0^3} - \frac{10B}{r_0^{11}} = 0 \implies 2A r_0^8 = 10B \implies A = \frac{5B}{r_0^8} \]
Second, use the binding energy information:
\[ U(r_0) = -\frac{A}{r_0^2} + \frac{B}{r_0^{10}} = -E_c \]
Step 3: Detailed Explanation:
We have the following given values:
\( E_c = 8.0 \) eV = \( 8.0 \times 1.602 \times 10^{-19} \) J \( \approx 1.28 \times 10^{-18} \) J
\( r_0 = 2.8 \) Å = \( 2.8 \times 10^{-10} \) m
Substitute the expression for A into the energy equation:
\[ U(r_0) = -\frac{1}{r_0^2}\left(\frac{5B}{r_0^8}\right) + \frac{B}{r_0^{10}} = -E_c \]
\[ -\frac{5B}{r_0^{10}} + \frac{B}{r_0^{10}} = -E_c \]
\[ -\frac{4B}{r_0^{10}} = -E_c \implies B = \frac{E_c r_0^{10}}{4} \]
Now, calculate B:
\[ B = \frac{(1.28 \times 10^{-18} \text{ J}) \times (2.8 \times 10^{-10} \text{ m})^{10}}{4} \]
\[ (2.8 \times 10^{-10})^{10} = (2.8)^{10} \times 10^{-100} \approx 2.97 \times 10^4 \times 10^{-100} = 2.97 \times 10^{-96} \]
\[ B \approx \frac{(1.28 \times 10^{-18}) \times (2.97 \times 10^{-96})}{4} \approx \frac{3.80 \times 10^{-114}}{4} \approx 0.95 \times 10^{-114} = 9.5 \times 10^{-115} \text{ J m}^{10} \]
This matches \( B = 9.52 \times 10^{-115} \) J m\(^{10}\).
Next, find A using the relation derived from the energy at equilibrium: \[ A = \frac{5B}{r_0^8} = \frac{5 E_c r_0^{10}}{4 r_0^8} = \frac{5}{4} E_c r_0^2 \]
Calculate A:
\[ A = \frac{5}{4} \times (1.28 \times 10^{-18} \text{ J}) \times (2.8 \times 10^{-10} \text{ m})^2 \]
\[ A = 1.25 \times (1.28 \times 10^{-18}) \times (7.84 \times 10^{-20}) \]
\[ A = 1.25 \times 10.035 \times 10^{-38} \approx 12.54 \times 10^{-38} = 1.254 \times 10^{-37} \text{ J m}^2 \]
Step 4: Final Answer:
The calculated values are A \( \approx 1.25 \times 10^{-37} \) J m\(^2\) and B \( \approx 9.52 \times 10^{-115} \) J m\(^{10}\). These values are in excellent agreement with option (A).