Question

Using Bohr's quantization condition, the rotational kinetic energy in the third orbit for a diatomic molecule is \( \left( h = \text{Planck's constant}, I = \text{moment of inertia of diatomic molecule} \right) \)

• A. \( \frac{9h^2}{7I} \)
• B. \( \frac{3h^2}{7I} \)
• C. \( \frac{6h^2}{7I} \)
• D. \( \frac{12h^2}{7I} \)

Hint:

In quantum mechanics, Bohr's quantization condition relates the angular momentum to integer multiples of Planck’s constant, and this is used to derive the rotational kinetic energy in the third orbit.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Bohr's quantization condition states that angular momentum \( L \) is quantized as \( n\frac{h}{2\pi} \). We need to find the rotational kinetic energy based on this condition for \( n = 3 \).
Step 2: Key Formula or Approach:
1. Angular momentum: \( L = \frac{nh}{2\pi} \).
2. Rotational Kinetic Energy: \( K_{rot} = \frac{L^2}{2I} \).
Step 3: Detailed Explanation:
Given \( n = 3 \), the angular momentum is:
\[ L = \frac{3h}{2\pi} \]
Substitute this into the kinetic energy formula:
\[ K_{rot} = \frac{(3h/2\pi)^2}{2I} \]
\[ K_{rot} = \frac{9h^2}{4\pi^2} \cdot \frac{1}{2I} = \frac{9h^2}{8\pi^2 I} \]
Step 4: Final Answer:
The rotational kinetic energy is \( \frac{9h^2}{8\pi^2 I} \).