Question

A molecule consists of two atoms each of mass \( m \) and separated by a distance \( d \). At room temperature the average rotational kinetic energy is \( E \), then its angular frequency is

• A. \( \frac{2}{d} \sqrt{\frac{E}{m}} \)
• B. \( \sqrt{\frac{m}{Ed}} \)
• C. \( \frac{d}{2} \sqrt{\frac{m}{E}} \)
• D. \( \sqrt{\frac{Ed}{m}} \)

Hint:

For rotational motion, use the relation \( E = \frac{1}{2} I \omega^2 \) to connect energy and angular frequency.

Answer 1

The Correct Option is A

Step 1: Understanding the relation.
The angular frequency \( \omega \) for a rotating molecule is related to the rotational kinetic energy by the formula: \[ E = \frac{1}{2} I \omega^2 \] where \( I \) is the moment of inertia. For a diatomic molecule, \( I = \mu r^2 \), where \( \mu \) is the reduced mass and \( r \) is the distance between the atoms.
Step 2: Using the given values.
Substitute \( I = \mu r^2 \) and \( \mu = \frac{m}{2} \), and solve for the angular frequency \( \omega \). After simplification, we obtain: \[ \omega = \frac{2}{d} \sqrt{\frac{E}{m}} \] Step 3: Conclusion.
Thus, the angular frequency is \( \frac{2}{d} \sqrt{\frac{E}{m}} \), which corresponds to option (A).