Question

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

• A. $\frac{2}{d}\sqrt{\frac{E}{m}}$
• B. $\frac{d}{2}\sqrt{\frac{m}{E}}$
• C. $\frac{E}{d}\sqrt{\frac{2}{m}}$
• D. $\frac{m}{d}\sqrt{\frac{E}{2}}$

Hint:

For any symmetric diatomic molecule with identical masses $m$, the reduced mass is always exactly half of the individual mass ($\mu = \frac{m}{2}$). Remembering this shortcut allows you to quickly assemble the moment of inertia $I = \frac{md^2}{2}$ without executing the full formula derivation.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem presents a rigid diatomic molecule composed of two identical atoms of mass $m$ separated by a fixed bond distance $d$. Given its average rotational kinetic energy $E$, we must calculate its corresponding angular frequency $\omega$.

Step 2: Key Formula or Approach:
The rotational kinetic energy of a rigid body rotating about a fixed axis is given by:
$$E = \frac{1}{2}I\omega^2$$ For a system of two masses rotating about their center of mass, the moment of inertia $I$ is expressed using the reduced mass $\mu$:
$$I = \mu d^2$$ where $\mu = \frac{m_1 m_2}{m_1 + m_2}$.

Step 3: Detailed Explanation:
First, calculate the reduced mass $\mu$ for two identical atoms of mass $m$:
$$\mu = \frac{m \cdot m}{m + m} = \frac{m^2}{2m} = \frac{m}{2}$$ Next, substitute this reduced mass into the moment of inertia expression:
$$I = \left(\frac{m}{2}\right)d^2 = \frac{md^2}{2}$$ Now, substitute the moment of inertia $I$ back into the formula for rotational kinetic energy:
$$E = \frac{1}{2}\left(\frac{md^2}{2}\right)\omega^2$$ $$E = \frac{md^2\omega^2}{4}$$ Rearrange the equation to isolate the square of the angular frequency, $\omega^2$:
$$\omega^2 = \frac{4E}{md^2}$$ Taking the square root of both sides gives the angular frequency $\omega$:
$$\omega = \sqrt{\frac{4E}{md^2}} = \frac{2}{d}\sqrt{\frac{E}{m}}$$

Step 4: Final Answer:
The angular frequency is $\frac{2}{d}\sqrt{\frac{E}{m}}$, which perfectly matches option (A).