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(A) The rotational kinetic energy $E$ of a rigid body is given by $E = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the

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$\frac{2}{d} \sqrt{\frac{E}{m}}$

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(A) The rotational kinetic energy $E$ of a rigid body is given by $E = \frac{1}{2} I \omega^2$,where $I$ is the moment of inertia and $\omega$ is the angular frequency.For a molecule consisting of two atoms of mass $m$ separated by a distance $d$,the moment of inertia about an axis passing through the center of mass and perpendicular to the line joining the atoms is $I = m(\frac{d}{2})^2 + m(\frac{d}{2})^2 = 2m(\frac{d^2}{4}) = \frac{md^2}{2}$.Substituting $I$ into the energy equation: $E = \frac{1}{2} (\frac{md^2}{2}) \omega^2 = \frac{md^2}{4} \omega^2$.Solving for $\omega^2$: $\omega^2 = \frac{4E}{md^2}$.Taking the square root: $\omega = \sqrt{\frac{4E}{md^2}} = \frac{2}{d} \sqrt{\frac{E}{m}}$.

$A$ 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:

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