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(A) The moment of inertia $I$ of a diatomic molecule consisting of two atoms of mass $m$ separated by a distance $d$ about an axis passing through its

Vedclass · Accepted Answer

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

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

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

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