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A diatomic gas consisting of rigid molecules is at a temperature of $87^{\circ} \mathrm{C}$. If the moment of inertia of the rotating diatomic rigid molecule is $2.76 \times 10^{-39} \mathrm{gcm}^2$, then the rms angular speed of the molecule is (Boltzmann constant $=1.38 \times 10^{-23} \mathrm{JK}^{-1}$ )
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Verified Answer
The correct answer is:
$6 \times 10^{12}$ rads $^{-1}$
Moment of inertia of diatomic molecule,
$$
I=\frac{2}{3} M r^2
$$
rms speed of molecule,
$$
v=\sqrt{\frac{3 k T}{M}}
$$
rms angular speed,
$$
\begin{aligned}
\omega=\frac{v}{r} & =\frac{\sqrt{3 k T / M}}{\sqrt{3 I / 2 M}}=\sqrt{\frac{2 k T}{I}} \\
& =\sqrt{\frac{2 \times 1.38 \times 10^{-23} \times(87+273)}{2.76 \times 10^{-46}}} \\
& =\sqrt{360 \times 10^{23}}=\sqrt{36 \times 10^{24}} \\
& =6 \times 10^{12} \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
$$
I=\frac{2}{3} M r^2
$$
rms speed of molecule,
$$
v=\sqrt{\frac{3 k T}{M}}
$$
rms angular speed,
$$
\begin{aligned}
\omega=\frac{v}{r} & =\frac{\sqrt{3 k T / M}}{\sqrt{3 I / 2 M}}=\sqrt{\frac{2 k T}{I}} \\
& =\sqrt{\frac{2 \times 1.38 \times 10^{-23} \times(87+273)}{2.76 \times 10^{-46}}} \\
& =\sqrt{360 \times 10^{23}}=\sqrt{36 \times 10^{24}} \\
& =6 \times 10^{12} \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
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