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The oxygen molecule has a mass of $5.3 \times 10^{-26} \mathrm{~kg}$ and a M.I of $1.94 \times 10^{-46} \mathrm{~kg} \mathrm{~m}^2$ about an axis through its centre perpendicular to the lines joining the two atoms. Suppose the mean speed of such a molecule in a gas is $500 \mathrm{~m} / \mathrm{s}$ and that its K.E. of rotation is two thirds of its K.E of translation. Find the average angular velocity of the molecule.
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Verified Answer
Let $m / 2=$ mass of each oxygen atom
$2 R=$ separation between them.
Total mass of the molecule
$=m=5.3 \times 10^{-26} \mathrm{~kg}$
M.I. $=I=1.94 \times 10^{-46} \mathrm{~kg} \mathrm{~m}{ }^2 ; v=500 \mathrm{~m} / \mathrm{s}$
$I_{X Y}=\frac{m}{2} r^2+\frac{m}{2} r^2=m r^2$
$\therefore r=\sqrt{\frac{I_{X Y}}{m}}=\sqrt{\frac{1.94 \times 10^{-46}}{5.3 \times 10^{-26}}}=0.61 \times 10^{-10} \mathrm{~m}$
Given that,
$$
\begin{aligned}
&\frac{1}{2} I \omega^2=\frac{2}{3} \times \frac{1}{2} m v^2 \Rightarrow \frac{1}{2}\left(m r^2\right) \omega^2=\frac{1}{3} m v^2 \\
&\omega=\sqrt{\frac{2}{3}} \cdot \frac{v}{r}=\sqrt{\frac{2}{3}} \times \frac{500}{0.61 \times 10^{-10}} \\
&=6.7 \times 10^{12} \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
$2 R=$ separation between them.
Total mass of the molecule
$=m=5.3 \times 10^{-26} \mathrm{~kg}$
M.I. $=I=1.94 \times 10^{-46} \mathrm{~kg} \mathrm{~m}{ }^2 ; v=500 \mathrm{~m} / \mathrm{s}$
$I_{X Y}=\frac{m}{2} r^2+\frac{m}{2} r^2=m r^2$
$\therefore r=\sqrt{\frac{I_{X Y}}{m}}=\sqrt{\frac{1.94 \times 10^{-46}}{5.3 \times 10^{-26}}}=0.61 \times 10^{-10} \mathrm{~m}$
Given that,
$$
\begin{aligned}
&\frac{1}{2} I \omega^2=\frac{2}{3} \times \frac{1}{2} m v^2 \Rightarrow \frac{1}{2}\left(m r^2\right) \omega^2=\frac{1}{3} m v^2 \\
&\omega=\sqrt{\frac{2}{3}} \cdot \frac{v}{r}=\sqrt{\frac{2}{3}} \times \frac{500}{0.61 \times 10^{-10}} \\
&=6.7 \times 10^{12} \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
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