The oxygen molecule has a mass of $5.30 \times 10^{-26}kg$ and a moment of inertia of $1.94 \times 10^{-46}kgm^2$ 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 $500m/s$ and that its kinetic energy of rotation is two thirds of its kinetic energy of translation. Find the average angular velocity of the molecule.
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Mass of an oxygen molecule, $m = 5.30 \times 10^{-26}kg$ Moment of inertia, $I = 1.94 \times 10^{-46}kgm^2$^ Velocity of the oxygen molecule, v = 500m/s The separation between the two atoms of the oxygen molecule = 2r Mass of each oxygen atom $=\frac{\text{m}}{2}$ Hence, moment of inertia I, is calculated as, $\Big(\frac{\text{m}}{2}\Big)\text{r}^2+\Big(\frac{\text{m}}{2}\Big)\text{r}^2=\text{mr}^2$
$\text{r}=\Big(\frac{\text{l}}{\text{m}}\Big)^{\frac{1}{2}}$
$\Big(\frac{1.94\times10^{-46}}{5.36\times10^{-26}}\Big)=0.60\times10^{-10}\text{m}$ It is given that, $\text{kE}_{\text{rot}}=\Big(\frac{2}{3}\Big)\times\Big(\frac{1}{2}\Big)\times\text{mv}^2$
$\text{m}\text{r}^2\omega^2=\Big(\frac{2}{3}\Big)\text{mv}^2$
$\omega=\Big(\frac{2}{3}\Big)^{\frac{1}{2}}\Big(\frac{\text{v}}{\text{r}}\Big)$
$\omega=\Big(\frac{2}{3}\Big)^{\frac{1}{2}}\Big(\frac{500}{0.6\times10^{-10}}\Big)$
$\omega=6.80\times10^{12}\text{rad/s}$
$\text{r}=\Big(\frac{\text{l}}{\text{m}}\Big)^{\frac{1}{2}}$
$\Big(\frac{1.94\times10^{-46}}{5.36\times10^{-26}}\Big)=0.60\times10^{-10}\text{m}$ It is given that, $\text{kE}_{\text{rot}}=\Big(\frac{2}{3}\Big)\times\Big(\frac{1}{2}\Big)\times\text{mv}^2$
$\text{m}\text{r}^2\omega^2=\Big(\frac{2}{3}\Big)\text{mv}^2$
$\omega=\Big(\frac{2}{3}\Big)^{\frac{1}{2}}\Big(\frac{\text{v}}{\text{r}}\Big)$
$\omega=\Big(\frac{2}{3}\Big)^{\frac{1}{2}}\Big(\frac{500}{0.6\times10^{-10}}\Big)$
$\omega=6.80\times10^{12}\text{rad/s}$
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