# Q7.20     The oxygen molecule has a mass of $5.30\times 10^{-26}kg$  and a moment of inertia of               $1.94\times 10^{-46}kg\: m^{2}$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 $500m/s$ and              that its kinetic energy of rotation is two thirds of its kinetic energy of translation.             Find the average angular velocity of the molecule.

D Devendra Khairwa

We are given the moment of inertia and the velocity of the molecule.

Let the mass of oxygen molecule be m.

So the mass of each oxygen atom is given by :      $\frac{m}{2}$

Moment of inertia is :

$I\ =\ \frac{m}{2}r^2\ +\ \frac{m}{2}r^2\ =\ mr^2$

or                                   $r\ =\ \sqrt{\frac{I}{m}}$

or                                  $r\ =\ \sqrt{\frac{1.94\times 10^{-46}}{5.36\times 10^{-26}}}\ =\ 0.60\times 10^{-10}\ m$

We are given that :

$E_{rot}\ =\ \frac{2}{3}E_{tra}$

or                                  $\frac{1}{2}I\omega ^2\ =\ \frac{2}{3}\times \frac{1}{2}mv^2$

or                                 $\omega \ =\ \sqrt{\frac{2}{3}}\times \frac{v}{r}$

or                                        $=\ 6.80\times 10^{12}\ rad/s$

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