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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.

Answers (1)

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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

Posted by

Devendra Khairwa

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