Question

The oxygen molecule has a mass of $5.4\times {10}^{-26}\text{kg}$ and moment of inertia is $1.8\times {10}^{-46}{\text{kg/m}}^2$ about an axis passing through its centre and perpendicular to the line joining the two atoms. Suppose the mean speed of the molecule in a sample of oxygen has is $500{\text{ms}}^{-1}$ and its kinetic energy of rotation is two third of its kinetic energy of translation. Its angular velocity is found to be approximately $n\times {10}^{12}\text{rad/sec}$ where $n=$

Answer 1

Given that rotational kinetic energy is two third of translational energy
$\frac{1}{2}I{\omega }^2=\frac{2}{3}\left(\frac{1}{2}m{v}^2\right)$
$\Rightarrow \omega =\sqrt{\frac{2m{v}^2}{3I}}=v\sqrt{\frac{2m}{3I}}$
or $\omega =500\times \sqrt{\frac{2\times 5.4\times {10}^{-26}}{3\times 1.8\times {10}^{-46}}}$
$=5\sqrt{2}\times {10}^{12}\approx 7\times {10}^{12}\text{rad/sec}$