### Q.20:- The oxygen molecule has a mass of 5.30 × 10^{–26} kg and a moment of inertia of 1.94×10^{–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 500 m/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.

**Answer:-**Mass of an oxygen molecule, *m* = 5.30 × 10^{–26} kg

Moment of inertia, *I* = 1.94 × 10^{–46} kg m^{2}

Velocity of the oxygen molecule, *v* = 500 m/s

The separation between the two atoms of the oxygen molecule = 2*r*

Mass of each oxygen atom = *m*/2

Hence, moment of inertia *I*, is calculated as:

(*m*/2)r^{2} + (*m*/2)*r*^{2} = *mr*^{2}

*r* = ( *I* / *m*)^{1/2}

(1.94 × 10^{-46} / 5.36 × 10^{-26} )^{1/2} = 0.60 × 10^{-10} m

It is given that:

KE_{rot} = (2/3)KE_{trans}

(1/2) I ω^{2} = (2/3) × (1/2) × *mv*^{2}

*mr*^{2}ω^{2} = (2/3)*mv*^{2}

ω = (2/3)^{1/2} (*v*/*r*)

= (2/3)^{1/2} (500 / 0.6 × 10^{-10}) = 6.80 × 10^{12} rad/s.