The oxygen molecule has a mass of `5.30 xx 10^(-26) kg` and a moment of inertia of `1.94 xx 10^(-46) kg m^(2)` about an axis through its centre perpendicular to the line 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.

Explanation

Here, `m=5.30xx10^(-26) kg`
`l=1.94xx10^(-46) kg m^(2)`
`v=500 m//s`
If `(m)/(2)` is mass of each atom of oxygen and 2r is distance between the two atom as shown in Fig, then
`I=(m)/(2) r^(2)+(m)/(2)r^(2)=mr^(2)`
`r=sqrt((l)/(m))=sqrt((1.94xx10^(-46))/(5.30xx10^(-26)))`
`=0.61xx10^(-10)m`
As K.E. or rotationa`=(2)/(3)` K.E. of translation
`therefore (1)/(2)I omega^(2)=(2)/(3)xx(1)/(3)momega^(2)`
`(1)/(2)(mr^(2))omega^(2)=(1)/(2)mv^(2)`
` omega=sqrt((2)/(3))(v)/(r )=sqrt((2)/(3))xx(500)/(0.61xx10^(-10))=6.7xx10^(12) rad//s`.