Determine the angular rotation velocity of an `S_(2)` molecule promoted to the first exccited rotational level if the distance between its nuclei is `d= 189p m`

Explanation

In the first excited raotaional level `J=1`
so `E_(J)=1xx2( ħ^(2))/(2I)=(1)/(2)I omega^(2)` classically
Thus `omega= sqrt(2)( ħ)/(I)`
Now `I=Sigma m_(I)r_(i)^(2)=(m)/(2)(d^(2))/(4)+(m)/(2)(d^(2))/(4)=m(d^(2))/(4)`
where `m` is the mass of the mole cub and `r_(i)` is the distance of the atom from the axis.
Thus
`omega=(4sqrt(2) ħ)/(md^(2))= 1.56xx10^(11) rad//s`