Question #ccb3c

1 Answer
Nov 5, 2017

Consider a diatomic gas molecule.

The rotational kinetic energy,

#E_(rot) = (Iomega^2)/2# where #I# is the moment of inertia.

It may be expressed in terms of total angular momentum of the molecule as,

#E_(rot) = J^2/(2I)#

But from Quantum mechanics,
#J = root(2)[j(j + 1)]barh#
Where #j# is angular momentum quantum number determined either by #LS# coupling or #jj# coupling.

Therefore,

#E_j = (j(j + 1)barh^2)/(2I)#

Therefore as the molecule makes a transition (selection rule #Deltaj = +1# or #-1#),

#DeltaE = E_(j+1) - E_j = (2(j + 1)barh^2)/(2I)#

Corresponding photon emitted has frequency, #nu = (DeltaE)/h = 2B(j + 1)#

Where #B = h/(8Ipi^2)# is some constant.

Therefore, #nu = 2B(j + 1)# which shows that lines are spaced equally (with varying #j#) in pure rotational spectra of the diatomic gas.