Physical Chemistry Third Edition

990 23 Optical Spectroscopy and Photochemistry

Since the selection rule requires that∆J±2, the Raman shift for the Stokes lines is

given by

∆ ̃ν ̃ν− ̃ν′ ̃B[(J+2)(J+3)−J(J+1)] ̃B[J^2 + 5 J+6)−J^2 +J)]

 ̃B[4J+6]

The^16 O nuclei are bosons (I0) and the electronic ground state is a sigma state, so only

even values ofJcan occur. The first spectral line should occur at 6 ̃B, and the splitting

between lines should equal 8 ̃B. The first full line that is visible in the figure is the second line,

corresponding to the transition fromJ2toJ4. The splitting is found to be 3.09 cm−^1

by measurement in the figure. Using the symbol∆ ̃vfor the splitting, we obtain

re^2 

h

2 π^2 mOc∆ ̃ν



6. 6261 × 10 −^34 Js

2 π^2 (2. 656 × 10 −^26 kg)(3.09 cm−^1 )(2. 9979 × 1010 cm s−^1 )

 1. 364 × 10 −^20 m^2

re 1. 17 × 10 −^10 m117 pm 1 .17 Å

This result agrees fairly well with the accepted value, 116.15 pm.

The vibrational selection rule for a diatomic molecule is that∆v0,±1 and the

rotational selection rule is∆J0,±2. A band in the vibrational Raman spectrum

of a diatomic molecule has three branches corresponding to the different changes in

the rotational quantum number. TheQbranch corresponds to∆J0, theObranch

90 80 70 60

58.2

50 40 30 20 10

8.9

Intensity

Raman shift/cm^21

CO 2

Figure 23.19 Rotational Raman Spectrum of Carbon Dioxide.The axis is the differ-

ence between the reciprocal wavelength of the incident and that of the scattered radiation.

The peak at zero Raman shift is the scattered radiation with no change in energy of the

molecules. From L. Claron Hoskins,J. Chem. Educ., 54 , 642 (1977).