Page 134 - Bruno Linder Elementary Physical Chemistry
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August 18, 2010 11:37      9in x 6in     b985-ch11     Elementary Physical Chemistry





                                             Elements of Molecular Spectroscopy           119























                                                                           1
                                                                             35
                                                 Vibration–rotation spectrum of H Cl.
                                       Fig. 11.2
                               The spacing between all adjacent lines is the same except the spacing
                            between the lines 2865.10 and 2906.25, which is twice as large. This gap
                            separates the R- and P-branches. The above information suggests that
                                                                       ∗
                                                            ∗
                                                J =2 → 3   ν = ν +2B × 3
                                                                 ∗
                                                            R    o
                                                                 ∗
                                             R J =1 → 2    ν = ν +2B × 2             (11.13a)
                                                            ∗
                                                                       ∗
                                                            R    o
                                                            ∗
                                                                       ∗
                                                J =0 → 1   ν = ν +2B × 1
                                                                 ∗
                                                            R    o
                                                                 ∗
                                                                       ∗
                                                J =1 → 0 ν = ν − 2B × 1
                                                            ∗
                                                            R
                                                                 o
                                                                 ∗
                                                                       ∗
                                                            ∗
                                             P J =2 → 1 ν = ν − 2B × 2               (11.13b)
                                                                 o
                                                            R
                                                                 ∗
                                                                       ∗
                                                            ∗
                                                J =3 → 2 ν = ν − 2B × 3
                                                            R
                                                                 R
                            It is clear that by adding the lowest R peak and the highest P peak, and
                            dividing the sum by 2 we can determine ν .Thus,
                                                                ∗
                                                                0
                                              ∗
                                             ν =1/2 (2906.25 + 2865.10) cm −1
                                              0
                                                = 2995.6cm −1                         (11.14)
                            This immediately gives B =10.34cm −1 , from which we can obtain the
                                                   ∗
                            moment of inertia, I, and the internuclear distance, d.
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