Question

For 1H
19F, the lowest wave number absorption line in its rotational spectrum occurs at
41.11 cm1
. Calculate the wave numbers corresponding to its second, third and fourth
absorption lines. Explain the term, rotational spacing using these values

Answer 1

the rotational spectrum of a diatomic molecule consists of a series of equally spaced lines. the energies of these lines are given as E = 2B(J+1), where J is the rotational quantum number.

energy spectrum is given as 

...................J = 3      E = 2B(3+1) = 8B

....................J = 2       E = 2B(2+1)=6B

.....................J = 1     E = 2B(1+1) = 4B

.......................J= 0   E = 2B
so that the transitions are equally spaced with a spacing of 2B.
now, given lowest wave absorption line is 41.11cm-1

=>    2B/hc = 41.11cm-1

=> wave number for second line = 4B = 2x41.11 = 82.22 cm-1
                            for third line = 6B = 3x41.11=123.33cm-1
                            Fourth line = 8B=4x41.11=164.44cm-1
  For allow transitions in rotational spectra, selection rule is difference in J corresponding to 2 levels should be +/-1 hence the spacing is 2B.From the above values the spacing comes out to be 41.11cm-1.