1. I = μr2 μ = m1m2/(m1+m2) I (uÅ2) = 16.863/ B(cm-1)
This is the far infra red spectrum of the diatomic molecule CO. It is due to absorption by pure rotational transitions of this molecule. Determine the B value as accurately as possible and thus the bond length in Å
I = μr2 μ = m1m2/(m1+m2) I (uÅ2) = / B(cm-1)
Assume C has mass and O mass 16.0
Harry Kroto 2004

2. Far infrared rotational spectrum of CO J= 12 15 20B10
Far infrared rotational spectrum of CO
J= 12 15
20B
23.0 cm-1
61.5 cm-1
Line separations 2B
Approximately 61.5 – 23 = 38.5 cm-1 = 20B
2B = B = cm-1 (
50/3.85 = = 13 so line at 50cm-1 is J=12
B = / I I = / B
I = 8.76 uA2 I =  r2
 = m1m2/(m1+m2)= 16x12/28 = 6.86
8.76/6.86 = = r2
r = 1.277½ = A ( acc B value 1.921)
Harry Kroto 2004

3. 5 10
J= 12 15
20B
23.0 cm-1
61.5 cm-1
Harry Kroto 2004

4. Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1
Line separations 2B
Approximately 61.5 – 23 = 38.5 cm-1 = 20B
2B = B = cm-1
Harry Kroto 2004

5. A Classical Description > E = T + V E = ½I2 V=0
B QM description > the Hamiltonian H J  = E J  H = J2/2I
C Solve the Hamiltonian > Energy Levels F (J) = BJ(J+1)
D Selection Rules > Allowed Transitions J = ±1
E Transition Frequencies > F B(J+1)
F Intensities > THE SPECTRUM
J Analysis > Pattern recognition; assign J numbers
H Experimental Details > microwave spectrometers
I More Advanced Details: Centrifugal distortion, spin effect
J Information obtainable: structures, dipole moments etc
Harry Kroto 2004

6. Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 (
Line separations 2B
Approximately 61.5 – 23 = 38.5 cm-1 = 20B
2B = B = cm-1 (
Harry Kroto 2004

7. Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1
J= 12
23.0 cm-1
61.5 cm-1
Line separations 2B
Approximately 61.5 – 23 = 38.5 cm-1 = 20B
2B = B = cm-1
50/3.85 = = 13 so line at 50cm-1 is J=12
Harry Kroto 2004

8. Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 ( 10 15
Line separations 2B
Approximately 61.5 – 23 = 38.5 cm-1 = 20B
2B = B = cm-1 (
50/3.85 = = 13 so line at 50cm-1 is J=12
Harry Kroto 2004

9. A Classical Description > E = T + V E = ½I2 V=0
B QM description > the Hamiltonian H J  = E J  H = J2/2I
C Solve the Hamiltonian > Energy Levels F (J) = BJ(J+1)
D Selection Rules > Allowed Transitions J = ±1
E Transition Frequencies > F B(J+1)
F Intensities > THE SPECTRUM
J Analysis > Pattern recognition; assign J numbers
H Experimental Details > microwave spectrometers
I More Advanced Details: Centrifugal distortion, spin effect
J Information obtainable: structures, dipole moments etc
Harry Kroto 2004

10. Radiotelescope in Canada
Harry Kroto 2004

11. A Classical Description > E = T + V E = ½I2 V=0
B QM description > the Hamiltonian H J  = E J  H = J2/2I
C Solve the Hamiltonian > Energy Levels F (J) = BJ(J+1)
D Selection Rules > Allowed Transitions J = ±1
E Transition Frequencies > F B(J+1)
F Intensities > THE SPECTRUM
J Analysis > Pattern recognition; assign J numbers
H Experimental Details > microwave spectrometers
I More Advanced Details: Centrifugal distortion, spin effect
J Information obtainable: structures, dipole moments etc
Harry Kroto 2004

12. A Classical Description > E = T + V E = ½I2 V=0
B QM description > the Hamiltonian H J  = E J  H = J2/2I
C Solve the Hamiltonian > Energy Levels F (J) = BJ(J+1)
D Selection Rules > Allowed Transitions J = ±1
E Transition Frequencies > F B(J+1)
F Intensities > THE SPECTRUM
J Analysis > Pattern recognition; assign J numbers
H Experimental Details > microwave spectrometers
I More Advanced Details: Centrifugal distortion, spin effect
J Information obtainable: structures, dipole moments etc
Harry Kroto 2004

13. Nuclear Energies H + H E(r) Chemical Energies r  v=3 2 1
r 
Harry Kroto 2004

14. Nuclear Energies H + H E(r) Chemical Energies Rotational levels r 
r 
Harry Kroto 2004