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Far Infrared Rotational Spectrum of CO: Determination of Bond Length

This article examines the far infrared rotational spectrum of CO to determine its bond length in angstroms and cm-1. It includes classical and quantum mechanical descriptions, selection rules, and experimental details.

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Far Infrared Rotational Spectrum of CO: Determination of Bond Length

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  1. Far infrared rotational spectrum of CO 5 10 J= 12 15 20B 61.5 cm-1 23.0 cm-1 Harry Kroto 2004

  2. 5 10 J= 12 15 20B Harry Kroto 2004

  3. 20B 61.5 cm-1 23.0 cm-1 Line separations 2B Harry Kroto 2004

  4. 20B 61.5 cm-1 23.0 cm-1 Line separations 2B 61.5 – 23 = 38.5 cm-1 = 20B Harry Kroto 2004

  5. 20B 61.5 cm-1 23.0 cm-1 Line separations 2B 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 Harry Kroto 2004

  6. 5 10 15 Homework Determine the bond length of the CO in A (Angstroms) and nm B (cm-1) = 16.863/ I (UA2) I =  r2 = m1m2/(m1+m2) Assume U of C =12 and O = 16 Note 1A = 0.1nm Harry Kroto 2004

  7. Harry Kroto 2004

  8. 5 10 15 Line separations 2B Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 ( 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 B = 16.863/ I I = 16.863/ B I = 8.76 uA2 I =  r2  = m1m2/(m1+m2)= 16x12/28 = 6.86 8.76/6.86 = 1.277 = r2 r = 1.277½ = 1.130 A (1.128 acc B value 1.921) Harry Kroto 2004

  9. J= 12 61.5 cm-1 23.0 cm-1 Line separations 2B Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 ( 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 B = 16.863/ I I = 16.863/ B I = 8.76 uA2 I =  r2  = m1m2/(m1+m2)= 16x12/28 = 6.86 8.76/6.86 = 1.277 = r2 r = 1.277½ = 1.130 A (1.128 acc B value 1.921) Harry Kroto 2004

  10. 5 10 15 Line separations 2B Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 ( 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 B = 16.863/ I I = 16.863/ B I = 8.76 uA2 I =  r2  = m1m2/(m1+m2)= 16x12/28 = 6.86 8.76/6.86 = 1.277 = r2 r = 1.277½ = 1.130 A (1.128 acc B value 1.921) Harry Kroto 2004

  11. 61.5 cm-1 23.0 cm-1 Line separations 2B Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 ( 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 B = 16.863/ I I = 16.863/ B I = 8.76 uA2 I =  r2  = m1m2/(m1+m2)= 16x12/28 = 6.86 8.76/6.86 = 1.277 = r2 r = 1.277½ = 1.130 A (1.128 acc B value 1.921) Harry Kroto 2004

  12. 20B 61.5 cm-1 23.0 cm-1 Line separations 2B Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 ( 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 B = 16.863/ I I = 16.863/ B I = 8.76 uA2 I =  r2  = m1m2/(m1+m2)= 16x12/28 = 6.86 8.76/6.86 = 1.277 = r2 r = 1.277½ = 1.130 A (1.128 acc B value 1.921) Harry Kroto 2004

  13. 5 10 15 Line separations 2B Approximately 61.5 – 23 = 38.5 cm-1 = 20B 2B = 3.85 B = 1.925 cm-1 ( 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 B = 16.863/ I I = 16.863/ B I = 8.76 uA2 I =  r2  = m1m2/(m1+m2)= 16x12/28 = 6.86 8.76/6.86 = 1.277 = r2 r = 1.277½ = 1.130 A (1.128 acc B value 1.921) Harry Kroto 2004

  14. 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

  15. Radiotelescope in Canada Harry Kroto 2004

  16. 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

  17. B(J+1)(J+2) – D(J+1)2(J+2)2 J+1 BJ(J+1) – DJ2(J+1)2 J F(J) = 2B(J+1) – 4D(J+1)3 Harry Kroto 2004

  18. 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

  19. 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

  20. 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 HExperimental Details > microwave spectrometers I More Advanced Details: Centrifugal distortion, spin effect J Information obtainable: structures, dipole moments etc Harry Kroto 2004

  21. Radiotelescope in Canada Harry Kroto 2004

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  23. Harry Kroto 2004

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