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) = 16.863/ B(cm-1) Assume C has mass 12.0 and O mass 16.0 Harry Kroto 2004
Far infrared rotational spectrum of CO J= 12 15 20B 10 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 = 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
5 10 J= 12 15 20B 23.0 cm-1 61.5 cm-1 Harry Kroto 2004
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 = 3.85 B = 1.925 cm-1 Harry Kroto 2004
A Classical Description > E = T + V E = ½I2 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
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 = 3.85 B = 1.925 cm-1 ( Harry Kroto 2004
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 = 3.85 B = 1.925 cm-1 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 Harry Kroto 2004
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 = 3.85 B = 1.925 cm-1 ( 50/3.85 = 12.99 = 13 so line at 50cm-1 is J=12 Harry Kroto 2004
A Classical Description > E = T + V E = ½I2 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
Radiotelescope in Canada Harry Kroto 2004
A Classical Description > E = T + V E = ½I2 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
A Classical Description > E = T + V E = ½I2 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
Nuclear Energies H + H E(r) Chemical Energies r v=3 2 1 r Harry Kroto 2004
Nuclear Energies H + H E(r) Chemical Energies Rotational levels r r Harry Kroto 2004