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H + H

Deuterium Isotope Fractionation. E(r). H + H. H 2. r . Harry Kroto 2004. Consider the zero point energy of a diatomic molecule Assuming SHO  = (1/2 ) ( k/ ). Harry Kroto 2004. Consider the zero point energy of a diatomic molecule Assuming SHO  = (1/2 ) ( k/ ).

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H + H

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  1. Deuterium Isotope Fractionation E(r) H + H H2 r  Harry Kroto 2004

  2. Consider the zero point energy of a diatomic molecule Assuming SHO  = (1/2) (k/) Harry Kroto 2004

  3. Consider the zero point energy of a diatomic molecule Assuming SHO  = (1/2) (k/) Harry Kroto 2004

  4.  = (1/2)(k/)   -1/2  = m1m2/(m1+ m2) HD = 2/3 H2 = 1/2 HD/H2 = 4/3 HD/H2 = (HD/H2)-1/2 = 3/2 • H2 = 4500 cm-1 • HD = 3000 x 1.732/2 = 2598 ½ HD – ½H2 = Harry Kroto 2004

  5.  = (1/2)(k/)   -1/2  = m1m2/(m1+ m2) HD = 2/3 H2 = 1/2 HD/H2 = 4/3 HD/H2 = (HD/H2)-1/2 = 3/2 • H2 = 4500 cm-1 • HD = 3000 x 1.732/2 = 2598 ½ HD – ½H2 = Harry Kroto 2004

  6. H2 = 4395 cm-1 • HD = 4395 x 1.732/2 = 3806 ½ HD – ½H2 = 2197 – 1909 = 288 cm-1 H2 + D+ HD + H+ + 288cm-1 Remember N = Noe-E/0.7T E(cm-1) and T(K) Harry Kroto 2004

  7. H2 = 4395 cm-1 • HD = 4395 x 1.732/2 = 3806 ½ HD – ½H2 = 2197 – 1909 = 288 cm-1 H2 + D+ HD + H+ + 288cm-1 Remember N = Noe-E/0.7T E(cm-1) and T(K) Harry Kroto 2004

  8. H2 = 4395 cm-1 • HD = 4395 x 1.732/2 = 3806 ½ HD – ½H2 = 2197 – 1909 = 288 cm-1 H2 + D+ HD + H+ + 288cm-1 Remember N = Noe-E/0.7T E(cm-1) and T(K) Harry Kroto 2004

  9. In 13C+ + 12CO ⇆13CO + 12C+ + H 13CO/12CO E = 24 cm-1 So significant in cold 5-10K clouds The excellent fit between theory and experiment is strong evidence for ion-molecule reactions Harry Kroto 2004

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