1. Harry Kroto 2004

2. Deuterium Isotope Fractionation
H + H H2
r 
Harry Kroto 2004

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

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

5. HD/H2 = (HD/H2)-1/2 = 3/2
 = (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. HD/H2 = (HD/H2)-1/2 = 3/2
 = (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

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 cm-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 cm-1
Remember N = Noe-E/0.7T
E(cm-1) and T(K)
Harry Kroto 2004

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

10. So significant in cold 5-10K clouds
13C CO ⇆ 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