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Bill Madden 559 2123.  = (4  /3ħc)  n  e  m  2  (N m -N n )  (  o -  ) 1 2 3 4 1.Square of the transition moment  n  e  m  2 2.Frequency.

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Presentation on theme: "Bill Madden 559 2123.  = (4  /3ħc)  n  e  m  2  (N m -N n )  (  o -  ) 1 2 3 4 1.Square of the transition moment  n  e  m  2 2.Frequency."— Presentation transcript:

1 Bill Madden 559 2123

2  = (4  /3ħc)  n  e  m  2  (N m -N n )  (  o -  ) 1 2 3 4 1.Square of the transition moment  n  e  m  2 2.Frequency of the light  3.Population difference (N m - N n ) 4.Resonance factor - Dirac delta function  (0) = 1

3  = (4  /3ħc)  n  e  m  2  (N m -N n )  (  o -  ) 1 Fermi’s Golden Rule

4   n  m n* mdn* md ~ d  /dq Quantum Mechanics Wave Mechanics Schrödinger Notation Quantum Mechanics Matrix Mechanics Dirac Notation Classical Analogue Dipole moment change over motion …coordinate q Take Home Message

5   n  m n* mdn* md ~ d  /dq Quantum Mechanics Wave Mechanics Schrödinger Notation Quantum Mechanics Matrix Mechanics Dirac Notation Classical Analogue Dipole moment change over motion …coordinate q Take Home Message  ~

6  = (4  /3ħc)  n  e  m  2  (N m -N n )  (  o -  ) 1 2 3 4 1.Square of the transition moment  n  e  m  2 2.Frequency of the light  3.Population difference (N m - N n ) 4.Resonance factor - Dirac delta function  (0) = 1

7 ABC Rotation of a Diatomic Molecule

8 For pure rotational transitions a molecule must have a permanent dipole moment

9 Observing the dipole change from the side i.e. the direction of propagation

10 dμ/dθ ≠ 0

11 Selection Rules Harry Kroto 2004 ∆N = ?

12 (N m - N n ) N m -N m

13 0 1 1.000 1.00 3.84 1 3 0.981 2.94 7.69 2 5 0.945 4.73 11.5 3 7 0.893 6.25 15.4 4 9 0.828 7.45 19.2 5 11 0.754 8.29 23.1 6 13 0.673 8.75 26.9 7 15 0.590 8.85 30.8 8 17 0.507 8.62 34.6 9 19 0.428 8.13 38.4 10 22 0.355 7.46 42.3 11 23 0.288 6.62 46.1 12 25 0.230 5.75 50.0 N m = N o e -∆E/kT In the case of degenerate levels such as rotational levels eacj J level is 2J+1 degenerate we get N m = N o e-∆E/kT J 2J+1 e -∆E/kT N m /N o F(J)

14 0 1 2 3 4 5 6 7 J 2B 4B 6B 8B 10B 12B 14B 16B0 Boltzmann

15 http://en.wiki pedia.org/wi ki/Boltzmann _constant Boltzmann

16 0 1 1.000 1.00 3.84 1 3 0.981 2.94 7.69 2 5 0.945 4.73 11.5 3 7 0.893 6.25 15.4 4 9 0.828 7.45 19.2 5 11 0.754 8.29 23.1 6 13 0.673 8.75 26.9 7 15 0.590 8.85 30.8 8 17 0.507 8.62 34.6 9 19 0.428 8.13 38.4 10 22 0.355 7.46 42.3 11 23 0.288 6.62 46.1 12 25 0.230 5.75 50.0 N m = N o e -∆E/kT In the case of degenerate levels such as rotational levels eacj J level is 2J+1 degenerate we get N m = N o e-∆E/kT J 2J+1 e -∆E/kT N m /N o F(J)

17 0 1 2 3 4 5 6 7 J 2B 4B 6B 8B 10B 12B 14B 16B0 Boltzmann

18 Boltzmann Population with Degeneracy

19 0 1 1.000 1.00 3.84 1 3 0.981 2.94 7.69 2 5 0.945 4.73 11.5 3 7 0.893 6.25 15.4 4 9 0.828 7.45 19.2 5 11 0.754 8.29 23.1 6 13 0.673 8.75 26.9 7 15 0.590 8.85 30.8 8 17 0.507 8.62 34.6 9 19 0.428 8.13 38.4 10 22 0.355 7.46 42.3 11 23 0.288 6.62 46.1 12 25 0.230 5.75 50.0 N m = N o e -∆E/kT In the case of degenerate levels such as rotational levels each J level is 2J+1 degenerate we get N m = N o (2J+1)e -∆E/kT J 2J+1 e -∆E/kT N m /N o F(J)

20 0 1 2 3 4 5 6 7 J 2B 4B 6B 8B 10B 12B 14B 16B0 Boltzmann

21 C≡O

22 CO Rotational Spectrum PROBLEM

23 Separation Vibration Rotation

24 ABC H Atom

25 H Atom Spectrum A

26 Positronium - +

27 Einstein Coefficients nn mm

28 Harry Kroto 2004 H 21 cm Line

29


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