Lecture 8: Case Study: UV - OH

Transcription

1 Lecture 8: Case Study: UV - OH UV absorption of OH: A Σ + X Π (~300nm). Introduction. OH energy leels Upper leel Lower leel 3. Allowed radiatie transitions Transition notations Allowed transitions 4. Working example - OH Allowed rotational transitions from "=3 in the A Σ + X Π system

2 . Introduction OH, a prominent flame emitter, absorber. Useful for T, X OH measurements. Selected region of A Σ + X Π(0,0) band at 000K

3 . Introduction Steps in analysis to obtain spectral absorption coefficient. Identify/calculate energy leels of upper + lower states. Establish allowed transitions 3. Introduce transition notation 4. Identify/characterize oscillator strengths using Hönl- London factors 5. Calculate Boltzmann fraction 6. Calculate lineshape function 7. Calculate absorption coefficient 3

4 . Energy leels Term energies elec. q. no. ib. q. no. ang. mom. q. no. Angular momentum energy (nuclei + electrons) n,, J T n G FJ E e Electronic energy Separation of terms: Born-Oppenheimer approximation G() = ω e ( + /) ω e x e ( + /) Sources of T e, ω e, ω e x e Herzberg Vibrational energy Oerall system : A Σ + X Π in [cm - ] A Σ + T e ω e ω e x e X Π T e ω e ω e x e Let s first look at the upper state Hund s case b! 4

5 . Energy leels Hund s case b (Λ=0, S 0) more standard, especially for hydrides Recall: Σ, Ω not rigorously defined = angular momentum without spin S = /-integer alues J = +S, +S-,, -S i =,, F i () = rotational term energy ow, specifically, for OH? 5

6 . Energy leels The upper state is A Σ + For OH: Λ = 0, Σ not defined use Hund s case b = 0,,, S = / J = ± / F denotes J = + / F denotes J = / Common to write either F () or F (J) 6

7 . Energy leels The upper state: A Σ + F B D B D for pure case b F splitting constant 0.cm the spin-splitting is γ (+) function of ; increases with for OH A γ (+) ~ 0.(5) ~ 0.5cm - for Compare with ν D (800K) = 0.3cm - otes: Progression for A Σ + + denotes positie parity for een [wae function symmetry] Importance? Selection rules require parity change in transition 7

8 . Energy leels The ground state: X Π (Λ=, S=/) Hund s case a Λ 0, S 0, Σ defined Hund s case b Λ = 0, S 0, Σ not defined ote:. Rules less strong for hydrides. OH behaes like Hund s low like Hund s large at large, L couples more to, Λ is less defined, S decouples from A-axis 3. Result? OH X Π is termed intermediate case 8

9 The ground state: X Π 9. Energy leels otes: 3. For intermediate/transition cases where Y A/B (< 0 for OH); A is effectiely the moment of inertia ote: F () < F () / / D Y Y B F D Y Y B F For large B F B F 0 B F F For small Behaes like Hund s a, i.e., symmetric top, with spin splitting ΛA Behaes like Hund s b, with small (declining) effect from spin

10 . Energy leels J 7/ 5/ 3/ The ground state: X Π otes: 4. Some similarity to symmetric top 3 Hund s a (A-B ) 30 F : J = + / F : J = / Ω = 3/ Ω = / Showed earlier that F < F J 5/ 3/ / T e = T 0 + AΛΣ For OH, A = -40 cm - T e = T 0 + (-40)()(/), Σ = / + (-40)()(-/), Σ = -/ T e = 40 cm - ot too far off the 30 cm - spacing for minimum J Recall: Hund s case a has constant difference of (A-B ) for same J F(J) = BJ(J+) + (A-B)Ω (A B)Ω -58.5Ω (A for OH~ -40, B ~ 8.5), Ω = 3/, / Ω = 3/ state lower by 36 cm - Actual spacing is only 88 cm -, reflects that hydrides quickly go to Hund s case b 0

11 . Energy leels The ground state: X Π otes: 5. Role of Λ-doubling J 7/ p + 3 p + J 5/ F F ic id Fi F i J cj J Fid Fic J J J d 5/ 3/ + + F : J = + / F : J = / Ω = 3/ Ω = / Showed earlier that F < F + + 3/ / F ic (J) F id (J) 0.04 cm - for typical J in OH c and d hae different parity (p) Splitting decreases with increasing ow let s proceed to draw transitions, but first let s gie a primer on transition notation.

12 3. Allowed radiatie transitions Transition notations Full description: A Σ + (') X Π (") Y X αβ (" or J") Strongest trans. e.g., R (7) or R 7 Y X " or J" General selection rules Parity must change + or + J = 0, otes:. Y suppressed when = J. β suppressed when α = β 3. Both " and J" are used where Y (O, P, Q, R, S for = - to +) X J (P, Q, R for J = -, 0, +) α = i in F i '; i.e., for F, for F β = i in F i "; i.e., for F, for F o Q (J = 0) transitions, J = 0 J = 0 not allowed Example: S R : J = +, = + F' = F (') F" = F (")

13 3. Allowed radiatie transitions Allowed transitions Allowed rotational transitions from "=3 in the A Σ + X Π system State or leel a specific ",J",",and Λ-coupling F c (3) F d (3) F (3) bands possible (3 originating from each lambda-doubled, spin-split X state) Main branches: α = β; Cross-branches: α β Cross-branches weaken as increases 3

14 3. Allowed radiatie transitions Allowed transitions Allowed rotational transitions from "=3 in the A Σ + X Π system otes: A gien J" (or ") has branches (6 are strong; J = ) + rule on parity F c F d 0.04(+) for OH for ~0, Λ-doubling is ~ 4cm -, giing clear separation If upper state has Λ-doubling, we get twice as many lines! 4

15 3. Allowed radiatie transitions Allowed transitions Allowed rotational transitions from "=3 in the A Σ + X Σ + system ote:. The effect of the parity selection rule in reducing the number of allowed main branches to 4. The simplification when Λ=0 in lower state, i.e., no Λ-doubling 5

16 4. Working example - OH Complete steps to calculate absorption coefficient. Identify/characterize oscillator strengths using Hönl-London factors. Calculate Boltzmann fraction 3. Calculate lineshape function (narrow-band s broad-band) 4. Calculate absorption coefficient k Absorption coefficient #/cm 3 of species (OH), #/cm 3 in state, h e, cm kt f exp 0 mec cm /s a a p A kt Fractional pop. in state ϕ[s] = (/c) ϕ[cm] To do: ealuate f, / a Step 4 Step 5 6

17 4.. Oscillator strengths Absorption oscillator strength f n",", ", J ", ", n', ', ', J ', ' or in shorthand notation elec. osc. strength f f f n" n' elec. ib. spin ang. mom. Λ-doubling F-C factor H-L factor n" n' q q " ' " ' f "' S J" S J" For OH A Σ + X Π (',") f '" (0,0) (,0) = band oscillator strength otes: q "' and S J"J' are normalized q " ' ' for Λ = 0 (Σ state), otherwise S J" S J ' g" el 4 for X this sum includes the S alues for all states with J" 7

18 4.. Oscillator strengths Is S J"J' = S J'J"? Yes, for our normalization scheme! From g f = g f, and recognizing that J+ is the ultimate (non remoable) degeneracy at the state leel, we can write, for a specific transition between single states In this way, there are no remaining electronic degeneracy and we require, for detailed balance, that f " f ', q q and Do we always enforce S J" for a state? o! But note we do enforce S J" S (4.7) and S J ' S (4.9) where, for OH A Σ X Π, (S+) = and δ =. When is there a problem? S S J ' J " el "' fel ' q' " J" J ' J" f " q J ' J ' S S J ' J " Eerything is okay for Σ-Σ and Π-Π, where there are equal elec. degeneracies, i,e., g" el = g' el. But for Σ-Π (as in OH), we hae an issue. In the X Π state, g el = 4 ( for spin and for Λ- doubling), meaning each J is split into 4 states. Inspection of our H-L tables for S J"J' for OH A Σ X Π (absorption) confirms ΣS J"J' from each state is J"+. All is well. But, in the upper state, Σ, we hae a degeneracy g' el of (for spin), not 4, and now we will find that the sum el el "' J ' J " '" of S J ' J " is twice J'+ for a single J' when we use the H-L alues for S J"J' for S J'J". Howeer, J " as there are states with J', the oerall sum S J ' 4 as required by (4.9) J " J ' J " 8

19 4.. Oscillator strengths Absorption oscillator strength for f 00 in OH A Σ + X Π Source f 00 Oldenberg, et al. (938) ± Dyne (958) ± Carrington (959) ± Lapp (96) ± Bennett, et al. (963) ± Golden, et al. (963) ± Engleman, et al. (973) Bennett, et al. (964) ± Anketell, et al. (967) ±

20 4.. Oscillator strengths Absorption oscillator strength Transition S J"J' /(J"+) ΣF (J) ΣF (J) Σ[F (J)+F (J)] Q (0.5) Q (0.5) R (0.5) R (0.5) P (.5) P (.5) P (.5) 0.39 P (.5) 0.75 Q (.5) 0.56 Q (.5) 0.37 Q (.5) 0.46 Q (.5) R (.5) 0.65 R (.5) 0.35 R (.5) R (.5) P (.5) P (.5) P (.5) 0.4 P (.5) Q (.5) Q (.5) 0.63 Q (.5) 0.4 Q (.5) R (.5) 0.56 R (.5) 0.73 R (.5) R (.5) Transition S J"J' /(J"+) ΣF (J) ΣF (J) Σ[F (J)+F (J)] P (3.5) P (3.5) P (3.5) 0.67 P (3.5) Q (3.5) Q (3.5) 0.95 Q (3.5) 0.70 Q (3.5) 0.84 R (3.5) 0.36 R (3.5) 0.3 R (3.5) R (3.5) 0.40 P (9.5) P (9.5) 0.06 P (9.5) P (9.5) Q (9.5) Q (9.5) Q (9.5) Q (9.5) R (9.5) 0.44 R (9.5) R (9.5) 0.04 R (9.5) 0.46 Hönl-London factors for selected OH transitions 0

21 4.. Boltzmann fraction. We seek the fraction of molecules in a single state for which. In general, 3. Electronic mode, S, 0 g e 0 g 4 e OH ge # of rot. leels produced by spin splitting & Λ-doubling = 4 for Π i J ' S / J Q Q " J ' J" g e e i Q / kt i Q r / Q c d c d A state ΣS J"J' =J"+ for each state J"="-/ J"= +/ " the sum of this oer all leels is Elec. leel Q n e S exp hct n n S exp hct n e e / kt / kt / Q e ote: hund s (a) includes AΩ

22 4.. Boltzmann fraction 4. Vibrational mode 5. Rotational mode (hund s (b)) n,, n, Q n,,, J n,, n,,, J, p n,,, J r n, n Q exp hcg exp hcg exp hcf exp hcf / kt / kt / kt / Q / kt r / Q Q r hcb r T r Again, each of these when summed / k for T ote: don t use F +F () here; until we add spin splitting ow what about fraction of those with in a gien J? J S Since # of states in is (+)(S+)ϕ, while # of states in J is (J+) ϕ ½ for OH as expected (fraction with spectral parity) r

23 6. Combining Boltzmann fraction ote:. The fraction in a gien state is /4 of that gien by rigid rotor!. Always know Σ( i /) =, both in total and for each mode separately. r e i e Q Q Q F G n T kt hc J J n p J n n J n n n n n n p J n exp,,,,,,,,,,,,,,,,,,,, Proper F i now! We hae loose end to deal with: narrow-band and broadband absorption measurement. (i.e., the Boltzmann fraction in state )

24 4.3. arrow-band s broad-band absorption measurement arrow-band absorption Measured quantity T I I S L with / 0 exp thus, if T ν (e.g. T ν0 ) is measured, and if L, p, γ, T, f are known S l e mec l f exp h kt Oscillator strength for transition tot. no.dens.of pl l kt Boltzmann fraction of species l = F ",J", (T) i bd. species i X i then can sole for l Quantity usually sought 4

25 4.3. arrow-band s broad-band absorption measurement Let s look at the classical (old-time) approach, pre 975 Broadband absorption line (for line from state) W W line Instrument broadening (no change in eq. width) Eq. width W J"J' (cm - ) W line ν A exp Transform ariables d K T L d d D exp K L V x, 0 ln D D ln ln x Integrated area is called: integrated absorbance, or eq. width K a dx S P i 5

26 4.3. arrow-band s broad-band absorption measurement Let s look at the classical (old-time) approach, pre 975 Broadband absorption Requires use of cures of growth D ln WJ " J ' exp K L V x, a 0 line ln D dx Cure of growth Eq. width W J"J' (cm - ) ν Procedure: measure W J"J', calculate ν D and a, infer K J"J', conert K J"J' to species ote:. Simple interpretation only in optically thin limit, W W K L e mc. Measured eq. width is indep. of instrument broadening! 3. Before lasers, use of absorption spectroscopy for species measurements require use of Cures of Growth! K f L L d 6

27 4.4. Example calculation (narrow-band) Consider spectral absorption coefficient of the (0,0)Q (9) line in the OH A Σ + X Π system, at line center. λ~309.6nm, ν~3300cm -, T = 000K, ν C = 0.05cm - Express k ν as a function of OH partial pressure k cm Pa n,,, J, cm.650 s kt a " f 0 a Pa / kt Oscillator strength (using tables) S fq J" 4 9 f"' Lineshape factor (narrow-band) D C 000K 0.5cm a 0 000K 0.05cm 3.3cm or.040 s 7

28 4.4. Example calculation (narrow-band) Consider spectral absorption coefficient of the (0,0)Q (9) line in the OH A Σ + X Π system, at line center. λ~309.6nm, ν~3300cm -, T = 000K, ν C = 0.05cm - Express k ν as a function of OH partial pressure k cm Pa n,,, J, " cm.650 f s kt a 0 a Pa / kt Population fraction in the absorbing state f c 0.5 hct 0/ kt exp hcg0/ kt J" exp hcf 9.5 a exp e Q Q Q exp e 0 exp 660K / T 0exp 33K / T T / 6.66K r / kt 8

29 4.4. Example calculation (narrow-band) Consider spectral absorption coefficient of the (0,0)Q (9) line in the OH A Σ + X Π system, at line center. λ~309.6nm, ν~3300cm -, T = 000K, ν C = 0.05cm - Express k ν as a function of OH partial pressure k cm Pa n,,, J, " cm.650 f s kt a 0 a Pa / kt k 3 cm 8 cm 4 0 cm.650 P atm % s Beer s Law cm 77 atm I I 0 P a s atm exp k L a atm 59% absorption for L = 5cm, X OH = 000ppm, T = 000K, P = atm 9

30 4.4. Example calculation (narrow-band) Selected region of OH A Σ + X Π (0,0) band at 000K otes: Lines belonging to a specific branch are connected with dashed or dotted cure Thicker dashed lines main branches; thin dotted lines cross branches Bandhead in R branches if B ' <B " ; Bandhead in P branches if B ' >B ote bandhead in R Q branch 30

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