1. HBr, angular distribution analysis E(0) Updated, 24.12.2013

2. Peak “A”

3. E0,J´=3

4. E0,J´=4

5. Rydb. H + Br/Br* r(HX) J´´ v´´= 0 HX HX** J´ v´ E H + X - /Ion-pair/V v´,J´ H* + Br H* + Br* H + + Br H + + Br* 2 h resonance excitation 3 h dissociation / H* formation 4 h ionization / H + formation

6. 2 h resonance excitation 3 h dissociation / H* formation According to https://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdfhttps://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdf :

7. where Unknown (variable in a fit) Unknown (variables in a fit)

8. BUT simpler for “non Q branches” (O, S) (i.e. for J´´  J´): 2 h resonance excitation 3 h dissociation / H* formation https://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdf …. i.e. independent of the R´s

9. An alternative way to analyse the angular distribution data is by the procedure given by Chichinin et al.: https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf -which involves the use of relative intensities of spectral lines I Q /I S and I Q /I O which could be derived from our REMPI spectra: i f ph HBr(Ji) HBr** H* + Br/Br*

10. 1. Determine „b“ factor, via mass resolved REMPI spectra (see p: 9 https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf ) from, https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf 2. Determine alignment parameter A 20 (<=„b“ ) (P: 9, https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf (43)(for J f = 1) (24) (25)

11. 4. Determine the angular distribution for one-photon photodissociation of the unpolarized „f“ state (i.e.  (n,n ph )) from for k = 1 (one photon), where  (n) is the photofragment angular distribution produced by a multiphoton excitation via the intermediate state (i.e. angular distribution derived from our experiments) (see p: 6 in https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf (1) ph 3. Determine angular distribution (  f (n)) via  –factor (  (f) ) determination for„f“(see slide 10): (P:6 (and 9), https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf (30) See: http://mathworld.wolfram.com/LegendrePolynomial.html ;J f = 1

12. 1. Determine „b“ factor, via mass resolved REMPI spectra (see p: 9 https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf ) from, https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf Detailed analysis, see : agust,heima, …./PPT-131219.pptx a´ c´ d´ a´´ c´´ d´´

13. HBr; E(0) angular distribution analysis; Areas a´c´d´a"c"d" b**2 IO IQIS IQ/IOIQ/IS IQ/IO SO J´´LorentzGaussLorentz GaussLorentz J´(S)LorentzJ´(O) 0732171029772947,1105181,666667301,4221042 1263790245711846010,735833,3333332,50,21,2082993 2427040381852869211,183454,1666672,3333330,13333316,666671,50,0857141,0931554 32341616781547620428573330823,3865712,777844,6666672,250,1211,666671,6666670,0888891,16360451,1494051 42717417033623810448973491922,9561313,8942552,20,114286101,750,0909091,21116561,2598312 53283121523607140375502765718,4928916,168845,2380952,1666670,1111119,1666671,80,0923081,37338571,0694993 64383728994558280314711656012,7353617,739515,4166672,1428570,1090918,6666671,8333330,0933331,47741880,7506174 75889541854447480255517977,597929175,13895,5555562,1250,1076928,3333331,8571430,09411814,7846290,4402645 853011419662812605,3056915,6666672,1111110,1066678,0952381,8750,0947370,2990256 93716630645693881,8669755,7575762,10,1058827,9166671,8888890,0952380,074437 1013743111935,8333332,0909090,1052637,7777781,90,095652 11288220195,8974362,0833330,1047627,6666671,9090910,096 OVERLAP av b**21,2103271,0573384 values with1,1595783 values V(m+4)av(av) 3 values1,188578 1,19 av1,278447S(J´=2-8) ATH Gaussian fit to Q linesav1,159578O(J´=1-3) ERGO, (b**2) ca.:1,2 b: 1,095445or-1,09545 https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx

14. 2. Determine alignment parameter A 20 (<=„b“ ) (P: 9, https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf (43)(for J f = 1) (24) (25)

15. Alignment parameter A 20 : http://www.ejournal.unam.mx/rmf/no503/RMF50315.pdfhttp://www.ejournal.unam.mx/rmf/no503/RMF50315.pdf, p:319 b A 20 (J f =1) A 20 (min) = -1 A 20 (max) =+0.5 b=1 Perpend. domin. b=-0.5 parallel. domin. https://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxphttps://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxp; Gr0,Lay0 <= https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsxhttps://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx

16. Alignment parameter A 20 : http://www.ejournal.unam.mx/rmf/no503/RMF50315.pdfhttp://www.ejournal.unam.mx/rmf/no503/RMF50315.pdf, p:319 b factor A 20 (J f ) A 20 (min;J f =1) = -1 A 20 (max;J f =1) =+0.5 b = 1 Perpendicular dominant b=-0.5 parallel dominant J f = 1 2 11 J f = 11 2 1 https://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxphttps://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxp; Gr0,Lay0 <= https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsxhttps://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx Our A 20 values = A 20 (b = 1.1; b 2 = 1.2)

17. f <-<- i transition dominantly perpendicular transition, i.e.  3. Determine angular distribution (  f (n)) via  –factor (  (f) ) determination for„f“(see slide 10): (P:6 (and 9), https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf )https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf (30) See: http://mathworld.wolfram.com/LegendrePolynomial.html  (f) = -0,62215 https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx ;J f = 1

18. https://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxphttps://notendur.hi.is/~agust/rannsoknir/Crete/PXP-131222.pxp; Gr1,Lay1<= https://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsxhttps://notendur.hi.is/~agust/rannsoknir/Crete/XLS-131221.xlsx   f (n; J f =1)

19. J´> 1: https://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdfhttps://notendur.hi.is/~agust/rannsoknir/papers/jcp125-034310-06.pdf: i.e.: (https://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdf)https://notendur.hi.is/~agust/rannsoknir/papers/jcp121-11802-04.pdf -but to a first approximation (?)  L = 0 for L = 4,6,…. ERGO: the angular distribution shown on slide 18 holds for all J f ´s ! -in which case the alteration in angular distribution vs. J observed (Slide 2) is due to the photofragmentation step(?!)

20. Now what?! How do we derive  (n,n ph ) ? (1) ph How about a „fitting procedure“:

21. i.e. something like:  0 180 f i ph f Experiment  (ph) (1) + A ~ -for red coloured parameters unknown => derive  (ph)