1. VMI images –fitting Helgi Rafn Hróðmarsson

2. The purpose of this fitting procedure is to check whether the assumption that the angular distribution data can be fitted with an expression of both the two-photon excitation transition as well as the ionization pathway. (see slides 9 – 21 in https://notendur.hi.is/~agust/rannsoknir/Crete/PPT-131219.pptx ) https://notendur.hi.is/~agust/rannsoknir/Crete/PPT-131219.pptx

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

4. Previous fit functional form used by Dimitris Zaouris was I(  )= A(1 +  2 *(1.5*cos 2 (  )-0.5) The new fit equation (named „Ice fit“) is based on a two-step excitation scheme as formulated in slides 9 – 21 in https://notendur.hi.is/~agust/rannsoknir/Crete/PPT-131219.pptx : https://notendur.hi.is/~agust/rannsoknir/Crete/PPT-131219.pptx I(  )=A*(1+  2 (f) *(1.5*cos 2 (  )-0.5))*(1+  2 (ph) *(1.5*cos 2 (  )-0.5)) where A is a normalizing constant.  2 (f) = -0.62215, obtained from REMPI data. (see slide 17 in https://notendur.hi.is/~agust/rannsoknir/Crete/PPT-131219.pptx) https://notendur.hi.is/~agust/rannsoknir/Crete/PPT-131219.pptx  2 (ph) is the “beta-factor” for the ionization pathway.

6. E(v=0,J=1) – peak A – Ice fit 1

7. E(v=0,J=1) – peak B – Ice fit 1

8. Peaks (J=1)GreekIcelandic 1 A (A;  1;  2) 0,76118±0,00925 0,45744±0,0283 0,86907±0,0127 -0,62215 1,1514±0,0345 B0,85873±0,00499 -0,37458±0,0126 0,89662±0,00382 -0,62215 0,33335±0,00962 C0,13704±0,0075 1,9112±0,173 0,17007±0,0112 -0,62215 2±0,17 D0,41475±0,00541 1,7632±0,0399 0,52983±0,0235 -0,62215 2±0,14 E0,27321±0,0106 1,8038±0,12 0,3591±0,0268 -0,62215 2±0,192 F0,36161±0,00817 1,5113±0,0651 0,46318±0,027 -0,62215 1,9759±0,148 G0,39409±0,00923 1,4896±0,0672 0,50614±0,0259 -0,62215 1,9828±0,131

9. E(v=0,J=2) – peak A – Ice Fit 1

10. E(v=0,J=2) – peak B – Ice fit 1

11. Peaks (J=2)GreekIcelandic 1 A (A;  1;  2) 0,76118±0,00925 0,45744±0,0283 0,81307±0,0134 -0,62215 1,2773±0,0392 B0,87669±0,00429 0,29162±0,0106 0,91825±0,00505 -0,62215 0,39804±0,0131 C0,17162±0,0121 2±0,227 0,23872±0,0199 -0,62215 2±0,45 D0,43901±0,0077 1,391±0,0492 0,46173±0,0246 -0,62215 2±0,30 E0,29792±0,0159 1,839±0,166 0,36734±0,0271 -0,62215 2±0,22 F0,15042±0,00463 1,4516±0,0875 0,20227±0,0123 -0,62215 2±0,18 G0,43901±0,0077 1,391±0,0492 0,53567±0,0212 -0,62215 1,8482±0,0971

12. E(v=0,J=3) – peak A – Ice fit 1

13. E(v=0,J=3) – peak B – Ice fit 1

14. Peaks (J=3)GreekIcelandic 1 A (A;  1;  2) 0,87109±0,0104 0,18036±0,0267 0,97163±0,0108 -0,62215 0,91082±0,0262 B0,90372±0,00754 -0,17687±0,0182 0,97059±0,0085 -0,62215 0,56026±0,02 C D0,22624±0,0106 1,3921±0,131 0,31474±0,00922 -0,62215 2±0,10 E0,48809±0,0241 -0,62215 2±0,15 F0,3953±0,0146 1,2594±0,1 0,49898±0,0216 -0,62215 1,8438±0108 G0,49621±0,0177 1,0946±0,0939 0,63018±0,0132 -0,62215 1,7876±0,0531

15. E(v=0,J=4) – peak A – Ice fit 1

16. E(v=0,J=4) – peak B – Ice fit 1

17. Peaks (J=4)GreekIcelandic 1 A (A;  1;  2) 0,8717±0,0101 0,33282±0,0252 0,92483±0,0146 -0,62215 0,41919±0,0355 B0,66393±0,00724 -0,036446±0,024 0,72084±0,00475 -0,62215 0,69182±0,0156 C D0,61987±0,0248 -0,62215 1,939±0,101 E0,35438±0,0194 1,3384±0,152 0,44253±0,0342 -0,62215 1,8133±0,192 F0,50083±0,0149 1,2284±0,0803 0,65381±0,0247 -0,62215 1,8808±0,0984 G0,25361±0,0101 0,80411±0,099 0,32169±0,0113 -0,62215 1,6199±0,0873

18. E(v=0,J=5) – peak A – Ice fit 1

19. E(v=0,J=5) – peak B – Ice fit 1

20. Peaks (J=5)GreekIcelandic 1 A (A;  1;  2) 0,56496±0,00739 -0,17113±0,0285 0,60009±0,00992 -0,62215 0,521±0,0397 B0,90547±0,00856 0,063685±0,0209 0,99436±0,0101 -0,62215 0,7879±0,0241 C D0,25092±0,0109 -0,62215 2±0,13 E0,1303±0,00748 1,3471±0,159 0,16249±0,0131 -0,62215 1,8091±0,201 F0,26804±0,00894 1,3276±0,0922 0,34558±0,0161 -0,62215 1,9056±0,121 G0,13687±0,00377 1,0828±0,0722 0,17482±0,00575 -0,62215 1,7706±0,083

21. E(v=0,J=6) – peak A – Ice fit 1

22. E(v=0,J=5) – peak B – Ice fit 1

23. Peaks (J=6)GreekIcelandic 1 A (A;  1;  2) ±±±± 0,82007±0,00725 -0,62215 0,3087±0,02 B0,85371±0,094 0,25965±0,0249 0,96185±0,0101 -0,62215 0,98997±0,0247 C D0,55652±0,0309 -0,62215 1,9759±0,147 E0,46686±0,0224 1,1077±0,127 0,59043±0,0284 -0,62215 1,7602±0,121 F0,1392±0,00437 1,0142±0,0811 0,17906±0,0073 -0,62215 1,7545±0,103 G0,41703±0,0184 0,97138±0,113 0,55327±0,026 -0,62215 1,803±0,121

24. E(v=0,J=7) – peak A – Ice fit 1

25. E(v=0,J=7) – peak B – Ice fit 1

26. Peaks (J=7)GreekIcelandic 1 A (A;  1;  2) 0,81874±0,00999 -0,1475±0,0266 0,8855±0,0118 -0,62215 0,59888±0,0303 B0,85244±0,0103 0,19325±0,027 0,96448±0,0113 -0,62215 0,95905±0,0272 C D0,35768±0,0169 -0,62215 1,8061±0,119 E0,20043±0,0121 1,3719±0,168 0,2557±0,0196 -0,62215 1,8678±0,196 F0,56079±0,0224 0,96569±0,102 0,71665±0,0184 -0,62215 1,7332±0,0651 G0,47816±0,0227 0,92885±0,121 0,61749±0,0223 -0,62215 1,7435±0,0918

27. E(v=0,J=8) – peak A – Ice fit 1

28. E(v=0,J=8) – peak B – Ice fit 1

29. Peaks (J=8)GreekIcelandic 1 A (A;  1;  2) 0,85434±0,00782 -0,32847±0,0199 0,89395±0,0102 -0,62215 0,36528±0,0272 B0,61819±0,00641 0,26281±0,0235 0,70103±0,0096 -0,62215 1,0059±0,032 C D0,48587±0,0314 -0,62215 1,8065±0,161 E0,050319±0,00362 0,81428±0,178 0,06375±0,00468 -0,62215 1,5693±0,18 F0,37184±0,00899 0,83385±0,0602 0,45401±0,0145 -0,62215 1,5277±0,0782 G0,53696±0,0225 0,98697±0,108 0,67354±0,0151 -0,62215 1,7095±0,0556

30. E(v=0,J=9) – peak A- Ice fit 1

31. E(v=0,J=9) – peak B- Ice fit 1

32. Peaks (J=9)GreekIcelandic 1 A (A;  1;  2) 0,7539±0,00573 -0,59057±0,0166 0,76278±0,00548 -0,62215 0,068452±0,0165 B0,8±0,00798 0,23631±0,0225 0,90479±0,0118 -0,62215 0,98217±0,0304 C D0,29856±0,0186 -0,62215 1,6688±0,155 E0,10717±0,00755 0,73928±0,172 0,12381±0,011 -0,62215 1,2622±0,21 F0,25428±0,102 0,55248±0,0946 0,30221±0,0139 -0,62215 1,2887±0,109 G0,21671±0,00926 0,78085±0,105 0,27046±0,0092 -0,62215 1,5698±0,0834

33. The Greek fit

34. The Icelandic fit  2 (ph)

35. Peak A angular distributions „Ice fit“:

37. Peak A – Intensity contour

38. Peak B angular distributions

40. Peak B – Intensity contour

41. Now to test another fit equation namely... I(  )=A*(1+  2 (f) *P 2 (cos(  4 (f) *P 4 (cos(  ))* (1+  2 (ph) *P 2 (cos(  4 (ph) *P 4 (cos(  )) I(  )=A*(1 +  2 (f) * (1.5 * cos 2 (  ) - 0.5) +  4 (f) * (1/8) * (35 * cos 4 (  ) – 30 * cos 2 (  ) + 3)) * (1 +  2 (ph) * (1.5 * cos 2 (  ) - 0.5) +  4 (ph) * (1/8) * (35 * cos 4 (  ) – 30 * cos 2 (  ) + 3)))...to try to get a better fit to the curves. Let’s try this fit of peak A and peak B. Again, as before, we keep  2 (ph) constant and we’ll be observant whether the  2 (f) constant stays about the same or it changes dramatically.

42. J=1 – Peak A

43. And comparison with the previous “Ice fit”.

44. E(v=0,J=1) – peak A – Ice fit 1

45. 1) We get a better fit than before with more points from the edges of the scattering. 2) The  2 (f) constant changes from 1.15 to 1.28, i.e. not a gargantuan increase. 3) We also get values for the  4 (f) constant as well as  4 (ph) constant.

46. J=1 – Peak B

47. J=2 – Peak A

48. J=2 – Peak B

49. J=3 – Peak A

50. J=3 – Peak B

51. J=4 – Peak A

52. J=4 – Peak B

53. J=5 – Peak A

54. J=5 – Peak B

55. J=6 – Peak A

56. J=6 – Peak B

57. J=7 – Peak A

58. J=7 – Peak B

59. J=8 – Peak A

60. J=8 – Peak B

61. J=9 – Peak A

62. J=9 – Peak B

63. Beta2 (fit 1)Beta2 (fit 2)Beta4ph (fit 2)Beta4f (fit 2) J=1 – peak A1.15±0.031.28±0.03-0.01±0.020.30±0.06 J=1 – peak B0,33±0,010.31±0,010.04±0,01-0.02±0,02 J=2 – peak A1,28±0,041.39±0,03-0.01±0,020.34±0,06 J=2 – peak B0,40±0,010.41±0,010.06±0,020.01±0,03 J=3 – peak A0,91±0,030.91±0,030.11±0,030.00±0,06 J=3 – peak B0,56±0,020.59±0,03-0.05±0,030.16±0,06 J=4 – peak A0,42±0,040.32±0,040.21±0,10-0.24±0,15 J=4 – peak B0,69±0,020.72±0,020.06±0,020.05±0,03 J=5 – peak A0,52±0,040,51±0,040.19±0,07-0.10±0,11 J=5 – peak B0,79±0,020.87±0,05-0.05±0,030.22±0,09 J=6 – peak A0,31±0,020.34±0,03-0.04±0,030.12±0,06 J=6 – peak B0,99±0,021.04±0,010.038±0,0080.14±0,02 J=7 – peak A0,60±0,030.56±0,030.09±0,04-0.01±0,08 J=7 – peak B0,96±0,030.98±0,030.02±0,020.15±0,05 J=8 – peak A0,37±0,030.35±0,050.11±0,07-0.04±0,12 J=8 – peak B1,01±0,031.05±0,020.05±0,010.18±0,03 J=9 – peak A0,07±0,020.02±0,020.13±0,07-0.16±0,09 J=9 – peak B0,98±0,030.99±0,020.12±0,020.08±0,05

64. A B Only small change in  2 (ph) compared to that for first „Ice fit“:  2 (ph)

65. The Icelandic fit  2 (ph)

66. All in all a)The vibrational peaks C – G involve almost purely Parallel photodissociation transition b)The H + + Br(1/2) formation (peak A) involves decreasing parallel/ increasing perpendicular photodissociation transition with J´. c) The H + + Br(3/2) formation (peak B) involves increasing parallel/ decreasing perpendicular photodissociation transition with J´. This can be compared with the predicted transitions, based on the comparison with HCl as shown on next slides (see also https://notendur.hi.is/~agust/rannsoknir/Crete/HBr%20REMPI%20work-Crete.pptx ) https://notendur.hi.is/~agust/rannsoknir/Crete/HBr%20REMPI%20work-Crete.pptx

67. H* + Br (3/2) H + + Br - Ry V/Ion-pair [ 4 ,5s ] 3   H* + Br*(1/2) [ B 2  ].. H + Br**(5s) H + + Br (3/2)/Br*(1/2) HBr + (v + )(3/2,1/2) E1+E1+ V/ 1  + Peaks C-G: [2 2  ] 1  0, 3  0 

68. H* + Br (3/2) H + + Br - Ry V/Ion-pair [ 4 ,5s ] 3   H* + Br*(1/2) [2 2  ] 1  0, 3  0 [ B 2  ].. H + Br**(5s) H + + Br*(1/2) E1+E1+ V/ 1  + Peak A:  and  states involved

69. H* + Br (3/2) Ry V/Ion-pair [ 4 ,5s ] 3   H* + Br*(1/2) [ B 2  ].. H + Br**(5s) H + + Br(3/2) H + + Br - [2 2  ] 1  0, 3  0 Peak B: E1+E1+ V/ 1  +  and  states involved