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H35Cl, j(0+) intensity ratio analysis and comparison of experimental data

H35Cl, j(0+) intensity ratio analysis and comparison of experimental data agust,www,....Jan11/PPT-210111ak.ppt agust,heima,...Jan11/Evaluation of coupling strength j state-2 i 0111kmak.xls agust,heima,....Jan11/PXP-210111ak.pxp The following holds for W12 = 25 cm-1:. g = 0.0171.

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H35Cl, j(0+) intensity ratio analysis and comparison of experimental data

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  1. H35Cl, j(0+) intensity ratio analysis and comparison of experimental data agust,www,....Jan11/PPT-210111ak.ppt agust,heima,...Jan11/Evaluation of coupling strength j state-2i0111kmak.xls agust,heima,....Jan11/PXP-210111ak.pxp The following holds for W12 = 25 cm-1:

  2. g = 0.0171 Least square minimization of I(35Cl+)/I(H35Cl+) vs J´ (for J´=0-5) with respect to a and g least sq. error (J=0-5) ag error 2 0.0171 0.000441 2.2 0.0149 0.000333 2.5 0.0124 0.000245 3 0.0092 0.000306 g = 0.0149 g = 0.0092 g = 0.0124 a Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0

  3. exp Calc.(v´=19-22) least sq. error (J=0-5) 0 3 5 J´ g = 0.0149 v´=19,20,21,22 and sum 0 3 5 J´ v´=18,19,20,21,22,23 and sum a Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0 agust,heima,...Jan11/Evaluation of coupling strength j state-210111kmak.xls

  4. exp Calc.(v´=19-22) least sq. error (J=0-5) 0 3 5 J´ v´=19,20,21,22 and sum 0 3 5 J´ g = 0.0124 a Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0 & agust,heima,...Jan11/Evaluation of coupling strength j state-210111kmak.xls

  5. exp Calc.(v´=19-22) least sq. error (J=0-5) 0 3 5 J´ v´=19,20,21,22 and sum 0 3 5 J´ g = 0.0092 a Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0 agust,heima,...Jan11/Evaluation of coupling strength j state-210111kmak.xls

  6. 1) NB!: contributions from v´< 20 and v´> 21 CLEARLYCAN NOT BE IGNORED!!! • This analysis assumes W12 to be constant and independent with v´(ip) and to be the • same value as that derived from shift analysis for v´(ip)=21. a and g are also • assumed to be constant and independent with v´(ip) : Thus least square analyses on • a and g (for W12 = 25 cm-1) resulted in • W12 = 25 cm-1 • a = 2.5 • = 0.0124 for j(0+) H35Cl The significantly larger g value, compared to that observed for other triplet states (g = 0.002 – 0.004) might be because of a large contribution to the dissocaiation Channels from photodissociation follwed by Cl ionization, i.e. 2hv + HCl ->-> HCl*(j(0+),v´=0, J´) HCl*(j(0+),v´=0, J´) + hv -> HCl** -> H + Cl* Cl* + hv -> Cl+ + e- Analogous analysis now need to be done for H37Cl!!!!!

  7. The J’ = 6 peak is problematic for H35Cl since the mass peaks for J’ = 6 and 8 overlap. Hence the experimentally evaluated ion ratio for J’ = 6 will be an underestimated value. Therefore it is acceptable that the calculated ratio is higher. This should not be the problem for H37Cl. Lets try to include v´=18 and 23 interactions:

  8. exp Calc.(v´=18-23) J´ v´=18,19,20,21,22,23 and sum J´ W12 = 25 cm-1; a = 1.7; g = 0.0135 agust,heima,...Jan11/Evaluation of coupling strength j state-220111kmak.xls

  9. It is interesting to see that the contribution falls down very slowly as DE(J´) • increases / v´ “moves further away” from the Rydberg state. • But what happens if W12 changes with v´, say W12 increases? • I tried • W12= 22,23,24,25,26,27 vs v´=18,19,20,21,22,23 & • W12= 19,21,23,25,27,29 vs v´=18,19,20,21,22,23 & • W12= 28,27,26,25,24,23 vs v´=18,19,20,21,22,23 & • W12= 31,29,27,25,23,21 vs v´=18,19,20,21,22,23 • No big change • Looking at calculations such as in the previous figure shows that contribution • from v´< 20 and v´>21 is close to a constant (f). Therefore the relevant expression for • I(Cl+)/I(HCl+) is • Is it perhaps possible to obtain good fit for the parameters a and f only • assuming g to be zero?

  10. No that does not seem to be the case. In other words gamma is an important parameter. Looking at: It is clear that c22 is very small and the ratio for v´<20 and v´> 21 is simply: NB! It is interesting to see that similar g values are obtained independet of the number of v´(V) contribution: g = 0.013 for v´=20-21 (KM) g = 0.0124 for v´= 19-22 g = 0.0135 for v´= 18-23 THIS IS IMPORTANT!

  11. Effect of g is clearly seen below:

  12. W12= 25, a = 1.7 How can I make the colors in the excel graph to stay unchanged? g = 0 J´ v´=18,19,20,21,22,23 and sum g = 0.006 J´ g = 0.013 J´ agust,heima,...Jan11/Evaluation of coupling strength j state-220111kmak.xls

  13. W12=25, f = 1 Minimize with respect to a and g=> a = 2.2, g = 0.0198, least sq. error(J´=0-5) = 0.000706 NB!: As a rough estimate I increased The experimental Ratio value to 0.5 We realy need to analogous test on H37Cl where the peak overlap problem does not exist. J´ agust,heima,...Jan11/Evaluation of coupling strength j state-220111kmak.xls

  14. Lets´ compare the calc. sum values for different optimizaed g values: Is the graph shape perhaps comparable?: Lets look at plots normalized to the largest peak (i.e. J´=6) See note from 230111: Fit of Irel(exp) = (I(Cl+)/I(HCl+)(J´;exp))/(I(Cl+)/I(HCl+)(J´max;exp)) vs J´ by Irel(calc)= (I(Cl+)/I(HCl+)(J´;calc))/(I(Cl+)/I(HCl+)(J´max;calc)) vs J´ Thus the a parameter drops out NB!: J´max = 6 All give equally good fit (see figure next slide) ERGO: 1) Use f = 0 (i.e. Neglect f*a*g) 2) Use only v´= 20 & 21 and perform fit on Irel (exp) vs J´ by varying g only!!! agust,heima,...Jan11/Evaluation of coupling strength j state-230111kmak.xls

  15. Fit of Irel(exp) = (I(Cl+)/I(HCl+)(J´;exp))/(I(Cl+)/I(HCl+)(J´max;exp)) vs J´ by Irel(calc)= (I(Cl+)/I(HCl+)(J´;calc))/(I(Cl+)/I(HCl+)(J´max;calc)) vs J´ : Relative Intensity ratios J´ agust,heima,...Jan11/Evaluation of coupling strength j state-230111kmak.xls

  16. Comparison of KM´s and JL´s ion ratios for j(0+), H35Cl: agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls

  17. JL 151210 JL 161210 KM J´ agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls

  18. JL 161210 JL 151210 KM J´ agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls

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