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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:
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0.0149 0.0124 0.0092 0.0171 error 20.01710.000441 2.20.01490.000333 2.50.01240.000245 30.00920.000306 Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0 Least square minimization of I(35Cl+)/I(H35Cl+) vs J´ (for J´=0-5) with respect to and least sq. error (J=0-5)
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least sq. error (J=0-5) 0.0149 Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0agust,heima,...Jan11/Evaluation of coupling strength j state-210111kmak.xls expCalc.(v´=19-22) v´=19,20,21,22 and sum 05 503 3J´ v´=18,19,20,21,22,23 and sum
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0.0124 Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0 &agust,heima,...Jan11/Evaluation of coupling strength j state-210111kmak.xls least sq. error (J=0-5) expCalc.(v´=19-22) v´=19,20,21,22 and sum 05 503 3J´
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0.0092 Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0agust,heima,...Jan11/Evaluation of coupling strength j state-210111kmak.xls least sq. error (J=0-5) expCalc.(v´=19-22) v´=19,20,21,22 and sum 05 503 3J´
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1) NB!: contributions from v´ 21 CLEARLY CAN NOT BE IGNORED!!! This analysis assumes W 12 to be constant and independent with v´(ip) and to be the same value as that derived from shift analysis for v´(ip)=21. and are also assumed to be constant and independent with v´(ip) : Thus least square analyses on and (for W 12 = 25 cm -1 ) resulted in W 12 = 25 cm -1 = 2.5 = 0.0124 for j(0 + ) H35Cl The significantly larger value, compared to that observed for other triplet states ( = 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 H 37 Cl!!!!!
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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:
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expCalc.(v´=18-23) v´=18,19,20,21,22,23 and sum W12 = 25 cm-1; = 1.7; = 0.0135 J´ agust,heima,...Jan11/Evaluation of coupling strength j state-220111kmak.xls
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It is interesting to see that the contribution falls down very slowly as E(J´) increases / v´ “moves further away” from the Rydberg state. But what happens if W 12 changes with v´, say W 12 increases? I tried 1)W 12 = 22,23,24,25,26,27 vs v´=18,19,20,21,22,23 & 2)W 12 = 19,21,23,25,27,29 vs v´=18,19,20,21,22,23 & 3)W 12 = 28,27,26,25,24,23 vs v´=18,19,20,21,22,23 & 4)W 12 = 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´ 21 is close to a constant ( ). Therefore the relevant expression for I(Cl+)/I(HCl+) is Is it perhaps possible to obtain good fit for the parameters and only assuming to be zero?
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No that does not seem to be the case. In other words gamma is an important parameter. Looking at: It is clear that c 2 2 is very small and the ratio for v´ 21 is simply: NB! It is interesting to see that similar values are obtained independet of the number of v´(V) contribution: = 0.013 for v´=20-21 (KM) = 0.0124 for v´= 19-22 = 0.0135 for v´= 18-23 THIS IS IMPORTANT!
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Effect of is clearly seen below:
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W 12 = 25, = 1.7 = 0 = 0.006 = 0.013 v´=18,19,20,21,22,23 and sum J´ agust,heima,...Jan11/Evaluation of coupling strength j state-220111kmak.xls How can I make the colors in the excel graph to stay unchanged?
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W 12 =25, f = 1 Minimize with respect to and => = 2.2, = 0.0198, least sq. error(J´=0-5) = 0.000706 agust,heima,...Jan11/Evaluation of coupling strength j state-220111kmak.xls J´ 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.
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Lets´ compare the calc. sum values for different optimizaed values: Is the graph shape perhaps comparable?: Lets look at plots normalized to the largest peak (i.e. J´=6) See note from 230111: ** least sq.(0-6)(rel) 0.0130.03710.002602391 0.0090.04460.002591518 0.0050.05260.002591518 00.062590.002591521 f (in f* * ) 0.0132.8160.002591523 0.0094.9540.002591524 0.00318.860.002591532 f (in f* * ) All same! 0.0312900.002591532 All same! agust,heima,...Jan11/Evaluation of coupling strength j state-230111kmak.xls Fit of I rel (exp) = (I(Cl + )/I(HCl + )(J´;exp))/(I(Cl + )/I(HCl + )(J´ max ;exp)) vs J´ by I rel (calc)= (I(Cl + )/I(HCl + )(J´;calc))/(I(Cl + )/I(HCl + )(J´ max ;calc)) vs J´ NB!: J´ max = 6 All give equally good fit (see figure next slide) ERGO: 1) Use f = 0 (i.e. Neglect f* * ) 2) Use only v´= 20 & 21 and perform fit on I rel (exp) vs J´ by varying only!!! Thus the parameter drops out
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agust,heima,...Jan11/Evaluation of coupling strength j state-230111kmak.xls Fit of I rel (exp) = (I(Cl + )/I(HCl + )(J´;exp))/(I(Cl + )/I(HCl + )(J´ max ;exp)) vs J´ by I rel (calc)= (I(Cl + )/I(HCl + )(J´;calc))/(I(Cl + )/I(HCl + )(J´ max ;calc)) vs J´ : Relative Intensity ratios J´
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Comparison of KM´s and JL´s ion ratios for j(0+), H35Cl: JCl/HCl(KM)(JL;15.12.10))JL(16.12.10) 00.17295960.171049682 10.187169960.203345849 20.175343650.188906718 30.181481750.229411338 40.187976060.201725744 50.199185490.2270846080.266582422 60.369315570.6029650360.485383285 70.70763260.913165678 8 90.149138678 suitable for low J´ suitable for high J´ agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls
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KM JL 151210 JL 161210 J´ agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls
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KM JL 151210 JL 161210 J´ agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls
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