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

      error Agust,heima,....Jan11/PXP ak.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)

least sq. error (J=0-5)  Agust,heima,....Jan11/PXP ak.pxp; Lay: 0, Gr:0agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls expCalc.(v´=19-22) v´=19,20,21,22 and sum J´  v´=18,19,20,21,22,23 and sum

 Agust,heima,....Jan11/PXP ak.pxp; Lay: 0, Gr:0 &agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls least sq. error (J=0-5) expCalc.(v´=19-22) v´=19,20,21,22 and sum J´ 

 Agust,heima,....Jan11/PXP ak.pxp; Lay: 0, Gr:0agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls least sq. error (J=0-5) expCalc.(v´=19-22) v´=19,20,21,22 and sum J´ 

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  = for j(0 + ) H35Cl The significantly larger  value, compared to that observed for other triplet states (  = – 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!!!!!

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:

expCalc.(v´=18-23) v´=18,19,20,21,22,23 and sum W12 = 25 cm-1;  = 1.7;  = J´ agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls

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?

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:  = for v´=20-21 (KM)  = for v´=  = for v´= THIS IS IMPORTANT!

Effect of  is clearly seen below:

W 12 = 25,  = 1.7  = 0  =  = v´=18,19,20,21,22,23 and sum J´ agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls How can I make the colors in the excel graph to stay unchanged?

W 12 =25, f = 1 Minimize with respect to  and  =>  = 2.2,  = , least sq. error(J´=0-5) = agust,heima,...Jan11/Evaluation of coupling strength j state kmak.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.

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 : ** least sq.(0-6)(rel) f (in f*  *  ) f (in f*  *  ) All same! All same! agust,heima,...Jan11/Evaluation of coupling strength j state kmak.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

agust,heima,...Jan11/Evaluation of coupling strength j state kmak.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´

Comparison of KM´s and JL´s ion ratios for j(0+), H35Cl: JCl/HCl(KM)(JL; ))JL( ) suitable for low J´ suitable for high J´ agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls

KM JL JL J´ agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls

KM JL JL J´ agust,heima,...Jan11/Evaluation of coupling strength j state kmak.xls