H35Cl, j(0+) intensity ratio analysis and comparison of experimental data
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Cl, j(0+) intensity ratio analysis and parison of experimental data st,www,....Jan11/PPT-210111ak.ppt st,heima,...Jan11/Evaluation of coupling strength j state-2i0111kmak st,heima,....Jan11/PXP-210111ak.pxp following holds for W12 = 25 cm-1:
description
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. - PowerPoint PPT Presentation
Transcript of 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.pptagust,heima,...Jan11/Evaluation of coupling strength j state-2i0111kmak.xlsagust,heima,....Jan11/PXP-210111ak.pxp
The following holds for W12 = 25 cm-1:
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x10-6
3.02.82.62.42.22.0
0.0149
0.0124
0.0092
0.0171
error2 0.0171 0.0004412.2 0.0149 0.0003332.5 0.0124 0.0002453 0.0092 0.000306
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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x10-6
3.02.82.62.42.22.0
least sq.error(J=0-5)
0.01490
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Agust,heima,....Jan11/PXP-210111ak.pxp; Lay: 0, Gr:0 agust,heima,...Jan11/Evaluation of coupling strength j state-210111kmak.xls
exp Calc.(v´=19-22)
v´=19,20,21,22 and sum
0 5
50 3
3 J´
J´
v´=18,19,20,21,22,23 and sum
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x10-6
3.02.82.62.42.22.0
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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)
exp Calc.(v´=19-22)
v´=19,20,21,22 and sum
0 5
50 3
3 J´
J´
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x10-6
3.02.82.62.42.22.0
0
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0.0092
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)
exp Calc.(v´=19-22)
v´=19,20,21,22 and sum
0 5
50 3
3 J´
J´
1) NB!: contributions from v´< 20 and v´> 21 CLEARLY CAN NOT BE IGNORED!!!
This analysis assumes W12 to be constant and independent with v´(ip) and to be thesame value as that derived from shift analysis for v´(ip)=21. and are alsoassumed to be constant and independent with v´(ip) : Thus least square analyses on and (for W12 = 25 cm-1) resulted in
W12 = 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 dissocaiationChannels 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!!!!!
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:
exp Calc.(v´=18-23)
v´=18,19,20,21,22,23 and sum
W12 = 25 cm-1; = 1.7; = 0.0135
J´
J´
agust,heima,...Jan11/Evaluation of coupling strength j state-220111kmak.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 W12 changes with v´, say W12 increases?
I tried 1)W12= 22,23,24,25,26,27 vs v´=18,19,20,21,22,23 &2)W12= 19,21,23,25,27,29 vs v´=18,19,20,21,22,23 &3)W12= 28,27,26,25,24,23 vs v´=18,19,20,21,22,23 &4)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 contributionfrom v´< 20 and 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 onlyassuming to be zero?
where
HClI
ClI
HClI
ClI
HClI
ClI
vv
21´)20´)(
)(
)(
)(
)(
)(
21,20´;
´))(1(
)1´)((
)(
)(22
22
´
vvc
vc
HClI
ClI
v
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 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!
21´,20´;
´))(1(
)1´)((
)(
)(22
22
´
vvvc
vc
HClI
ClI
v
21´,20´..
21´,20´;1)(
)(
´
vvforei
vvHClI
ClI
v
´´21´20
´´21´2021´20´
;)(
)(
)(
)(
)(
)(
vvv
vvvvv
HClI
ClI
HClI
ClI
HClI
ClI
Effect of is clearly seen below:
W12= 25, = 1.7
= 0
= 0.006
= 0.013
v´=18,19,20,21,22,23 and sum
J´
J´
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?
;)(
)(
)(
)(
)(
)(
21´20´
fHClI
ClI
HClI
ClI
HClI
ClI
vv
W12=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 increasedThe experimentalRatio value to 0.5
We realy need to analogous test onH37Cl where the peak overlapproblem does notexist.
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)0.013 0.0371 0.0026023910.009 0.0446 0.0025915180.005 0.0526 0.002591518
0 0.06259 0.002591521f (in f**)
0.013 2.816 0.0025915230.009 4.954 0.0025915240.003 18.86 0.002591532
f (in f**)All same!0.03129 0 0.002591532
All same!
agust,heima,...Jan11/Evaluation of coupling strength j state-230111kmak.xls
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´
´**
))22´(1(
)1)(22´(
))20´(1(
)1)(20´()´;)((/)(
22
22
22
22 Jvsf
vc
vc
vc
vccalcJHClIClI
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 & 21and perform fit on Irel (exp) vs J´by varying only!!!
Thus the parameter drops out
agust,heima,...Jan11/Evaluation of coupling strength j state-230111kmak.xls
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´
Comparison of KM´s and JL´s ion ratios for j(0+), H35Cl:
J Cl/HCl(KM) (JL;15.12.10)) JL(16.12.10)
0 0.1729596 0.171049682
1 0.18716996 0.203345849
2 0.17534365 0.188906718
3 0.18148175 0.229411338
4 0.18797606 0.201725744
5 0.19918549 0.227084608 0.266582422
6 0.36931557 0.602965036 0.485383285
7 0.7076326 0.913165678
8
9 0.149138678
suitable for low J´
suitable for high J´
agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls
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KM
JL151210
JL161210
)(
)(35
35
ClHI
ClI
J´
agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls
KM
JL151210
JL161210
)(
)(35
35
ClHI
ClI
J´
agust,heima,...Jan11/Evaluation of coupling strength j state-240111kmak.xls
