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HCl,  =0, H 3 7Cl and H35Cl analysis. agust,www,.....Sept10/PPT-211210aak.ppt agust,heima,...Sept10/XLS-201210ak.xls agust,heima,...Sept10/Look for J7-211210ak.pxp.

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Presentation on theme: "HCl,  =0, H 3 7Cl and H35Cl analysis. agust,www,.....Sept10/PPT-211210aak.ppt agust,heima,...Sept10/XLS-201210ak.xls agust,heima,...Sept10/Look for J7-211210ak.pxp."— Presentation transcript:

1 HCl,  =0, H 3 7Cl and H35Cl analysis. agust,www,.....Sept10/PPT-211210aak.ppt agust,heima,...Sept10/XLS-201210ak.xls agust,heima,...Sept10/Look for J7-211210ak.pxp agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-211210ak.pxp agust,heima,...Sept10/XLS-221210ak.xls agust,heima,...Jan11/XLS-010111ak.xls http://www3.hi.is/~agust/rannsoknir/rempi/hcl/Jan11/PDF-020111ak.pdf

2 AK(3) AK3 E(J´)=nju(Q)+E(J´´)DE(J´,J´-1)DE(J´,J´-1; 37)E(J´;37)=nju+E(J´´,37) 89282 89301.3776219.3776169219.3460075989301.34601 89339.9206438.5430233238.4798416389339.82585 89397.3046557.3840087157.2893285489397.11518 89475.3930178.0883625777.9622947789475.07747 89571.7368996.3438743996.1865667789571.26404 89683.87522112.1383337 89844.13475160.2595299 89986.13141.9952525 90162.65329176.5232911 H35Cl H37Cl Very small difference in energies and DE for the j(0+) state For 35 and 37; How about the V, v´=21 state? agust,heima,...Sept10/XLS-201210ak.xls

3 New assignment (AK4)! 5 6 7 8 9 10 j(0+), Q; H35Cl j(1); S 5 5 6 7 89 6 and 8 overlap j(0+), Q; H37Cl H35Cl+ 35Cl+ H+ 37Cl+ H37Cl+ agust,heima,...Sept10/Look for J7-211210ak.pxp; Lay:0, Gr: 1

4 = 19.586 J´ = 2B´J´=> B´= 9.794 cm-1  E J´,J´-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-211210ak.pxp; Lay:14, Gr: 18 j(0+), Q, H37Cl

5 = 19.657 J´ = 2B´J´=> B´= 9.8285 cm-1  E J´,J´-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:15, Gr: 19 j(0+), Q, H35Cl

6 j(0+), Q, H37Cl j(0+), Q, H35Cl  E J´,J´-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:15, Gr: 19

7 (Re)evaluation of W12 for j(0+) V,v´=21 interaction

8 E 1 (J´) E 1 0 (J´) (E10(J´) +E20(J´))/2 E 2 0 (J´) (E 2 (J´) agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 8, Gr:12 J´(V, v´=21) J´(j, v´=0) ) W12 derived E 1 0 (J´)E 1 (J´) andE 2 0 (J´) needed:

9 See 221210; 1-2 Calculated both for j and V,v´=21 states

10 j(0+) B´9.8285cm-1 J2´=6578 J1´:E10(J1´)E10(J2´) 08928289694.79789576.85589832.39689989.652 189301.3776289694.5176289576.5756289832.1166289989.37262 289339.9206489693.7466489575.8046489831.3456489988.60164 389397.3046589692.1596589574.2176589829.7586589987.01465 89693.8052389575.8632389831.4042389988.66023 V,v´=21 B´3.4389cm-1 J2´=678 J1´:E20(J1´)E20(J2´) 089607.389751.733889799.878489854.9008 189613.9776289751.5336289799.6782289854.70062 289627.2206489751.0210489799.1656489854.18804 389647.2046589750.3716589798.5162589853.53865 489674.2930189749.9488189798.0934189853.11581 589708.4368989749.7036989797.8482989852.87069 89750.7187789798.8633789853.88577 agust,heima,...Sept10/XLS-221210ak.xls H35Cl Based on Green´s data Need to reevaluate this from Longs data

11 AK4AK(4) J=6J=5J=7J=8 E1(J)89683.8752289571.7368989844.1347589995.33 E10(J)89693.8052389575.8632389831.4042389988.66023 E20(J)89750.7187789708.4368989798.8633789853.88577 (1/2)(E10(J)+E20(J))89722.26289642.1500689815.133889921.273 E20(J)-E10(J)56.91354073132.5736594-32.54085927-134.7744593 W12(J)25.7634798523.7501739724.0068400330.71483816 W´12(J)3.9753912484.3361686763.208048943.619778391 These values are much closer to KM´s evaluation base on intensity ratios NB: See efect of slight increase in B´ for the j state from 9.8285 to 9.8385 on next slide agust,heima,...Sept10/XLS-221210ak.xls H35Cl

12 AK4AK(4) J=6J=5J=7J=8 E1(J)89683.8752289571.7368989844.1347589995.33 E10(J)89694.1752389576.1132389831.9142389989.33023 E20(J)89750.7187789708.4368989798.8633789853.88577 (1/2)(E10(J)+E20(J))89722.44789642.2750689815.388889921.608 E20(J)-E10(J)56.54354073132.3236594-33.05085927-135.4444593 W12(J)26.2390740524.4590630223.5210536229.13131727 W´12(J)4.0487770254.4655935173.1431330013.433158665 agust,heima,...Sept10/XLS-221210ak.xls H35Cl B´(j)= 9.8385 B´(V)=3.4389

13 = 5.8919 J´ = 2B´J´=> B´= 2.94595 cm-1  E J´,J´-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:16, Gr: 20 and...Sept10/XLS-211210ak.xls V,v´=21, H37Cl a = -1.0729 ± 2.03 b = 6.8079 ± 0.613 V,v´=21 Long E(J´;37)=nju+E(J´´,37)DE(37) 89603.7454 89610.531416.786008 89622.9453612.41395 89641.5150818.56972 89665.4182323.90315 89700.4989135.08068

14 = 6.8778 J´ = 2B´J´=> B´= 3.4389 cm-1 H35Cl Based on Greens data: 24.95 cm-1  E J´,J´-1 J´ DE(J´,J´+1) 6.677617 13.24302 19.98401 27.08836 34.14387 66.21833 J´ 1 2 3 4 5 6 agust,heima,...Sept10/XLS-151210ak.xls agust,heima,...Sept10/22320_22356-161210ak.pxp, Lay:1, Gr:2

15  E J´,J´-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:17, Gr: 21 and...Sept10/XLS-211210ak.xls V,v´=21, H35Cl and H37Cl a = -1.0729 ± 2.03 b = 6.8079 ± 0.613 B´(37)= 3.40395 a = -0.42488 ± 2.37 b = 6.604 ± 0.714 B´(35) = 3.302

16 AK4AK(4) J=6J=5J=7J=8 E1(J)89683.8752289571.7368989844.1347589995.33 E10(J)89693.8052389575.8632389831.4042389988.66023 E20(J)89746.5661389708.4368989792.7941389845.62613 (1/2)(E10(J)+E20(J))89720.1856889642.1500689812.0991889917.14318 E20(J)-E10(J)52.76090739132.5736594-38.61009261-143.0340926 W12(J)24.9503750723.7501739725.5654626831.5989111 W´12(J)3.8499264564.3361686763.4163286543.723967387 H35Cl B´(j)= 9.8285 B´(V)=3.302 For E20(J´) levels based on Green Not a big difference in W12 evaluations agust,heima,...Sept10/XLS-221210ak.xls

17 AK4AK(4) J=6J=5J=7J=8 E1(J)89683.8752289571.7368989844.1347589995.33 E10(J)89693.8052389575.8632389831.4042389988.66023 E20(J)89748.7133889708.0275189794.9413889847.77338 (1/2)(E10(J)+E20(J))89721.259389641.9453789813.172889918.2168 E20(J)-E10(J)54.90815406132.1642794-36.46284594-140.8868459 W12(J)25.3740692623.7145845925.0251335531.37147637 W´12(J)3.9153038884.3296709743.3441241363.697163947 H35Cl B´(j)= 9.8285 B´(V)=3.302 For E20(J´) levels based on Long agust,heima,...Sept10/XLS-221210ak.xls

18 NOW comes H37Cl:

19 j(0+) H37Cl: B´=9.794cm-1 J2´=6578 J1´:E10(J1´)E10(J2´) 08928289693.34889575.8289830.46489987.168 189301.34689693.1060189575.5780189830.22289986.92601 289339.82689692.4098589574.8818589829.52689986.22985 389397.11589690.9351889573.4071889828.05189984.75518 489475.07789690.5454789573.0174789827.66189984.36547 89692.4497689574.9217689829.56689986.26976 V,v´=21 H37Cl B´3.40395cm-1 J2´=6578 J1´:E20(J1´)E20(J2´) 089603.74589746.711389705.863989794.36789848.8298 189610.53189746.6894189705.8420189794.34589848.80791 289622.94589745.4875689704.6401689793.14389847.60606 89746.2960989705.4486989793.95189848.41459 agust,heima,...Sept10/XLS-221210ak.xls H37Cl Based on Long´s data

20 AK(4) J=6J=5J=7J=8 89682.40489571.2640489843.189994.77807 89692.4589574.9217689829.56689986.26976 89746.29689705.4486989793.95189848.41459 89719.37389640.1852289811.75989917.34217 53.84633130.5269303-35.61437-137.8551697 25.33463522.1542273125.79109635.28888522 2.98571542.6109007273.03950984.158835006 H37Cl B´(j)= 9.794 B´(V)= 3.40395 For E 2 0 (J´) levels based on Long E1(J) E10(J) E20(J) (1/2)(E10(J)+E20(J)) E20(J)-E10(J) W12(J) W´12(J) agust,heima,...Sept10/XLS-221210ak.xls

21 NB! The W12 value for J´=8 is very sensitive to slight changes in E(J ´=8)

22 Different ways to evaluted W12 (W12´) from E0(J´), E(J´) and EV(J´) via  (J´) (= E(J´)-E0(J´)), see: http://www3.hi.is/~agust/rannsoknir/rempi/hcl/Jan11/PDF-020111ak.pdf j(0+)V, v´=21 H35Cl AK(4) assignmentj(0+) V,v´=21W12 J´E(J´)=nju(Q)+E(J´´)EV(J´)=nju(Q)+E(J´´)commentDE0DEE0D(E-E0)EV0 08928289607.3 189301.3776289613.9776219.65719.37762 289339.9206489627.2206439.31438.54302 389397.3046589647.2046558.97157.38401 J´-3489475.3930189674.2930178.62878.0883689475.39089674.290 J´-2589571.7368989708.4368998.28596.3438789573.681.94112689706.516.17356 J´-1689683.8752289755.59086Diffuse peak117.942112.138389691.627.74479289747.8522.2585 J´789844.13475137.599160.2595 889995.33157.256151.1953 990172.56329176.913177.2333 1090369.41473196.57196.8514 from XLS-221210ak.xls agust,heima,...Jan11/XLS-010111ak.xls The above calculation for E0 evaluations is based on E(J´=4) = E0(J´=4) I will now try to evaluate E0´s by same method (E(J´) = E0(J´)) for J´ different from J´=4 and J´ further away from the interaction region. Try J´= 1 (see next page)

23 j(0+)V, v´=21 H35Cl AK(4) assignmentj(0+) V,v´=21W12 J´E(J´)=nju(Q)+E(J´´)EV(J´)=nju(Q)+E(J´´)commentDE0DEE0D(E-E0)EV0 08928289607.3 J´-6189301.3776289613.9776219.65719.3776289301.38089613.980 J´-5289339.9206489627.2206439.31438.5430289340.690.77097789626.4514.86295 J´-4389397.3046589647.2046558.97157.3840189399.662.35796889644.8524.15981 J´-3489475.3930189674.2930178.62878.0883689478.292.89760589671.423.83144 J´-2589571.7368989708.4368998.28596.3438789576.584.83873189703.625.25948 J´-1689683.8752289755.59086Diffuse peak117.942112.138389694.5210.642489744.9525.49443 J´789844.13475137.599160.2595 889995.33157.256151.1953 990172.56329176.913177.2333 1090369.41473196.57196.8514 Good consistancy In W12 values! I need to perform analogous calculations for - H37Cl, J(0+) V, v´=21 and - HiCl, J(1) V, v´=20: i = 35 & 37 see PPT-211210ak.ppt agust,heima,...Jan11/XLS-010111ak.xls

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