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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 aak.ppt agust,heima,...Sept10/XLS ak.xls agust,heima,...Sept10/Look for J ak.pxp agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc ak.pxp agust,heima,...Sept10/XLS ak.xls agust,heima,...Jan11/XLS ak.xls

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) 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 ak.xls

3 New assignment (AK4)! j(0+), Q; H35Cl j(1); S and 8 overlap j(0+), Q; H37Cl H35Cl+ 35Cl+ H+ 37Cl+ H37Cl+ agust,heima,...Sept10/Look for J ak.pxp; Lay:0, Gr: 1

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

5 = J´ = 2B´J´=> B´= cm-1  E J´,J´-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc ak.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 ak.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 ak.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 ; 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´) V,v´=21 B´3.4389cm-1 J2´=678 J1´:E20(J1´)E20(J2´) agust,heima,...Sept10/XLS ak.xls H35Cl Based on Green´s data Need to reevaluate this from Longs data

11 AK4AK(4) J=6J=5J=7J=8 E1(J) E10(J) E20(J) (1/2)(E10(J)+E20(J)) E20(J)-E10(J) W12(J) W´12(J) 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 to on next slide agust,heima,...Sept10/XLS ak.xls H35Cl

12 AK4AK(4) J=6J=5J=7J=8 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 ak.xls H35Cl B´(j)= B´(V)=3.4389

13 = J´ = 2B´J´=> B´= cm-1  E J´,J´-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc ak.pxp; Lay:16, Gr: 20 and...Sept10/XLS ak.xls V,v´=21, H37Cl a = ± 2.03 b = ± V,v´=21 Long E(J´;37)=nju+E(J´´,37)DE(37)

14 = J´ = 2B´J´=> B´= cm-1 H35Cl Based on Greens data: cm-1  E J´,J´-1 J´ DE(J´,J´+1) J´ agust,heima,...Sept10/XLS ak.xls agust,heima,...Sept10/22320_ ak.pxp, Lay:1, Gr:2

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

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

17 AK4AK(4) J=6J=5J=7J=8 E1(J) E10(J) E20(J) (1/2)(E10(J)+E20(J)) E20(J)-E10(J) W12(J) W´12(J) H35Cl B´(j)= B´(V)=3.302 For E20(J´) levels based on Long agust,heima,...Sept10/XLS ak.xls

18 NOW comes H37Cl:

19 j(0+) H37Cl: B´=9.794cm-1 J2´=6578 J1´:E10(J1´)E10(J2´) V,v´=21 H37Cl B´ cm-1 J2´=6578 J1´:E20(J1´)E20(J2´) agust,heima,...Sept10/XLS ak.xls H37Cl Based on Long´s data

20 AK(4) J=6J=5J=7J= H37Cl B´(j)= B´(V)= 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 ak.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: 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)EV J´ J´ J´ Diffuse peak J´ from XLS ak.xls agust,heima,...Jan11/XLS ak.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)EV J´ J´ J´ J´ J´ J´ Diffuse peak J´ 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 ak.ppt agust,heima,...Jan11/XLS ak.xls


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