230 likes | 240 Views
HCl, W =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
E N D
HCl, W=0, H37Cl 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.xlsagust,heima,...Jan11/XLS-010111ak.xls http://www3.hi.is/~agust/rannsoknir/rempi/hcl/Jan11/PDF-020111ak.pdf
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
New assignment (AK4)! j(0+), Q; H35Cl 5 j(1); S j(0+), Q; H37Cl 6 and 8 overlap 6 8 10 7 9 H35Cl+ 5 35Cl+ H+ 5 H37Cl+ 6 7 37Cl+ 9 8 agust,heima,...Sept10/Look for J7-211210ak.pxp; Lay:0, Gr: 1
DE J´,J´-1 j(0+), Q, H37Cl = 19.586 J´ = 2B´J´=> B´= 9.794 cm-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-211210ak.pxp; Lay:14, Gr: 18
DE J´,J´-1 j(0+), Q, H35Cl = 19.657 J´ = 2B´J´=> B´= 9.8285 cm-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:15, Gr: 19
DE J´,J´-1 j(0+), Q, H37Cl j(0+), Q, H35Cl J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:15, Gr: 19
W12 derived ) (E2 (J´) J´(V, v´=21) E20(J´) (E10(J´) +E20(J´))/2 E10(J´) J´(j, v´=0) E1 (J´) E10(J´) E1 (J´) and E20(J´) needed: agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 8, Gr:12
See 221210; 1-2 Calculated both for j and V,v´=21 states
H35Cl Based on Green´s data Need to reevaluate this from Longs data agust,heima,...Sept10/XLS-221210ak.xls
H35Cl 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 B´(j)= 9.8385 B´(V)=3.4389 agust,heima,...Sept10/XLS-221210ak.xls
DE J´,J´-1 V,v´=21, H37Cl a = -1.0729 ± 2.03 b = 6.8079 ± 0.613 = 5.8919 J´ = 2B´J´=> B´= 2.94595 cm-1 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:16, Gr: 20 and ...Sept10/XLS-211210ak.xls
H35Cl Based on Greens data: DE J´,J´-1 24.95 cm-1 agust,heima,...Sept10/XLS-151210ak.xls = 6.8778 J´ = 2B´J´=> B´= 3.4389 cm-1 J´ agust,heima,...Sept10/22320_22356-161210ak.pxp, Lay:1, Gr:2
DE J´,J´-1 V,v´=21, H35Cl andH37Cl 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 J´ agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-221210ak.pxp; Lay:17, Gr: 21 and ...Sept10/XLS-211210ak.xls
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
H35Cl B´(j)= 9.8285 B´(V)=3.302 For E20(J´) levels based on Long agust,heima,...Sept10/XLS-221210ak.xls
H37Cl Based on Long´s data agust,heima,...Sept10/XLS-221210ak.xls
H37Cl B´(j)= 9.794 B´(V)= 3.40395 For E20(J´) levels based on Long agust,heima,...Sept10/XLS-221210ak.xls
NB! The W12 value for J´=8 is very sensitive to slight changes in E(J´=8)
Different ways to evaluted W12 (W12´) from E0(J´), E(J´) and EV(J´) via D(J´) (= E(J´)-E0(J´)), see: http://www3.hi.is/~agust/rannsoknir/rempi/hcl/Jan11/PDF-020111ak.pdf 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) agust,heima,...Jan11/XLS-010111ak.xls
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