1. HCl agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp (JMS paper) agust,www,....Sept10/PPT-141210ak.ppt agust,heima,...Sept10/HCl(3+1)j3Sigma(0) Calc-141210ak.pxp (R-C spectrum) agust,heima,...Sept10/XLS-141210ak.xls agust,heima,...Sept10/XLS-151210ak.xls agust,heima,...Sept10/XLS-161210ak.xls agust,heima,...Sept10/22320_22356-161210ak.pxp (new R-T spectrum) agust, heima,...Sept10/J3S_V1S21-171210ak.pxp (new R-T spectrum) agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc-201210ak.pxp agust,heima,...Sept10/XLS-201210ak.xls agust,heima,...Sept10/Look for J7-201210.pxp

2. agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 5, Gr:2_1 KM´s assignment JMS, 228,143,(2004)-assignment REMPI-current

3. agust,heima,...Sept10/HCl(3+1)j3Sigma(0) Calc-141210ak.pxp (R-C spectrum), Lay:2, Gr:1 1234 5 6 7 890 KM´s assignment JMS, 228,143,(2004)-assignment

4. J´ (JMS, 2004)J´(KM) 10 21 32 43 54 85 96 67 78 109 11Not seen NB!: KM´s (REMPI-TOF; R-T) spectrum is a lot colder than the JMS (20014)-REMPI- current (R-C) spectrum

5. Now lets try 1)Derive Energy level values both from R-T and R-C spectra using KM´s assignment and check shifts. 2) Insert Wangs and Longs´newest spectra uncluding “newly observed “ peaks for comparison. DE(J´,J´-1) 19.37762 38.54302 57.38401 78.08836 96.34387 112.1383 160.2595 153.4953 174.9333 J´ 1 2 3 4 5 6 7 8 9 Jnju(Q, j(0+)<-<-X 0 89282 1 89280.5 2 89277.3 3 89272.1 4 89266.8 5 89259 6 89246.3 7 89261.1 8 89248.6 9 89237.1 from KM: agust,heima,...Sept10/XLS-151210ak.xls E(J´)=nju(Q)+E(J´´) 89282 89301.37762 89339.92064 89397.30465 89475.39301 89571.73689 89683.87522 89844.13475 89997.63 90172.56329 Slide 6 Slide 7

6. agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 7, Gr:11 E(J´) J´(J´+1) Coefficient values ± one standard deviation K0=C = 89281 ± 3.56 K1=B´= 9.7179 ± 0.229 K2=D´= 0.0026284 ± 0.0026 J´=7 J´=9 J´=6 From REMPI-TOF (KM):

7. agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 6, Gr:9  E J´,J´-1 J´ DE(J´,J´-1) 19.37762 38.54302 57.38401 78.08836 96.34387 112.1383 160.2595 153.4953 174.9333 J´ 1 2 3 4 5 6 7 8 9 = 19.497 J´ = 2B´J´=> B´= 9.7485 cm-1 agust,heima,...Sept10/XLS-151210ak.xls From REMPI-TOF (KM): 4.8436663 cm-1 23.78

8. Now let´s look at the latest data for V, v´=21 from Long and Wang: I will wait for Longs analysis and summaries of mass spectra vs REMPI peaks

9. Let´s look at the REMPI-current data Let´s look at energy levels or the V, v´=21 state: Is there a shift of levels to be found?

10. agust,heima,...Sept10/22320_22356-161210ak.pxp, Lay:0, Gr:1 H+ i=35 i=37 22334.2789337.08 cm-1 V,v´=21, J´=6

11. V, v´=21 J´1hv(Laser)2hvE(J´´)E(J´)=nju(Q)+E(J´´)DE(J´,J´+1) 022401.82589607.3Green089607.3 122398.27589593.1Green20.8776289613.977626.677617 222391.1589564.6Green62.6206489627.2206413.24302 322380.589522Green125.204689647.2046519.98401 422366.42589465.7Green208.59389674.2930127.08836 522348.92589395.7Green312.736989708.4368934.14387 622334.2789337.08Wang437.575289774.6552266.21833 622327.57589310.3KM prediction89747.87522 90.78 agust,heima,...Sept10/XLS-161210ak.xls

12. = 6.8778 J´ = 2B´J´=> B´= 3.4389 cm-1 From REMPI-TOF (VHW&JL): 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

13. KM´s prediction, assuming (J’=6;V)- (J´=6,j)=64cm-1 (W&L´s observation for J´=6 (V)(?): (J’=6;V)- (J´=6,j)=90.8 cm-1) agust,heima,...Sept10/22320_22356-161210ak.pxp, Lay:0, Gr:1 NB!: According to KM (211210) the peak at 22334.27 89337.08 is a S line peak for V!! The broad peaks However might be V,v´=21, J´=6

14. E1(6)89683.87522see 161210;2 E10(6)89688.71889 DE10(5,6)-DE1(5,6) = D(DE1(5,6))4.8436663 E2(6)89774.65522 E20(6)89749.70522 D(DE2(5,6))24.95 (1/2)(E10(6)+E20(6))89719.21205 E20(6)-E10(6)60.9863337 W12(6)17.85661089cm-1 agust,heima,...Sept10/XLS-161210ak.xls

15. 89683.9 = E 1 (6) 89688.7 = E 1 0 (6) 89719.2 cm -1 E 2 0 (6) = 89749.7 (E 2 (6) = 89774.7 4.84 cm -1 24.95cm-1 agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 8, Gr:12 ? J´(V, v´=21) = 6 J´(j, v´=0) = 6 ) NB! Not needed in calculation of W 12 (see below) W12 dervid from J´=6

16. W 12 derived from J´=7 : J´ agust,heima,...Sept10/22320_22356-171210ak.pxp, Lay:1, Gr:2  E J´,J´-1 J=6J=7 E1(J)89683.87522see 161210;289844.13475 E10(J)89688.7188989820.35475 DE10(J-1,J)-DE1(J- 1,J) = D(DE1(J-1,J))4.843666323.78 E2(J)89774.65522? E20(J)89749.7052289756.58149 D(DE2(J-1,J))24.95? (1/2)(E10(J)+E20(J))89719.2120589788.46812 E20(J)-E10(J)60.9863337 - 63.77326354 W12(J)17.8566108945.62912016 V,v´=21 state data This is largely overestimated value This is slightly underestimated value

17. The large difference in W12 obtained for J´= 6 and 7 suggests that the Line assignment for j<-<-X is wrong(?). Let´s check other possibilities. 1) OLD assignment: NO evaluation method is no good!

18. Old assignment OLD assignmentOLD J´ nju(Q, j(0+)<-<-XE(J´)=nju(Q)+E(J´´) DE(J´,J´+1 ) 89282 89302.877621 89280.5 89343.1206440.2430232 89277.3 89402.5046559.3840093 89272.1 89480.6930178.1883634 89266.8 89579.5368998.8438745 89261.1 89698.67522119.138336 89248.6 89831.63475132.959537 89259 90008.03176.395258 89246.3 90181.76329173.733299 89237.1 90379.32473197.5614410 20.707 3.27 j,v´=0 state data

19. OLD assignment gives largest gap for J´=7 J´=8 whereas KM´s assignment gives largest gap for J´=6 J´=7 Is it possible to find largest gap for J´=5 J´=6? That would fit with the V, v´= 21 gap / shifts! = j state peaks KM: 9 6 8 5 7 4 3 2 1 0 OLD: 10 11 9 7 8 6 5 4 3 2 1 AK(1) 9 5 7 8 6 4 3 2 1 0 AK(2) 9 5 8 7 6 4 3 2 1 0 AK(3) 8 - 6 - 5 7 4 3 2 1 0 AK(4) 9 6 6/8 5 5 7 4 3 2 1 0 Larger Cl+/HCl+ ratios according to KM Large gap: 6 7 7 8 5 6 6 7 ATH! NO! 9 10 i=37 J(1);S overlap

20. AK(1) assignment: agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 10, Gr:14 NO!

21. agust, heima,...Sept10/J3S_V1S21-171210ak.pxp; Lay: 2, Gr: 1 1hv H+ H35Cl+ 35Cl+ H37Cl+ J, v´=0 Could be 37 peak from V, v´=21

22. AK(2) assignment (see slide 19): agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-141210ak.pxp, Lay: 11 Gr:15 NO!

23. agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-201210ak.pxp, Lay: 12, Gr:16 and.....Sept10/XLS-201210ak.xls Energy levels (E(J´): J(0+)-state <= KM J´=7 J´=6 89828.476 89844.13475 89799.8784 Diff = 15.65875 Prediction for E(J´=7): 89784.21965 89201.1849 J´´=7 -> J´(V) = 7): 89201.1849= 4 x 22300.296 i.e. difference between peaks for J´=7: 59.5 cm-1 13.018 cm-1 V,v´=21 state

24. AK(3) assignment (see slide 19): agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-201210ak.pxp, Lay: 13 Gr:17

25. J(0+)-state <= AK(3) V,v´=21 state 12.298 cm-1 J´=7 J´=6 agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-201210ak.pxp, Lay: 12, Gr:16 and.....Sept10/XLS-201210ak.xls

26. Now it is possible to predict the position of the J´=9 band nju = 89224.802 E(J´) = 90160.26529 (see next slide) Now let´s check position on a spectrum: It could indeed be the peak which Wang thinks that is J´=7 for the V state which is found at 89227.19!!! Lets try that agust,heima.....Sept10/XLS-201210ak.xls

27. Fits nicely J´=7 J´=6 J(0+)-state <= AK(3) agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc-201210ak.pxp, Lay: 12, Gr:16 and.....Sept10/XLS-201210ak.xls

28. agust,heima,...Sept10/Look for J7-201210.pxp 9 8 6 5 7 This could be V, v´=21, J´=7(?) 4.02 6 7 8 9 5

29. See also figs in KM´s manuscript and my comments there. It is worth performing some isotopomer calculations NB! According to Longs estimate ( agust,heima,..../Sept10/V state V21q energy diference between H35Cl and H37Cl-141210jl-201210ak.xls ) The isotope separation for J´= could be < 5. Now let´s check where exactly j(1), S serie peak is to be found.