HCl agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc ak.pxp (JMS paper) agust,www,....Sept10/PPT ak.ppt agust,heima,...Sept10/HCl(3+1)j3Sigma(0) Calc ak.pxp (R-C spectrum) agust,heima,...Sept10/XLS ak.xls agust,heima,...Sept10/XLS ak.xls agust,heima,...Sept10/XLS ak.xls agust,heima,...Sept10/22320_ ak.pxp (new R-T spectrum) agust, heima,...Sept10/J3S_V1S ak.pxp (new R-T spectrum) agust,heima,...Sept10/aHCl(3+1)j3S(0)Calc ak.pxp agust,heima,...Sept10/XLS ak.xls agust,heima,...Sept10/Look for J pxp

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

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

J´ (JMS, 2004)J´(KM) Not seen NB!: KM´s (REMPI-TOF; R-T) spectrum is a lot colder than the JMS (20014)-REMPI- current (R-C) spectrum

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) J´ Jnju(Q, j(0+)<-<-X from KM: agust,heima,...Sept10/XLS ak.xls E(J´)=nju(Q)+E(J´´) Slide 6 Slide 7

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

agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc ak.pxp, Lay: 6, Gr:9  E J´,J´-1 J´ DE(J´,J´-1) J´ = J´ = 2B´J´=> B´= cm-1 agust,heima,...Sept10/XLS ak.xls From REMPI-TOF (KM): cm

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

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?

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

V, v´=21 J´1hv(Laser)2hvE(J´´)E(J´)=nju(Q)+E(J´´)DE(J´,J´+1) Green Green Green Green Green Green Wang KM prediction agust,heima,...Sept10/XLS ak.xls

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

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_ ak.pxp, Lay:0, Gr:1 NB!: According to KM (211210) the peak at is a S line peak for V!! The broad peaks However might be V,v´=21, J´=6

E1(6) see ;2 E10(6) DE10(5,6)-DE1(5,6) = D(DE1(5,6)) E2(6) E20(6) D(DE2(5,6))24.95 (1/2)(E10(6)+E20(6)) E20(6)-E10(6) W12(6) cm-1 agust,heima,...Sept10/XLS ak.xls

= E 1 (6) = E 1 0 (6) cm -1 E 2 0 (6) = (E 2 (6) = cm cm-1 agust,heima,....Sept10/aHCl(3+1)j3S(0)Calc ak.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

W 12 derived from J´=7 : J´ agust,heima,...Sept10/22320_ ak.pxp, Lay:1, Gr:2  E J´,J´-1 J=6J=7 E1(J) see ; E10(J) DE10(J-1,J)-DE1(J- 1,J) = D(DE1(J-1,J)) E2(J) ? E20(J) D(DE2(J-1,J))24.95? (1/2)(E10(J)+E20(J)) E20(J)-E10(J) W12(J) V,v´=21 state data This is largely overestimated value This is slightly underestimated value

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!

Old assignment OLD assignmentOLD J´ nju(Q, j(0+)<-<-XE(J´)=nju(Q)+E(J´´) DE(J´,J´+1 ) j,v´=0 state data

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: OLD: AK(1) AK(2) AK(3) AK(4) 9 6 6/ Larger Cl+/HCl+ ratios according to KM Large gap: ATH! NO! 9 10 i=37 J(1);S overlap

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

agust, heima,...Sept10/J3S_V1S ak.pxp; Lay: 2, Gr: 1 1hv H+ H35Cl+ 35Cl+ H37Cl+ J, v´=0 Could be 37 peak from V, v´=21

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

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

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

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

Now it is possible to predict the position of the J´=9 band nju = E(J´) = (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 !!! Lets try that agust,heima.....Sept10/XLS ak.xls

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

agust,heima,...Sept10/Look for J pxp This could be V, v´=21, J´=7(?)

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 jl ak.xls ) The isotope separation for J´= could be < 5. Now let´s check where exactly j(1), S serie peak is to be found.