1. HCl, j(  =1) agust,www,...Sept10/PPT-211210ak.ppt agust,heima,....Sept10/XLS-211210ak.xls agust,heima,....Sept10/PXP-211210ak.pxp agust,heima,...Sept10/Look for J7-211210.pxp agust,heima,....Sept10/PXP-311210ak.pxp agust,heima,....juni09/grof fyrir J astond-150609hrh.xls

2. agust,heima,....Sept10/PXP-211210ak.pxp; Lay:0, Gr:2 j(  =1; S) Green  E(J´,J´-1) J´

3. S 234 5 agust,heima,...Sept10/Look for J7-211210.pxp 2? 345 AK assignment H35Cl+ 35Cl+ H+ S j(  =1; S)Green / j(  =1; S)AK NB: 35Cl+/H35Cl+ is very high according to Q line according to KM; see slide 6 below NB: actually This could be J´=2 (35Cl+)

4. agust,heima,....Sept10/PXP-211210ak.pxp; Lay:1, Gr:1 S j(  =1; P,Q,R,S) Green

5. Probably not correct since this point should be larger due to Interaction with V, v´=20; see slide 3 above Fits nicely with the R serie data i.e.  points agust,heima,....Sept10/PXP-211210ak.pxp; Lay:1, Gr:1 &...XLS-211210ak.xls S, AK

6. KM´s paper manuscript (TS(II) paper, version 12.12.10, fig. 2a

7.  E(J´,J´-1) J´ agust,heima,....Sept10/PXP-311210ak.pxp; Lay:2, Gr:3 &.. j(  =1; Q) Green a = 21.667 ± 3.02 b = 18.358 ± 0.498 Green E(J´) Q 89030.28 89087.02 89166.7 89261.19 89374.94 89506.18 89656.63 J´ 0 1 2 3 4 5 6 7 agust,heima,....Sept10/XLS-211210ak.xls

8.  E(J´,J´-1) J´ agust,heima,....Sept10/PXP-311210ak.pxp; Lay:3, Gr:4 &.. V(  =1; Q), v´=20; Green a = 1.128 ± 1.84 b = 6.446 ± 0.434 agust,heima,....juni09/grof fyrir J astond-150609hrh.xls Energy level 89085.6 89092.47 89107.61 89126.47 89152.94 89187.54 89227.03 J' 0 1 2 3 4 5 6 Orkumun ur á þrepum 6.87 15.13 18.87 26.46 34.61 39.49 agust,heima,....juni09/grof fyrir J astond-150609hrh.xls

9. For 1 = Rydberg state and 2 0 ion-pair state Here Calculated both for j and V,v´=20 states

10. 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:

11. 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 agust,heima,...Jan11/XLS-020111ak.xls The above calculation for E0 evaluations is based on E(J´=7) = E0(J´=7) j(1)V, v´=20 H35Cl j(1) V,v´=20W12W12´ J´E(J´)=nju(Q)+E(J´´)EV(J´)=nju(Q)+E(J´´)commentDE0DEE0D(E-E0)EV0 089085.6 189030.2776289092.4740.02589030.970.68913252289091.78087lower6.5102914.603471 289087.0206489107.6158.38356.74302389089.352.32910920189105.28089lower6.5215122.662396 389166.7046589126.4776.74179.68400989166.090.61389950789127.0839higher4.9318511.423703 489261.1930189152.9495.09994.48836389261.190.00326207389152.94326higher0.5942380.132876 589374.9368989187.54113.46113.7438789374.650.2901364689187.83014higher7.3679371.345195 689506.1752289227.03131.82131.2383389506.460.28652987989227.31653higher#NUM! 789656.63475150.17150.4595389656.630