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agust,www,...januar09/Profun-190209ak

agust,www,...januar09/Profun-190209ak.ppt. 4). 2). 3). 1). 1) Ed(0), J´= 0 – EVv´=8,J´= 0 = 45.553 cm-1. 2) EVv´=9, J´= 5 – EDv´=0,J´= 5 = 164.24 cm-1. 3) Ef(2), J´= 5 – EVv´=8,J´= 5 = 17.14 cm-1. 4) EVv=10, J´= 3 – Egv=0 J´= 3 = 106.64 cm-1.

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agust,www,...januar09/Profun-190209ak

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  1. agust,www,...januar09/Profun-190209ak.ppt

  2. 4) 2) 3) 1) 1) Ed(0), J´=0 – EVv´=8,J´=0 = 45.553 cm-1 2) EVv´=9, J´=5 – EDv´=0,J´=5 = 164.24 cm-1 3) Ef(2), J´=5 – EVv´=8,J´=5 = 17.14 cm-1 4) EVv=10, J´=3 – Egv=0 J´=3 = 106.64 cm-1 agust,heima,...january09/Profun-080209ak.pxp <= agust, heima, ...january09Term values for triplet paper-080209kmak.xls

  3. DW=0 homogeneous triplet triplet D1P1(v´=0) DW=0 homogeneous triplet singlet g3Sm1 DW=0 homogeneous triplet triplet DW=1 / heterogeneous triplet triplet DW=1 / heterogeneous triplet singlet agust, heima, ...january09/Term values for triplet paper-110209kmak.xls

  4. agust,heima,...january09/Profun-100209ak.pxp <= agust, heima, ...january09Term values for triplet paper-100209kmak.xls

  5. DE f3D1 – D1P1 J´ agust,heima,...january09/Profun-100209ak.pxp <= agust, heima, ...january09Term values for triplet paper-100209kmak.xls

  6. DE f3D1 – g3S+1 J´ agust,heima,...january09/Profun-100209ak.pxp <= agust, heima, ...january09Term values for triplet paper-100209kmak.xls

  7. DE f3D1 - V,v´=9 J´ agust,heima,...january09/Profun-100209ak.pxp <= agust, heima, ...january09Term values for triplet paper-100209kmak.xls

  8. DE g3S+1 - V,v´=9 J´ agust,heima,...january09/Profun-110209ak.pxp <= agust, heima, ...january09Term values for triplet paper-110209kmak.xls

  9. DE f3D1- g3S-0 J´ agust,heima,...january09/Profun-100209ak.pxp <= agust, heima, ...january09Term values for triplet paper-100209kmak.xls

  10. DE f3D1- g3S-1 J´ agust,heima,...january09/Profun-100209ak.pxp <= agust, heima, ...january09 / Term values for triplet paper-100209kmak.xls

  11. The question arose whether the “New state” (assigned as g3S+(1) from Q lines) could simply be Q lines for the f3D1 <-<-X1S+ ??? Factors which favour that are: 1) Term values for “New state” (derived from Q lines; Term values for triplet paper-110209kmak.xls) are close to that for f3D1 derived from S lines (see slides 3, 7 And 8 above) 2) B´s are similar: B´(“New state” ) 10.26 cm-1; B´(f3D1) = 10.293 cm-1 3) n0´s are similar: n0(New state) = 82521.2 cm-1; f3D1) = 82523.65 cm-1 Arguments agains it (from KM): 1) Although difference in term values is small it is significant and simultaneous simulation of line positions for Q lines in the “New state” spectrum and line positions for S lines in the f3D1 <-<-X1S+ spectrum can not be done for a unique set of B´(and D´) values: Thus if the S lines are fitted the position of the Q lines will be at higher cm-1 and close to the Q line near 82523.65 cm-1 which Green et al assigned as the Q line peak for the f3D1 <-<-X1S+ spectrum. 2) The single peak at 82523.65 cm-1 which Green et al. assigned as the Q line peak can not be assigned to any other nearby system which favours the Greens assignment.

  12. 3) A single peak for a Q line serie is obtained for DW = 1 (i.e. For W´ = 1 (W´´=0), whereas different shapes are obtained for DW = 0 and DW = 2, roughly: a) DW=0 DW=1 DW=2 Hunds case c DL=0 DL=1 DL=2 Hunds cas a-b 3S+(1) assuming Hunds case (b) 3D(1) assuming Hunds case (c) 3D(1) assuming Hunds case (a) 3S+(1) assuming Hunds case (c) Most likely a) Shape closest to that observed for “new state” 4) Good fit was obtained for P and R lines using “the other Set” Of B´and D´values derived by Green et al. 5) The intensity of the Q-line peaks is way to small for them to belong to the f 3D(1) state. Simply put, the intensity of the known R and S lines is too high compared to the proposed Q-lines to fit. However, the intensity fits nicely assuming the single peak (82523.65 cm-1 )being the Q-line in the f 3D(1) state.

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