____________________________________________________________________________ | | | QSTARK - Fully diagonalizing fit of Stark shifts in a rotor | | with zero or one quadrupolar nuclei | |____________________________________________________________________________| version 4.X.2017 Zbigniew KISIEL ------------------------------------------------------------------------------ Ar3...H35Cl ------------------------------------------------------------------------------ Calculation for J and F exceeding the value in data by at least 1 The fit will be made to TRANSITION FREQUENCIES 27 Lines read in 26 Lines fitted ---> 5 field free and 21 at non-zero field 1 Constants fitted TRANSITIONS (F and MF in units of 1/2): J K K <- J K K F MF<- F MF Volts Obs Obs-Calc Calc Field Field**2 -1 +1 -1 +1 1. 5 0 5 4 0 4 13 1 11 1 0.0 8437.96890 -0.00115 8437.9700 0.0000 0.00 2. 5 0 5 4 0 4 13 1 11 1 10032.0 8437.89690 0.00105 8437.8958 372.5214 138772.16 3. 5 0 5 4 0 4 13 1 11 1 12632.0 8437.85060 -0.00165 8437.8523 469.0680 220024.75 4. 5 0 5 4 0 4 13 3 11 3 12632.0 8437.88610 0.00295 8437.8832 469.0680 220024.75 5. 5 0 5 4 0 4 13 7 11 7 12632.0 8438.03830 -0.00155 8438.0399 469.0680 220024.75 6. 5 0 5 4 0 4 13 7 11 7 10032.0 8438.01570 0.00077 8438.0149 372.5214 138772.16 7. 5 0 5 4 0 4 13 9 11 9 10032.0 8438.09510 -0.00043 8438.0955 372.5214 138772.16 8. 5 0 5 4 0 4 11 1 9 1 0.0 8437.96890 -0.00090 8437.9698 0.0000 0.00 9. 5 0 5 4 0 4 11 1 9 1 10032.0 8437.91340 -0.00434 8437.9177 372.5214 138772.16 10. 5 0 5 4 0 4 11 1 9 1 12632.0 8437.88610 -0.00164 8437.8877 469.0680 220024.75 11. 5 0 5 4 0 4 11 5 9 5 10032.0 8437.99780 0.00090 8437.9969 372.5214 138772.16 12. 5 0 5 4 0 4 11 5 9 5 12632.0 8438.01940 -0.00018 8438.0196 469.0680 220024.75 13. 5 0 5 4 0 4 11 7 9 7 10032.0 8438.06770 -0.00229 8438.0700 372.5214 138772.16 14. 5 0 5 4 0 4 9 1 7 1 0.0 8437.43630 -0.00040 8437.4367 0.0000 0.00 15. 5 0 5 4 0 4 9 1 7 1 12632.0 8437.36550 -0.00374 8437.3692 469.0680 220024.75 16. 5 0 5 4 0 4 9 3 7 3 10032.0 8437.42330 -0.00717 8437.4305 372.5214 138772.16 17. 5 0 5 4 0 4 9 5 7 5 10032.0 8437.50250 -0.00051 8437.5030 372.5214 138772.16 18. 5 0 5 4 0 4 9 5 7 5 12632.0 8437.53670 -0.00195 8437.5387 469.0680 220024.75 19. 5 0 5 4 0 4 7 1 5 1 0.0 8437.43630 -0.00126 8437.4376 0.0000 0.00 20. 5 0 5 4 0 4 7 1 5 1 12632.0 8437.34700 -0.00063 8437.3476 469.0680 220024.75 21. 5 0 5 4 0 4 7 3 5 3 10032.0 8437.47580 0.00177 8437.4740 372.5214 138772.16 22. 5 0 5 4 0 4 7 3 5 3 12632.0 8437.49340 -0.00037 8437.4938 469.0680 220024.75 23. 5 3 3 4 3 2 13 1 11 1 0.0 8438.93270 0.00215 8438.9305 0.0000 0.00 24. 5 3 2 4 3 1 13 1 11 1 0.0 8438.93270 0.00215-- 8438.9305 0.0000 0.00 25. 5 3 3 4 3 2 13 1 11 1 8466.4 8438.64760 -0.00030 8438.6479 314.3854 98838.21 26. 5 3 2 4 3 1 13 1 11 1 8466.4 8438.60240 -0.00121 8438.6036 314.3854 98838.21 27. 5 3 3 4 3 2 13 1 11 1 12632.0 8438.73670 0.00059 8438.7361 469.0680 220024.75 Standard deviation = 0.002252 ITERATION NO = 1 CONSTANTS, deviations and changes: Mu.a = 0.982814241 +- 0.002907114 0.000074241 TRANSITIONS (F and MF in units of 1/2): J K K <- J K K F MF<- F MF Volts Obs Obs-Calc Calc Field Field**2 -1 +1 -1 +1 1. 5 0 5 4 0 4 13 1 11 1 0.0 8437.96890 -0.00115 8437.9700 0.0000 0.00 2. 5 0 5 4 0 4 13 1 11 1 10032.0 8437.89690 0.00106 8437.8958 372.5214 138772.16 3. 5 0 5 4 0 4 13 1 11 1 12632.0 8437.85060 -0.00164 8437.8522 469.0680 220024.75 4. 5 0 5 4 0 4 13 3 11 3 12632.0 8437.88610 0.00296 8437.8831 469.0680 220024.75 5. 5 0 5 4 0 4 13 7 11 7 12632.0 8438.03830 -0.00156 8438.0399 469.0680 220024.75 6. 5 0 5 4 0 4 13 7 11 7 10032.0 8438.01570 0.00076 8438.0149 372.5214 138772.16 7. 5 0 5 4 0 4 13 9 11 9 10032.0 8438.09510 -0.00044 8438.0955 372.5214 138772.16 8. 5 0 5 4 0 4 11 1 9 1 0.0 8437.96890 -0.00090 8437.9698 0.0000 0.00 9. 5 0 5 4 0 4 11 1 9 1 10032.0 8437.91340 -0.00433 8437.9177 372.5214 138772.16 10. 5 0 5 4 0 4 11 1 9 1 12632.0 8437.88610 -0.00163 8437.8877 469.0680 220024.75 11. 5 0 5 4 0 4 11 5 9 5 10032.0 8437.99780 0.00089 8437.9969 372.5214 138772.16 12. 5 0 5 4 0 4 11 5 9 5 12632.0 8438.01940 -0.00019 8438.0196 469.0680 220024.75 13. 5 0 5 4 0 4 11 7 9 7 10032.0 8438.06770 -0.00231 8438.0700 372.5214 138772.16 14. 5 0 5 4 0 4 9 1 7 1 0.0 8437.43630 -0.00040 8437.4367 0.0000 0.00 15. 5 0 5 4 0 4 9 1 7 1 12632.0 8437.36550 -0.00373 8437.3692 469.0680 220024.75 16. 5 0 5 4 0 4 9 3 7 3 10032.0 8437.42330 -0.00717 8437.4305 372.5214 138772.16 17. 5 0 5 4 0 4 9 5 7 5 10032.0 8437.50250 -0.00052 8437.5030 372.5214 138772.16 18. 5 0 5 4 0 4 9 5 7 5 12632.0 8437.53670 -0.00197 8437.5387 469.0680 220024.75 19. 5 0 5 4 0 4 7 1 5 1 0.0 8437.43630 -0.00126 8437.4376 0.0000 0.00 20. 5 0 5 4 0 4 7 1 5 1 12632.0 8437.34700 -0.00062 8437.3476 469.0680 220024.75 21. 5 0 5 4 0 4 7 3 5 3 10032.0 8437.47580 0.00176 8437.4740 372.5214 138772.16 22. 5 0 5 4 0 4 7 3 5 3 12632.0 8437.49340 -0.00038 8437.4938 469.0680 220024.75 23. 5 3 3 4 3 2 13 1 11 1 0.0 8438.93270 0.00215 8438.9305 0.0000 0.00 24. 5 3 2 4 3 1 13 1 11 1 0.0 8438.93270 0.00215-- 8438.9305 0.0000 0.00 25. 5 3 3 4 3 2 13 1 11 1 8466.4 8438.64760 -0.00031 8438.6479 314.3854 98838.21 26. 5 3 2 4 3 1 13 1 11 1 8466.4 8438.60240 -0.00122 8438.6036 314.3854 98838.21 27. 5 3 3 4 3 2 13 1 11 1 12632.0 8438.73670 0.00056 8438.7361 469.0680 220024.75 Standard deviation = 0.002252 ITERATION NO = 2 CONSTANTS, deviations and changes: Mu.a = 0.982814259 +- 0.002906848 0.000000018 FINAL RESULTS OF LEAST SQUARES FITTING PROCEDURE ================================================ FITTED CONSTANTS: A /MHz 852.8 1:Xab /MHz 0. B /MHz 843.897726 1:XJ.a/kHz 0. C /MHz 0. 1:XK.a/kHz 0. DJ /kHz 2.3406 1:XJbc/kHz 0. DJK /kHz 1.802 1:Ma /MHz 0. DK /kHz 0. 1:Mb-c/MHz 0. dJ /kHz 0. 1:Tr /MHz 0. dK /kHz 0. 1:Xd /kHz 0. HJ / Hz 0. 1:Xbc /MHz 0. HJK / Hz 0. 1:Xac /MHz 0. HKJ / Hz 0. HK / Hz 0. hJ / Hz 0. hJK / Hz 0. Mu.a /D 0.9828(29) hK / Hz 0. Mu.b /D 0. LKKJ /mHz 0. Mu.c /D 0. 1:Xa /MHz -31.0048 d /cm 26.93 1:Xb-c/MHz 0. k /cm 0. 1:X.bb /MHz 15.502400000000 1:X.cc /MHz 15.502400000000 ------------------------------------------------------------------------------