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solution-probset1 - Chm Qtlo SON-n ~ M M at I l 573:9 4kg...

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Unformatted text preview: Chm Qtlo SON-n +~ M. M at I l 573:9 4kg mdn’vhc 47¢“ and 1M, ”me “A" We we 1w) [incarflfl {,myfini, 7,4113 J'mmflcw ,1~L¢_4e rUmwc/o Wm? , 7, 7, war/m2 x194 —.v. we Hz 7 3 TM: 3c :Ibnlfimfla 14!! 4%.th ({Pd’c‘ SUMNEM ’f MHZ, so 1k "‘d‘““k umdh., .2.Cl1.J—) Ox). 'C -; (A). x (141»!!4'630.) =, 'L .. _._____7° = or "J” W' " ' 'me'xu‘é sec [’37 X! ”h’é‘c‘ Ma. % g: 207.2 k”! , Udna f]; 02.1) 0” F‘U‘f CRWW- :90” = “32031. SW 1) , 75);)" a fi aflmwahon, :;. 2.0.[Aa—u2Lr—3_= la ,LHE =£~lo.lié: WIN? d 20. .1 km 9.. (m) [me up M3~ _p?vw%_gg_4/zsw - ? fem 29-7 fwéo. 1:: y'fff‘j' "7 ~ {‘5’ =VQ:‘~=M a w ‘1; 2—62: fig 6qu6.—~—~2r/«:~~;r =* a» m In. A 7 77 777W 7—0 2. e , Am%:[email protected]’<-Cw)= Th5: i:+« mvdayfAvlern—sslmmflwwhe robs}: 49.!!an “to fab. 4": ”ML P‘Mr; :rhett V mm“ k 9» H95 rix’tf rnk aru'ra_ 4 :(IUF) Mrficg 7%} me Join Pink. 6Y9,M.a,.favt;£\»% H YQfVUEnh-hwof 7‘19 Snooflx curve, Lem +‘ta an WM *0 0°?“de +¢ one '10 A: ‘fuur £0.5ij Er' (LL?) ’5‘ PPM {J'd‘ooMflz 9 (Px’odlx (3x196 ¢ [1000 H2 7/): W 4 _fi¢;1.w:£esr I Hagan? 7M eth ,rfreinrra/‘em {reimbt . 79 (haul-— I‘m-Wmflh, is M ”-6oaoHé . to féoaoHe. 'Armm' him/«1,43: , 7,. _J__.= “3 7L3 ’ 24:; mm”; [email protected] . , W, 7V“ / , a; A A 7n. (3.4). £114,747».er “194.00 z. ”-0441431431 In) (MO) 7 ,AWA _, 2 W U}: 3938‘ + Vééj‘z‘lfiifi, z , £3!ny +%V511%fi (£2; sill-2%, + '21: V31 3 1““ > AW, flees)Esuég'waiacfinacgml“pgfilm. We um yam VJ“ ”Jy-Jengviéufl): , . if H ,, f : W9 fixgwgwmad'wflflepvifl'wwwm ,- ' Joa— Mw-k] fit »J#LvC1u--$i»3-ACJu-§§é 7?“ SM ‘25 55.1%?th ' 7 , ‘*H¢‘r3fl=méldo®9 :‘Qa EMOT gmyflisguwgeg,» W . —-3 3', lm)kmenfln—f£fiw IMH‘ enerya (8% 3.5 Reproducing Table 3.2 on p. 40 for 110,1 = —'10() Hz, v03 = ~200 Hz and .112 = 5 Hz: ______—________—_________.__———————-———-————— number m, m; spin states eigenfunction eigenvalue/Hz l +% +-£- (m l/lmldlmg +1§U0J + %UO,3 + 44-1113 = —l48.75 2 4% ‘i 013 $11.11/I/m +%110.1 — %U0,2 - i112 = 48-75 3 *3: 4% .50 WAWM £110.] + #112 — i112 = -51-25 4 —% —% 16.6 mil/1.2 ‘lUm — ém + 1112 : 151.25 ___,_________-_..—__———-———————-————Af__. The set of allowed transitions is: transition spin states frequency/Hz 1—9 2 (m —9 (113 E2 — E1 = 197.50 3 —+ 4 Ba —>fiB E4 - E3 = 202.50 1—9 3 cm —-> [311/ E3 — E1 = 97.50 2—>4 aB—afifl E4—E2 : 102.50 spin 1 flips spin 1 (1 B spin 2 01 [3 spin 2 flips 13 24 12 34 W 80 100 120 140 160 180 200 220 ——~—.—1> frequency I Hz If J12 = —5 Hz, the table of energies becomes: number m] m; s in states ei enfunction ei envalue/Hz P g g 1 +% +% at! wmilfiag +%U0,1+%U0_2 + $112 2 ~151.25 . 1 1 1 1 1 2 +5 ‘3 0/3 Want/1m +§llo,1 — 5110.2 - 1J12 = 51.25 1 1 1 1 1 3 ‘5 +3 ,3“ 11/111111"; ”5110.1 + 51102 - 3-712 = ‘43-75 1 1 1 1 1 4 ‘5 *3 13/3 Wank/32 ~§v0.1 - 5110.2 + 2112 = 148-75 spin 1 flips spin 1 B (1 spin 2 l3 1): spin 2 flips 24 13 34 12 WM 80 100 120 140 160 180 200 220 —> frequency / Hz The spectrum in unchanged in appearance. However, the labels of the lines have changed; the spi state of the passive spin for each line of the doublet has swapped over. M 7 ,7 ...
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