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LEC5 - “v.=.4 JW 40 JLW 3 ‘ w:wW mm a ¢-¢W §~ Wmhm W...

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Unformatted text preview: “v .= .4 JW% 40 +JLW+3 ‘ w:wW+mm+a ¢-¢W- §~ Wmhm W U WWI-$.52 (4.5; . " (M 1.3- Mon—T W) 714. M cm 1-; _ W W fl : W M 54,414.... __ Lit: “HOPE .L+K=HLE+A \hf‘} A Z Cf . 5;}! SI *1. '-'-' 3(5“) ‘31”; $1.11.: ”flint! m mu— [‘1 1$ “NH—I a .— _ \ W = “0051-0 - fl. 5 L=o a: a J I ' No. a"- N5 Slat-.15 I W ' ‘ (_ '15 +1 _ “id!"‘hp'hbi‘fi u 4" _ [5" W M I 5:0 i-wu. i— _ ‘525 T L20 "‘5‘: 17}. msa=t‘/,, “S“ L123- .-.—7Ms=10,0,l on S: 335;, T15’l,0,-l =5 S=| 5.45; I, .. l5\-5;' N5 ‘3 0 7?. 5:0 ' 3 Tm NW 8 S 1%“ W q Mm ..£ :7 W W m ,a.( :1 =3 7~ fl ‘3 Fr» W - W SCI: W W +441. MW 7i m ‘) 50 W 5‘34 5 / y%( liz+7t2:+> J— uznra:—7+l1:—7u-+ [4.z-712:—7 -* ("5:10 -1) 7 wwwmw¢c 5(35) 1 E(’S)‘ “’ Gum Vila: W (’m WWW-1M! '0’? 1% (Let/it) m fi('s)¢o Amt 13(35): 0 (S W 1‘ M‘Mfl’féfi; 4/» W m‘w ‘ 3:57am 4M H’W' IS #1. I? p 37 (D 36 II I .5 —»- ‘Now/wMLwaWWW/t‘wz 2,9" “WW. S=l.0 fluff}: . “‘1' ti. 0,-l .'. . U6 W“ M Ml; '5 Ila)". .+M "L: 2-,‘30, Ila.) —' l0} ‘IJ‘Z 3 3 3 l g} L: 1(l‘ly: 23'10. -'l"1) D< P< S 4 D<|P<IS L31 (“1.7-3 1.01...) - _ L=o C"...:o) m (5,2.- : :l: 1 51.1.4”. :0 AK: A£,=o 52‘2'0 59‘: Afllu- A31: 1 AL=| 5.5-: AS-él AL‘J'I AS :0 Physics 780.04 Dr. Herbst ‘ ENUMERATING THE STATES OF COMPLEX ATOMS 1. Build atomic configurations from the single-electron energy level scheme using the ordering ls < 2s < 2p < 3s < 3p < 4s < 3d and the Pauli Exclusion Principle. Example: Li (3 electrons) Lowest Configuration 1s22sl 1st Excited Configuration 1s7‘2p1 2. Configurations may contain degeneracy. Enumerate the possible values of L and S in a given configuration. Sets of L, S values correspond to states of different energy with symbols known as term symbols. 3. The possible values of L and S within a given configuration can be deduced in the following manner: a) ignore closed shell electrons (e. g. ls2 or 1s22s2 or 1s22s22p6) since they contribute only L = S = 0 b) find the number of diagrams or wA involving the open—shell electrons: W = no. of diagrams = N!/(M![N-M]!) N = 2(21+1) for an unfilled shell of angular momentum l M = ncelectrons in that shell EXAMPLES: 2p2 N = 2x3 = 6 (1:1) M = 2 W = 6!/2!4! = 15 2p3p M = 1 for each shell W=W1xW2 = 6!/1!5! x 6!/1!5! = 36 c) group the diagrams according to ML and MS and deduce the possible L and S values. EXAMPLE 1s22s12p1 — only consider 2s2p W = 6 x 2 = 12 1 _¢ ;1 _1 _ ”:§I:§::EI:EI 1311111111111 ML 1 o -1 1 o -1 1 o -1 1 0 -1 Ms 1 1 1 o o 0 -1 -1 -1 o o 0 3 1 L=1 8:1 ==> P L=1 S=O P For complex cases, try to enumerate the highest L term first followed by the term of highest multiplicity. The other terms are then easier to deduce. (d) If the Pauli Exclusion Principle need not be invoked, a simpler method is to add the 1 and s of open shell electrons. EXAMPLE: 2s2p 11 = 0 12 = 1 ===> L = 1 (L =1,+1,,11+12—1,...|1,—12|) s1: 1/2 s2 =1/2 ==> s = 1,0 L=lS=1===> 3P whileL=lS=0 ===>1P 4. If L and S are both non—zero, spin-orbit coupling occurs. The total angular momentum quantum number J = L+S, L+S-l,... |L-S| and EJ = N2 [J(J+l) — L(L+1) — S(S+1)] EXAMPLE: 3P L = 1 S = l J = 2,1,0 A > 0 3P0 < 3P1 < 3P2 (ascending energy) A < 0 3P2< 3P1< 3Po 5. States can be ordered by Hund’s Rules (based on actual calculations). Hund’s Rules state that for a given configuration, the energy ordering of the individual states is given by 3 rules (rule 1 being most important): 1. The higher the spin multiplicity, the lower the energy (e. g. 3P < 1P) 2. The higher the L, the lower the energy (e.g. 1D < 1S) 3. For open shells less than half-full, the lower the J the lower the energy (A > 0), while for open shells more than half-full, the higher the J the lower the energy (A < 0). For exactly half—full shells (e.g. p3, d5), ignore rule 3. ...
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