Many-Electron AtomsThornton and Rex, Ch. 8In principle, can now solve Sch. Eqn for anyatom.In practice, -> Complicated!Goal--To explain properties of elements fromprinciples of quantum theory(without exact solutions)• Elements distinguished by nuclearcharge Z (= number of electrons)• To first approx., each electron moves inelectric field of nucleus + remainingelectrons:electron state : (n, l, ml, ms)• Principles for filling electron states:1) Always fill lowest energy state first.2) No two electrons can have samequantum numbers (n, l, ml, ms).Pauli Exclusion PrincipleNo two electrons can occupy the samequantum state.shellsubshelln labels energy, but no simple formula.l subshells no longer completely degenerate.Building up atomic structure of atomsn l mlmsHydrogen 1 0 0 ±1/2Helium 1 0 0 +1/21 0 0 -1/2Helium has a closed shell.For Lithium, now add n=2 electron,but l =0 or l =1?Smaller l always has lower energy.Lithium 1 0 0 +1/21 0 0 -1/22 0 0 ±1/2n l mlmsHydrogen 1 0 0 +1/2Helium 1 0 0 -1/2Lithium 2 0 0 +1/2Beryllium 2 0 0 -1/2Boron 2 1 -1 +1/2Carbon 2 1 0 +1/2Nitrogen 2 1 +1 +1/2Oxygen 2 1 -1 -1/2Flourine 2 1 0 -1/2Neon 2 1 +1 -1/2Sodium 3 0 0 +1/2Magnesium 3 0 0 -1/2Aluminum 3 1 -1 +1/2Silicon 3 1 0 +1/2Phosporus 3 1 +1 +1/2Sulfur 3 1 -1 -1/2Chlorine 3 1 0 -1/2Argon 3 1 +1 -1/2Potassium 4 0 0 +1/2(Last Electron Added)Chemical properties of elementsElectrons in outermost, largest n orbitsare most weakly bound. They determinethe chemical properties of the elements.Elements with similar electron structurehave similar properties.• Inert or Noble GasesClosed p subshell (s for He).He (1s2), Ne (2s22p6), Ar (3s23p6)• AlkalisHave single electron electron outsideclosed shell.Li (2s1), Na (3s1), K (4s1)• HalogensAre one electron short of a closed shell.F (2s22p5), Cl (3s23p5)Total Angular MomentumConsider a 1-electron atom (or with just1 electron outside closed shell).It has Orbital Angular momentum L and Spin Angular momentum S.These can be combined to giveTotal Angular momentum J = L + S .J is quantized withJ = √j(j+1) handJz = mj hwhere j = l ± sor j = l ± 1/2 (since s = 1/2)j will be half-integral (1/2, 3/2, 5/2, ...)mj will also be half-integral, ranging from-j to j.Example: l =1, s=1/2ml = (1,0,-1) ms = (-1/2,+1/2)3•2 = 6 statesCan combine intoj = 3/2 = 1 + 1/2mj = (-3/2,-1/2,+1/2,+3/2) (4 states)orj = 1/2 = 1 - 1/2mj = (-1/2,+1/2) (2 states)Total number of j-states is 6 = 4 + 2.Spectroscopic notationnLjExamples: 2S1/2 3P3/2 etc.Principle QuantumNumberOrbital Angular Momentum LetterTotal Angular Momentum NumberSpin-Orbit Coupling• Recall, coupling of spin to a magneticfield shifts the energy (VB = -ms• B).• Motion of electron produces an“internal” magnetic field.So there is an additional contribution tothe energy: VSL = -ms• Bint VSL µ S • LProportionalto -SProportionalto LThis is the Spin-Orbit Coupling: VSL µ S • LNow states with definite energy do nothave unique L and S quantum numbers(ml, ms). We must use J quantumnumbers (j, mj).States with j = l - 1/2 have slightly lessenergy than states with j = l + 1/2 .2P3/22P2P1/2(States with different mj are stilldegenerate for each j.)Selection RulesAllowed transitions:• lifetimes t ~ 10-9 secDn = anything, Dl = ±1,Dj = 0, ±1, Dmj = 0, ±1Forbidden transitions:• lifetimes much longerEx. 2s Æ 1s, t ~ 1/7 secn l j mjn’ l‘ j’ mj’emittedphotonMany-Electron AtomsA careful analysis involving L and S inmulti-electron atoms is very complicated.Hund’s Rules(Empirical rules for filling a subshell,while minimizing the energy)1) The total Spin should be maximized(without violating Pauli ExclusionPrinciple).2) Without violating Rule 1, the OrbitalAngular momentum should also bemaximized.Handwaving explanation:Electrons repel each other, so we wantthem as far from each other as possible.1) If spins of two electrons are aligned(for maximum S), then Pauli ExclusionPrinciple says they must havedifferent L orbits. They will tend tobe farther apart.2) If the L orbits are aligned (althoughwith different magnitudes), then theelectrons will travel around thenucleus in the same direction, so theydon’t pass each other as often.Example:A d subshell (l =2) can contain 10 electrons.ms=+1/2ml =+2 -1/2 +1/2ml =+1 -1/2 +1/2ml =0 -1/2 +1/2ml =-1 -1/2 +1/2ml =-2 -1/2Example:A d subshell (l =2) can contain 10 electrons.ms=+1/2ml =+2 -1/2 +1/2ml =+1 -1/2 +1/2ml =0 -1/2 +1/2ml =-1 -1/2 +1/2ml =-2 -1/254321Put first 5 electrons all with spin in same direction. First 2 electrons have ml =-2 and -1, etc.Many-Electron AtomsFor many-electron atoms there is noworbit-orbit and spin-spin interactions,in addition to spin-orbit interactions.Consider simplest case of 2 electronswith L1, S1 and L2, S2.Only “good” quantum number isassociated with total angularmomentumJ = L1+L2+S1+S2 .(By “good”, I mean states with definiteenergy have definite j and mj. )How can we describe atom to bestunderstand energy levels?LS, or Russell-Saunders, CouplingFor most atoms the spin-orbit coupling isrelatively weak. Then it makes sense toadd the angular momentum in steps:First, L = L1 + L2S = S1 + S2Then J = L + SFor 2 electrons the Total Spin QuantumNumber S is = 0 (spins anti-parallel) or = 1 (spins parallel).The Total Orbital Angular MomentumQuantum Number L is an integer in therange between |l1 - l2| and |l1 + l2|.The Total Angular MomentumQuantum Number J is an integer in therange between |L - S| and |L + S|.Note that for S=0, there is 1 value ofJ, given by J=L. This state is called aSinglet.For S=1, there are 3 values of J, given byJ=L-1, J=L, J=L+1. These states arecalled a Triplet.In general, the multiplicity of the statesis given by (2S+1).The Spectroscopic notation isn(2S+1)LJExample:2 electrons, one in 4p, other in 4d.I.e., n=4, l1 =1, s1=1/2 l2 =2, s2=1/2Possible values of S:S=0 or S=1Possible values of L:L=1, 2, or 3Possible values of J:for singlet (S=0): J=Lfor triplet (S=1): J=L-1or J=Lor J=L+1Use Hund’s rules to order the energies.4p4dUse Hund’s rules to order the energies.4p4dS=0S=11P1D1F3P3D3F1P11D21F33P23P13P03D33D23D13F43F33F2Maximize SMaximize LMinimize J(Spin-Orbit ~S•L)Example:Helium 1s2l1 =0, s1=1/2l2 =0, s2=1/2Possible values of S=0,1Possible values of L=0Possible values of J=0,1States: 1S0, 3S1not allowed by Pauli Exclusion(requires both electrons all same QN’s)If one electron