Atomic Physics VI. Greg Anderson Department of Physics & Astronomy. December Northeastern Illinois University

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1 Atomic Physics VI Greg Anderson Department of Physics & Astronomy Northeastern December 2010 c G. Anderson Modern Physics I slide 1 / 30

2 Overview c G. Anderson Modern Physics I slide 2 / 30

3 Atomic Hydrogen Hydrogen Energy Levels Rydberg Formula c G. Anderson Modern Physics I slide 3 / 30

4 Atomic Hydrogen Atomic Hydrogen Hydrogen Energy Levels Rydberg Formula Schrödinger s Equation 2 2m 2 ψ nlml m s k Ze2 r ψ nlm l m s = E n ψ nlml m s Hydrogen Wave Functions ψ nlml m s (r, θ, φ) = R nl (r)y lml (θ, φ)χ ms Lowest Order Bound State Energies E n = 1 2 mc2z2 α 2 n 2 = Z2 (13.6 ev) n 2 c G. Anderson Modern Physics I slide 4 / 30

5 Hydrogen Energy Levels (n, l, m l, m s ) l = 0 l = 1 l = 2 l = 3 l = 4 l = 5 n = 4 n = 3 2(1) 2(3) 2(5) n = 2 2(1) 2(3) l = 0, 1,...,n 1. (n) m l = 0, ±1,...,±l. (2l + 1) m s = ± 1. (2) 2 Allowed transitions: m = 0, ±1, l = ±1 n = 1 2(1) Hydrogen Applet c G. Anderson Modern Physics I slide 5 / 30

6 The Rydberg-Ritz Formula (1908) Atomic Hydrogen Hydrogen Energy Levels Rydberg Formula m n ( 1 1 = R λ mn m 1 ), (m < n < ) 2 n 2 R = m Rydberg constant 3 5 n m E γ = hf = hc λ = E n E m ( 1 E n E m = 13.6 ev m 1 ) 2 n 2 c G. Anderson Modern Physics I slide 6 / 30

7 J L + S J 1 + J 2 Spectroscopic Notation c G. Anderson Modern Physics I slide 7 / 30

8 Total J J L + S J 1 + J 2 Spectroscopic Notation Total J = L + S Total angular momentum quantum numbers: where J 2 = j(j + 1) 2, J z = m j m j = j, j + 1,...,j 1, j Q: Given l and s, what are the possible values of j? c G. Anderson Modern Physics I slide 8 / 30

9 Addition of J L + S J 1 + J 2 Spectroscopic Notation Total J = L + S For the quantum mechanical addition of two angular momentum vectors: j = l + s, l + s 1,..., l s For s = 1 2 j = l + s, l s c G. Anderson Modern Physics I slide 9 / 30

10 Addition of (II) J L + S J 1 + J 2 Spectroscopic Notation Adding J = J 1 + J 2 J 2 1 = j 1 (j 1 + 1) 2 J 2 2 = j 2 (j 2 + 1) 2 J 2 = j(j + 1) 2 The resulting quantum number j may take on several possible values: j = j 1 + j 2, j 1 + j 2 1,..., j 1 j 2 c G. Anderson Modern Physics I slide 10 / 30

11 Spectroscopic Notation J L + S J 1 + J 2 Spectroscopic Notation Shells n K L M N O Subshells l s p d f g h Historical spectroscopic naming: s = sharp, p = principal, d = diffuse, f = fundamental For a l = 1 state: Hydrogen States: n 2s+1 P j 1 2 S 1/2, 2 2 S 1/2, 2 2 P 1/2, 2 2 P 3/2,... c G. Anderson Modern Physics I slide 11 / 30

12 Spin-Orbit Coupling (Fine Structure) Relativistic Corrections Hydrogen Fine Structure c G. Anderson Modern Physics I slide 12 / 30

13 Spin-Orbit Coupling () B L Spin-Orbit Coupling (Fine Structure) Relativistic Corrections Hydrogen Fine Structure r p v Electron Rest Frame: Spin magnetic moment interacts with magnetic field produced by proton. U = µ B = µ z B = g s µ B Bm s 2P 3/2 B L S µ 2P B L S µ 2P 1/2 Hydrogen 2P splitting: E = ev c G. Anderson Modern Physics I slide 13 / 30

14 Relativistic Corrections Spin-Orbit Coupling (Fine Structure) Relativistic Corrections Hydrogen Fine Structure Recall the Lorentz factor: γ = 1/ 1 v 2 /c 2 ) The relativistic kinetic energy is: ( E k = (γ 1)mc 2 = In terms of the momentum p: v 2 c ) v 4 c E k = p2 2m p4 8mc 2 + c G. Anderson Modern Physics I slide 14 / 30

15 Hydrogen Spin-Orbit Coupling (Fine Structure) Relativistic Corrections Hydrogen Fine Structure Fine structure constant Bound state energies α = ke2 c = E n = 1 2 m ec 2Z2 α 2 n 2 e2 4πǫ 0 c = Z2 (13.6 ev) n 2 + Spin-Orbit + Relativistic + Darwin [ ( E nj = E n 1 + (Zα)2 n n 2 j )] c G. Anderson Modern Physics I slide 15 / 30

16 Spin-Orbit Coupling (Fine Structure) Relativistic Corrections Hydrogen Fine Structure At high resolution, hydrogen spectral lines contain closely-spaced doublets. U = µ z B 2P 1S 2P 3/2 2P 1/2 B L S µ B L S µ U = 2µ B B e c G. Anderson Modern Physics I slide 16 / 30

17 Schrödinger s Eq. for 2 or more ptls. Pauli Exclusion Spin & Exchange Symmetry c G. Anderson Modern Physics I slide 17 / 30

18 Schrödinger s Eq. for 2 or more ptls. Schrödinger s Eq. for 2 or more ptls. Pauli Exclusion Spin & Exchange Symmetry Two ψ nm (x 1, x 2 ) = ψ n (x 1 )ψ m (x 2 ) For identical particles ψ(x 1, x 2 ) 2 = ψ(x 2, x 1 ) 2 Two Possibilities ψ S = 1 2 [ψ n (x 1 )ψ m (x 2 ) + ψ n (x 2 )ψ m (x 1 )] ψ A = 1 2 [ψ n (x 1 )ψ m (x 2 ) ψ n (x 2 )ψ m (x 1 )] c G. Anderson Modern Physics I slide 18 / 30

19 Pauli Exclusion Principle Schrödinger s Eq. for 2 or more ptls. Pauli Exclusion Spin & Exchange Symmetry Pauli Exclusion Principle: No two fermions can occupy the same quantum state The electron is a fermion: s = 1/2. c G. Anderson Modern Physics I slide 19 / 30

20 Spin & Exchange Symmetry Schrödinger s Eq. for 2 or more ptls. Pauli Exclusion Spin & Exchange Symmetry Bosons: ψ(r 1,r 2 ) = +ψ(r 2,r 1 ) Fermions: ψ(r 1,r 2 ) = ψ(r 2,r 1 ) Pauli exclusion principle: No more that one electron (identical fermion) may occupy a quantum state specified by the same set of single particle quantum numbers n, l, m l, m s Active Learning: Prove the Pauli exclusion principle given that the total wave function for fermions is antisymmetric under particle exchange. c G. Anderson Modern Physics I slide 20 / 30

21 Hydrogen Energy Levels Relative Energy Levels Multi-Electron Atoms Radial Probability Filling Shells & Subshells Periodic Periodic Ionization Energy Atomic Radii Rough Estimates c G. Anderson Modern Physics I slide 21 / 30

22 Hydrogen Energy Levels (n, l, m l, m s ) l = 0 l = 1 l = 2 l = 3 l = 4 l = 5 n = 4 n = 3 2(1) 2(3) 2(5) n = 2 2(1) 2(3) l = 0, 1,...,n 1. (n) m l = 0, ±1,...,±l. (2l + 1) m s = ± 1. (2) 2 Allowed transitions: m = 0, ±1, l = ±1 n = 1 2(1) Hydrogen Applet c G. Anderson Modern Physics I slide 22 / 30

23 Relative Energy Levels Multi-Electron Atoms Hydrogen Energy Levels Relative Energy Levels Multi-Electron Atoms Radial Probability Filling Shells & Subshells Periodic Periodic Ionization Energy Atomic Radii Rough Estimates 4p 3d 4s 1s 3p 3s 2p 2s c G. Anderson Modern Physics I slide 23 / 30

24 Radial Probability: 4πr 2 ψ ψ n = 1, l = 0 Bohr Model: r n = n 2 a 0 Z Probability n = 2, l = 1 n = 2, l = r/a 0 c G. Anderson Modern Physics I slide 24 / 30

25 Filling Shells & Subshells Hydrogen Energy Levels Relative Energy Levels Multi-Electron Atoms Radial Probability Filling Shells & Subshells Periodic Periodic Ionization Energy Atomic Radii Rough Estimates 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 4d... 6s... Mnemonic Z < 56 (Ba) (1s) (2s) (2p) (3s) (3p) (4s) (3d) (4p) (5s) (4d) (5p) (6s) Z Elem configuration 1 H (1s) 2 He (1s) 2 3 Li [He](2s)... 9 F [He](2s) 2 (2p) 5 10 Ne [He](2p) 6 11 Na [Ne](4p)... c G. Anderson Modern Physics I slide 25 / 30

26 c G. Anderson Modern Physics I slide 26 / 30 H 1 He 2 Li 3 Be 4 B 5 C 6 N 7 O 8 F 9 Ne 10 Na 11 Mg 12 Al 13 Si 14 P 15 S 16 Cl 17 Ar 18 K 19 Ca 20 Sc 21 Ti 22 V 23 Cr 24 Mn 25 Fe 26 Co 27 Ni 28 Cu 29 Zn 30 Ga 31 Ge 32 As 33 Se 34 Br 35 Kr 36 Rb 37 Sr 38 Y 39 Zr 40 Nb 41 Mo 42 Tc 43 Ru 44 Rh 45 Pd 46 Ag 47 Cd 48 In 49 Sn 50 Sb 51 Te 52 I 53 Xe 54 Cs 55 Ba Hf 72 Ta 73 W 74 Re 75 Os 76 Ir 77 Pt 78 Au 79 Hg 80 Tl 81 Pb 82 Bi 83 Po 84 At 85 Rn 86 Fr 87 Ra Unq 104 Unp 105 Unh 106 Uns 107 Uno 108 Une 109 Uun 110 La 57 Ce 58 Pr 59 Nd 60 Pm 61 Sm 62 Eu 63 Gd 64 Tb 65 Dy 66 Ho 67 Er 68 Tm 69 Yb 70 Lu 71 Ac 89 Th 90 Pa 91 U 92 Np 93 Pu 94 Am 95 Cm 96 Bk 97 Cf 98 Es 99 Fm 100 Md 101 No 102 Lr 103

27 Electron configuration, nl, of outer most elecrons. 1s 1 1s 2 2s 1 2s 2 2p 1 2p 2 2p 3 2p 4 2p 5 2p 6 3s 1 3s 2 3p 1 3p 2 3p 3 3p 4 3p 5 3p 6 4s 1 4s 2 fill ten 3d states: 4s a 3d b 4p 1 4p 2 4p 3 4p 4 4p 5 4p 6 5s 1 5s 2 fill ten 4d states: 5s a 4d b 5p 1 5p 2 5p 3 5p 4 5p 5 5p 6 6s 1 6s 2 fill ten 5d states, fourteen 4f states 6p 1 6p 2 6p 3 6p 4 6p 5 6p 6 Subshell contains 2(2l + 1) states Lanthanide Series 4f states. Actinide Series 5f states c G. Anderson Modern Physics I slide 27 / 30

28 Northeastern Ionization Energy (ev) vs. Z He Ne H Li Na Ar Kr Xe K Rb Cs Rn c G. Anderson Modern Physics I slide 28 / 30

29 Northeastern Atomic Radii (pm) H Li Na K Rb c G. Anderson Modern Physics I slide 29 / 30

30 Rough Estimates For Single electron atoms E n = (13.6 ev)z 2 /n 2 Z = 2 Helium, 1s 2, E I = 24.5 ev. Screened to Z = 1 half the time: n = 1, Z eff 1.5 vs E I 13.6 ev (1.5)2 1 2 = 30.6 ev Z = 3 Lithium, 1s 2 2s 1, E I = 5.4 ev. For 2s electron, two protons well screened by the 1s 2 electrons: n = 2, Z eff 1 vs E I 13.6 ev (1)2 2 2 = 3.4 ev Z = 4 Beryllium, 1s 2 2s 2, E I = 9.3 ev. For 2s 2 electrons, two protons screened by 1s 2 electrons, a third proton is screened half the time by other 2s electron. n = 2, Z eff = 1.5 vs E I 13.6 ev (1.5)2 2 2 = 7.6 ev c G. Anderson Modern Physics I slide 30 / 30

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