4πε. me 1,2,3,... 1 n. H atom 4. in a.u. atomic units. energy: 1 a.u. = ev distance 1 a.u. = Å

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Spencer Summers
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1 H atom 4 E a me =, n=,,3,... 8ε 0 0 π me e e 0 hn ε h = = 0.59Å E = me (4 πε ) 4 e 0 n n in a.u. atomic units E = r = Z n nao Z = e = me = 4πε = 0 energy: a.u. = 7. ev distance a.u. = 0.59 Å General results for hydrogenic atoms (H +, He +, Li +, U 9+ ), Z = nuclear charge

2 Note: We really should be using μ instead of m e. H, D, T have slightly different μ and thus can be distinguished spectroscopically Radial distribution function integrate over angular degrees of freedom [ ] Prdr () = r Rr () dr

3 Chapter 0 Many e - atoms He atom: + ψ = r r r r cannot separated due to term Eψ in a.u. ψ = φ φ However, it is still useful to use an approximate wavefunction that does separate ( r ) ( r ) orbital approximation Simplest approach: neglect the r term poor approximation Better approach: each e - experiences a potential from the average charge distribution of the other e - V r r r dr ( ) = + φ ( ) φ ( ) eff r r Hartree model N electrons N one-electron hamiltonians {φ I, ε i }

4 h i φ i = ε i φ i Effective one electron hamiltonian To proceed further, we must consider e - spin spin of e - = ½; two components m s = +/, m s = -/ spin eigenfunctions: α ms = β ms = sˆ α = ()( s s+ ) α = 3 α 4 sˆ β = 3 β 4

5 sˆ ; ˆ zα = ms α = α szβ = β * * * ααdσ ββdσ αβdσ = = ; = 0 Indistinguishability of electrons wavefunction must be antisymmetric wrt exchange of two e - ψ = φ He: () () [ () () () () s φ s α β β α ] r r σ σ Pauli exclusion principle ψ = () s α() () s β () () s α() () s β() = ()() s s () () () () = ss( αβ βα ) [ α β β α ]

6 Two electrons cannot have all quantum #s the same. In general: ψ (,,... n) = m= n/, if n even n + =, if n odd n! ( ) φ α() φ () β()... φ () β() φ() α() φ() β()... φm() β() φ ( n) α( n) φ ( n) β( n)... φ ( n) β( n) m Slater determinant m excited states s s singlet + triplet ψ ψ ψ 3 ψ 4 = = = = sα sβ sα sβ sα sα sα sα sβ sβ sβ sβ sβ sα sβ sα

7 s s {s()s()+s()s()}(αβ-βα) Singlet, S = 0, M s =0 {s()s()-(s()s()}αα {s()s()-s()s()}(αβ+βα) Triplet, S =, M s =, 0, - {s()s()-s()s()}ββ Normally, the three components of the triplet give the same energy Excited triple is energetically below excited singlet

Variational method: approximate wavefunction Φ HΦ EΦ E = Φ * HΦdτ * ΦΦ dτ If Φ has a parameter, b, solve E/ b= 0 example: 3 5 7 9 x x x x x α

8 Variational method: approximate wavefunction Φ HΦ EΦ E = Φ * HΦdτ * ΦΦ dτ If Φ has a parameter, b, solve E/ b= 0 example: x x x x x α Φ= a a a a a for particle-in-box problem with 0 < x < a The variational parameter is α Plot of the exact and approximate wavefunctions

9 if α = 0 E = 0.03 h ma E α 0 = 0 α = E = 0.7 α E exact h = 0.5 ma h ma The energy can never fall below the exact energy Hartree-Fock Self-Consistent Field method Φ is taken to be a Slater determinant parameters in orbitals are varied eff i + Vi ( ri) φi( ri) = εi( ri) depends on orbitals that we are trying to solve for

10 Guess a set of orbitals construct eff V i solve for orbitals + energies Iterate until energies and orbitals are converged. E 0 ε 3 ε ε = ε ( ) E J K total i ij ij i j> i empty filled exchange Coulomb to remove double counting in IP i = -ε I EA j = - ε j for filled orbitals for empty orbitals εi i Koopmans theorem

11 many electron atoms H atom ε ns < ε np < ε nd, ε ns = ε np = ε nd, deshielding s has more weight near nucleus than does p which has more weight near nucleus than d.

12 He s Ne s s p + = Ar s s p s p =8 Sc[ Ar]4s 3 d Mn[ Ar] 4s 3 d Cu[ Ar] 4s3d 5 0 Ti[ Ar]4s 3 d Fe[ Ar]4s 3 d Zn[ Ar] s 3d 6 0 V[ Ar]4s 3 d Co[ Ar]4s 3d 3 7 Energy Cr[ Ar]4s3 d Ni[ Ar]4s 3d 5 8

13 Cr 4s3 d vs. 4s 3d 5 4 Cu 4s3 d vs. 4s 3d 0 9 d 5 Especially stable Electronegativity: X = IP + EA definition due to Mulliken Does Na transfer an e - to Cl? Δ E = IPNa EACl = =.5eV

14 Many electron atoms: n, l, m l, m s not good quantum #s L=, S = s i i Lˆ = ˆ, Sˆ = sˆ z z, i z z, i i i i Lower case: individual orbitals Upper case: many electron state L, S, M L, M S are the good quantum #s Actually, for heavy elements, one needs to add a spin-orbit coupling term to Ĥ. Lˆ, Lˆ, Sˆ, Sˆ no longer commute with Ĥ z z

15 with spin-orbit coupling, need to use the good quantum #s are J, M J J = L+ S z > 40

16 Terms + States H s L= S = S 0; Term symbol S+ L He s L = S = S 0; 0 B s s p L= ; S = P multiplicity 3 C s s p L=,,0; S =,0 P, D, S N s s p 3 3,,0;,,, 4 L= S = S P D 4 3 O s s p L= S = D S,,0;,0 P,, 5 F s s p L= S = P ; 6 Ne s s p L = 0; S = 0 S Ti Ar s d L [ ]4 3 = 4,3,,,0; 0,,,,, 3 3 S = S D G P F

17 A closer look at spin He ss α β M s = 0 β α M s = 0 α α M s = β β M s = There must be a S = state. S = M s = -, 0, There must also be an S = 0 state M s = 0 ψ T = ( s s s s ) αα T T S, T ψ S = ( s s+ s s )( αβ βα ) ψ T = ( s s s s ))( αβ + βα ) ψ T = ( s s s s ) ββ

18 In the absence of a magnetic field, the three triplet components are degenerate. The T and S states are different energy E T < E s term S+ L (S+)(L+) degeneracy filled shells S : 3 3,,,, 0 C s p d L= S = F, F, D, D, P, P = 60 states p: 6 choices 3d: 0 choices 60 states

C s s p p L= S = : 3,, 0;, 0 D, D, P, P, S, S 3 3 3 3 C s p P S D

19 C s s p p L= S = : 3,, 0;, 0 D, D, P, P, S, S C s p P S D :,, what happens to 3 D, P, 3 S? violate the Pauli exclusion principle.

5 4 8 3 7 Note p, p ; p, p ; d, d ; d, d etc. give the same states for each pair. Hund s Rules:.

21 Note p, p ; p, p ; d, d ; d, d etc. give the same states for each pair. Hund s Rules:. The lowest energy term is that with the highest spin. For terms that have the same spin, that with the greatest L value lies lowest in energy 3 C: s p P < D < S

22 S. O. Coupling P P P P J Energy level diagram for C atom

23 Spin-orbit coupling adds a term LS i to H generally can ignore for light atoms 3 P 3 P 3 P 3 P 0 Hund s rule 3: If a subshell is > half full, the level with the highest J is lowest in energy. If it is < half full, the level with the lowest J is lowest in energy 3 P splits into 3 levels upon application of magnetic field

24 Return to the ss singlet/triplet problem Esinglet = ( ) ˆ ( ) s s+ s s H s s+ s s dτ dτ = E + E + (s s+ s s) (s s+ s s) dτ dτ s s r = E + E + J + K s s J = s s s sdτ dτ r coulomb exchange K = s s s sdτdτ r E ( ) ˆ triplet = s s s s H(ss s s) dτdτ ss = E + E + J K J s s S T, K J, K are positive K arises from the antisymmetrization of ψ.

Chemistry 120A 2nd Midterm. 1. (36 pts) For this question, recall the energy levels of the Hydrogenic Hamiltonian (1-electron):

Chemistry 120A 2nd Midterm. 1. (36 pts) For this question, recall the energy levels of the Hydrogenic Hamiltonian (1-electron): April 6th, 24 Chemistry 2A 2nd Midterm. (36 pts) For this question, recall the energy levels of the Hydrogenic Hamiltonian (-electron): E n = m e Z 2 e 4 /2 2 n 2 = E Z 2 /n 2, n =, 2, 3,... where Ze is