Atoms 09 update-- start with single electron: H-atom

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1 Atoms 09 update-- start with single electron: H-atom VII 33 x z φ θ e -1 y 3-D problem - free move in x, y, z - handy to change systems: Cartesian Spherical Coordinate (x, y, z) (r, θ, φ) Reason: V(r) = -Ze 2 / r only depends on separation electrostatic potential not orientation Note: (if proton fixed at 0,0,0 then r =x 2 +y 2 +z 2 ) r = (x e x p )i + (y e y p )j + (z e z p )k --vector r = [(x e x p ) 2 + (y e y p ) 2 + (z e z p ) 2 ] 1/2 --length (scalar) Goal separate variables V(r) x,y,z mixed no problem for K.E. T already separated First step formal: reduce from 6-coord: x e,y e,z e & x p,y p,z p to 3-internal coordinates. Eliminate center of mass whole atom: R = X+Y+Z X = (m e x e +m p x p )/(m e +m p ).. etc. this normalizes the position correct for mass for H-atom these are almost equal to: x p, y p, z p but process is general move equal mass issue other 3 coord: relative x = x e x p etc. ideal for V(r) Problem separates, V = V(r) only depend on internal coord. Hψ = [-h 2 /2M R 2 + -h 2 /2μ r 2 + V(r)] Ψ(R,r) = EΨ M = m e + m p μ = m e. m p /( m e + m p ) 33

2 Ψ(R,r) = Ξ(R) ψ(r) product wave function separates summed Η like before (3-D p.i.b.) i.e. Operate H on Ψ(R,r) and operators pass through R and r dependent terms, Ξ(R) and ψ(r), to give: left side: right side = E {Ξ(R) ψ(r)} (-h 2 /2M)ψ(r) 2 R Ξ(R) and Ξ(R)[(-h 2 /2μ) 2 r +V(r)]ψ(r) so divide through by Ψ(R,r) = Ξ(R) ψ(r) and results are independent (R and r) sum equal constant, E VII 34 R-dependent equation: (-h 2 /2M)(1/Ξ(R)) R 2 Ξ(R) = E T Motion (T) of whole atom free particle -not quantized r-dependent equation: (1/ψ(r))[(-h 2 /2μ) 2 r +V(r)]ψ(r) = E int relative (internal) coord. here we let E = E T +E int internal equation simplified by convert: x,y,z r,θ,φ result internal: H(r)ψ(r) = [(-h 2 /2μ) r 2 Ze 2 /r]ψ(r) = E int ψ(r) (idea potential only depends on r, so other two coordinates, φ,θ only contribute K.E.) r,θ,φ 2 = 1/r 2 { / r(r 2 / r)+[1/sinθ] / θ(sinθ / θ) + [1/sin 2 θ] 2 / φ 2 } This easy to separate φ dependence set up Hψ = Eψ, multiply r 2 sin 2 θ, φ only in 1 term [(-h 2 /2μ) r,θ,φ 2 Ze 2 /r] ψ(r,θ,φ) = (-h 2 /2μr 2 ){ / r(r 2 / r)+ [1/sinθ] / θ(sinθ / θ) + [1/sin 2 θ] 2 / φ 2 }ψ(r,θ,φ) Ze 2 /r ψ(r,θ,φ) = E int ψ(r,θ,φ) 34

3 VII 35 (-h 2 /2μ) sin 2 θ{ / r(r 2 / r)+ [1/sinθ] / θ(sinθ / θ) + [1/sin 2 θ] 2 / φ 2 }ψ(r,θ,φ) [r 2 sin 2 θze 2 /r]ψ(r,θ,φ) = r 2 sin 2 θe int ψ(r,θ,φ) (-h 2 /2μ) {sin 2 θ / r(r 2 / r)+ sinθ / θ(sinθ / θ) r 2 sin 2 θ Ze 2 /r}ψ(r,θ,φ) - r 2 sin 2 θe int ψ(r,θ,φ) = - 2 / φ 2 ψ(r,θ,φ) (-h 2 /2μ) {sin 2 θ / r(r 2 / r)+ sinθ / θ(sinθ / θ) r 2 sin 2 θ (E int + Ze 2 /r)} R(r)Θ(θ)Φ(φ) = - 2 / φ 2 R(r)Θ(θ)Φ(φ) Use ψ(r, θ, φ) = R(r) Θ(θ) Φ(φ) divide through as before Recall 2 / φ 2 only operate on Φ(φ) part, rest pass through Φ(φ) (-h 2 /2μ) {sin 2 θ / r(r 2 / r)+ sinθ / θ(sinθ / θ) r 2 sin 2 θ (E int + Ze 2 /r)} R(r)Θ(θ) = - R(r)Θ(θ) 2 / φ 2 Φ(φ) divide by Ψ(r,θ,φ) [R(r)Θ(θ)] 1 (-h 2 /2μ){sin 2 θ / r(r 2 / r)+sinθ / θ(sinθ / θ) r 2 sin 2 θ (E int + Ze 2 /r)} R(r)Θ(θ) = - [Φ(φ)] 1 2 / φ 2 Φ(φ) = const. right side (red) independent of left side, so must be constant a) Let Φ part equal constant, m 2 : 2 / φ 2 Φ(φ) = -m 2 Φ(φ) y x +m -m rotation about z-axis Φ(φ) = e imφ m = 0,±1,±

4 b) Can similarly separate Θ(θ) but arithmetic messier 1 st divide through by sin 2 θ [R(r)Θ(θ)] 1 (-h 2 /2μ){ / r(r 2 / r)+(sinθ) 1 / θ(sinθ / θ) r 2 (E int + Ze 2 /r)} R(r)Θ(θ) = m 2 / sin 2 θ VII 36 Note 1 st and 3 rd terms r-dependent but middle only θ, separate [Θ(θ)] 1 (-h 2 /2μ){ (sinθ) 1 / θ(sinθ / θ)-m 2 / sin 2 θ}θ(θ) = [R(r) ] 1 (-h 2 /2μ){ / r(r 2 / r) r 2 (E int + Ze 2 /r)} R(r) = const. = l(l+1) LeGendre polynomial: Θ lm (θ)= P l m (cos θ) l = 0,1,2, l = 0 2 /2 l = 1 m = 0 (3/2) 1/2 cos θ l = 1 m = ±1 - (3/4) 1/2 sin θ l = 2 m = 0 (5/8) 1/2 (3 cos 2 θ 1) l = 2 m = ±1 (15/4) 1/2 (sin θ cos θ) l = 2 m = ±1 (15/16) 1/2 (sin 2 θ).... c) Radial function messier yet but must fit B.C. r R nl (r) 0 (must be integrable) exponential decay, penetrate potential ~ e -αr damp (r always +) Must be orthogonal this works when fct. oscillate (wave-like) power series will do 36

5 VII 37 Associated LaGuerre Polynomial σ = Zr/a 0 R nl = [const] (2σ/n) l 2l 1 L + -σ/n n + l (2σ/n)e n = 1, 2, 3, l = 0, 1, 2, n 1, l m n = 1, l = 0 ~ e -σ n = 2 l = 0 ~ (2 σ)e -σ/2 n = 2 l = 1 ~ σe -σ/2 n = 3 l = 0 ~ (1 2σ/3 + 2σ 2 /27)e -σ/3 n = 3 l = 1 ~ (σ σ 2 /6)e -σ/3 n = 3 l = 2 ~ σ 2 e -σ/3.... Comparison of potentials when potential not infinite, levels collapse, when sides not infinite and vertical w/f penetrates the potential wall Particle in box; Stubby box; 37

6 VII 38 Harmonic oscillator; Anharmonic oscillator H-atom Solutions drawn are for l = 0 If rotate this around r = 0, get a symmetric well and shapes look a little like harmonic oscillator shapes since potential bends over, V=0 at r=, get collapse of Levels as n Energy vary with nodes and curvature as before 38

7 Review 1. Talked about the H-atom problem all on handouts see Web page a. Separation of C of M from relative coord. interested in relative position e vs. p b. Separation into spherical coordinates idea V = -Ze 2 /r ; 1 coord. Spher., 3 Cartes. c. Method get all of one variable on one side must be constant use product wavefunction: ψ = R(r) Θ(θ) Φ(φ) VII 39 Φ(φ) = e imφ Θ(θ) = P m l (cos θ) m = 0, ±1, ±2, l = 0, 1, 2, ; l m LaGuerre Polynomial n = 1, 2, and l n 1 2l 1 R nl (r) = (constant)(2zr/na 0 ) L + n + l (2Zr/na0)e -Zr/na 0 (only r dependent equation has energy in it) E n = -Z 2 e 2 μ/2h 2 n 2 same as Bohr energy levels collapse with increasing n 39

8 These functions can be combined ψ(r,θ,φ) = R nl (r) Θ l m (θ) Φ m (φ) VII 40 Note: only R nl depend on r as does V(r) Energy will not depend on θ,φ Often separate as Y l m (θ,φ) = Θ l m (θ) Φ m (φ) these are eigenfunctions of Angular Momentum recall : L=r x p, L 2 ~-h 2 r 2, L z =xp y -yp x, L z =-ih / φ L 2 Y l m (θ,φ) = l(l + 1) h 2 Y l m L z Y l m (θ φ) = mhy l m This is source of familiar orbitals, l = 0,1,2,3.. or s,p,d,f Solving R nl equation E n = -Z 2 e 2 μ/2h 2 n 2 exactly Bohr solution (must be, since works) Familiar: n = 1 l = 0 m = 0 1s n = 2 l = 0 m = 0 2s l = 1 m = 0,±1 2p (2p 0 +2p ±1 ) n = 3 l = 0 m = 0 3s l = 1 m = 0,±1 3p l = 2 m=0,±1,±2 3d (3d 0,3d ±1,3d ±2 ) Note: 1s decays, 2s has node (2-σ) term, 2p starts at 0 40

9 VII 41 Note: 2s node makes dip Note: # radial nodes (in R nl ) in probability (e- density) decrease with l, i.e. #=n-l-1 Angular functions have no radial value --> a surface, combine with radial function to get e- density, represented as a contour map or probability surface 41

10 Real orbitals, take linear combinations of ±m values, eliminate i dependent terms, get x,y,z functions Cartesian form: e imφ +e -imφ = cosφ+isinφ + cosφ isinφ ~x VII 42 42

11 43 VII 43

12 Energy level diagram H-atom E n VII 44 Allowed any n change: Δn 0 l, m l as before: Δl = ±1, Δm l = 0, ±1 Spectral transitions match Balmer series but also must account for Θ,Φ functions Allowed selection rules (see box) n n' = 1 Lyman must start p orbital --> end 1s n n' = 2 Balmer must start d or s orbital end 2p or start in p orbital end in 2s etc. Test with Zeeman effect m l βh = E' add E from field 44

13 Extend this to more than 1 electron: VII 45 H-atom sol'n use for N-elect., assume product wavefct. n ψ = φni li mi where: ψ multi electron w/fct φ n i li mi one electron w/fct we would expect the lowest energy state like H-atom with n for all electrons --> N but this is not an allowed multielectron wavefunction Election Spin changes how things work for electrons: Pauli Principle: a. Every wavefunction for fermion (spin 1/2 particle) must be anti-symmetric with respect to exchange of identical particles b. For electrons in atoms this turns out to mean each electron has different set of quantum numbers But there is also spin quantum number so for each n l m l 2 electrons maximum Spin intrinsic magnetic moment or angular momentum no physical picture no functional form - represent as: α, β S z α = 1/2 hα S z β = -1/2 hβ 45

14 Multi-electron Atoms -- Simplest idea if H-atom describes how electrons are arranged around nucleus use them to describe multi-electron atom VII 46 Problem potential now has new term -- electron repel N V(r) = -Ze 2 N N /r i + e 2 /r ij j = 1 r ij = [(x i x j ) 2 + (y i y j ) 2 + (z i z j ) 2 ] 1/2 distance between electrons N H = T + V = -h 2 /2m 2 N i + -Ze 2 /r i + ½ i j e 2 /r ij sum over e - attraction repulsion assume C of M if ignore 3rd term H 0 ~ N h i (r i ) - separable each h i (r i ) is H-atom problem with solution that we know N N E 0 = E ψ 0 = ψ i (r i ) i sum of orbital E i product H-atom solution Which orbitals to use? a) Could put all e - in 1s lowest energy, but Pauli prevent that b) Put 2e - each orbital (opposite spin), fill in order of increasing energy 46

15 Order of filling Aufbau (build up) in order of increasing n (idea: low to high energy) and increasing l (skip one n for d and again for f) VII 47 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s 5f Why this order? relates back to the d and f orbitals being smaller because they have fewer nodes Added electrons shield outer electron from attraction to nucleus (3rd term left out) i.e. as Z increases 1s to more negative energy Same for n=2 etc. but each shielded by 2e - in 1s and 2e - in 2s, 6 in 2p, etc. Due to different s and p radii n-level splits with l: E E nl But d, f abnormal do not fill until fill (s + p) higher n Seems counter-intuitive, but goes like nodes more nodes e - get sucked in close to nucleus 47

16 Approximation -- Where does this come from? Basically these orbitals for multielectron atoms must adjust H-atom like solution to account for e 2 /r ij term,ij This term not separable Central field approximation: V(r) = [-Ze 2 /r i + V(r i )] + [e 2 /r ij V(r i )] i i,j pull out of repulsion that part depends on r i Think of as what is average potential for electron i a) attracted to nucleus b) repelled by all other electron j (average) Still a problem with central force, now separable and include average repulsion Miss out on correlation instant e e motion Solution ψ(r 1, r 2, r 3, ) = N get product wavefunction N get summed energy (orbitals): E = ψ r i li mi (r i, θ i, φ i ) ε i VII 48 Method underlying this is Variation Principle if use exact H, approximate (guess) ψ a then H = ψ a *Hψ a dτ / ψ a *ψ a dτ E 0 {true ground state energy if guess w/f with a parameter λ choose form 48

17 then H / λ = 0 will give best value λ (minimum E) VII 49 improve ψ alter form example He-atom 2 electrons ψ ~ φ 1s (r 1 ) φ 1s (r 2 ) if shield then Z Z' (less attraction to nucleus) ψ a ~ e -Z'r 1/a0 e -Z'r 2/a0 solve H / Z' = 0 Z' = z 5/16 = 27/16 for best function E 0 = 2 4 E H = 8 E H E' = ev ~ ev E exp ~ ev To get better add more variation eg ψ'' = (1 + br 12 ) e -Z'r 1/a0 e -Z'r 2/a0 get: Z' ~ 1.85 E'' ~ ev b ~ 0.364/a 0 error ~ 0.5% could go on and get E calc more precise than E exp!! For atoms represent orbital as sum of function φ nl (r i ) = ckƒ k (r i ) ƒ k could be various exponent on other k do optimization: H / c k = 0 find best c k linear combination solve problem Actual work these days uses Hartree-Fock method underlying Variation Principle is same but in Hartree-Fock optimize form of V(r i ) by use orbitals to calculate average repulsion then solve for improved orbitals until self-consistent 49