Atoms – start with single electron: H-atom



Atoms – start with single electron: H-atom

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) = -Ze2/(r( only depends on separation

not orientation

Note:

r = (xe – xp)i + (ye – yp)j + (ze – zp)k --vector

|r( = [(xe – xp)2 + (ye – yp)2 + (ze – zp)2]1/2 --length

Goal – separate variables

– no problem for K.E. – T already separated

First step reduce from 6-coord: xe, ye, ze & xp, yp, zp

to 3-internal coordinate. Eliminate center of mass

whole atom: R = X+Y +Z X = (mexe+mpxp)/(me+mp) . . .

for H-atom these are almost equal to: xp, yp, zp

but process is general – move equal mass – issue

other 3 coord: relative x = xe – xp etc. ideal for V(r)

Problem separates, V = V(r) only depend on internal

Hψ = [-(2/2M (R2 + -(2/2μ (r2 + V(r)] Ψ(R,r) = EΨ

M = me++mp μ= me.+mp /( me++mp)

Ψ(R,r) = Ξ(R) ψ(r) ( separates like before

i.e. Operate H on Ψ(R,r) and operators pass through

R and r dependent terms, Ξ(R) and ψ(r) to give:

(-(2/2M)ψ(r)(R2Ξ(R) and Ξ(R)[(-(2/2μ)(r2+V(r)]ψ(r)

so divide through by Ψ(R,r) = Ξ(R) ψ(r) and result

are independent (R and r) - correspond to constant

R-dependent equation: (-(2/2M)(1/ Ξ(R))(R2Ξ(R) = ET

Motion of whole atom – free particle not quantized

r-dependent equation: (1/ψ(r))[(-(2/2μ)(r2+V(r)]ψ(r)=Eint

here we let E=ET+Eint

internal equation simplified by convert: x, y, z ( r, θ, φ

result internal: H(r)ψ(r) = [(-(2/2μ)(r2–Ze2/r]ψ(r) = Eψ(r)

(idea – potential only depends on r,

so other two coordinates –only contribute to K.E.)

(r,θ,φ2 = 1/r2{(/(r(r2(/(r)+[1/sinθ](/(θ(sinθ (/(θ) + [1/sin2θ](2/(φ2}

This easy to separate φ dependence

set up Hψ = Eψ , multiply r2sin2θ , φ only in 1 term

Use ψ(r, θ, φ) = R(r) Θ(θ) Φ(φ) divide through as before

a) Let Φ part equal constant, m2: (2/(φ2 Φ(φ) = -m2 Φ(φ)

[pic] rotation about z-axis Φ(φ) = eimφ m = 0,(1,(2

b) Can similarly separate Θ(θ) ( but arithmetic messier

LeGendre polynomial: Θℓm(θ)= Pℓ(m((cos θ) ℓ = 0,1,2, …

ℓ = 0 – [pic]/2

ℓ = 1 m = 0 – (3/2)1/2 cos θ m = (1 (3/4)1/2 sin θ

ℓ = 2 m = 0 – (5/8)1/2 (3 cos2 – 1) …

c) Radial function messier yet but must fit B.C.

r ( ( ( Rnℓ(r) ( 0 (must be integrable)

~ e-αr damp (r always +)

Must be orthogonal

this works when fct. oscillate (wave-like)

power series will do

Associated LaGuerre Polynomial

σ = Zr/a0

Rnℓ = [const] (2σ/n)ℓ [pic] (2σ/n)e-σ/n

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

Comparison of potentials – when potential not infinite, levels collapse, when sides not infinite and verticals w/f penetrates the potential wall

Particle in box; Stubby box;

[pic]

Harmonic oscillator; Anharmonic oscillator

[pic]

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

Not edited

Review

1. Talked about the H-atom problem – all on handouts

a. Separation of C of M from relative

( not interested in position / relativity of whole atom, only e vs. p

b. Separation into spherical coordinates

( idea V = -Ze2/r ; one coordinate if spherical, 3 if Cartesian

c. Method – get all of one variable on one side

must be constant ( use product

wavefunction: ψ = R(r) Θ(θ) Φ(φ)

Φ = eimφ m = 0, (1, (2, … ℓ = 0, 1, 2, …; ℓ ( (m(

Θ = Pe(m((cos θ) LaGuerre Polynomial n = 1, 2, …

R = (constant) (2Zr/na0) [pic] (2Zr/na0) e-Zr/na0 ℓ ( n – 1

(only equation with energy in it)

En = -Z2e2μ/2K2n2 – same as Bohr

energy levels collapse with increasing n

These functions can be combined

ψ(r,θ,φ) = Rnℓ(r) Θe(m( (θ) Φm(φ)

Note: only Rnℓ depend on r as does V(r)

Energy will not depend on θ,φ

Often separate as Yem(θ,φ) = Θℓ(m((θ) Φm(φ)

( these are eigenfunctions of Angular Momentum

L2 Yem (θ,φ) = ℓ(ℓ + 1) (2 Yem

L2 Yem (θ φ) = m(Yem

Solving Rnℓ equation En = -Z2e2μ/2(2n2 ( exactly Bohr solution (must be)

Familiar: n = 0 ℓ = 0 m = 0 – 1s

n = 1 ℓ = 0 – 2s

ℓ = 1 m = 0, (1 – 2p (2p0 + 2p(1)

n = 2 ℓ = 0 – 3s

ℓ = 1 m = 0, (1 – 3p

ℓ = 2 m = 0, (1, (2 – 3d (3d0 , 3d(1 , 3d(2)

[INSERT r/Bohr plots]

Energy level diagram – H-atom

[INSERT diagram]

Spectral transitions match Balmen series but also must account for Θ,Φ functions

Allowed – any n change: Δn ( 0

– ℓ, mℓ as before: Δℓ = (1 , Δmℓ = 0, (1

n ( n' = 1 – Lyman must start p orbital

n ( n' = 2 – Balmen must start d or s orbital ( p or p orbital ( s

etc.

Test with Zeeman effect – mℓ βH = E' added E due to field

[INSERT H-atom & atomic orbitals plots and drawings]

[INSERT Atomic Orbitals – Table 7.1]

[INSERT Reduced mass – Figure 7.10]

Election Spin

if do this technique n-electrons

ψ = [pic]φni ℓi mℓi

ψ – multi electron

φ – one electron

we would expect the lowest energy state to be ni = 1 i = 1 – h

but this is not an allowed multielectron wavefunction

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 – set of quantum numbers must be different for each one electron. But also spin quantum number so for each n ℓ me – 2 electrons maximum

“Spin” – intrinsic magnetic moment or angular momentum – no physical picture

no function form α, β

Szα = 1/2 (α

Szβ = -1/2 (β

Multi-electron Atoms

Simplest idea – if H-atom describes how electrons are arranged around nucleus ( use them to describe multi-electron atom

Problem – potential no has new term

V(r) = [pic]-Ze2/ri + [pic][pic]e2/rij

rij = [(xi – xj)2 + (yi – yj)2 + (zi – zj)2]1/2

distance between electrons

H = T + V = [pic]-(2/2m (E2 + [pic]-Ze2/ri + [pic]e2/rij

sum over e- attraction repulsion

assume C of M

if ignore 3rd term ( H0 ~ [pic]hi (ri)

each hi (ri) is H-atom problem with solution that we know

E0 = [pic]Ei ψ0 = [pic]ψi(ri)

H-atom solution

product

sum of orbital E:

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

Order of filling – Aufbau (build up)

increased n –

increased ℓ – (skip one n for d and again for f)

1s ( 2s – 2p ( 3s – 3p ( 4s – 3d – 4p (

5s – 4d – 5p ( 6s – 4f – 5d – 6p (

7s – 5f …

Why this order ( this relates back to the d and f orbitals being smaller

each added electron shields outer electron from attraction to nucleus (3rd term left out)

i.e. As Z increases ( 1s plunge to more negative energy

Same for rest but each will be shielded by 2e- in 1s and 2 in 2s, 6 in 2p, etc.

Due to different s and p radii each get splitting E ( Enℓ

But d, f abnormal – do not fill until (s + p) higher n

Seems counterintuitive, but goes like nodes – more nodes get sucked in close

Approximation

Where does this come from?

Basically these orbitals for multielectron atoms are adjusting for [pic]e2/rij term

This term not separable

Central field approximation: V(r) = [pic]-Ze2/ri + V(ri) + [pic][e2/rij – V(rij)]

pull out of repulsion

that part depends on r

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(r1, r2, r3, …) = [pic]ψri ℓi mi (ri, θi, φi)

get product wavefunction

get summed energy (orbitals): E = [pic]εi

Method – underlie is Variation Principle

if use exact H, approximate (guess) ψa

then (H( = (ψa*Hψadτ / (ψa*ψadτ ( E0 {true ground state energy

if guess w/f with a parameter λ ( choose form

then ((H(/(x = 0 will give best value λ (minimum E)

improve ψ – alter for m

example He-atom 2 electrons ψ ~ θ1s(r1) φ1s(r2)

if shield then Z ( Z' (less attraction to nucleus)

ψa ~ e-Z'r1/a0 e-Z'r2/a0

solve ((H(/(Z' = 0 ( Z' = z – 5/16 = 27/16 for best function

E0 = 2(4 EH( = 8 EH E' = -77.4 eV

~ -108.8 eV Eexp ~ -78.9 eV

To get better – add more variation

eg ψ'' = (1 + br12) e-Z'r1/a0 e-Z'r2/a0

get: Z' ~ 1.85 E'' ~ -78.6 eV

b ~ 0.364/a0 error ~ 0.5%

( could go on and get Ecalc more precise than Eexp!!

For atoms – represent orbital as sum of function

φnℓ(ri) = [pic]ck(k (ri) (k could be various exponent on other

do optimization: ((H(/(ck = 0

find best ck ( 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(ri) by use orbitals to calculate average repulsion then solve for improved orbitals until self-consistent

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