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
-----------------------
[pic]
................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related searches
- easiest business to start with no money
- words that start with c
- ted talk simon sinek start with why
- simon sinek start with why
- adjectives that start with c
- business to start with little money
- start with why statement samples
- businesses you can start with no money
- descriptive words that start with o
- inspirational words start with n
- descriptive positive words start with o
- characteristics that start with o