• Hydrogen Atom: 3D Spherical Coordinates

– Ψ = (spherical harmonics)(radial) and probability density P

– E, L2, Lz operators and resulting eigenvalues

• Angular momenta: Orbital L and Spin S

– Addition of angular momenta

– Magnetic moments and Zeeman effect

– Spin-orbit coupling and Stern-Gerlach (Proof of electron

spin s)

• Periodic table

– Relationship to quantum numbers n, l, m

– Trends in radii and ionization energies

Zeeman Effect

( ) ( ) ( ) ( )xExxVdx

xd

mψψ

ψ=+−

2

22

2

ℏ

Kinetic energyPotential

energyTotal energy

Schrödinger Equation: Coordinate Systems

( ) ( ) ( ) ( )xExzyxVxzyxm

ψψψ =+

∂

∂+

∂

∂+

∂

∂− ,,2 2

2

2

2

2

22ℏ

( )2 2 2

2

2 2 2 2

1 1 1sin

2 2 sin sinr V r

r r r r

E

ψ ψ ψθ ψ

µ µ θ θ θ θ φ

ψ

∂ ∂ ∂ ∂ ∂ − − + + ∂ ∂ ∂ ∂ ∂ =

ℏ ℏ

1D Cartesian

3D Cartesian

3D Spherical

Kinetic energy Potential energy Total energy

Kinetic energy Potential

energy

Total energy

Convert to spherical using: sin cos , sin sin , cosx r y r z rθ φ θ φ θ= = =

Hydrogen Atom: 3D Spherical Schrödinger Equation

( ) ( ) ( )φθφθψ ,,, lmnnlm YrRr =

Laguerre

Polynomials

Spherical

Harmonics

( ) ( ) ( )QrEQrVQrp

neff ,,,,,,2

ˆ 2θψθψθψ

µ=+

∂∂

∂∂

−=r

rrr

p 2

2

22 1ˆ where ℏ

22

0

2

0

2where

1 13.6

2 n

keE eVE

z

n

Eµ= ≈

− =

ℏEigenvalues:

Eigenfunctions:

“Rewritten” Schrodinger Eqn.:

Hydrogen Atom: 3D Spherical Schrödinger Equation

3 Quantum Numbers (3-dimensions)

n = energy level value (average radius of orbit)

n = 1, 2, 3 …

l = angular momentum value (shape of orbit)

l = 0, 1, 2, … (n – 1)

m = z component of l (orientation of orbit)

m = -l,(-l + 1) ..0,1,2,..+l

How many quantum states (n, l, m) exist for n = 3? Is there a general

formula?

Wave Functions : Formulas

n = 1

n = 2

l = 1

l = 0

m = –2 m = –1 m = 0 m = 1 m = 2

Wave Functions: Angular Component

cosθsinθ sinφ sinθ cosφ

3cos2θ–1sinθcosθsinφ

sinθcosθcosφ

sin2θ sin2φ sin2θ cos2φ

s-orbital

p-orbitals

d-orbitals

l = 0

l = 1

l = 2

Wave Function: Radial

Component

What is the relationship between the number of zero crossings for the radial

component of the wave function and the quantum numbers n and l?

http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter40/multiple3/deluxe-content.html

Wave Functions: Angular & Radial Components

200 211 211210

100

300 311 311310 321 321320322 322

400 411 411410 421 421420422 422

431 431430432 432 433433

From: http://pcgate.thch.uni-bonn.de/tc/people/

hanrath.michael/hanrath/HAtomGifs.htmll = 1 p-orbitals

l = 2 d-orbitals

l = 0 s-orbitals

l = 3 f-orbitals

Probability Density: Formula

2

2

0 0 0

( , , )

sin

P r dV

r d d dr

π π

ψ ψ θ φ

ψ ψ θ ϕ θ

∗

∞∗

=

=

∫

∫ ∫ ∫

ψψψπ symmetricy sphericallfor 4 2

0

drrP ∫∞

∗=

⇒For small ∆r, can use P = P(r) ∆ r (analogous to 1D case)

( ) ( )20

( ) where 4P P r dr P r rπ ψ ψ∞

∗= =∫

Volume Element dV

Probability Density: “Density” Plots

p-orbitals

200

2ψ 210

2ψ 21 1

2ψ ±

“Dumbbells”“Spherical Shells”

n = 2 n = 2

l = 1

s-orbital

l = 0

m = 0

m = 1

Probability Density: Cross Sections

Can you draw the radial probability functions for the 2s to 3d wave functions?

http://webphysics.davidson.edu/faculty/dmb/hydrogen/default.html

Probability Density: Cross Sections

Rank the states (1s to 3d) from smallest to largest for the electron’s most

PROBABLE radial position.

For which state(s) do(es) the most probable value(s) of the electron's position

agree with the Bohr model?

http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/chapter40/multiple3/deluxe-content.html

Probability Density: Problem

[ ]

[ ]

22100 100

22

100 1001.5 3

2100 3

10

10

0

01

( ) (4 ) ( )

( ) ( )

( ) ( ) (0.05 )

, ,

(( ) 0.00.05 8) 1

o o

o o

r a r a

o o

a a

o o

o

o

o

P r r r r r

e ewhere r and r

a a

eP a r a a

a

after substitution o

P a

f r and r

er

π

π π

ππ

− −

−

−

∆ = Ψ ∆

Ψ = Ψ =

∆ =

Ψ ∆

=∆ =

For the ground state n = 1, l = 0, m = 0 of hydrogen, calculate the

probability P(r)∆r of finding the electron in the range ∆r = 0.05aoat r = a

o/2

Orbital Angular Momentum L:

Related to Orbital “Shape”

• Magnitude of Orbital Angular Momentum L

( ) ( )

( )

2 2 2

2 2

1 1ˆ , , sin 1sin sin

Eigenvalues: 1

L r l l

L l l

ψ θ φ θ ψ ψθ θ θ θ φ

∂ ∂ ∂ = − + = + ∂ ∂ ∂

= +

ℏ ℏ

ℏ

( )ˆ , ,

Eigenvalues:

z

z

L r i m

L m

ψ θ φ ψ ψφ∂

= − =∂

=

ℏ ℏ

ℏ

• Z-component of L

Orbital Momentum L: Vector Diagram

For l = 2, find the magnitude of the angular momentum L and the possible m

values. Draw a vector diagram showing the orientations of L with the z axis.

( +1) 2(2+1)

6 2.45

2

0, 1

0, 1 , 2

2

z

L l l

L or

l to l

L

m

m

l

= =

=

= − =

= = ± ±

=

± ±

ℏ ℏ

ℏ ℏ

ℏ ℏ ℏ

z

2h

24º

L = √√√√6h = 2.45h

m = 1

m = 0

m = −−−−1

−−−−2h

h

−−−−h

55º

m = 2

m = −−−−2

Can you draw the vector diagram for l = 3? For j = 3/2?

Spin Angular Momentum S:

Property of Electron

• Magnitude of Spin Angular Momentum S

( )Eigenvalues:

1 1 3For one electron: 1

2 2 4

1

S

S s s

= + =

= +

ℏ ℏ

ℏ

Eigenvalues:

For one electron: 2

z

z sS m

S =

=ℏ

ℏ

• Z-component of SExample for s = 1/2

z s = √√√√0.75h

m = 1/2

m = −−−−1/2

h/2

-h/2

• Orbital angular momentum L and spin angular momentum s of

electrons result in magnetic moments µl and µs.

Angular Momentum: Link to Magnetic Moments

( )

z-compone

1

nt

l

l

L B

L

z l L

B

B

L

m g

gl l gµ

µµ

µ µ

=

= −

−= +

�

ℏ

�

Orbital l = 0, 1, 2, …

( )

z-component

1s

sz s s B

s B

s B

B

gss

m g

s gµ

µ µ

µ

µ

µ

= ±

−= = +

− ≈ℏ

� �

Spin: s = ½

where eV55.79 10 and , gyromagnetic ratios

2 TB L s

e

eg g

mµ −= = × =

ℏ�

( )2Remember that where

2 2 2

qv q qr vr L mvriA

r mπ

π → = = =

=

Lµ���

Zeeman Effect: Splits m values

• Orbital magnetic moment µL interacts with an external magnetic field B and separates degenerate energy levels.

ext lzU B U Bµ µ= − ⋅ ⇒ = −��

assume z

direction

Different energies for

different ml values!

NO B

FieldB Field

l = 1

l = 0

m = 0

m = 1

m = –1

m = 0

“Anomalous” Zeeman Effect: More Lines??

l = 0

ml = 0

ml = 1

ml = –1

l = 1

????Zeeman

ml = 0

Why are there more energy levels than expected from the Zeeman effect?

Add external

magnetic field

Spin-Orbit Coupling: Splits j values

BBBU Bszs µµµ ±=−=⋅= int

��BU Bµ2=∆

• Electron’s spin magnetic moment µs interacts with internal B field caused by its orbital magnetic moment µl and separates energy levels.

Spin up:

High Energy

vp+

L, BlS

µs

e-

vp+

L, Bl

µs

e-

Spin down:

Low Energy

S

l = 1

l = 0

j = 3/2

j = 1/2

j = 1/2

s = 1/2

s = 1/2

Angular Momentum Addition: L + S gives J

• Special Case:

Example: l = 1, s = ½

( )ℏ�

���

1+=

+=

jjJ

SLJ

Quantum Numbers

,

, 1,... 1,j

j l s l s

m j j j j

= + −

= − − + −

Vectors

SL��

+

21

21

23

21

21

23

21

21

23

21

, ,,,

11

and

and

−=−−=

=−==+=

jj mm

jj

j = 3/2 j = 1/2

“Anomalous” Zeeman Effect: Spin-Orbit + Zeeman

• Quantum numbers mj (j-j coupling) for HIGHER Z elements.

j = 3/2

j = 1/2

j = 1/2

mj = 1/2

mj = 3/2

mj = –1/2

l = 1mj = –3/2

mj = 1/2

mj = –1/2

mj = 1/2

mj = –1/2

Spin-Orbit Zeeman

s = 1/2

l = 0 s = 1/2

Angular Momentum Addition: General Rules

21 JJ��

+

( ) ( )jjjjm

jjjjjjj

j ,1,...1,

,...1, 212121

−+−−=

−−++=

( )ℏ�

���

1

21

+=

+=

jjJ

JJJ

tot

tot

Example: j1 = 3/2, j2 = 3/2

• General Case:Quantum NumbersVectors

3 3 3 3max min2 2 2 2

and

for

3 0

3, 2, 1, 0

3, 2, 1, 0,1, 2, 3 3j

j j

j

m j

= + = = − =

=

= − − − =

Stern-Gerlach Experiment: ALSO splits m values

• A magnetic force deflects atoms up or down by an amount that depends on its magnet moment and the B field gradient.

• For hydrogen (mlz = 0), two lines are observed (spin up, spin down).

– Since l = 0, this experiment gave direct evidence for the existence of spin.

=dz

dBF zz µ

Case of ml = -1, 0, 1

What is the “story” of this experiment? Did Stern & Gerlach know about

The angular momentum of the yttrium atom in the ground state is

characterized by the quantum number j = 5/2. How many lines would

you expect to see if you could do a Stern-Gerlach experiment with

yttrium atoms?

Remember that in the Stern-Gerlach experiment all of the atoms with

different mj values are separated when passing through an

inhomogeneous magnetic field, resulting in the presence of distinct

lines.

How many lines would you expect to see if the beam consisted of

atoms with l = 1 and s = 1/2?

Stern-Gerlach Experiment: Problem

Multi-electron Atoms

• Schrödinger equation cannot be solved exactly for multi-electron

atoms because Coulombic repulsion “mixes” variables.

– Estimate energies using single-electron wave functions and

“correcting” energies with 1st-order perturbation theory.

• Orbitals “fill” in table as follows: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p . .

– Only ONE electron per state (n, l, ml , ms ) - Pauli Exclusion Rule!

– Why is 4s filled before 3d? ⇒ 4s orbital has a small bump near

origin and “penetrates” shielding of core electrons better than 3d

orbital, resulting in a larger effective nuclear charge and lower

energy.

12

2

intrr

keV ��

−=

Periodic Table

l = 1 (p)

l = 0 (s)

l = 2 (d)

l = 3 (f)

1

2

3

4

5

6

7

Noble Gas

Halogen

Alkali

Group VI

Group IV

Group V

Group III

n

• Effective atomic radii decrease

across each row of table.

– Why? Effective nuclear charge

increases and more strongly

attracts outer electrons,

decreasing their radius.

Periodic Table: Trends for Radii and Ionization Energies

• Ionization energies increaseacross each row of table until the complete “shell” is filled.

– Alkali atoms easily give up s-orbital electrons.

– Halogens have strong affinity for outer electrons.

Maxima = noble gases

Maxima = alkali metals

APPENDIX: Wave Functions Formulas

Complete Wave Function ψn,l,m

l = 1m = ±1

m = ±1

s-orbital

p-orbital

d-orbital

f-orbital

Spherical Component Yl,m

l = 0n = 1

n = 2

n = 3

l = 0,1

l = 0,1,2

l = 1

l = 0

l = 2

l = 3