Section 2Atomic Spectra(Lectures 2-3 ish)

Atomic Structure:IC yr 1

Quantum theoryof atoms / molecules

Previously:

H-Atom: Wavefunctions, quantum numbers and energy levels Spectral transitions: Rydberg series and selection rules Spin and spin-orbit coupling

Alkali Metal Atoms (e.g., Na): Penetration and shielding, quantum defects Larger spin-orbit coupling

He Atom Singlets and triplets Electron correlation

General Russell-Saunders and j-j coupling Atoms in external fields

Lecture 2: Atomic Spectroscopy

Solutions: n,l,m (r, q, j) = Rn,l(r)Yl,m(, )

2.1 Revision: The H-atom (& other 1e- atoms)2

2

2

22

22 11

rr

rr

sin

sin

1

sin

12

2

2

2

2

0

Quantum numbers:Principal quantum number, nOrbital angular momentum quantum number, lOrbital angular momentum projection quantum number, ml

Permitted values:n = 1, 2, 3,...l = 0, 1, 2, 3, 4,... n-1ml = -l, -l + 1,... +l

e.g., n = 3: l = 0: ml = 0 one of 3s orbitall = 1: ml = 0, ±1 three of 3p orbitalsl = 2: ml = 0, ±1, ±2 five of 3d orbitals

Spherical harmonics (complex)Laguerre functions

(or “magnetic”)

2.1.1 Atomic Quantum Numbers

The principal quantum number, n:

determines the energies of orbitals:n.b., Ens = Enp = End ...

Ionizationpotential

2

2

y

Leads to radial wavefunctions / distribution functions:

and the mean radius22

0

0 0 4

e

1s2s2p

1s

2s

2p

y y2P(r)

y(r) has n - l -1 nodes

4

1 1

3 2

0

y

2.1.1 Atomic Quantum Numbers

Orbital angular momentum quantum number, l& orb. ang. mom. projection (or magnetic) quantum number, ml

The magnitude of the angular momentum come from l:

|l| = ħ {l(l+1)}1/2

and the projection of l on the z-axis is provided by ml:

lz = mlħ

Lead to the angular wavefunctions and shapes of orbitals (1e- wavefunctions):

llz

1s

2snp

nd

(n.b. Real linear combinations plotted)

2.2 H-Atom Spectrum: Grotrian Diagrams

absorption

emission

Selection Rules:

Dn unrestricted

Dl = ±1 (Laporte)

Dml = 0, ±1

2

2 2

1 2

y

Transitions observedat wavenumbers:

Multiple series of lines (different n1) Graph of n vs 1/n2 is straight with

intercept = ionization energy

UVIR

2. 3 Origin of selection rules I Selection Rules:

Dn unrestrictedDl = ±1

Dml = 0, ±1

A physical picture: Conservation of momentum

The photon has intrinsic angular momentum ofphl

Total angular momentum must be conservedin the absorption/ emission process

F I l l

But |lF| is also quantised,

only specific relative orientations are permitted

F F F l l l

It is easily shown that the maximum and minimumlF are given by

F max I

F min I

l l i.e. l

l l i.e. l

I

F

I I

F

F

l lWhich is not quite our selection rule...

vectorially:

An atomic orbital with quantum number l has parity (-1)l

Since i changes the parity, yF and yI must have opposite parity

Hence Dl ≠ 0 for reasons of symmetry

Dml = 0, ±1 arises due to the helicity (s = ±1) of the photon

2.3 Origin of selection rules II Selection Rules:

Dn unrestrictedDl = ±1

Dml = 0, ±1

All “dipole allowed” selection rules ultimately originatefrom the need for a non-zero transition dipole moment:

F I F I i i

i

To be non-zero the integrand must be totally symmetric under the symmetryoperations of the group (the full rotation group, R3 for atoms).

Consider the operation i:

2.5 More than one electron

For any atom other than the H-atom, the Schrödinger equation becomes:

The total energy, E, is the energy of all electrons and all interactions

The electron repulsion term, Vij, renders the Schrödinger equation insoluble.

For one electron: y =y(x, y, z)yspin = y(x, y, z, ms)

For two electrons: y =y(x1, y1, z1, (ms)1, x2, y2, z2, (ms)2)

In general we need: 3N spatial and N spin coordinates

Electron K.E Nuclear attractions Electron repulsions

2

2

i iN iji i i je

2

0

iN

i

2

0

ij

i j

The wavefunction, y:

2.6 The orbital approximation

In order to proceed, assume Yspace is the product of n, one-electron wavefunctions(or orbitals)

Yspace= ja(r1)jb(r2)jc(r3)jd(r4)... jz(rn)

For each orbital

i.e.., electron i experiences the mean field of all other electrons

Solve using numerically by e.g., Self Consistent Field approach (Hartree) toyield orbital energies. Including exchange yields Hartree-Fock orbitals.

This method is flawed because in order to satisfy the Pauli principle we must reallywrite Yspace as linear combinations of orbitals with defined symmetry with respect toelectron permutation.Oh, and it also neglects spin and correlation (see 2.18, 2.19). But apart from that.....

2

2

i iN ij i ii je

e- – e- repulsion averaged overpositions of all other electrons

2.7 The spectra of alkali metal atoms

Ground state electron configurations [RG] ns1 i.e., closed shell cores

We are only interested here in valence excitations (core excitations at higher energies).Within the orbital approximation only one electron can change orbital per photon.

Energy Levels: The effects of penetration and shielding

3p

3s

Consider the radial distributionfunctions of 3s, 3p

A 3s electron spends more of its timepenetrating the core region andexperiencing the full nuclear attraction(or equivalently, 3p is more shielded)

3s

3p3d

Na atomn = 3H atom

The Energy of a level now depends on nand l quantum numbers with Ens< Enp< End

2.8 The spectrum of the Na atom

2S 2P 2D 2F

H-Atomenergylevels

“sharp”series

“principal”series

“diffuse”series

“fundamental”series

We account for penetration, shieldingwith a quantum defect, dnl for each level:

2 2 2

22 2

y y eff y

nl

nlnl

Absorption: 3s npEmission: ns np “sharp”

np ns “principal”nd np “diffuse”nf nd “fundamental”

Selection Rules:

Dn unrestrictedDl = ±1

Dml = 0, ±1Dj = 0, ±1

For a given l, dnl is only weaklydependent on n

2.9 Spin-Orbit Coupling I

A minor effect for the H-atom SOC becomes increasingly important with atomic number

The electron has spin quantum number s = ½and spin projection quantum number ms = ±½

Associated with any angular momentum of a charged particle is a magnetic momentvia which orbital and spin angular momenta interact leading to spectral fine structure.

The orbital angular momentum l has (if l > 0) an associated magnetic moment towhich the spin angular momentum s can couple, yielding a total angular momentum, j.

Vectorially:

The permitted range of quantum numbers, j, are givenby a Clebsch-Gordon series:

j = l+s, l+s-1, l+s-2,....... |l–s|

Hence, for a single electron, j = l ± ½:

j

2.9.1 Spin-Orbit Coupling II: Quantitative

The spin-orbit Hamiltonian,

But

SO

2 22 2 j l j l l l

Hence: 2 22 l l

Substituting for eigenvalues: l l l

The energy of a given |l, s, j> level is l ,s , j l l

Where A is the spin-orbit coupling constant which, forone electron atoms is given by

2 4

3 12

n.b. spin-orbit couplingincreases rapidly with

atomic number

Spin-Orbit Coupling constants:H(2p) = 0.243 cm-1

H(3p) = 0.072 cm-1

Li(2p) = 0.3 cm-1

Na(3p) = 17.2 cm-1

K(4p) = 57.7 cm-1

Rb(5p) = 237.6 cm-1

Cs(6p) = 554.1cm-1

2.10 Spectral Fine Structure

Spin-orbit coupling gives rise to splittings in spectra called fine structure .

l = 1, s = ½

j = 3/2

j = 1/2

mj = ± 1/2, ± 3/2

mj = ± 1/2

Selection Rule Dj = 0, ± 1 (but not 0 ↔ 0) arises from conservation of angular momentum.e.g.,

Na D-lines

3p

3s

2P3/2

2P1/2

2S1/2

17 cm-1

e.g.,24d

3p

Na Na

2P3/2

2P1/2

2D5/2

2D3/2

2.11 Atomic Term Symbols

Just a succint notation for the angular momentum coupling in an atomTerm symbols contain 3 pieces of information:

23/2

“Spin multiplicity”= 2S+1

(Where S is the total spinquantum number for the atom)

Gives L, the total orbital angular

momentum quantum number

for the atom:S: L = 0P: L = 1D: L = 2F: L = 3, etc.

J, the total angular

momentum quantumnumber for the atom.i.e., how L and S are

coupled

2.12 Atomic Term Symbols: one unpaired electron

For atoms with a single unpaired electron:

A. s = ½ and therefore S = ½2S+1 = 2 and all terms are “doublets”

B. l ( and therefore L) is dependent on the orbital the electron is in:l = 0(s), 1(p), 2(d), give rise to S, P and D terms, respectively

C. As we’ve seen, J depends on how L and S are coupled but the possiblequantum numbers are given by the Clebsch-Gordan series:

J = L+S, L+S-1, L+S-2,... |L-S|

e.g., So, the [Ne]3s1 ground state configuration of Na is 2S1/2

The excited [Ne]3p1 confguration yields 2P3/2 and 2P1/2 terms.

n.b., In the absence of external fields, each J levelhas (2J+1) degeneracy arising from mJ states:

1/2

-1/2

3/2

-3/2

mJ

2.13 Atomic Term Symbols: General Russell-Saunders Coupling

Term symbols really become useful when dealing with many-electron atoms.In Russell-Saunders Coupling:

A. The spins of all electrons, si couple to yield a total spin angular momentum S.The possible values of the corresponding S quantum number arise from aClebsch-Gordan Series.

S = s1 + s2, s1 + s2 -1,..... |s1 – s2| for two electrons

B. The orbital angular momenta of all electrons, li couple similarly to yield a totalorbital angular momentum L.The possible values of the corresponding L quantum number arise from aClebsch-Gordan Series.

L = l1 + l2, l1 + l2 -1,..... |l1 – l2| for two electrons

e.g., coupling of orbitalangular momenta l = 1,2:

e.g., coupling of twoelectron spins to give S=0, 1

2.13 Atomic Term Symbols: General Russell-Saunders Coupling

C. L and S can then couple to yield the total angular momentum, JThe possible values of the corresponding J quantum number arise from aClebsch-Gordan Series.

J = L+S, L+S-1,..... |L-S|

n.b., the above coupling scheme is appropriate in the limit of weak spin-orbitcoupling (small Z) and is known as Russell-Saunders Coupling

L L L

SS

S

JJ

J

e.g., 3D terms

L = 2, S = 1

3D33D2

3D1

2.14 The spectrum of the He atom

The simplest example of a 2 valence electron atom:

The ground state is simple: configuration 1s2, electrons spin-paired, hence a 1S0 term.

This satisfies the Pauli Exclusion Principle as formulated:

“No two electrons may have identical quantum numbers”

- Since n, l, ml and s are the same for both electrons they must have different ms. Thislies at the heart of the aufbau principle of electronic configurations.

Singly excited configurations, 1s1nl1 (n 2) are more interesting:

Now the electrons may be spin-paired (antiparallel) or spins-aligned (parallel) These differ in their overall spin angular momentum (S = 0 and S = 1,

respectively) denoted singlet and triplet states.

Why singlet and triplet? We need to understand the Spin wavefunctions and to dothat we must consider the more general form of the Pauli Principle:

2.15 The Pauli Principle

Each electron can have either ms = ½ (spin up, or a state) or ms = -½ (spin-down, b)

Four combinations can arise: a(1)a(2), b(1)b(2), a(1)b(2), a(2)b(1)

The first two are clearly symmetric with respect to exchange but the latter two areneither symmetric, a(1)b(2)≠ a(2)b(1), nor anti-symmetric, a(1)b(2)≠ – a(2)b(1).

“Any acceptable wavefunction must be anti-symmetric with respect tothe exchange of two identical fermions and totally symmetric with

respect to the exchange of identical bosons”

Applies to the total wavefunction. We will see several manifestations of thisthroughout this course but for the exchange of two electrons (Fermions, s = ½):

Ytot = YspaceYspin. Consider Yspin for He:

2.16 The Pauli Principle

The two electrons are, however, indistinguishable and we can take linearcombinations of these to form eigenfunctions of the S2 and Sz operators:

1 11 2 1 2 2 1 and 1 2 1 2 2 1

2 2, ,

s+(1,2) is clearly symmetric with respect to exchange, s-(1,2) anti-symmetric

So our four spin wavefunctions are:

1 2

1 2

11 2 1 2 2 1

21

1 2 1 2 2 12

,

,

Of which the 1st three are symmetric wrt exchange and the final one anti-symmetric.

These need pairing with Yspace to give anti-symmetric total wavefunctions Ytot.

2.17 Picturing spin states

SINGLET (spin paired):Complete cancellation of s1,s2 to give

S = 0, S = 0, Ms = 0 a single arrangement

S

Ms=0 Ms=+1 Ms=-1

TRIPLET: three arrangements with S = 1 [|S| = (2)1/2ħ]

n.b., ms1, ms2 are no longer well-defined

S

S

2.18 The Energy of Singlet and Triplet states: Electron Correlation

For states arising from the same configuration, the triplet state lies lower in energythan the singlet. To understand why, consider the space wavefunctions:

Consider the 1s12s1 configuration: Yspace= j1s(1) j2s(2) or j1s(2) j2s(1)

Which, again are neither symmetric nor anti-symmetric wrt exchange.Form linear combinations of these:

1 2 1 2space s s s s

Of which Y+ is symmetric under exchange and Y- anti-symmetric.

As the spin wavefunctions for the triplet states are always symmetric wrt exchangethese must be paired with Y-. Consider exchange for Y-:

When r1 = r2

1 2 1 2space s s s s space

space space

which can only be true if space

2.19 The Energy of Singlet and Triplets: Fermi Heaps & Fermi Holes

There is zero probability of finding two electrons in the same region of space if theyare described by Y- (and triplet states are). Conversely there is a slight maximum inthe probability of finding two electrons described by Y+ in exactly the same place.

|Y-|2

r1-r20

0

|Y+|2

r1-r20

0

Fermi Hole

Fermi Heap

The electrons are correlated – the location of each depends intimately on the other.

Triplet

Lower energy

Singlet

Higher Energyr1 = r2 is maximum electronrepulsion (i.e., high energy)

The degeneracy of singlet and triplet is lifted by the electron repulsion. A moredetailed treatment (e.g., MQM) shows the energy difference to be DE = 2K where Kis the exchange integral:

2

1 2 1 2

0 12

s s s s

2.20 The Grotrian diagram of the He atom (finally)

1s3d

1S 3S1P 1D 3P 3D

Singlet Triplet

1s2s

1s3s1s3p

1s2p

1s2

1s2s

1s3s1s3p

1s2p

1s3d

Emission:1s1ns1 (1S) 1s1n’p1 (1P) 1s1ns1 (3S) 1s1n’p1 (3P)1s1np1 (1P) 1s1n’s1 (1S) 1s1np1 (3P) 1s1n’s1 (3S)1s1np1 (1P) 1s1n’d1 (1D) 1s1np1 (3P) 1s1n’d1 (3D)etc.,

Absorption:1s2 (1S) 1s1np1 (1P)

Selection Rules:

Single e– changesDn unrestricted

Dl = ±1 (= DL)DS= 0

NEW!Singlet ↔ Triplet

forbidden

For each configuration, everytriplet state is lower in energythan the corresponding singlet

2.21 Configurations, Terms and Levels

A given electron configuration may give rise, within Russell-Saunders coupling, toseveral Terms (different L, S combinations). These in turn are split by spin-orbitcoupling into Levels (unique J) which comprise 2J+1 degenerate quantum states (mJ).

For example:

ns np

configuration terms

1P

3P

levels

1P1

3P0

3P2

3P1

2J+1=

3

1

5

states(12 in total)

3

0

1

-1

2

-2

J

Jz = mJħ

mJ =

2J+1 components(degenerate in

zero field)Electrostatic /

Spin correlation magneticInteractions:

2.22 But beware the Pauli principle......

Consider the C atom ground state, 1s22s22p2 and which terms arise.

Naively: s1 = ½, s2 = ½ and thus S = 0, 1 (i.e., singlet and triplet terms)

l1 = 1, l2 = 1 and thus L = 0, 1, 2 (i.e., S, P and D terms)

1 3 1 3 1 3i.e., we expect

But think carefully about the 3D term:

L = 2 hence it must have ML = 2 component which must mean ml1 = 1 and ml2 = 1S = 1 hence it must have MS = 1 component which must mean ms1 = ½ and ms2 = ½But l1 = l2 = 1 and n1 = n2 = 2

These two electrons would have to have identical quantum numbers which wouldviolate the Pauli Exclusion Principle.

It is possible to invoke group theory to tell if a given term is allowed (MQM p238) orwe can consider all possible combinations of ML and MS

2.23 Microstate tables l1=l2=1 ml = -1, 0, 1 s1 = s2 = ½ ms= -½ , ½

ml

1

0

-1

Sml 2 0 -2 1 0 1 -1 0 -1 1 0 -1 1 0 1

Sms 0 0 0 0 0 0 0 0 0 1 1 1 -1 -1 -1

Draw up a table of every possible combination of ms and ml permitted by Pauli

The largest value of Sml = 2 which must correspond to a D term. This only occurswith Sms = 0. Hence a 1D term with all it’s 5 ML states (ML = ±2, ±1, 0) must exist.

P2:

Strike 5 such states from the table:

ml

1

0

-1

Sml 2 0 -2 1 0 1 -1 0 -1 1 0 -1 1 0 1

Sms 0 0 0 0 0 0 0 0 0 1 1 1 -1 -1 -1

2.23 Microstate tables

ml

1

0

-1

Sml 2 0 -2 1 0 1 -1 0 -1 1 0 -1 1 0 1

Sms 0 0 0 0 0 0 0 0 0 1 1 1 -1 -1 -1

The largest remaining Sml = 1 which must correspond to a P term. This occurs withSms =0, ±1, i.e., triplet terms. Hence a 3P term with all it’s 9 states must exist.

ml

1

0

-1

Sml 2 0 -2 1 0 1 -1 0 -1 1 0 -1 1 0 1

Sms 0 0 0 0 0 0 0 0 0 1 1 1 -1 -1 -1

Leaving only a 1S term remaining. Hence a p2 configuration gives rise to 1D, 3P and 1S.Interestingly, a p4 configuration gives the same terms.

2.24 Energetic ordering: Hund’s Rules Friedrich Hund1896-1997

Within Russell-Saunders (LS) Coupling, for a given configuration:

1. The term with largest S is lowest in Energy2. For a given S the term with largest L is lowest in Energy3. For a term with several levels:

if the sub-shell is less than half full the lowest J level is lowest in Energy if the sub-shell is more than half full the highest J level is lowest in Energy

Strictly applies only to finding ground states of atoms.

As we’ve seen, the C atom ground state p2 configuration yields 1D, 3P, 1S terms.

3P0 ,3P1, 3P2

Lowest energy term(Rules 1, 2)

Lowest energy level(Rule 3)

Assumes spin-correlation >> orbital angular momm coupling >> spin-orbit coupling

2.25 LS (Russell-Saunders) versus j-j coupling

l1 + l2 + l3 ... L

+

s1 + s2 + s3 ... S

J

Russell-Saunders (LS) Coupling:

Good quantum numbers: L, S, J

Selection Rules:Dn unrestricted

Dl = ±1DL = 0, ±1

DJ = 0, ±1 (not 0↔0)DS= 0

2.25 LS (Russell-Saunders) versus j-j coupling

But if spin-orbit coupling is strong (large Z) then the spin, s, of an electron prefers tocouple to its own orbital angular momentum, l, to give j:

Good quantum numbers:

ji = li ± ½

J = j1 + j2 +..., j1 + j2 + ... -1, etc..

n.b. L, S undefined

Selection Rules:Dn unrestricted

Dl = ±1, Dj = 0, ±1 for one electronDl = 0, Dj = 0 for other electrons

DJ = 0, ±1 (not 0↔0)

l1 l2 l3

+ + +

s1 s2 s3

j1 + j2 + j3 ... J

m

agn

etic

electrostatic

e.g., down Group IV:C Si Ge Sn Pb

Coupling: LS intermediate j-j

j-j Coupling:

= the Bohr magneton

2.26 Atoms in External Fields: lifting degeneracy

In the absence of external fields the 3 ML components of an 1P term are degenerate.

But the magnetic moment associated with the orbital angular momentum caninteract with any external B field to lift this degeneracy.

2.26.1 The Normal Zeeman Effect (when S = 0)

Classically, the interaction energy:

E = – m.B = – geLzB

[ ge is the gyromagnetic (or magnetogyric) ratio]

0

1

-1

ML =B

Hence, quantum mechanically:

E = – geML ħB = mBMLB

e B e

e

LLz

lifts the field-free degeneracy resulting in splittings in the spectrum

As we’ve seen S and L give rise to magnetic moments. These can interact with anexternally applied B field:

2.26.1 The Normal Zeeman Effect (when S = 0) cont.

0

1

-1

ML =B

LLz

Selection Rule: DML = 0, ±1

ML

2

-2-101

-101

1P

1D

DML = -1 0 1

B = 0 B > 0 E = mBMLB

0

1

-1

2

-2

B ML =

L

2.26.2 The Anomalous Zeeman Effect (when S ≠ 0)

Of course the spin angular momentum also gives rise to a magnetic moment andwhen both L and S are non-zero the splittings are more complex.

B

J

L

S

Now, E = – ge(L + 2S).B

But we need consider only projections of L and S on B.

e.g., 2

J J

J

1 1 1

12 1

J

S S L Lg L ,S

J J

J JWhere the Landé g-factor,

n.b., when S = 0, J = L and gJ = 1 Normal Zeeman Effect

e J e J J

e.g., gJ = 1.5 for a 3P2 level= 2.00 for 3S1 level (normal spin g-factor)

2.26.2 The Anomalous Zeeman Effect cont.

Selection Rule: DMJ = 0, ±1

MJ

3/2

-3/2-1/21/2

2P1/2

2D3/2

DMJ = -1 0 1

B = 0 B > 0

-1/2

1/2

So the levels are split into 2J + 1 components but the splitting itself is dependenton J, L, and S.

i.e., splitting proportional to mJ, MJ & B

E = -gegJħMJB