Lecture on-line 14 Molecular structure The Born-Oppenheimer approximation

Term Symbols-II (PowerPoint) Term Symbols-II (PDF) The Born-Oppenheimer Approximation (PowerPoint) The Born-Oppenheimer Approximation (PDF)

Handout for Term Symbols-II Handout for Born-Oppenheimer Approximation

The Term Symbol

( ) ( ) ,...,( )nn ll mm nn ll mm nn ll mm1 1 1 2 2 2n n

m m mn1 2 m

For a configuration

We have a number of different states (eigenfunctions to the Schrödinger equations)

They are characterized by differentTERM SYMBOLS :

L(llll T)

2ssssT+ 1

jjjjT

Total orbital angularmomentum quantumnumber Tll

Total spin angular quantum number with spin - multiplicity 2

T

T

ssss + 1

Total angular momentum quantumnumber Tjj

As an example 2p d1 13The Term Symbol

Total orbital angularmomentum quantumnumber

: 0 1 2 3 4 S P D F G

T

T

llll

Total spin angular quantum number with spin - multiplicity 2

T

T

ssss + 1

Total angular momentum quantumnumber Tjj

( , L( (2 (2 Number of States

T T2S

T T

T

TLL SS LL SSLL

) ) ))

+ + ×+

1 11

(3,1) F 21 3

(2,1) D 15 3

(2, 0) D 51

(1,1) P 93

(1, 0) P 31

Total 60

(3, 0) F 71

The Term Symbol States with different spin -multiplicity will differ in energy. The state withthe higher spin - multiplicity willbe lower in energy. The energy willdecrease with increasing spin - multiplicity

States with different quantum numbers will have differentenergies. The higher the quantum number the lower the energy

T

T

LL

LL

The Term Symbol

Example He : 2p d1 13

We have (1) = 1; 1) =

12

ll ss(

We have (2) = 2; 2) =

12

ll ss(

ll T = +2 1;3

ll T = + −2 1 1;2

ll T = −2 1;1

Next combining spin -angular momenta

ssT = +1

212

; ssT = −1

212

;

1 0

Thus combining orbitalangular momenta

P3

D3

F3

D

P1

1

F1

2p 3d1 1

1r12

The Term Symbol

Example He : 2p d1 13

We have (1) = 1; 1) =

12

ll ss(

We have (2) = 2; 2) =

12

ll ss(

ll T = +2 1;3

ll T = + −2 1 1;2

ll T = −2 1;1

Next combining spin -angular momenta

ssT = +1

212

; ssT = −1

212

;

1 0

Thus combining orbitalangular momenta

H = -

2m-

2mZr

Zr r

2

e

2

e 1 2 12

h h∇ ∇ − − +12

22 1

Adding to the Hamiltonian

An interaction between theorbital angular momentum

L

S

L S

T

T

T T

r

r

r r

and the spin - angular

momentum in termsof the spin - orbit interaction

term Hso = ⋅aaGives rise to the HamiltonianH1 = ⋅H+ aL S

With this Hamiltonian states with the same and but different

will have different energiesT T

T

L SJ

E hc J J L L

S SL S J, , { ( ) ( )

( )}

= + − +

− +

12

1 1

1

α

P3

D3

F3

D

P1

1

F1

P11

1D 2

F13

P3

P3

P3

2

1

0D3

D3

D3

321

F3F3

F3

43

2

2p 3d1 1

1r12

Spin-orbitinteraction

The Term Symbol

Example He : 2p d1 13

We have (1) = 1; 1) =

12

ll ss(

We have (2) = 2; 2) =

12

ll ss(

ll T = +2 1;3

ll T = + −2 1 1;2

ll T = −2 1;1

Next combining spin -angular momenta

ssT = +1

212

; ssT = −1

212

;

1 0

Thus combining orbitalangular momenta

The Term Symbol

Example He : 2p d1 13

We have (1) = 1; 1) =

12

ll ss(

We have (2) = 2; 2) =

12

ll ss(

Thus combining orbitalangular momenta

ll T = +2 1;

3 ll T = + −2 1 1;

2 ll T = −2 1;

1

Next combining spin -angular momenta

ssT = +1

212

; ssT = −1

212

;

1 0

Finally combining and to construct

T T

T

SS LLJJ

LL SST T 3 , = = 1

JJ T 3 +1, 3 +1- 1, 3 - 1= 4 3 2

Term Symbols :

F ; F ; F34

33

32

LL SST T 3 , = = 0

JJ T 3 + 0= 3

Term Symbols :

F13

LL SST T 2 , = = 1

JJ T 2 +1, 2 +1- 1, 2 - 1= 3 2 1

Term Symbols :

D ; D ; D33

32

31

The Term Symbol

Example He : 2p d1 13

We have (1) = 1; 1) =

12

ll ss(

We have (2) = 2; 2) =

12

ll ss(

Thus combining orbitalangular momenta

3 ll T = + −2 1 1;

Next combining spin -angular momenta

ssT = +1

212

; ssT = −1

212

;

1 0

Finally combining and to construct

T T

T

SS LLJJ

LL SST T 2 , = = 0

JJ T 2 + 0= 2

Term Symbols :

D12

ll T = +2 1; ll T = + −2 1 1;

2 ll T = −2 1;

1

LL SST T 1 , = = 1

JJ T 1+1, 1+1- 1 , 1- 1= 2 1 0

Term Symbols :

P ; P ; P32

31

30

The Term Symbol

Example He : 2p d1 13

We have (1) = 1; 1) =

12

ll ss(

We have (2) = 2; 2) =

12

ll ss(

Thus combining orbitalangular momenta

3 ll T = + −2 1 1;

Next combining spin -angular momenta

ssT = +1

212

; ssT = −1

212

;

1 0

Finally combining and to construct

T T

T

SS LLJJ

LL SST T 1 , = = 0

JJ T 1+ 0= 1

Term Symbols :

P12

ll T = +2 1; ll T = + −2 1 1;

2 ll T = −2 1;

1

P3

D3

F3

D

P1

1

F1

P11

1D2

F13

P3

P3

P3

2

10

D3

D3

D3

321

F3F3

F3

43

2

1r12

Spin-orbitinteractio

The Term Symbol

( , )ll SS jj ll jjSSjjT T

2T

34

33

32

13

33

32

31

12

32

31

3

, ) L( 2

(3,1,4) F 9

(3,1,3) F 7

(3,1,2) F 5

(3,0,2) F 7

(2,1,3) D 7

(2,1,2) D 5

(2,1,1) D 3

(2,0,2) D 5

(1,1,2) P 5

(1,1,1) P 3

(1,1,0) P

T + +1 1

003

1

1

(1,0,1) P 3 Total 60

P3

D3

F3

D

P1

1

F1

P11

1D2

F13

P3

P3

P3

2

10

D3

D3

D3

321

F3F3

F3

43

2

1r12

Spin-orbitinteractio

The Term Symbol

For less than half filled sub shells,states (within a TERM of equal and with lowest has lowest energy

T T

T

LL SSJJ

)

For more than half filled sub shells,states (within a TERM of equal and with higest has lowest energy

T T

T

LL SSJJ

)

P3

D3

F3

D

P1

1

F1

P11

1D2

F13

P3

P3

P3

2

10

D3

D3

D3

321

F3F3

F3

43

2

1r12

Spin-orbitinteractio

The Term Symbol

Ex. He : 1s ; He 1s3d;

O 1s 2s 2p

Cl : 1s 2s 2p

2

2 2 4

2 2 6

;

3 32 5ss pp

All filled shells or subshells give 0and They can be disregarded

T

T

LLSS

== 0.

The Term Symbol

In electronic transitionswe start with theelectron configuration (A)

( ( (

( (

1 1n

2 2n

i in

k kn

m mn

1

m

nn ll mm nn ll mm nn ll mm

nn ll mm nn ll mm

jjii

kkii

mm

1 22) ) . )

) ........ )

of TERM symbol

L( (A)(A)(A)

2SSTT JJTT

TT LL )

we next move an electron from shell (i) to shell (k)to produce the newconfiguration (B)

( ( (

( (

1 1n

2 2n

i in

k kn

m mn

1

m

nn ll mm nn ll mm nn ll mm

nn ll mm nn ll mm

jjii

kkkk

mm

1 22 1

1

) ) . )

) ........ )

−

+

of TERM symbol

L( (B)(B)(B)

2SSTT JJTT

TT LL )

For elements 1- 80Such a transition is onlyallowed if

1. ll lli k l = 1since absorbed photon has l = 1

− = ±∆

2. (A) - (B) = = 0a photon has no spin

T TSS SS SS∆

3. (A) - (B) = = 0, 1Note we can still have

= 0 and (i) - (k) = 1

T TLL LL LL

LL ll ll

∆

∆

±

The Term Symbol

S s ns

P s np

s nd

s nf

==

( ) )

( ) )

( ) )

( ) )

1

1

1

1

1 1

1 1

1 1

1 1

(

(

D = (

F = (

sin glet triplet

1 21 1s s

1 31 1s s

1. ll lli k l = 1since absorbed photon has l = 1

− = ±∆

2. (A) - (B) = = 0a photon has no spin

T TSS SS SS∆

3. (A) - (B) = = 0, 1Note we can still have

= 0 and (i) - (k) = 1

T TLL LL LL

LL ll ll

∆

∆

±

The Term Symbol

For a many - electron atom with the quantum number the possible quantum numbers are

...., -

LLmm

mm LL LL LL LL

TT

TT TT TT TT= − −, , ,1 2

For each m we have an angular momentumalong the z - axis of L = m and amagnetic moment

lz l

z B l

hµ µ= − mm

The Term Symbol

For each m we have an angular momentumalong the z - axis of L = m and amagnetic moment

lz l

z B l

hµ µ= − mm

The magnetic moment can interact with an externalmagnetic field along thez - axis to give

E = - z B l

BB

BB mm BBµ µ=

The Term Symbol

The normal Zeeman effect. On the left, when the field is off, asingle spectral line is observed. When the field is on, the linesplits into three, with different polarizations. The circularlypolarized lines are calledthe (-lines; the plane-polarized linesare called (-lines. Which line is observed depends on theorientation of the observer.

What you should recall from this lecture

Selection rules for electronic transitions S = 0 , L = 0, 1, l = 1J = 0, 1; but J = 0 to J = 0 not allowed

∆ ∆ ∆∆

± ±±

You should recallspliting in and external magnetic field of levels with same L quantum as

E m

m L,L - 1,...., -L - 1, -L

m l

l

l

BB

BBBB=

=

µ

The Term Symbol