Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3...

24
The one-electron atom H(Z =1), He + (Z =2), Li 2+ (Z =3), ...,U 91+ (Z = 92), ... Electronic Hamilton operator (for point-like clamped nucleus) * : c H el = b T e + b V en = - ~ 2 2m e 2 - Ze 2 κ 0 r (139) Determination of stationary bound states, i.e. solutions h r | ψ i = ψ(r) with E< 0, to the time-independent Schr¨ odinger equation (with c H el from above): c H el - E ψ(r)= - ~ 2 2m e 2 - Ze 2 κ 0 r - E ! ψ(r)=0 (140) Firstly, we remove the fundamental constants by switching to ‘atomic units’. This reduces the mathematical work to pure numbers, and eliminates quantities which have experimental uncertainties. * The finite mass of the nucleus can be taken into account by switching from the electron mass m e to a reduced mass μ, where μ -1 = m e -1 + m N -1 and m N is the nuclear mass. FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 2/ 48-5 1. Historical introduction 2. The Schr¨ odinger equation for one-particle problems 3. Mathematical tools for quantum chemistry 4. The postulates of quantum mechanics 5. Atoms and the ‘periodic’ table of chemical elements 6. Diatomic molecules 7. Ten-electron systems from the second row 8. More complicated molecules FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 1/ 48-5

Transcript of Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3...

Page 1: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

The one-electron atom

H (Z = 1), He+ (Z = 2), Li2+ (Z = 3), . . ., U91+ (Z = 92), . . .

Electronic Hamilton operator (for point-like clamped nucleus)∗:

Hel = Te + Ven = − ~2

2me∇2 − Z e

2

κ0 r(139)

Determination of stationary bound states, i.e. solutions 〈 r |ψ 〉 = ψ(r)

with E < 0, to the time-independent Schrodinger equation (with Helfrom above):

(Hel −E

)ψ(r) =

(− ~2

2me∇2 − Z e

2

κ0 r−E

)ψ(r) = 0 (140)

Firstly, we remove the fundamental constants by switching to ‘atomic

units’. This reduces the mathematical work to pure numbers, and

eliminates quantities which have experimental uncertainties.

∗The finite mass of the nucleus can be taken into account by switching from the electron mass me

to a reduced mass µ, where µ−1 = me−1 +mN

−1 and mN is the nuclear mass.

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 2/ 48-5

1. Historical introduction

2. The Schrodinger equation for one-particle problems

3. Mathematical tools for quantum chemistry

4. The postulates of quantum mechanics

5. Atoms and the ‘periodic’ table of chemical elements

6. Diatomic molecules

7. Ten-electron systems from the second row

8. More complicated molecules

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 1/ 48-5

Page 2: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Now the Schrodinger equation reads

(Hel − E

)ψ(r) =

(− 1

2∇2 − Z

r− E

)ψ(r) = 0 (141)

which is transformed from cartesian coordinates (x, y, z) to spherical

coordinates (r, θ, φ), with 0 ≤ r <∞, 0 ≤ θ ≤ π, and 0 ≤ φ ≤ 2π.

Laplace operator ∆ and squared angular momentum operator l2 (in

atomic units) in spherical coordinates:

∆ = ∇2 =1

r2∂

∂rr2∂

∂r− l2

r2=

1

r

∂2

∂r2r − l2

r2(142)

l2 = −

1

sin θ

∂θsin θ

∂θ+

1

sin2 θ

∂2

∂φ2

(143)

Separation ansatz for the state function:

ψ(r) = ψ(r, θ, φ) = R(r)Y (θ, φ) (144)

(this is always possible for ‘central fields’, i.e. V = V (r)).

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 4/ 48-5

Atomic unitsa

Physical quantity Symbol (name) Value in SI unitsb

mass me 9.1093826(16) × 10−31 kgcharge e 1.60217653(14) × 10−19 Cangular momentum, action ~ 1.05457168(18) × 10−34 J sel. permittivity 4πε0 κ0 1.112650056 . . . × 10−10 F m−1

length κ0~2/(mee2) = ~/(αmec) a0 (bohr) 5.291772 × 10−11 m

time ~/Eh = ~/(α2mec2) 2.418884 × 10−17 s

velocity a0Eh/~ = αc 2.187691 × 106 m s−1

linear momentum ~/a0 = αmec 1.992852 × 10−24 kgm s−1

force Eh/a0 8.238723 × 10−8 Nenergy e2/(κ0a0) = α2mec

2 Eh (hartree) 4.359744 × 10−18 J

power Eh2/~ 1.802378 × 10−1 W

charge density e/a03 1.081202 × 1012 C m−3

el. current eEh/~ 6.623618 × 10−3 Ael. potential Eh/e 2.721138 × 101 Vel. capacitance κ0a0 5.887891 × 10−21 Fel. resistance ~/e2 4.108236 × 103 Ωel. field strength (E) Eh/(ea0) 5.142206 × 1011 V m−1

el. displacement (D) e/a02 5.721476 × 101 C m−2

el. dipole moment ea0 8.478353 × 10−30 C mel. quadrupole moment ea0

2 4.486551 × 10−40 C m2

el. polarizability (ea0)2/Eh 1.648777 × 10−41 C2 m2 J−1

magn. flux ~/e 6.582119 × 10−16 Wbmagn. flux density (B) ~/(ea0

2) 2.350517 × 105 Tmagnetizing force (H) eEh/(a0~) 1.251682 × 108 A m−1

magn. dipole moment e~/me = 2µB 1.854802 × 10−23 J T−1

aBased on CODATA recommended values 2002 (http://physics.nist.gov/constants/).

bThe standard deviation uncertainty in the least significant digits is given in parentheses.

Page 3: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Notation for single-particle eigenfunctions of angular momentum:

l2 |ψαlm 〉 = l(l+ 1) ~2 |ψαlm 〉lz |ψαlm 〉 = m ~ |ψαlm 〉

l 0 1 2 3 4 5 6 7 8 9 10 ...

s p d f g h i k l m n ...

Notation for many-particle eigenfunctions of total angular momen-

tum:

L2 |ΨαLM 〉 = L(L+ 1) ~2 |ΨαLM 〉Lz |ΨαLM 〉 = M ~ |ΨαLM 〉

Lz =n∑

i=1

lz,i , L=n∑

i=1

li , L2 = L · L

L 0 1 2 3 4 5 6 7 8 9 10 ...

S P D F G H I K L M N ...

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 6/ 48-5

The angular part is found to be a spherical harmonic, Y (θ, φ) =

Ylm(θ, φ), which is an ‘angular momentum eigenfunction’, i.e. a si-

multaneous eigenfunction of l2 and lz (in atomic units):

l2 Ylm(θ, φ) = l(l+ 1)Ylm(θ, φ) (145)

lz Ylm(θ, φ) = mYlm(θ, φ) (146)

Orbital angular momentum quantum number l: l = 0,1,2,3, . . .,

Magnetic quantum number m: −l ≤ m ≤ l.

Explicit form for the spherical harmonics (with Condon-Shortley† phaseconvention), −l ≤ m ≤ l:

Ylm(θ, φ) = Θlm(θ)Φm(φ) = (−1)mNlmPml (cos θ) eimφ (147)

Nlm =

√2l+ 1

(l−m)!

(l+m)!, Yl,−m(θ, φ) = (−1)mYlm

∗(θ, φ)

ΩYlm∗(θ, φ)Yl′m′(θ, φ) dΩ = δll′ δmm′

† E. U. Condon (1902-1974), G. H. Shortley (∗1910)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 5/ 48-5

Page 4: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

The first Legendre polynomials

l Pl(x)

0 1

1 x

2 12 (3x2 − 1)

3 12 (5x3 − 3x)

4 18 (35x4 − 30x2 + 3)

5 18 (63x5 − 70x3 + 15x)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 8/ 48-5

The associated Legendre‡ functions Pml (cos θ) are related to the Le-

gendre polynomials Pl(x) = P0l (x) (x = cos θ). The following relations

hold (for −l ≤ m ≤ l, where applicable):

Pml (x) =1

2ll!(1− x2)m/2 dl+m

dxl+m(x2 − 1)l (148)

Pl(x) =1

2ll!

dl

dxl(x2 − 1)l = 2F1(−l, l+ 1; 1; (1− x)/2) (149)

P−ml (x) = (−1)m(l−m)!

(l+m)!Pml (x) (150)

Pml+1(x) =2l+ 1

l −m+ 1xPml (x)− l+m

l−m+ 1Pml−1(x) (151)

Pm+1l (x) =

2m√1− x2

xPml (x)− (l+m)(l −m+ 1)Pm−1l (x) (152)

‡ A.-M. Legendre (1752-1833)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 7/ 48-5

Page 5: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

The first spherical harmonics

(Condon-Shortley phase convention)

l m Ylm(θ, φ) l m Ylm(θ, φ)

0 0√

14π 3 +3 −15

√7

2880π sin3 θ e3iφ

1 +1 −√

38π sin θ eiφ 3 +2 15

√7

480π sin2 θ cos θ e2iφ

1 0√

34π cos θ 3 +1 −3

2

√7

48π sin θ (5 cos2 θ − 1) eiφ

1 −1√

38π sin θ e−iφ 3 0 1

2

√74π(5 cos3 θ − 3cos θ)

2 +2 3√

596π sin2 θ e2iφ 3 −1 3

2

√7

48π sin θ (5 cos2 θ − 1) e−iφ

2 +1 −3√

524π sin θ cos θ eiφ 3 −2 15

√7

480π sin2 θ cos θ e−2iφ

2 0 12

√54π(3 cos2 θ − 1) 3 −3 15

√7

2880π sin3 θ e−3iφ

2 −1 3√

524π sin θ cos θ e−iφ

2 −2 3√

596π sin2 θ e−2iφ

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 10/ 48-5

The first associated Legendre functions

l m Pml (cos θ) l m Pml (cos θ)

0 0 1 3 0 12 (5 cos3 θ − 3cos θ)

1 0 cos θ 3 +1 32 sin θ (5 cos2 θ − 1)

1 +1 sin θ 3 −1 −18 sin θ (5 cos2 θ − 1)

1 −1 −12 sin θ 3 +2 15 sin2 θ cos θ

2 0 12 (3 cos2 θ − 1) 3 −2 1

8 sin2 θ cos θ

2 +1 3 sin θ cos θ 3 +3 15 sin3 θ

2 −1 −12 sin θ cos θ 3 −3 − 1

48 sin3 θ

2 +2 3 sin2 θ

2 −2 18 sin2 θ

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 9/ 48-5

Page 6: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Energy eigenvalues (in atomic units):

En = Enlm = − 1

2

Z2

n2(155)

These are degenerate for n > 1 (further details below).

Principal quantum number n: n = 1,2,3, . . .

This result for En is also an important hint for the understanding of

the stability of matter: En > −∞, i.e. the electron does not collapse

into the nucleus, despite the singular attractive potential, due to a

balance between kinetic and potential energy.

Generalized Laguerre§ polynomials:

L(α)k (x) =

(α+ 1)kk! 1F1(−k;α+ 1;x) (156)

L(α)k (x) =

2k+ α− 1− xk

L(α)k−1(x)−

k+ α− 1

kL

(α)k−2(x) (k ≥ 2)

(157)

§ E. N. Laguerre (1834-1886)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 12/ 48-5

Two-point boundary value problem for P (r) = r R(r) (as resulting

from the separation ansatz):(

d2

dr2+ 2 [ E − Vl(r) ]

)P (r) = 0 (153)

Vl(r) =l(l+ 1)

2 r2− Zr, P (0) = 0 , lim

r→∞P (r) = 0

Physically acceptable (i.e. normalizable) solutions exist only for a

discrete set of energies: Quantization due to the boundary condi-

tions.

Radial functions (eigenfunctions):

Pnl(r) = r Rnl(r) = Nnl xl+1L

(2l+1)nr (x) e−x/2 (154)

x =2Z

nr , Nnl =

1

n

√Z

nr!

(n+ l)!, nr = n− l− 1 ≥ 0

∫ ∞0

Rnl(r)Rn′l(r) r2dr =

∫ ∞0

Pnl(r)Pn′l(r) dr = δnn′

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 11/ 48-5

Page 7: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

The first radial functions of one-electron atoms

n l nr Pnl(r) (x = 2Zr/n)

1s 1 0 0√Z x e−x/2

2s 2 0 1 12

√Z2 x (2− x) e−x/2

2p 2 1 0 12

√Z6 x

2 e−x/2

3s 3 0 2 13

√Z3 x (3− 3x+ x2/2) e−x/2

3p 3 1 1 13

√Z24 x

2 (4− x) e−x/2

3d 3 2 0 13

√Z

120 x3 e−x/2

4s 4 0 3 14

√Z4 x (4− 6x+ 2x2 − x3/6) e−x/2

4p 4 1 2 14

√Z60 x

2 (10− 5x+ x2/2) e−x/2

4d 4 2 1 14

√Z

720 x3 (6− x) e−x/2

4f 4 3 0 14

√Z

5040 x4 e−x/2

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 14/ 48-5

The first generalized Laguerre polynomials

k L(α)k (x)

0 1

1 (α+ 1)− x2 1

2

[(α+ 1)(α+ 2)− 2(α+ 2)x+ x2

]

3 16

[(α+ 1)(α+ 2)(α+ 3)− 3(α+ 2)(α+ 3)x+ 3(α+ 3)x2 − x3

]

4 124

[(α+ 1)(α+ 2)(α+ 3)(α+ 4)− 4(α+ 2)(α+ 3)(α+ 4)x

+ 6(α+ 3)(α+ 4)x2 − 4(α+ 4)x3 + x4]

5 1120

[(α+ 1)(α+ 2)(α+ 3)(α+ 4)(α+ 5)

− 5(α+ 2)(α+ 3)(α+ 4)(α+ 5)x+ 10(α+ 3)(α+ 4)(α+ 5)x2

− 10(α+ 4)(α+ 5)x3 + 5(α+ 5)x4 − x5]

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 13/ 48-5

Page 8: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Eigenfunctions and energy eigenvalues (in atomic units) of the one-

electron atom:

ψnlm(r) =1

rPnl(r)Ylm(θ, φ) , Enlm = En = − 1

2

Z2

n2(158)

n = 1,2,3,4, . . . , l = 0,1, . . . , n− 1 , m = −l,−l+ 1, . . . , l

〈nlm |n′l′m′ 〉 =∫ψnlm(r)ψn′l′m′(r) dr = δnn′ δll′ δmm′

Ground state and corresponding energy:

ψ100(r) =

√Z3

πe−Zr =

1

r

√Z 2Zr e−Zr

1√4π

, E1 = − Z2

2(159)

For isovalue plots — i.e. representations of all points r with ψnlm(r) =

|c| for chosen c ∈ R — of the eigenfunctions of the one-electron atom,

see J. Brickmann, M. Kloffler, H.-U. Raab, Chemie in unserer Zeit 12

(1978) 23-26.

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 16/ 48-5

Radial functions Pnl(r) of the one-electron atom (for n = 1, 2, 3, 4)

n = 1 (1s)

0 10 Zr

0.0

0.2

0.4

0.6

0.8

Pnl(r)

Z1/2

...........

...........

..........

..........

..........

.........

.........

.........

.........

.........

........

........

........

........

........

........

........

........

........

.........

.........

.........

.........

..........

..........

........

......

.......

.......

........

.........

..........

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

n = 3 (3s, 3p, 3d)

0 10 20 30 Zr

-0.4

-0.2

0.0

0.2

0.4Pnl(r)

Z1/2

..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

...............................................................................................................................................................................................................................................................

...........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..........................................

.....................................

.....................................

.........................................

................................................

...................................................................

...............................................................................................

........................................................................................................................................................................................................................

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

n = 2 (2s, 2p)

0 10 20 Zr

-0.4

-0.2

0.0

0.2

0.4Pnl(r)

Z1/2

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.............................................................................................................................................................................................................

.................................................

.......................................................................................................................

.....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

n = 4 (4s, 4p, 4d, 4f)

0 10 20 30 40 Zr

-0.4

-0.2

0.0

0.2

0.4Pnl(r)

Z1/2

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

....................................................................

.............................................................

..................................................................

.........................................................................

...............................................................................

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......................................................

...........................................................................................................................................................................................................................................................................................................................................................................................................

....................................................................................................................................................

..............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..............................................................

..............................................................

........................................................................

...................................................................................................

.................................................................................................

.............................................

....................................

....................................................

......................................

...................................................

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 15/ 48-5

Page 9: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

The ‘periodic’ table of the chemical elements

(2004)

1

2

3

4

5

6

7

1

1

H

3

Li

11

Na

19

K

37

Rb

55

Cs

87

Fr

2

4

Be12

Mg

20

Ca

38

Sr

56

Ba

88

Ra

3

21

Sc

39

Y

∗∗

∗ 57

La58

Ce59

Pr60

Nd61

Pm62

Sm63

Eu64

Gd65

Tb66

Dy67

Ho68

Er69

Tm70

Yb71

Lu∗∗ 89

Ac90

Th91

Pa92

U93

Np94

Pu95

Am96

Cm97

Bk98

Cf99

Es100

Fm101

Md102

No103

Lr

4

22

Ti

40

Zr

72

Hf

104

Rf

5

23

V

41

Nb

73

Ta

105

Db

6

24

Cr

42

Mo

74

W106

Sg

7

25

Mn

43

Tc

75

Re

107

Bh

8

26

Fe

44

Ru

76

Os

108

Hs

9

27

Co

45

Rh

77

Ir

109

Mt

10

28

Ni

46

Pd

78

Pt

110

Ds

11

29

Cu47

Ag

79

Au

111

X

12

30

Zn

48

Cd80

Hg112

(X)

13

5

B

13

Al

31

Ga

49

In

81

Tl

113

14

6

C

14

Si

32

Ge

50

Sn

82

Pb114

(X)

15

7

N

15

P

33

As

51

Sb

83

Bi

115

16

8

O

16

S

34

Se

52

Te

84

Po116

(X)

17

9

F

17

Cl

35

Br

53

I

85

At

117

18

2

He

10

Ne

18

Ar

36

Kr

54

Xe

86

Rn118

(?)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 18/ 48-5

Degree of degeneracy¶:

m degeneracy: gl =l∑

m=−l1 = 2l+ 1 (160)

l degeneracy (without spin): gn =n−1∑

l=0

gl = n2 (161)

l degeneracy (spin included): gsn = 2 gn = 2n2 (162)

The value gsn essentially determines the length of the rows (‘periods’)

in the table of chemical elements (2, 8, 18, 32), whereas the value

2 gl = 2(2l+1) determines the block structure of the ‘periodic’ table

(s-, p-, d-, and f-block for l = 0,1,2,3, respectively).

¶ The degeneracy with respect to l, eq. (161), is a special property of the one-electron atom (withpoint-like nucleus), and is not present in many-electron atoms. For example, the 2s and 2p statesof a one-electron atom are degenerate, i.e. they have the same energy, but the ‘orbital energies’for the 2s and 2p orbitals in any state of a many-electron atom are always different.

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 17/ 48-5

Page 10: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Trial functions (unnormalized)

for the ground state of the one-electron atom

0

1

f(x)

0 1 2 3x

S: exp(−x), x = ζr

G: exp(−x2), x =√αr

L: 1/(1 + x2), x = ar

P: 1/(1 + x)2, x = cr

...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 20/ 48-5

The variation principle at work (I)

The ground state 1s1 2S of the one-electron atom

Definition of the system under consideration:

H = T + V , T = − 1

2∆ , V = Vnuc(r) = − Z

r

A set of trial functions‖:

1. Slater function (ζ > 0):

ψS = NS exp (−ζr)

2. Gauß function (α > 0):

ψG = NG exp (−αr2)

3. Lorentz function (a > 0):

ψL = NL

[1 + (ar)2

]−1

4. Preuß function (c > 0):

ψP = NP

[1 + cr

]−2

‖Refs.: C. Zener, Phys. Rev. 36 (1930) 51, J. C. Slater, Phys. Rev. 36 (1930)57 (Slater function) — S. F. Boys, Proc. R. Soc. London A 200 (1950) 542,H. Preuß, Z. Naturforsch. A 11 (1956) 823 (Gauß function) — H. Preuß, Z.Naturforsch. A 13 (1958) 439 (Preuß function).

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 19/ 48-5

Page 11: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

1. The Slater function ψS = NS exp (−ζr), ζopt =?:

〈rk〉 = N2S 4π

∫ ∞0

dr rk+2e−2ζr = N2S 4π

Γ(k+ 3)

(2ζ)k+3(k > −3)

〈1〉 = 〈r0〉 = N2Sπ

ζ3≡ 1 ⇒ NS =

√ζ3/π

〈rk〉 = Γ(k+ 3)

2k+1

1

ζk

〈T 〉 = +1

2N2

S4π

ζ

∫ ∞0

dx

(d

dxx e−x

)2

=1

2N2

S4π

ζ

∫ ∞0

dx (1− x)2e−2x =1

2ζ2

〈V 〉 = − Z ζ

〈E 〉 = 1

2ζ2 − Z ζ ⇒ d〈E 〉

dζ= ζ − Z = 0 ⇒ ζopt = Z

〈E 〉Z2

= x

(1

2x− 1

)(x = ζ/Z) ⇒ Emin = − 1

2Z2

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 22/ 48-5

Estimate for the ground state energy:

〈E 〉 = 〈H 〉 = 〈ψ | H |ψ 〉 =∫ψ(r) H ψ(r) dr = 〈T 〉+ 〈V 〉

〈T 〉 = − 1

2〈ψ |∆ |ψ 〉 = +

1

2〈∇ψ |∇ψ 〉 , 〈V 〉 = − Z 〈r−1〉

Mathematical preliminaries:

• The beta function:

B(a, b) =Γ(a) Γ(b)

Γ(a+ b)= B(b, a)

B(a, b) =∫ 1

0ta−1(1− t)b−1 dt =

∫ ∞0

ta−1

(1 + t)a+bdt (a > 0, b > 0)

• Useful integral formulas:∫ ∞0

xs−1 exp (−p xn) dx =1

n

Γ(s/n)

ps/n(p > 0, s/n > 0)

∫ ∞0

xs−1

(1 + xn)pdx =

1

nB(p− s/n, s/n) (s/n > 0, np− s > 0)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 21/ 48-5

Page 12: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

3. The Lorentz function ψL = NL

[1 + (ar)2

]−1, aopt =?:

〈rk〉 = N2L 4π

∫ ∞0

rk+2 dr

(1 + (ar)2)2= N2

L4π

ak+3

1

2B(

1− k2,k+ 3

2)

〈1〉 = 〈r0〉 = N2Lπ2

a3≡ 1 ⇒ NL =

√a3/π

〈rk〉 = 2

π

1

akB(

1− k2,k+ 3

2) (−3 < k < 1)

〈T 〉 = +1

2N2

L4π

a

∫ ∞0

dx

(d

dx

x

1 + x2

)2

=1

2N2

L4π

a

∫ ∞0

dx(1− x2)2(1 + x2)4

=1

4a2

〈V 〉 = − 2Z a/π

〈E 〉 = 1

4a2 − 2Z

a

π⇒ d〈E 〉

da=

1

2a− Z 2

π= 0 ⇒ aopt =

4Z

π〈E 〉Z2

= x

(1

4x− 2

π

)(x = a/Z) ⇒ Emin = − 4

π2Z2

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 24/ 48-5

2. The Gauß function ψG = NG exp (−αr2), αopt =?:

〈rk〉 = N2G 4π

∫ ∞0

dr rk+2e−2αr2 = N2G 4π

1

2

Γ(k+32 )

(2α)k+32

(k > −3)

〈1〉 = 〈r0〉 = N2G

)3/2≡ 1 ⇒ NG =

(2α

π

)3/4

〈rk〉 = 2√π

Γ(k+32 )

(2α)k/2

〈T 〉 = +1

2N2

G4π√α

∫ ∞0

dx

(d

dxx e−x

2)2

=1

2N2

G4π√α

∫ ∞0

dx (1− 2x2)2e−2x2 =3

〈V 〉 = − 2Z√

2α/π

〈E 〉 = 3

2α− 2Z

√2α

π⇒ d〈E 〉

dα=

3

2− Z

√2

πα= 0 ⇒ αopt =

8Z2

〈E 〉Z2

= x

(3

2x− 2

√2

π

)(x =

√α/Z) ⇒ Emin = − 4

3πZ2

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 23/ 48-5

Page 13: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Variation of the ground state energy 〈E 〉 of the one-electron atom

with the trial function parameter

〈E 〉Z2

= f(x) = a2 x2 + a1 x

−0.5

0

f(x)

1 2x

S: x2/2− x, x = ζ/Z

G: 3x2/2−2√

2/π x, x =√α/Z

L: x2/4− 2x/π, x = a/Z

P: x2/5− x/2, x = c/Z

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

.............................................................................................................................................................................................................................................................................................................................................................................................................................

............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

..................................................

.......................................

...............................................................................

.............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

......................................................

............................................

.......................................

...................

For optimal choice of the parameter x (i.e. at the minima), the

quantum mechanical virial theorem, 〈V 〉 / 〈T 〉 = −2, is fulfilled, and

thus 〈E 〉 = 〈V 〉 /2 = − 〈T 〉 in all four cases.

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 26/ 48-5

4. The Preuß function ψP = NP

[1 + cr

]−2, copt =?:

〈rk〉 = N2P 4π

∫ ∞0

rk+2 dr

(1 + cr)4= N2

P4π

ck+3B(1− k, k+ 3)

〈1〉 = 〈r0〉 = N2P

3c3≡ 1 ⇒ NP =

√3c3

〈rk〉 = 3

ckB(1− k, k+ 3) (−3 < k < 1)

〈T 〉 = +1

2N2

P4π

c

∫ ∞0

dx

(d

dx

x

(1 + x)2

)2

=1

2N2

P4π

c

∫ ∞0

dx(1− x)2(1 + x)6

=1

5c2

〈V 〉 = − Z c/2

〈E 〉 = 1

5c2 − Z c

2⇒ d〈E 〉

dc=

2

5c− Z

2= 0 ⇒ copt =

5Z

4〈E 〉Z2

= x

(1

5x− 1

2

)(x = c/Z) ⇒ Emin = − 5

16Z2

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-06-30 25/ 48-5

Page 14: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

into eq. (163):

0 =

H

(0) − E(0) +∞∑

k=1

(H(k) − ε(k)

)λk

ψ(0) +

∞∑

l=1

χ(l) λl

=(H(0) − E(0)

)ψ(0) +

∞∑

m=1

λm(

H(0) −E(0))χ(m)

+m−1∑

k=1

(H(k) − ε(k)

)χ(m−k) +

(H(m) − ε(m)

)ψ(0)

Thus we obtain from the coefficient of λm:

m = 0 :(H(0) −E(0)

)ψ(0) = 0

m = 1 :(H(0) −E(0)

)χ(1) +

(H(1) − E(1)

)ψ(0) = 0

m = 2 :(H(0) −E(0)

)χ(2) +

(H(1) − E(1)

)χ(1)

+(H(2) −E(2)

)ψ(0) = 0

...

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 28/ 48-5

Perturbation theory

Time-independent Rayleigh-Schrodinger perturbation theory for non-

degenerate states

Find a solution for the Schrodinger equation

Hλψλ = Eλψλ ⇔(Hλ −Eλ

)ψλ = 0 (163)

with the Hamilton operator

Hλ = H(0) +∞∑

k=1

λk H(k) , (164)

where λ is a (natural or artificial) perturbation parameter (|λ| < λmax),

and assume that the solutions of the unperturbed problem

H(0)ψ(0) = E(0)ψ(0) (165)

are completely known (with all E(0) non-degenerate). Then put

Eλ = E(0) +∞∑

k=1

ε(k) λk and ψλ = ψ(0) +∞∑

l=1

χ(l) λl (166)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 27/ 48-5

Page 15: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

The perturbation theory may be used advantageously

• to determine (non-variational) approximations to the solutions of

eq. (163):

Eλ ≈ E(n) = E(0) +n∑

k=1

ε(k) λk

ψλ ≈ ψ(n) = ψ(0) +n∑

l=1

χ(l) λl

• to obtain exact values for derivatives of the energy E or the state

function ψ to various orders in λ at λ = 0, e.g.:

∂nE

∂λn

∣∣∣∣∣λ=0

= n! ε(n)

The perturbation theory presented above can be extended to include

the case of degenerate states.

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 30/ 48-5

The ‘intermediate normalization’ condition

〈ψ(0) |ψλ 〉 = 1 with 〈ψ(0) |ψ(0) 〉 = 1

implies orthogonality between ψ(0) and all χ(l):

〈ψ(0) |χ(l) 〉 = 0 (for l = 1,2, . . .)

Resulting expressions for ε(1) and ε(2):

ε(1)= 〈ψ(0) | H(1) |ψ(0) 〉 (167)

ε(2)= 〈ψ(0) | H(1) |χ(1) 〉+ 〈ψ(0) | H(2) |ψ(0) 〉 (168)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 29/ 48-5

Page 16: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

For the spatial part f(r1, r2) a suitable choice must be made. For S

states we can simplify further to f(r1, r2, r12), or equivalently f(s, t, u)

with s = r1 + r2, t = r1 − r2, and u = r12.

Variational results for the helium ground state 1s2 1S a.

f(s, t, u) −Eopt/a.u.

e−ζr1e−ζr2 = e−ζs 2.8477

ϕ(r1)ϕ(r2) 2.8617 b

e−ζr1e−ηr2 + e−ηr1e−ζr2 2.8757 c

e−ζs+cu 2.8896

e−ζs (1 + cu) 2.8911 d

e−ζs+cu cosh (at) 2.8994e−ζs

(c0 + c1u+ c2t2 + c3s+ c4s2 + c5u2

)2.9032

exact 2.9037 e

a E. A. Hylleraas, Z. Phys. 54 (1929) 347

b C. Froese Fischer: The Hartree-Fock method for atoms. Wiley, New York, 1977

c C. Eckart, Phys. Rev. 36 (1930) 878

d W.-K. Li, J. Chem. Educ. 64 (1987) 128

e K. Frankowski, C. L. Pekeris, Phys. Rev. 146 (1966) 46

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 32/ 48-5

The two-electron atom

H− (Z = 1), He (Z = 2), Li+ (Z = 3), . . ., U90+ (Z = 92), . . .

Hamilton operator (in atomic units):

Hel = − 1

2∇2

1 −1

2∇2

2 −Z

r1− Z

r2+

1

r12= h1 + h2 +

1

r12(169)

General structure of state functions for two-electron systems∗∗:

Φ(x1,x2) = f(r1, r2)Θ(σ1, σ2)

Spin part Θ = |SMS 〉: Singlet (S = 0, para-He) or triplet (S = 1,

ortho-He)

|00 〉 = (αβ − βα)/√

2

|11 〉 = αα |10 〉 = (αβ + βα)/√

2 |1− 1 〉 = ββ

∗∗ . . . as long as the Hamilton operator does not act on the spin of the particles.

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 31/ 48-5

Page 17: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Electron configuration of neutral atoms in the ground state (designated as 2S+1LJ)

1 H 1s12S1/2

2 He 1s21S0

3 Li [He] 2s1 2S1/2

4 Be [He] 2s2 1S0

5 B [He] 2s2 2p1 2P1/2

6 C [He] 2s2 2p2 3P0

7 N [He] 2s2 2p3 4S3/2

8 O [He] 2s2 2p4 3P2

9 F [He] 2s2 2p5 2P3/2

10 Ne [He] 2s2 2p6 1S0

11 Na [Ne] 3s1 2S1/2

12 Mg [Ne] 3s2 1S0

13 Al [Ne] 3s2 3p1 2P1/2

14 Si [Ne] 3s2 3p2 3P0

15 P [Ne] 3s2 3p3 4S3/2

16 S [Ne] 3s2 3p4 3P2

17 Cl [Ne] 3s2 3p5 2P3/2

18 Ar [Ne] 3s2 3p6 1S0

19 K [Ar] 4s1 2S1/2

20 Ca [Ar] 4s2 1S0

21 Sc [Ar] 3d1 4s2 2D3/2

22 Ti [Ar] 3d2 4s2 3F2

23 V [Ar] 3d3 4s2 4F3/2

24 Cr [Ar] 3d5 4s1 7S3

25 Mn [Ar] 3d5 4s2 6S5/2

26 Fe [Ar] 3d6 4s2 5D4

27 Co [Ar] 3d7 4s2 4F9/2

28 Ni [Ar] 3d8 4s2 3F4

29 Cu [Ar] 3d10 4s1 2S1/2

30 Zn [Ar] 3d10 4s2 1S0

31 Ga [Ar] 3d10 4s2 4p1 2P1/2

32 Ge [Ar] 3d10 4s2 4p2 3P0

33 As [Ar] 3d10 4s2 4p3 4S3/2

34 Se [Ar] 3d10 4s2 4p4 3P2

35 Br [Ar] 3d10 4s2 4p5 2P3/2

36 Kr [Ar] 3d10 4s2 4p6 1S0

37 Rb [Kr] 5s1 2S1/2

38 Sr [Kr] 5s2 1S0

39 Y [Kr] 4d1 5s2 2D3/2

40 Zr [Kr] 4d2 5s2 3F2

41 Nb [Kr] 4d4 5s1 6D1/2

42 Mo [Kr] 4d5 5s1 7S3

43 Tc [Kr] 4d5 5s2 6S5/2

44 Ru [Kr] 4d7 5s1 5F5

45 Rh [Kr] 4d8 5s1 4F9/2

46 Pd [Kr] 4d10 1S0

47 Ag [Kr] 4d10 5s1 2S1/2

48 Cd [Kr] 4d10 5s2 1S0

49 In [Kr] 4d10 5s2 5p1 2P1/2

50 Sn [Kr] 4d10 5s2 5p2 3P0

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 34/ 48-5

The many-electron atom

A part of the knowledge of the state functions of the one-electron

atom can be transferred to the many-electron atom, if the following

assumptions are made††:1. Central field approximation: The electrons in the many-electron

atom are assumed to move in an effective central field Veff,l(r),

so that the orbitals can be written as ψ(r) = R(r) Y (θ, φ), with

Y (θ, φ) = Ylm(θ, φ).

2. Equivalence restriction: The radial parts are assumed to be

independent of the magnetic quantum number m: R(r) = Rnl(r).

The resulting set of radial functions Pnl(r) = r Rnl(r) has to be de-

termined for every state of the many-electron atom, e.g.

- He ground state (singlet): 1s2 1S → P10(r)

- He excited states (singlet or triplet): 1s1 2s1 1,3S → P10(r), P20(r)

- Li ground state (doublet): 1s2 2s1 2S → P10(r), P20(r)

†† In addition to the approximation of the many-electron state function as a Slater determinant (anantisymmetrized product of spin orbitals), or a linear combination thereof.

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 33/ 48-5

Page 18: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

LS terms for electron configurations p2 and p3

p2: 3P (9), 1D (5), 1S (1)(62

)= 15 = 9 + 5 + 1

p3: 4S (4), 2D (10), 2P (6)(63

)= 20 = 4 + 10 + 6

Energy levels of neutral tetravalent atoms from the p-, d-, and f-block

(C, Ti, Ce), within an energy range above lowest ground state level:

∆E = 1eV ≈ 8065.55 cm−1 (Eh = 2h cR∞)

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 36/ 48-5

51 Sb [Kr] 4d10 5s2 5p3 4S3/2

52 Te [Kr] 4d10 5s2 5p4 3P2

53 I [Kr] 4d10 5s2 5p5 2P3/2

54 Xe [Kr] 4d10 5s2 5p6 1S0

55 Cs [Xe] 6s1 2S1/2

56 Ba [Xe] 6s2 1S0

57 La [Xe] 5d1 6s2 2D3/2

58 Ce [Xe] 4f1 5d1 6s2 1G4

59 Pr [Xe] 4f3 6s2 4I9/260 Nd [Xe] 4f4 6s2 5I461 Pm [Xe] 4f5 6s2 6H5/2

62 Sm [Xe] 4f6 6s2 7F0

63 Eu [Xe] 4f7 6s2 8S7/2

64 Gd [Xe] 4f7 5d1 6s2 9D2

65 Tb [Xe] 4f9 6s2 6H15/2

66 Dy [Xe] 4f10 6s2 5I867 Ho [Xe] 4f11 6s2 4I15/2

68 Er [Xe] 4f12 6s2 3H6

69 Tm [Xe] 4f13 6s2 2F7/2

70 Yb [Xe] 4f14 6s2 1S0

71 Lu [Xe] 4f14 5d1 6s2 2D3/2

72 Hf [Xe] 4f14 5d2 6s2 3F2

73 Ta [Xe] 4f14 5d3 6s2 4F3/2

74 W [Xe] 4f14 5d4 6s2 5D0

75 Re [Xe] 4f14 5d5 6s2 6S5/2

76 Os [Xe] 4f14 5d6 6s2 5D4

77 Ir [Xe] 4f14 5d7 6s2 4F9/2

78 Pt [Xe] 4f14 5d9 6s1 3D3

79 Au [Xe] 4f14 5d10 6s1 2S1/2

80 Hg [Xe] 4f14 5d10 6s2 1S0

81 Tl [Xe] 4f14 5d10 6s2 6p1 2P1/2

82 Pb [Xe] 4f14 5d10 6s2 6p2 3P0

83 Bi [Xe] 4f14 5d10 6s2 6p3 4S3/2

84 Po [Xe] 4f14 5d10 6s2 6p4 3P2

85 At [Xe] 4f14 5d10 6s2 6p5 2P3/2

86 Rn [Xe] 4f14 5d10 6s2 6p6 1S0

87 Fr [Rn] 7s1 2S1/2

88 Ra [Rn] 7s2 1S0

89 Ac [Rn] 6d1 7s2 2D3/2

90 Th [Rn] 6d2 7s2 3F2

91 Pa [Rn] 5f2 6d1 7s2 4K11/2

92 U [Rn] 5f3 6d1 7s2 5L6

93 Np [Rn] 5f4 6d1 7s2 6L11/2

94 Pu [Rn] 5f6 7s2 7F0

95 Am [Rn] 5f7 7s2 8S7/2

96 Cm [Rn] 5f7 6d1 7s2 9D2

97 Bk [Rn] 5f8 6d1 7s2 8H17/2

98 Cf [Rn] 5f10 7s2 5I899 Es [Rn] 5f11 7s2 4I15/2

100 Fm [Rn] 5f12 7s2 3H6

101 Md [Rn] 5f13 7s2 2F7/2

102 No [Rn] 5f14 7s2 1S0

103 Lr [Rn] 5f14 6d1 7s2

104 Rf [Rn] 5f14 6d2 7s2

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 35/ 48-5

Page 19: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Energy levels of Ti I‡‡ (380 lines of data) — IP = 6.820 eV

-----------------------------------------------------------------Configuration | Term | J | Level | Lande | Leading

| | | (cm-1) | g | Percentages

------------------|-------|----|------------|-------|------------

3d2.4s2 | a 3F | 2 | 0.000 | 0.66 | 100

| | 3 | 170.132 | 1.08 | 100

| | 4 | 386.874 | 1.25 | 100| | | | |

3d3(4F)4s | a 5F | 1 | 6556.828 | 0.00 | 100

| | 2 | 6598.749 | 0.99 | 100

| | 3 | 6661.003 | 1.25 | 100

| | 4 | 6742.757 | 1.35 | 100

| | 5 | 6842.964 | 1.41 | 100| | | | |

3d2.4s2 | a 1D | 2 | 7255.369 | 1.02 | 96 2 3d3.(2D2).4s 1D

| | | | |

3d2.4s2 | a 3P | 0 | 8436.618 | | 92 7 3d3.(2P).4s 3P

| | 1 | 8492.421 | 1.50 | 92 7 3d3.(2P).4s 3P

| | 2 | 8602.340 | 1.49 | 90 7 3d3.(2P).4s 3P| | | | |

3d3(4F)4s | b 3F | 2 | 11531.760 | 0.67 | 100

| | 3 | 11639.804 | 1.08 | 100

| | 4 | 11776.806 | 1.26 | 98 1 3d2.3s2 1G

| | | | |

‡‡ Source: Atomic Spectra Database, http://physics.nist.gov/PhysRefData/contents.html

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 38/ 48-5

Energy levels of C I‡‡ (280 lines of data) — IP = 11.26030 eV

-----------------------------------------------Configuration | Term | J | Level

| | | (cm-1)

-------------------|--------|-----|------------

2s2.2p2 | 3P | 0 | 0.00

| | 1 | 16.40

| | 2 | 43.40| | |

2s2.2p2 | 1D | 2 | 10192.63

| | |

2s2.2p2 | 1S | 0 | 21648.01

| | |

2s.2p3 | 5S* | 2 | 33735.20| | |

2s2.2p(2P*)3s | 3P* | 0 | 60333.43

| | 1 | 60352.63

| | 2 | 60393.14

| | |

2s2.2p(2P*)3s | 1P* | 1 | 61981.82| | |

2s.2p3 | 3D* | 3 | 64086.92

| | 1 | 64089.85

| | 2 | 64090.95

| | |

......

...

......

...

| | |

2s2.2p(2P*)26d | 1F* | 3 | 90721.0

| | |

2s2.2p(2P*)27d | 1F* | 3 | 90732.7

| | |

2s2.2p(2P*)28d | 1F* | 3 | 90742.2

| | |

2s2.2p(2P*)29d | 1F* | 3 | 90753.8

| | |

-------------------|--------|-----|------------

| | |

C II (2P*<1/2>) | Limit | | 90820.42

| | |

C II (2P*<3/2>) | Limit | | 90883.84

| | |

2s.2p2(4P)3s | 5P | 1 | 103541.8

| | 2 | 103562.5

| | 3 | 103587.3

| | |

2s.2p3 | 3S* | 1 | 105798.7

| | |

2s.2p3 | 1P* | 1 | [119878]

‡‡ Source: Atomic Spectra Database, http://physics.nist.gov/PhysRefData/contents.html

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 37/ 48-5

Page 20: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Energy levels of Ti I (contd.)

| f 3D | 2 | 49571.69 | |

| | 3 | 49619.72 | |

| | | | |

| f 1D | 2 | 50128.08 | || | | | |

| f 1G | 4 | 52125.98 | |

| | | | |

| e 1P | 1 | 53663.32 | |

| | | | |

------------------|-------|----|------------| || | | | |

Ti II (4F<3/2>) | Limit | | 55010 | |

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 40/ 48-5

Energy levels of Ti I (contd.)

| | | | |

3d2.4s2 | a 1G | 4 | 12118.394 | 0.98 | 90 8 3d3.(2G).4s 1G

| | | | |

......

...

| | | | |3d4 | a 5D | 0 | 28772.86 | | 100

| | 1 | 28791.62 | | 100

| | 2 | 28828.51 | | 100

| | 3 | 28882.44 | | 100

| | 4 | 28952.10 | | 100

| | | | |

......

...

| | | | |3d2.4p2 | e 3D | 2 | 48724.34 | |

| | 1 | 48724.83 | |

| | 3 | 48839.74 | |

| | | | |

3d2.4p2 | h 5D | 0 | 48802.32 | |

| | 1 | 48859.51 | || | 2 | 48915.07 | |

| | 3 | 49024.43 | |

| | 4 | 49036.46 | |

| | | | |

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 39/ 48-5

Page 21: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Energy levels of Ce I (contd.)

| | | | |

4f(2F*) 5d2(3F)6s (4F) | 5I* | 4 | 3196.607 | 0.66612 | 67 11 4f.5d.6s2 3F*

| | 5 | 3764.008 | 0.90691 | 90 3 4f.5d.6s2 3H*| | 6 | 4455.756 | 1.11714 | 64 26 (2F*) (3F)(4F) 5H*

| | 7 | 5315.803 | 1.21625 | 56 33 (2F*) (3F)(4F) 5H*

| | 8 | 6809.128 | 1.250 | 100

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 5 | 3210.583 | 1.16277 | 41 3G* 37 (2F*) (3F)(4F) 5H*

| | | | |4f.5d 6s2 | * | 4 | 3312.240* | 1.08582 | 29 3F* 26 3G*

| | | | |

4f.5d 6s2 | 3D* | 1 | 3710.513 | 0.61549 | 64 10 4f(2F*).5d2(1D).6s(2D) 3D*

| | 2 | 4766.323 | 1.14945 | 67 12 4f(2F*).5d2(1D).6s(2D) 3D*

| | 3 | 5006.719 | 1.23674 | 58 20 1F*

| | | | |4f(2F*) 5d2(3F)6s (4F) | * | 0 | 3974.503 | | 29 5D* 23 (2F*) (1D)(2D) 3P*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 1 | 4020.954 | 1.49404 | 17 3S* 15 (2F*) (1D)(2D) 3P*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | 3G* | 3 | 4160.283 | 0.72933 | 47 23 (2F*) (3F)(4F) 5H*

| | | | |4f(2F*) 5d2(3F)6s (4F) | * | 4 | 4173.494 | 1.02948 | 41 3G* 30 (2F*) (3F)(4F) 5H*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 5 | 4417.618 | 1.17790 | 29 3G* 28 4f.5d.6s2 3G*

| | | | |

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 42/ 48-5

Energy levels of Ce I‡‡ (953 lines of data) — IP = 5.5386 eV

-------------------------------------------------------------------------------Configuration | Term | J | Level | Lande | Leading

| | | (cm-1) | g | Percentages

-------------------------|---------|-------|------------|---------|------------

4f.5d 6s2 | 1G* | 4 | 0.000 | 0.94543 | 55 29 3H*

| | | | |

4f.5d 6s2 | 3F* | 2 | 228.849 | 0.76515 | 66 24 1D*| | 3 | 1663.120 | 1.07736 | 85 7 4f(2F*).5d2(1D).6s(2D) 3F*

| | 4 | 3100.151* | 1.07703 | 34 27 4f(2F*).5d2(3F).6s(4F) 5I*

| | | | |

4f.5d 6s2 | 3H* | 4 | 1279.424 | 0.88979 | 54 20 1G*

| | 5 | 2208.657 | 1.03212 | 80 12 4f(2F*).5d2(1D).6s(2D) 3H*

| | 6 | 3976.104 | 1.16032 | 78 13 4f(2F*).5d2(1D).6s(2D) 3H*| | | | |

4f.5d 6s2 | 3G* | 3 | 1388.941 | 0.73494 | 66 11 4f(2F*).5d2(1D).6s(2D) 3G*

| | 5 | 4199.367 | 1.15021 | 48 34 4f(2F*).5d2(3F).6s(4F) 5H*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | 5H* | 3 | 2369.068 | 0.59978 | 66 16 4f.5d.6s2 3G*

| | 4 | 2437.629 | 0.98592 | 38 25 (2F*) (3F)(4F) 3G*| | 6 | 4746.627 | 1.16593 | 67 24 (2F*) (3F)(4F) 5I*

| | 7 | 5802.108 | 1.237 | 62 38 (2F*) (3F)(4F) 5I*

| | | | |

4f.5d 6s2 | 1D* | 2 | 2378.827 | 0.93654 | 54 23 3F*

| | | | |

‡‡ Source: Atomic Spectra Database, http://physics.nist.gov/PhysRefData/contents.html

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 41/ 48-5

Page 22: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Energy levels of Ce I (contd.)

| | | | |

4f.5d 6s2 | * | 2 | 6303.984 | 1.419 | 36 3P* 16 4f(2F*).5d2(1D).6s(2D) 3P*

| | | | |4f(2F*) 5d2(3F)6s (2F) | * | 4 | 6475.540 | 0.897 | 43 3H* 17 (2F*)(3F)(2F) 1G*

| | | | |

4f.5d 6s2 | * | 3 | 6621.892 | 1.147 | 35 1F* 17 3D*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 5 | 6663.226 | 0.953 | 19 3I* 18 (2F*)(3F)(2F) 3I*

| | | | |4f(2F*) 5d2(3F)6s (4F) | * | 2 | 6836.628 | 0.68078 | 33 3F* 18 (2F*)(3F)(4F) 5G*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 3 | 7169.751 | 1.13 | 28 3F* 26 (2F*)(3F)(4F) 5F*

| | | | |

4f(2F*) 5d2(3F)6s (2F) | * | 4 | 7348.299 | 0.964 | 22 1G* 21 (2F*)(3F)(4F) 3H*

| | | | |4f(2F*) 5d2(3F)6s (4F) | * | 6 | 7696.210 | 1.076 | 39 3I* 19 (2F*)(1G)(2G) 3I*

| | | | |

4f(2F*) 5d2(3F)6s (4F)? | * | 5 | 7715.236 | 0.934 | 22 3I* 14 (2F*)(3F)(4F) 5G*

| | | | |

4f(2F*) 5d2(3F)6s (2F)? | * | 5 | 7841.955 | 1.063 | 24 3H* 21 (2F*)(3F)(2F) 3I*

| | | | |4f(2F*) 5d2(1D)6s (2D) | * | 1 | 7853.119 | 0.983 | 27 1P* 16 (2F*)(3F)(2F) 3D*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 4 | 7890.429 | 1.242 | 25 3F* 18 (2F*)(3F)(4F) 5D*

| | | | |

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 44/ 48-5

Energy levels of Ce I (contd.)

| | | | |

4f2.6s2 | 3H | 4 | 4762.718 | 0.80508 | 89 4 4f.5d(3G*).6s.6p(1P*) 3H

| | 5 | 6238.934 | 1.035 | 92 4 4f.5d(3G*).6s.6p(1P*) 3H

| | 6 | 7780.202 | 1.169 | 92 4 4f.5d(3G*).6s.6p(1P*) 3H

| | | | |

4f(2F*) 5d2(3F)6s (4F) | 3S* | 1 | 5097.777 | 1.88257 | 49 25 (2F*) (3F)(4F) 5D*| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 2 | 5210.906 | 1.22694 | 28 5D* 27 (2F*) (3F)(4F) 5G*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | 5G* | 2 | 5409.236 | 0.773 | 35 24 (2F*) (3F)(4F) 5D*

| | 3 | 6234.792 | 1.04965 | 69 13 (2F*) (3F)(4F) 5D*

| | 4 | 6856.559 | 1.150 | 72 6 (2F*) (3F)(2F) 1G*| | 5 | 7467.160 | 1.179 | 69 7 (2F*) (3F)(4F) 3H*

| | 6 | 8055.526 | 1.207 | 50 18 (2F*) (3F)(4F) 3H*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 3 | 5519.751 | 1.24530 | 32 5D* 21 (2F*) (3F)(4F) 3F*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 0 | 5571.156 | | 44 5D* 35 4f.5d.6s2 3P*| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 4 | 5572.074 | 1.31658 | 29 5D* 27 (2F*) (3F)(4F) 3F*

| | | | |

4f.5d 6s2 | * | 1 | 5637.233 | 1.389 | 30 3P* 26 4f(2F*).5d2(3F).6s(4F) 5D*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | 5F* | 1 | 5674.829 | 0.140 | 73 6 (2F*) (3F)(4F) 5D*

| | 2 | 5904.006 | 0.905 | 66 13 (2F*) (3F)(4F) 5G*| | 3 | 6337.061 | 1.232 | 42 16 (2F*) (3F)(4F) 5D*

| | 4 | 7174.156 | 1.373 | 67 19 (2F*) (3F)(4F) 5D*

| | 5 | 7933.558 | 1.345 | 81 5 (2F*) (3P)(4P) 5F*

| | | | |

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 43/ 48-5

Page 23: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

Energy levels of Ce I (contd.)

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 1 | 8695.201 | 1.285 | 22 3P* 12 (2F*)(3F)(4F) 5P*

| | | | |

4f(2F*) 5d2(3F)6s (2F) | * | 3 | 8902.306 | 1.128 | 20 1F* 14 4f.5d.6s2 1F*

| | | | |

4f(2F*) 5d2(3F)6s (2F) | * | 5 | 8991.451 | 1.067 | 31 3H* 17 (2F*)(3P)(4P) 5G*| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 0 | 9119.094 | | 40 3P* 32 (2F*)(3F)(2F) 1S*

| | | | |

4f(2F*) 5d2(3P)6s (4P) | 3D* | 3 | 9135.099 | 1.274 | 50 12 (2F*)(3F)(4F) 5P*

| | | | |

4f(2F*) 5d2(3F)6s (2F) | * | 2 | 9200.707 | 1.376 | 24 3D* 21 (2F*)(3P)(2P) 3D*| | | | |

4f(2F*) 5d2(3F)6s (2F) | 3I* | 6 | 9333.222 | 1.047 | 61 20 (2F*)(1G)(2G) 3I*

| | 7 | 11061.551 | 1.141 | 79 16 (2F*)(1G)(2G) 3I*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 1 | 9369.628 | 1.065 | 21 3P* 17 (2F*)(3F)(2F) 3D*

| | | | |

4f(2F*) 5d2(3P)6s (4P) | 3D* | 2 | 9425.529 | 1.207 | 46 8 (2F*)(3F)(4F) 5S*| | | | |

4f(2F*) 5d2(3F)6s (2F) | 3F* | 2 | 9709.012 | 0.799 | 51 13 (2F*)(3P)(2P) 3F*

| | | | |

4f(2F*) 5d2(3F)6s (2F)? | 3G* | 3 | 9787.220 | 0.868 | 42 26 (2F*)(3P)(2P) 3G*

| | | | |

......

...

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 46/ 48-5

Energy levels of Ce I (contd.)

| | | | |

4f(2F*) 5d2(3P)6s (4P) | 5G* | 2 | 8088.912 | 0.403 | 95 3 (2F*)(3F)(4F) 5G*

| | 3 | 8307.309 | 0.957 | 81 6 (2F*)(3P)(4P) 3G*

| | 4 | 8762.126 | 1.054 | 50 9 (2F*)(3F)(2F) 3H*

| | 5 | 9462.705 | 1.19 | 46 12 (2F*)(3P)(4P) 3G*

| | 6 | 11030.470 | 1.193 | 51 21 (2F*)(3F)(2F) 3H*| | | | |

4f(2F*) 5d2(3F)6s (4F) | 5P* | 2 | 8101.187 | 1.735 | 85 5 (2F*)(3F)(4F) 5S*

| | 3 | 8270.249 | 1.504 | 74 11 (2F*)(3F)(4F) 3D*

| | 1 | 8430.846 | 2.04 | 62 10 (2F*)(3F)(4F) 3P*

| | | | |

4f2.6s2 | 3F | 2 | 8235.605 | 0.680 | 88 4 4f.5d(3F*).6s.6p(1P*) 3F| | 3 | 9206.305 | 1.083 | 90 4 4f.5d(3F*).6s.6p(1P*) 3F

| | 4 | 9379.148 | 1.139 | 55 32 1G

| | | | |

4f(2F*) 5d2(3F)6s (2F) | 1S* | 0 | 8351.167 | | 44 17 4f.5d.6s2 3P*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 2 | 8366.098 | 1.525 | 33 3P* 16 (2F*)(3F)(4F) 5S*| | | | |

4f(2F*) 5d2(3F)6s (2F) | * | 5 | 8400.730 | 0.917 | 30 3I* 19 4f.5d.6s2 1H*

| | | | |

4f(2F*) 5d2(3F)6s (2F) | * | 4 | 8509.209 | 0.954 | 20 3H* 19 (2F*)(3P)(4P) 5G*

| | | | |

4f(2F*) 5d2(3F)6s (4F) | 3I* | 7 | 8587.973 | 1.155 | 59 24 (2F*)(1G)(2G) 3I*| | | | |

4f(2F*) 5d2(3F)6s (4F) | * | 6 | 8603.531 | 1.225 | 36 5G* 25 (2F*)(3F)(4F) 3H*

| | | | |

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 45/ 48-5

Page 24: Atoms and the ‘periodic’ table of chemical elements · 2p 2 1 0 1 2 q Z 6 x 2e x=2 3s 3 0 2 1 3 q Z 3 x(3 3x+x2=2)e x=2 3p 3 1 1 1 3 q Z 24x 2(4 x)e x=2 3d 3 2 0 1 3 q Z 120x

H atom in a weak homogeneous electric field (Stark effect) — per-

turbation theory

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 48/ 48-5

Energy levels of Ce I (contd.)

......

...

| | | | |

| * | 6 | 30603.359 | |

| | | | |

| * | 4,3 | 30624.042 | || | | | |

| * | 2 | 30706.313 | |

| | | | |

| * | 4 | 30739.270 | |

| | | | |

| * | 3 | 30740.884 | |

| | | | || * | 5 | 30767.047 | |

| | | | |

| * | 4 | 30787.639 | 0.695? |

| | | | |

| * | 3 | 30854.052 | 0.737 |

| | | | || * | 7 | 30876.234 | |

| | | | |

| * | 2 | 30991.841 | |

| | | | |

-------------------------|---------|-------|------------| |

| | | | |Ce II (4H*<7/2>) | Limit | | 44672 | |

FAQC — D. Andrae, Theoretical Chemistry, U Bielefeld — 2004-07-14 47/ 48-5