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3. Atomic structure and atomic spectra
3.1. Structure and spectra of hydrogenic atoms
3.1.1 The structure of hydrogenic atoms
3.1.2 Atomic orbitals and their energies
3.1.3 Spectroscopic transitions and selection rules
3.2. The structure of many electron atoms
3.2.1 The orbital approximation
3.2.2 The spectra of complex atoms
3.2.3 Singlet and triplet states
3.2.4 Spin-orbit coupling
H
Xe
3.1 Structure and spectra of hydrogenic atoms
 Hydrogenic atoms: atoms having only one electron (H, He+, Li2+, …)  the Schrödinger
equation can be solved
 Many-electron atoms  impossible to solve the Schrödinger equation  approximations
are needed to solve the SE (Hartree-Fock Theory, Density Functional Theory, etc..).
3.1.1 The structure of hydrogenic atoms
Rydberg observed that the lines in the emission spectrum of H fit the expression:
 1
1
~

  H  2  2 
 n1 n2 
where H is the Rydberg constant for H (H= 109677cm-1). One series is characterized by
n1, within this series the lines are specified with n2 that can take the values n1+1, n1+2, …
For n1=1  Lyman series in the ultraviolet region
For n1=2  Balmer series in the visible region
For n1=3  Paschen series in the infrared region
Hydrogen spectrum
The gas inside the tube; that emits blue light is mercury, but the same can be
done with H2. The gas pressure in the tubes are low, because of natural
radioactivity and cosmic rays there are always few free electrons and ions. When
something around 2000-3000 volts is applied to these tubes under low pressure,
free electrons and ions are accelerated and collide with the gas molecule H2 and
dissociate them in excited H atoms. These collisions generate an avalanche of
charged particles. As a result, the gas in the discharge tube contains a lot of
ionized and highly excited atoms. The excited atoms give back this energy in the
form of light, while they are turning back to their ground states. (NB: Neon tubes
are typical examples of discharge tubes) The electromagnetic radiation emitted is
then analyzed by a spectrometer and various emission lines of hydrogen can be
observed in the visible (Balmer series), the ultraviolet (Lyman series) and infrared (Paschen series).
From the experimental law by Rydberg, a general principle is found: the Ritz
combination principle. It states that the wavenumber of any spectral line is the
difference between two terms. NB: it’s only for hydrogenic atoms that the terms have the
form:
R
Tn  H2
1 1
~  T1  T2  RH  2  2 
n
 n1 n2 
Bohr frequency condition: when an atom changes its
energy by E, the difference is carried away as a photon
of frequency : E = h
 In the emission spectrum of an atom, a line appearing
at a frequency  corresponds to a change in energy
E=E2-E1= hc (T1-T2) of an atom that is characterized by
the difference between two terms T.
 The atom can only have specific energies:
quantization.
 What is the origin of these permitted discrete values of energy?
A. The Schrödinger equation for the electron of a hydrogenic atom
 Atom = 1e- + 1 nucleus of mass mN. Both move and have a kinetic energy Te and TN.
Because of the electron charge and the positive charge |Ze| of the nucleus, the Coulomb
potential energy V(r) has to be taken into account in the energetics of the system.
Ze 2
V(r) = 40 r
 The Hamiltonian of the Schrödinger equation H tot = E tot
describing the system is:
2
2
2


Ze
2
2
H  Tˆe  TˆN  Vˆ = 
e 
N 
2 me
2 mN
40 r
B. Separation of internal motions
The atom is a 2-body system: both the nucleus and e- move with respect to each other (relative
motion) and both move with respect to an outside reference point.
However, it’s possible to separate the relative motion of two particles from the motion of the
center of mass. Moreover, since me<< mN, we’ll show that the simplified picture of an electron
moving in a spherical potential created by the nucleus can be envisaged.
In classical mechanics: let’s try to express total energy of two bodies (electron and
nucleus) not with respect to their coordinates x1 and x2, but with respect to the position of
the center of mass X (of the atom) and the distance x between the two bodies (i.e; the
distance between the electron and the nucleus).
p12
p22
E

V
2m1 2m2
X
m1
m
x1  2 x2
m
m
m  m1  m2
m m 
p1  m1 x1  m1 X   2 1  x
 m 
m m 
p2  m2 x2  m2 X   2 1  x
 m 
  mX ;
p   x
x  x1  x2

m2 m1
m1  m2
m 
x1  X   2  x
 m
m 
x2  X   1  x
 m
p12
p22
1
1
E

 V  mX 2   x 2  V
2m1 2m2
2
2
m= total mass of the system
2
2 p
E

V
2m 2 
P= linear momentum of the system as a whole
= reduced mass
p= “internal” linear momentum
 Via the correspondence principles, the Hamiltonian of the system (hydrogenic atom) can
then be expressed in 3D with respect to the coordinates of the center of mass (c.m.) and the
relative coordinate (distance between the e- and the nucleus).
2
2 p
E

V
2m 2 
2 2
2 2
H 
 c . m. 
 V
2m
2
 The total wavefunction tot appears as the product of a wavefunction c.m.describing the
motion of the center of mass of the system (e- + N) and an electronic wavefunction 
characterizing the motion of the e- around the nucleus: tot = c.m. 
 The total Schrödinger equation is then split in two equations:
H tot = Etot tot
(1)
(2)
2 2

 c.m.  c.m.  Ec.m.  c.m.
2m
2 2

   V   E
2
Ec.m.  E  Etot
c.m.  =tot
NB1: Concerning equation (1), we concluded in chap 2 that the translational energy of an
atom in gas phase is not quantized and can be described in classical physics.
 We now focus on Equation (2)
C. Separation of the variables in the electronic wave function 
In Equation (2),   me because mN>>me  Equation (2) describes the electron moving in
the spherical potential of the nucleus. It should be solved to rationalize the presence of lines
in the emission spectrum of the atoms.
z
2 2
In the rest of this chapter, we’ll only consider

   V   E
this electronic Schrödinger equation:
2
From chap 2 (rotation in 3D), the SE in spherical coordinates is:

2  2
2 
1
 2 

 2 2 ,   V (r )  E
2   r
r r r

Since the potential is spherical (does not depend of angles,  and ):
 (r , ,  )  R(r ) Y ( ,  )
 2   2 R 2Y R R 2 
 Y 2 

 2  , Y   V (r ) RY  E R Y
2   r
r r r


2
(r /RY)
C-C
r
x
2 
1
2
1  
 
 sin 


2
2
sin    
  
sin   
2  2 2R
R 
2 2
2
2
 r



2
r

V
r


Y

E
r

,

2R   r 2
 r 
2Y
 2  2 2R
  2 2 
R 
2
 r
  V r  C   
 2r
 , Y   E r 2  C

2
r 

 2R   r
  2Y
y
NB: For the separation of the SE equation C is arbitrary. However, the angular equation is an eigenvalue equation, which has the
eigenvalues: -l(l+1). Hence, for the sake of simplicity in the mathematical development C is chosen to be directly:
C
2
l (l  1)
2
 2   2 R 2 R 
(1’) 

  Veff R  ER

2
2   r
r r 
and
(2’) 2 , Y  l (l  1)Y
 As seen in Chap 2, the solutions Y(,) of (2’) are the spherical harmonics that are
specified by the quantum numbers l and ml.
 Let’s focus now on the radial wave equation (1’) that describes the motion of a particle of
masse  in a one-dimensional region where the potential is Veff(r).
D. The radial solutions R(r)
Veff
Ze 2
l (l  1) 2


40 r
2 r 2
Coulombic potential
created by the nucleus
Potential coming from the centrifugal
force that arises from the angular
momentum of the e- turning around the
nucleus
For l=0: e- attracted by the nucleus; r (purely Coulombic)
For l0: e- has an angular momentum that gives a positive
contribution to Veff  repulsion from the nucleus.
 Solving the radial equation (1’) gives eigenvalues En characterized by a quantum number:
e 4
Z2
En  
32 2 02 2 n 2
n= 1, 2, 3, …
 The solutions of (1’) are the Hydrogenic radial wavefunctions Rn,l(r), which are specified
by 2 quantum numbers n and l. They have the form:



R n , l ( r )  N n , l   Ln , l e 2 n
n
l
L= polynomial in : the associated Laguerre polynomial.
2 Zr

(no dimension)
a0
40  2
a0 
me e 2
Bohr radius = radius of the electron orbit of
the lowest energy; a0= 0.529 Å
The general form of these solutions are submitted to 2 constrains:
(i) At the nucleus (i.e.; r=0 or =0), the solution should never be infinite: R(0)  , because
the probability density cannot be infinite.
(ii) Far away from the nucleus (i.e.; r): R()  
 In order to fulfill those conditions, the values of n and l must respect that criterion:
l  (n-1) and n=1, 2, 3, ...
For l0: R(0)= 0, the wavefunction is zero
at the nucleus because of the centrifugal
force pushes away the electron
For l= 0: R(0)= a finite value, because the
attractive Coulombic potential localizes the
electron close to the nucleus
3.1.2 Atomic orbitals and their energies
Atomic orbital is a oneelectron wavefunction
defined by 3 quantum
numbers (n, l, ml)
=
Associated Laguerre
polynomials multiplied by
an exponential, and
characterized by n and l
 (r , ,  )  R(r ) Y ( ,  )

Spherical harmonics
specified with
l and ml
 Principal quantum number: n= 1, 2, 3, …
In an orbital characterized by n, the energy
of the electron is :
hc
Z 2  atom e 4
 atom
E


 atom
n
2
atom 

2 2 2
n
32  0 
me
 Angular momentum quantum number: l = 0, 1, 2, …, n-1
In an orbital characterized
by l, the angular momentum due to the rotation of the electron around the nucleus is:
L   l (l  1)
 Magnetic quantum number: ml= l , l-1, l-2 , … , - l
In an orbital characterized by ml, the
plane of rotation of the electron has a specific orientation characterized by the z-component of L :
Lz  ml 
 NB: A spin function should be added in the wavefunction. Hence, a spin quantum number, ms
= ½, should also be specified in order to characterize totally the electron.
A. The energy levels and the ionization energies
En  
hc
 atom
n2
 The energies given by the formula are negative, they
correspond to a bound state of the electron.
 The zero energy, reference level for the energy, corresponds to
a situation where the electron is not bound to the proton H+ and
has a zero kinetic energy. It is characterized by n = 
 The positive energies correspond to unbound states of the
electron. The wavefunction describing such an electron is also
solution of the SE, and it is the wavefunction of a free electron.
The energy of the unbound electron are not quantized and form
the continuum states of the atom.
 The ionization energy, I, of an atom is the minimum energy
required to remove an electron from the atom that is in its
ground state (the state of lowest energy of the atom).
e.g: For H, there is one electron. The atom is in its ground state
when the electron is in the state characterized by n=1.
I=-En=1= hc H = 2.179 10-18J= 13.6 eV
B. Shells and subshells  In hydrogenic atoms, all orbitals in the same shell have the
same energy (characterized by n). NB: In many-electron atoms,
we’ll see that the subshells have different energies.
ml  |l |
ml
l
Number of orbitals
with different ml
l  (n-1)
n=
l=
n
1
2
3
4
K
L
M
N
0
1
2
3
s
p
d
f
C. s orbitals
 For Hydrogen (Z=1), the 1s orbital (n= 1, l= 0, ml= 0) has
a spherical symmetry. The wavefunction decays
exponentially from a maximum value, 1/(a03)1/2, at the
nucleus.
r

1
 1Hs 
e a0
1
/
2
a03
T<<, V>>
 
Radial node
1s
2s
 Virial theorem: for V(r)= rb  2 <T> = b <V>. For n=1,
l=0, V(r) is Coulombic= r -1  <T> = -1/2 <V>.
<T> = -1/2 <V>
Actual ground state
wave-function has to
fulfill the Virial theorem
T>>, V<<
 All the s orbitals are spherical, but differ by the number of
radial nodes (where the probability to find the electron at the
node is zero). The higher the “n”, the higher the number of High curvature  T >> and V <<
nodes. Because: the higher is “n”, the more there are
solutions for the associated Laguerre polynomial Ln,l. The High amplitude  e is close to
position rnode of the nodes are found by solving Ln,l= 0, such the nucleus
that R(rnode)= 0  2(rnode)=0.
D. The mean radius <r>
This is the expectation value of the operator “r”
r   * rˆ d   r |  |2 d
Spherical coordinates
d  r 2 dr sin  d d
  2
r     r Rn2,l | Yn,l |2 r 2 dr sin  d d
0 0 0
r
n ,l
 1  l (l  1)  a0
 n 2 1  1 

2
2
n

 Z

For a given n, the mean radius
follows the order d < p < s
E. Radial distribution functions
For a s orbital, the probability density ||2 to find the electron is
higher close to the nucleus: ||2  exp(-2Zr/a0)
But, close to the nucleus, there is not a lot of room for the electron
to be. Hence, the probability to find the electron in a spherical shell
of thickness dr is lower. Because the volume of the shell d= 4r2
dr (surface of a sphere  thickness) is smaller.
 The most probable radius at which the electron will be found is
a compromise between the volume of the spherical shell and the
probability density ||2.
 The probability P(r)dr of finding the electron at any angle
(dd) at a constant radius r is the integral of the probability
density ||2 over the surface of a sphere of radius r:
 2
P(r )dr    Rn2,l (r ) | Yn ,l ( ,  ) |2 r 2 dr sin  d d
0 0
 2
 r R (r ) dr   | Yn ,l ( ,  ) |2 sin  d d  r 2 R 2 (r ) dr
2
2
n ,l
0 0
 For the 1s orbital of H, the maximum of P(r) occurs at r=a0, the Bohr radius.
http://www3.adnc.com/~topquark/quantum/hydrogenmain.html
For the stationary states nlm(x,y,z) of the hydrogen atom, the probability densities
|nlm(x,0,z)|2 are plotted in a plane containing the z axis.
spherical coordinates:
F. p orbitals
p0  R2,1 ( r )Y1,0 ( ,  ) 
x= r sincos; y= r sin sin; z= r cos
Z
 
1/ 2  a 
4( 2 )  0 
1
1 Z
p1  R2,1 (r )Y1, 1 ( ,  )   1/ 2  
8  a0 
1
  1/ 2 r sin  e i f (r )
2
5/ 2
r cos  e
5/ 2

re
Zr
2 a0

Zr
2 a0
 r cos  f ( r )  z f ( r )  pz
sin  e i
Different rotations: clockwise,
counter-clockwise
z
r
y
x
1
p x   1 / 2  p 1  p 1   r sin  cos  f (r )  x f (r ) These linear combinations are real
functions. They are also solutions of
2
the SE for the atom, but have no net
i
p y  1 / 2  p 1  p 1   r sin  sin  f (r )  y f (r ) angular momentum vs. the z-axis
2
compared to p1.
Nodal plane
+
-
Demonstration:
“if 1 and 2 are solutions of the Schrödinger Equation,
i.e., H1= E 1 and H2= E 2,
then the linear combination  = c11+ c22 is also a solution”
H  = H (c11+ c22)= c1 H1+ c2 H2
H  = c1 E1+ c2 E2= E(c11+ c22)
H=E
Hence,  is indeed also a solution.
It means that px and py are solutions
of the SE for a hydrogenic atom
1
 p1  p1 
1/ 2
2
i
p y  1/ 2  p1  p1 
2
px  
F. d orbitals In the shell n=3, there are 3 subshells:  l= 0  ml= 0
 1 s-orbital
 l= 1  ml= -1, 0, 1  3 p-orbitals
 l= 2  ml= -2, -1, 0, 1, 2 5 d-orbitals
From a linear combination of these
orbitals, we can build real orbitals that
are also solutions of the SE for the
hydrogenic atom: dxy, dyz, dzx, dx2-y2, dz2
http://www.albany.net/~cprimus/orb/
The
electron
orbitals
presented here represent a
volume of space within which
an electron would have a
certain probability of being
based on particular energy
states and atoms.
3.1.3 Spectroscopic transitions and selection rules
 From the Bohr frequency condition, the excess of energy produced when the atom undergoes
a transition from an excited state (characterized by an electron in the orbital n1,l1,ml1) to a lower
energy state (for which the electron is in the orbital n2,l2,ml2) is discarded as a photon of
frequency .
 All the possible transitions, (n1,l1,ml1)[(n2,l2,ml2)+h], are not permissible because the
global angular momentum of the system should be conserved. The photon has an intrinsic spin
angular momentum corresponding to s=1. The change in angular momentum of the electron
must compensate for the angular momentum carried away by the photon.
d orbital (l=2)  s orbital (l=0) + h (s=1)
Transition forbidden because 2 0 + 1
 The probability for a transition to occur, i.e. the intensity of the different lines in the
spectrum, is given by the transition dipole moment fi between an initial state i and a final
state f :
 fi  e  f r  i d  e f r i
The use of transition dipole moment to evaluate
the intensity of a line is a general method that
can be applied to any atoms or molecules. For a
hydrogenic atom, the introduction of the final
and initial states, characterized by two different
sets of numbers (n, l, ml), leads to the selection
rules indicating which transition is allowed:
l= 1
ml= 0, 1
Note: There is no specific rule for the principal
quantum number, because it does not carry
information about the angular momentum.
3.2 Structure of many-electron atoms
The Schrödinger equation for a three-body system cannot be solved analytically (e.g.: He= 1
nucleus, 2 electrons)  approximations required.
3.2.1 The orbital approximation
The spatial electronic wavefunction of an atom with n electrons is written: (r1, r2,..., rn),
where the ri are the vectors between the nucleus and the electron i. Because the electrons
interact with each other via a Coulombic potential Vee, (r1, r2,..., rn) is exactly unknown. The
electronic Hamiltonian of the atom is:
n
n

1
Z
i
Hˆ elect = Te  Ven  Vee   
  + 
i=1 2mi
i=1 ri
i=1 i>i ri  ri '
n
2
A first approximation is to neglect the interaction between electrons. Consequently, the
Hamiltonian can be divided in a sum of terms, each acting on one electron: the monoelectronic
Hamiltonians hi. The Schrödinger equation can be split in n monoelectronic equations:
hi (ri)=i (ri)
i=1, … n
 The wavefunction of the atom is then the product of monoelectronic wavefunctions (ri).
This is the orbital approximation: each electron occupies its “own” orbital (ri) :
(r1, r2,..., rn)= (r1) (r2) ... (rn)
In order to associate an orbital (ri) to each electron in a many-electron atom, we can think to
use the solutions of the SE for a hydrogenic atom (1e- + 1 nucleus of charge Z).
He: 2e- + nucleus (Z=2)  electron configuration: 1s2….but is it  or  ?
 The answer is given by the Pauli exclusion principle.
Li: 3e- + nucleus (Z=3)  using the hydrogenic orbitals, the possible electron configuration
are: 1s2 2p1, 1s2 2s1 or 1s3. The last proposal can be directly rejected because electrons are
Fermions and they follow the Pauli exclusion principle:
“No more than 2 electrons may occupy any given orbital. If
2 electrons occupy one orbital, then their spin must be paired
(antiparallel: ).”
 The spin of the electron must be taken into account to
understand the electronic configuration of atoms. A system of
two paired electrons has a resultant spin of zero.
A. Pauli exclusion principle
 A system of Fermions (particles with half integral spin) is described by a Total
wavefunction (x1, x2,..., xn) (where xi= ri, i. i is the spin coordinate of the electron i:  or
), which is antisymmetric: when the labels of any two identical fermions are exchanged, the
total wavefunction changes sign:
(x1, x2,..., xn)= - (x2, x1,..., xn)
 In the orbital approximation: each electron is associated to a spinorbital that is the product
between a spatial orbital and a spin function:
(xi)= (ri) s(i)
 Let’s consider a system of 2 electrons that occupy the same spatial orbital (r). The total
wave function of the system is (1,2)= [(1)(2)] S(1,2). Where the spatial part of the total
wavefunction is (1)(2) and it is symmetric: (1)(2)= +(2)(1). Hence, in order to have
the total wavefunction (1,2) antisymmetric, the total spinfunction S(1,2) must be
antisymmetric: S(1,2)= -S(2,1).
What is this antisymmetric spinfunction S(1,2)?
 Let’s look at the different possible association of 2 spins:
- The two spins are “up” ( ) : S1(1,2)= (1)(2)  symmetric
- The two spins are “down” ( ) : S2(1,2)= (1) (2)  symmetric
- one “up” and one “down” () …. but as the particle are indiscernible, we cannot
distinguish between (1)(2) or (1)(2). Two linear combinations are then used in order
to represent the situation where the spins are different:
S3+(1,2)= 1/21/2 {(1)(2) + (1)(2)}  symmetric
S4-(1,2)= 1/21/2 {(1)(2) - (1)(2)}  antisymmetric
 Only the spinfunction S4-(1,2) is antisymmetric and characterized by two different spins.
That proofs the Pauli exclusion principle: 2 electrons in the same spatial orbital must be
paired, i.e. must have different spins. The total wavefunction of a system of 2 electrons in
the same orbital is:
(1,2)= [(1)(2)] 1/21/2 {(1)(2) - (1)(2)}
 He: 2e- + nucleus (Z=2)  electron configuration: 1s2….and it is  and not  ?
B. Penetration and shielding
 For hydrogenic atoms: 2s and 2p are degenerated
 For many-electron atoms: the 2s orbital lies lower in energy than
the 2p orbitals….
Why?
If the nucleus has a charge Z, the 2s electrons do not feel the potential
created by Z, because the 1s electrons shield this nuclear charge. The
2s electrons feel an effective nuclear charge Zeff:
Zeff= Z - 
 is the shielding constant. This shielding constant is not the same for
an electron in 2s and 2p, because these orbitals have different radial
distributions (see Figure). A s orbital has a greater penetration
through inner shells than a p orbital of the same shell.
 the 2s electrons feel a less shielded nuclear charge, i.e. more
positive, than the 2p electrons. The 2s electrons will be more tightly
bound to the nucleus.
 Usually the energies of subshells in a many-electron atom lie in
the order:
s<p<d<f
Li: 3e- + nucleus (Z=2)  1s2 2p1 is rejected, 1s2 2s1 is the correct
configuration.
2s
1s
For heavy atoms, the rules for the energy order
of the orbitals (s<p<d<f) fails.
Example: the d subshells have 5 orbitals, a
maximum of 10 electrons can be there…. These
electrons are close to each other in space.
 For heavy atoms, the electron-electron
repulsion energy is important and comparable to
the energy difference between the 4s and 3d
orbitals.
 That can lead to deviation of usual energy
order of the orbitals (s<p<d<f)
C. Hund’s rule
C: 6e- + nucleus (Z=6)  1s2 2s2 2p2 …. The 2 last electrons in the 2p orbitals are the valence
electrons.
But is it 2px12py1 or 2px2?
 The 2 electrons repel each other by Coulombic repulsion,
they prefer to be spatially far from each other. The
configuration 2px2 is less stable than when the 2 electrons
are on two different spatial orbitals: 2px12py1 is favorable.
 When adding electrons in the hydrogenic orbitals in order to estimate the electron
configuration of atom (building-up principle), this rule has to be followed:
“electrons occupy different orbitals of a given subshell before doubly occupying any one of
them”
But is it 2px1 () 2py1() or 2px1 () 2py1()?
 This question is resolved by the empirical Hund’s rule:
“An atom in its ground state adopts a configuration with the greatest number of unpaired
electrons”
It is then 2px1 () 2py1(). This is due to a quantum mechanical property: the spin
correlation
 Let’s consider the situation where 2 electrons are in different orbitals a(1) and b(2)
(like for C: 2px, 2py). The spatial part of the total wavefunction (1,2) cannot be the
simple product: a(1)b(2), because it suggests that we know which electron is in which
orbitals, whereas we cannot keep track of electrons.
The correct description is a linear combination, either of the two following wavefunctions:
+(1,2)= 1/21/2 {a(1)b(2) + b(1)a(2)}  symmetric
-(1,2)= 1/21/2 {a(1)b(2) - b(1)a(2)}  antisymmetric
In order to have an antisymmetric total wave function (1,2), +(1,2) should be multiplied
by an antisymmetric spinfunction S(1,2) and -(1,2) by a symmetric spinfunction.
Antisymmetric spinfunction: Singlet
S1(1,2)= +(1,2) S4-(1,2)= 1/2 {a(1)b(2) + b(1)a(2)} {(1)(2) - (1)(2)}
Symmetric spinfunction: Triplet
T2(1,2)= -(1,2) S1(1,2)= 1/21/2 {a(1)b(2) - b(1)a(2)} {(1)(2)}
T3 (1,2)= -(1,2) S2(1,2)= 1/21/2 {a(1)b(2) - b(1)a(2)} {(1) (2)}
T4 (1,2)= -(1,2) S3+(1,2)= 1/2 {a(1)b(2) - b(1)a(2)} {(1)(2)+ (1)(2)}
What happens if the electron 1 approaches the electron 2: r1=r2 ?
+(r1,r2=r1)= 1/21/2 {a(r1)b(r1) + b(r1)a(r1)} 0
-(r1,r2=r1)= 1/21/2 {a(r1)b(r1) - b(r1)a(r1)} = 0
 When the two electrons are close to each other, the total wavefunction with an
antisymmetric spinfunction, S1(r1,r2=r1) 0 : two electrons with different spin (antiparallel)
have a certain probability to be at the same position.
 The electrostatic repulsion is increased: the system in its singlet state is destabilized
 For two electrons having a symmetric spinfunction, the total wavefunction is zero:
T2(r1,r2=r1)= T3(r1,r2=r1)= T4(r1,r2=r1)=0 : when 2 electrons in 2 different orbitals have
the same spin (parallel), they are repelled to each other thanks to their spin.
 that prevents an additional electrostatic repulsion. The Triplet state (2 spins parallel) is
more stable than the singlet state (2 spins antiparallel): this is the explanation for Hund’s
rule.
This quantum effect is called the spin correlation.
 The carbon atom has a valence configuration: 2px1 () 2py1() and not 2px1 () 2py1() .
3.2.2 Singlet and triplet states
Triplet State : Symmetric spinfunction
T2(1,2)= 1/21/2 {a(1)b(2) - b(1)a(2)} {(1)(2)}
T3 (1,2)= 1/21/2 {a(1)b(2) - b(1)a(2)} {(1) (2)}
T4 (1,2)= 1/2 {a(1)b(2) - b(1)a(2)} {(1)(2) + (1)(2)}
 When two electrons have “parallel” spins, they have a
nonzero total spin angular momentum.
 Spin Multiplicity g = number of states having the same
energy: g= 2S+1. Here, MS=1/2+1/2=1, there are 3 Triplet states
with the same energy.
Singlet State: Antisymmetric spinfunction
S1(1,2)= 1/2 {a(1)b(2) + b(1)a(2)} {(1)(2) - (1)(2)}
 When two electrons have “antiparallel” spins, they have a
zero total spin angular momentum.
g= 2S+1. Here, S=1/2-1/2=0, there is only one Singlet state.
Spectrum of the He atom:
Z=2, 2 s electrons.
During the excitation, only one electron is excited. When the atom comes back to a lower
energy state, two types of transitions are observed: Between triplet states: T  T, Between
singlet states: S  S. But not: T  S or S  T
 The probability for a transition is given by the
transition dipole moment fi between an initial state
i and a final state f :
Both states of the atom are characterized by a spatial
function  and a spinfunction S.
 fi  e   f r i d  e  f r  i dr  S f Si d
 If the initial and final states have both a
spinfunction of the same symmetry, the transition
dipole moment is non-zero: the transition is allowed.
 If the two states have different spinfunction
symmetries, the transition is forbidden.
3.2.3 Spin-orbit coupling
A moving charge creates a magnetic fields:
 e- moves around it-self  a magnetic moment arises from its
spin: s= es
 e- moves around the nucleus in an orbital  a magnetic moment
arises from its orbital angular momentum: l= el
 The two magnetic moments (antiparallel to the corresponding
angular momentum), s and l interacts with each other, like two
magnets: this is the spin-orbit coupling.
 When the angular momenta (s and l) are parallel, as in (a), the
magnetic moments are aligned unfavorably, the energy of the
system is high. When they are opposed, as in (b), the interaction is
favorable, the energy of the electron system is stabilized.
 This magnetic coupling is the cause of the splitting of a
configuration into levels.
The strength of the coupling depends on the relative orientation
of the orbital angular momentum and the spin angular momentum.
A. The total angular momentum
The relative orientation of the orbital angular momentum, l, and the spin angular momentum,
s, is characterized by their vector sum giving the total angular momentum: j=l+s
 The strength of the coupling depends on the total angular momentum j.
Example: The configuration with 1e- in a p orbital (l=1) can be split in 2 levels.
A) if “s” and “l” are parallel (s and l are parallel), j is high: j= 1+1/2= 3/2. This is a
situation where the magnetic interaction is unfavorable. The energy of this level is destabilized.
B) if “s” and “l” are antiparallel (s and l are antiparallel), j is low: j= 1-1/2= 1/2. This is a
situation where the magnetic interaction is favorable. The energy of this level is stabilized.
 The energy of the two levels, characterized by
(j, l, s) are given by the expression:
Ej,l,s= ½hcA {j(j+1)-l(l+1)-s(s+1)}
A= spin-orbit coupling constant
 In spectroscopy, new lines and new selection
rules appear due to this spin-orbit coupling, when
the electron involved in the transition is in an
orbital with l0.
Absorption spectrum of Saturnus
The light emitted from the sun is absorbed by atoms contained in the
ring of the planet. The absorption spectrum gives us the composition
of the ring: ice, mineral, iron ...
The shift in frequency observed for an absorption line appearing on
both sides of the ring is explained like this: the ring is composed of
particles (from 10 cm to 10 m) rotating around the planet at a high
velocities. The spectral line appears shifted from the earth because of
the speed of rotation of the particles: the Doppler effect modifies the
apparent frequency of the line. From the measured frequency shift,
the speed of the particles can be deduced (20km/s).