Ph235
Autumn 2008
F. Merritt
Oct 16, 2008
Lecture 6 (October 10, 2008): (version 1.1; 16-Oct-08)
Atomic systems and the periodic table. Part 2
Questions from Lecture 5:
Where did the formula for hydrogen energy levels come from? I don’t remember
 , but you wrote
mc 2  Z 2 2
(1.1)
En  
2n 2
Answer: We might not have actually written it in this form in class, but we certainly
derived
me4 1
1
(1.2)
En   2 2   Ry 2 where Ry  =Rydberg  =13.6 ev
2 n
n
from this, it’s easy to derive (1.1) using the definition   e2 /  c  . Also recall
2
a0 
me e
e
2

c
me
  a0
 Bohr radius (=.529 10-8 cm)
 Compton wavelength of electron
(1.3)
re   a0
 classical electron radius
This would be a good time for you to review Shankar 13.3 and 13.4, where all these
formulas are discussed and are related to our present discussion of multielectron atoms
and the periodic table.
2
Question on HW#2:
Why doesn’t the electron just emit a photon when it drops to the ground state?
That’s what usually happens in atomic transitions.
Answer: Yes, but this case is a bit different. Consider the He atom as a two-particle
system, first particle= He  ion (containing the first electron), and second particle =
second electron. When the first electron drops to the ground state, that changes the mass
of the He  ion. The two particles are now no longer in a bound state; they were initially,
when the first electron was in the n=2 shell, but not after it drops to the n=1 shell. So the
2 particles simply fly apart. (See comment on HW#2 sheet on web).
=========================
Helium (Z=2)
The helium ion  He  has only one electron, but Z  2 . This means the ground
state energy is a factor of 4 times that of hydrogen, and in general
EnHe   4  EnH
(1.4)

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Autumn 2008
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Oct 16, 2008
We might expect the helium atom (with 2 electrons) to have energy of twice this, so a
ground state of E1He  8E1H  108.8 eV . But experimentally the ground state energy
has -79.0 eV.
It’s not hard to identify the problem. Equation (1.4) assumes that the Hamiltonian
for helium has a coulomb potential like hydrogen. But for all atoms with N electrons, the
inter-electron interactions are quite large. The Hamiltonian for atomic systems of N
electrons is
2
N 
Ze2 
e2

 N N
2
(1.5)
H   
j 
  
rj 
j 1 
 2m
 j 1 k  j 1 rj  ri
It is best to explicitly include the effect of the second term – but it is a lot easier to
think of the inner electrons as providing “shielding” for the jth electron. In the case of
helium, the ground state electrons are clearly equivalent, so we might imagine that each
one provides 50% shielding for the other (since each one must be the “inner” electron
half the time). That would mean that the effective Z of the helium nucleus is about 1.5,
and that would give a ground state energy of -61.2 eV. This at least moves us in the right
direction (but overcorrects substantially). The effective Z that gives the correct groundstate energy in the absence of ee interactions is Z eff  1.70 . [We will derive this result
next week using the variational methods (Shankar 16.1)].
But this picture ignores the variation of effective Z with r, and also treats the two
electrons as independent, uncorrelated states. Recall that a) the overall electron wf must
be antisymmetric under exchange of electrons (since they are fermions), which means
that the product of space wf and spin wf for the 2-electron system must be odd. We
know that the spin wf can be either symmetric (triplet) or antisymmetric (singlet). The
triplet spin function must be associated with an antisymmetric space wf for the the 2e
system, and the singlet must be associated with a symmetric space function.
But we saw in Ph234 (Griffiths Section 5.1.2, or Eisberg) that the rms separation
between two particles in a symmetric wf is less than for antisymmetric. This is what we
would expect, since the exclusion principle prevents 2 electrons from occupying the same
spatial state (if they are in the same spin state). But this means the ee electrostatic term in
the Hamiltonian should be smaller for the triplet spin state (antisymmetric space) than for
the singlet spin state (symmetric space). The difference can be reasonably large
( 1 eV for the 2S state ).
[Suppose 2 electrons occupy states spatial a and b (orthogonal). Then the triplet
wf with Sz=0 is
 abtrip  12 (ab  ba)  12     
(1.6)
while the same state with singlet spin (S=Sz=0) is
 absing  12 (ab  ba)  12     
(1.7)
In this equation, “a” and “b” mean the spatial part of the wave-functions of the first and
second particles, e.g.
a  RnlYlm
and
b  Rnl Yl m
(1.8)
ab  ba   RnlYlm 1  Rnl Yl m 2   Rnl Yl m 1  RnlYlm 2
Note that both states [equations (1.6) and (1.7)] are antisymmetric under total exchange.
The triplet has S=1, since applying S  will give a new state with S z  1 , whereas the
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Oct 16, 2008
singlet state has S=0, and applying S  gives zero (kills the state). Also note that the two
states have identical expectation values for x1 and x2 , but different expectation values for
 x1  x2 
2
. There is no classical analogue for this.
[Note also that when the 2 spatial states are the same, then the spatial part is simply aa,
which is obviously symmetrical. Then overall symmetry of the wavefunction requires
that the spin function be antisymmetric. In this case, there is no triplet state.].
But we know that correlations are very important. We have already (from
Griffiths section 5.1.2) that the expectation value of the separation of two particles in
states a and b is greater if the spatial wf is antisymmetric than if the spatial wf is
symmetric. This would mean that the ee interaction is smaller for particles that are in a
spatially antisymmetric state. This means that the negative interaction energy of the state
should be smaller in magnitude, and consequently the total energy of the sytem should be
lower in absolute magnitude, for identical electrons that are in a spatially antisymmetric
state.
Indeed, there are different energy levels for helium eigenstates that have S=1
(triplet) and S=0 (singlet). The former are called orthohelium, and have antisymmetric
spatial wavefunctions. The latter are called parahelium, and have symmetric spatial
wavefunctions. Because of the exclusion principle, orthohelium states (spatially
antisymmetric) have lower energy than parahelium states (spatially symmetric). This is
shown in Griffiths Figure 5.2 (page 213).
The parahelium and orthohelium states are also shown on the physics website of
Georgia State University:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/atomstructcon.html#c1
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http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/helium.html .
In fact, this website seems quite good to me, and is worth exploring. It covers most of
the topics that we’ll be addressing in the next few lectures. To find the helium levels, go
to the GSU website above and click on the “Helium” bubble in the upper center region,
just below “Atomic Structure” (which gives the map shown above). Any reference that I
give to GSU will refer to one of the links on this page. For example, if you click on the
“Helium” bubble, you’ll find the diagram and discussion of orthohelium and parahelium
states that we have just been looking at. Click on the links on this page to find a graph of
the radial wavefunctions, which shows why shielding is smaller for s-wave electrons than
for p-wave, and smaller for p-wave than d-wave. To see this effect in higher-Z atoms,
find the page showing the lithium and sodium energy levels; these are compared to the
corresponding energy levels of hydrogen..
We will return to Helium later in the course (soon!) and look at different ways of
dealing with the Hamiltonian in a more quantitative manner (Shankar Ch 16 and 17). But
first we will review spectroscopic notation and the general structure of the Periodic
Table.
Spectroscopic notation for atomic systems
There are two ways of representing the eigenstates of an atomic system with
nuclear charge Z. The subshell occupancy is given by writing the state as a product of
terms (nl ) factors, where n is the principal quantum number and l is the orbital angular
momentum of an electron; but for largely historical reason, l is written as a letter, coded
as follows: s (sharp) means l=0, p (principal) means l=1, d(diffuse) means l=2, and f
(fundamental) means l=3. Above this, the coding just follows the alphabet (g, h, …).
The electron configuration of the ground state for the first 12 elements is thus:
Z
1
2
3
4
5
6
7
8
9
10
Sym
Name
H Hydrogen
He
Helium
Li
Lithium
Be Beryllium
B
Boron
C
Carbon
N
Nitrogen
O
Oxygen
F
Flourine
Ne
Neon
Electron configuration (nl notation).
(1s)
(1s)2
(1s)2 (2s)
(1s)2 (2s)2
(1s)2 (2s)2 (2p)
(1s)2 (2s)2 (2p)2
(1s)2 (2s)2 (2p)3
(1s)2 (2s)2 (2p)4
(1s)2 (2s)2 (2p)5
(1s)2 (2s)2 (2p)6
But this notation does not completely specify the atomic state. One can also
choose the mutually commuting operators H , L2 , S 2 , J 2 , J z where these variables now
refer to the sums over all electrons. The quantum numbers L, S, and J are incorporated
into the standard spectroscopic notation of atomic states, as
2 S 1
LJ
(1.9)
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Oct 16, 2008
The total angular momentum L is written using the SPDFG notation described above for
electron states. So the ground state of hydrogen is written as (S= 12 , L=0, J= 12 )
2
Hydrogen ground state:
S1/ 2
(1.10)
while that of helium is written as (S=0, L=0, J=0)
Helium ground state: 1S0
(1.11)
Lithium (Z=3)
Here the ground state must have a filled n=1 shell with a single electron in the
n=2 shell. We would expect the Lithium ion L (with 2 electrons) to be just like the He
2
ground state, but with energies higher by a factor of  32  . Then we would expect the
lithium energy levels to be similar to the n=2 state in hydrogen, which has an ionization
energy of about 3 eV. But because of the shielding effect, the s-wave state has a lower
energy for lithium than for hydrogen.
This is shown on the GSU website: Go from “Atomic structure” to “Helium” to
“Orbital Dependence”; read all of this, then click on the link “What is the origin of the
orbital quantum number dependence?”.}
For the ground state of lithium, then, l3  0 (s-wave), so S 12 , L  0, J  12 and
Lithium ground state:
2
S1/2
(1.12)
Beryllium (Z=4):
The 4th electron completes the (2s) subshell, so S  0, L  0, J  0
Beryllium ground state (1s) 2  2s  : 1S0
2
(1.13)
So far, the (nl) configuration of the valence electrons has uniquely specified S, L, and J of
the system in the ground state. Either S=0 or L=0 (or both = 0) for the first 4 elements,
and the determination of the ground state has been trivial.
==========================================
Higher Z atoms (Z>4) and the Periodic Table:
As soon as we enter the (2p) shell, things become more interesting. In order to
specify the ground state completely, we need to know S,L, and J of the system, and these
are no longer uniquely determined by the electron (nl) orbitals. The prescription for
determining these is given by Hund’s Rules 1, 2, and 3 (developed below and in next
lecture).
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Autumn 2008
F. Merritt
Oct 16, 2008
But even more important than these rules are the relative energies of the different
subshells. This ordering, which determines how the subshells are filled as we go to
higher Z, is sometimes called “Hund’s Rule #0”.
We have already seen that the S-wave states (L=0) are lower in energy than the
P,D,F… states because of the reduced shielding for these states. We might have expected
all subshells to fill in order of increasing n, and for a given n to fill in order of increasing
l. That is how the periodic table starts out, but when Z becomes greater than 18 we find
that the (n+1,s) subshell is even lower in energy than the (n,p) (n,d) subshell. When we
get above Z=54 we find (n+1,s) is lower than both (n,p) (n,d) and (n,d) (n-1,f). (And note
that for a given n, the lower-l subshells are always lower in energy than the higher-l
ones.)
The order in which (n,l) subshells is filled is shown in the diagram below, which
is a kind of schematic picture of the Periodic Table (which is shown in full on the next
page).
The ordering of subshells shown in this figure is what I will call “Hund’s Rule
Zero”:
Learn how to reproduce this without looking at it, and relate it to the ordering to the plots
in GSU “Orbital Dependance” plots. That’s the best way to understand the structure of
the Periodic Table.
When a particular subshell is being filled (like the (2p) subshell), the inner shells
shield the nucleus so that the valence energy levels are similar to those of hydrogen.
Because of this, one sees a periodic structure in the atomic radius, plotted as a function of
Z, and also in the ionization energy plotted as a function of Z (see plots below).
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Oct 16, 2008
** The Periodic Table (from WebElements.com) **
Atomic radius as a function of Z
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** Ionization energy as a function of Z (from Wikipedia): **
Boron (Z=5) through Neon (Z=10):
With Boron, we enter the (2p) subshell. Since there is only one valence electron
(valence electrons are those beyond the filled subshells), we know that S=1/2 and L=1.
The sum of these is J, and from our rules for adding angular momentum we know that the
only possible values are J=1/2 and J=3/2. From the rules we have developed so far, we
cannot tell which of these is the ground state.
The next element is Carbon (Z=6), and the electron configuration is
2
2
2
1s   2s   2 p  . Here, the two valence electrons could be in either a singlet or triplet
spin state so S=0 and S=1 are both possible. Also, the sum of the two p-waves could give
a total L of 0, 1, or 2. We don’t know which of these will give the lowest energy. The
possible values of J would appear to range from 0 to 3. [But note that some of these will
be excluded by the antisymmetry requirement on the total wavefunction; e.g., S=1 and
L=2 would give a wf which is symmetric under particle exchange.]
For the higher Z states, there are even more possibilities. In general, these are all
possible states, but we don’t know what their relative energies are or which will
correspond to the ground state. But we can figure it out! That is what Hund’s Rules do
for us. See Griffiths page 218, or GSU “Term Symbols” and GSU “Hund’s Rules”.
First I will state the rules, and then go through the reasons for them:
Hund #1: The lowest energy state will have the maximum value of S (total spin)
consistent with the electron configuration specified by Hund Zero.
Hund#2: The lowest state will have the maximum value of L (total orbital angular
momentum consistent with the previous rules (Hund#0 and Hund#1).
Hund#3: For shells that are half-filled or less, the lowest state is the one with
J  L  S ; for shells that are more than half filled, the minimum energy state
is the one with J  L  S .
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