Download N - University of St Andrews

Many-electron atoms
• We were able to solve exactly the one-electron atom
( H , He + , Li + + ...)
• For any other atom with more than one electron:
 p
Ze 
e
 + ∑
H = ∑ 
−
4πε 0 ri  i > j 4πε 0 rij
i =1  2m0
N
2
i
2
2
for N electrons
Bad news: too complicated, even for only 2 electrons.
We can’t calculate the eigenvalues and functions exactly
Good news: it is still possible to solve the problem,
to a VERY good approximation.
It turns out that:
• even when several electrons are bound to the nucleus, their
individual electron states can be characterised by the three
quantum numbers
n, l, m
so we can still think in terms of orbitals.
• however the corresponding energies are strongly modified,
with respect to the one-electron problem, by the interactions
of the electrons with each other.
Spectroscopic notation for orbitals:
l= 0, 1, 2 , 3, 4….
s, p, d, f, g... state
E.g. (n,l)=3,1 is a 3p state
Let’s order these orbitals starting from the ground state:
1s , 2s , 2p , 3s , 3p , 3d , …
E1 p E2 p
E3
Let’s consider now the ground state of a N-electron atom:
will the electrons all sit in 1s ? …
Of course not, because of the
Pauli exclusion principle:
No two electrons have the same set of quantum numbers.
This means that we have to occupy several orbitals.
Electron configuration:
state of occupation of the energy levels
(or terms) of an atom by electrons
For example, the electron configuration
for the ground state of Lithium (3 electrons) is
We can allocate two electrons in each
orbital because the electron has spin ½,
i.e. its spin state can be either up or down.
So we have two spin states for each orbital.
2
1s 2 s
2 electrons
in 1s orbital
1 electron
in 2s orbital
The Pauli principle then tells us what is the maximum number
of electrons that we can allocate to each orbital:
orbital s (l=0, m=0):
up to 2 electrons
orbital p (l=1, m=0, ±1):
up to 6 electrons
orbital d (l=0, m=0, ±1, ±2):
up to 10 electrons
e.g: the ground state of Na (Z=11) is
2
2
6
1s 2s 2 p 3s
These ideas apply to all atoms. Let’s consider now the case
of alkali atoms.
H&W chapter 11
Alkali atoms
• They belong to the first column of the periodic table:
Li Na K Rb Cs Fr
Z= 3 11 19 37 55 87
• Next simplest spectra after 1-electron atoms
• The reason for this is that for the alkalis
electron configuration = closed shell + 1 electron with
next quantum number n
(known as valence electron)
e.g. Lithium ground state is
• Ionisation potentials:
2
1s 2 s
24.5eV for He
5.40 eV for Li
closed shell
(He - atom)
closed shells
are especially stable
Our aim is to have a qualitative understanding of the energy levels of
alkali atoms. For this it is useful to start from a “Bohr orbit” picture:
+Ze
-(Z-1)e
closed shell
valence electron
(new n)
Because of its larger quantum
number n, the valence electron
spends all of its time outside the
closed shell.
The valence electron “sees” a net
charge +e at the nuclear site.
Hence according to this model, the valence electron experiences
the same potential V(r) as the H-atom, and has the same spectrum:
V(r)
r
− e2
4πε 0 r
Ry
En = − 2
n
The valence electron is perfectly screened from the
nucleus by the closed shell.
Let’s now generalise to elliptical orbits (Bohr-Sommerfeld model):
+Ze
-(Z-1)e
closed shell
valence electron
When the electron gets closer to the nucleus, it experiences unscreened
nuclear potential:
V(r)
r
− e2
4πε 0 r
− Ze 2
4πε 0 r
almost perfect screening at large r
almost no screening at small r
The screening now depends on the eccentricity of the orbit,
i.e. on the angular momentum :
Small angular momentum (small l)
more eccentric orbit
electron gets closer to nucleus – less screening
stronger, i.e. more negative, binding energy
For instance in Lithium the valence electron has n=2
and based on our reasoning the electron configuration of the
2
ground state is 1s 2 2s and NOT 1s 2 p , because the orbital
2s is more strongly bound than 2p.
More generally, we expect all the energy levels to depend on
the quantum number l, as well as n:
En for hydrogen ⇒ En ,l for many - electron atoms
Degeneracy of levels
• l degeneracy is lifted:
states with same n but different l have different energies.
• m degeneracy is still true:
states with same l but different m have the same energy.
(This makes sense because m is the projection of the angular momentum along the
z-axis, which is arbitrarily defined.)
We can arrive at the same conclusions if we use radial probability
densities instead of classical orbits:
lithium
sodium
Due to “imperfect screening”, the energies of Li and Na
always lie lower than the corresponding energies of the H-atom.
We can quantify screening effects by introducing quantum defects:
RH hc
En = − 2
n
E n ⇒ E n ,l
for hydrogen
RH = Rydberg constant
for H-atom
for many-electron atoms because we know
different shapes of orbital (different ls )
mean different energies.
E.g. for sodium:
E n ,l


1
1

= − RNa hc 2 = − RNa hc
2 
neff
 [n − ∆ (n, l )] 
quantum defect
Quantum defects
e.g. for Na atom:
l
0 s
1 p
2 d
∆ ( n, l )
are determined empirically,
n=3
4
5
1.373
0.883
0.010
1.357
0.867
0.011
1.352
0.862
0.013
These quantum defects are
•large for s electrons
•decrease with increasing orbital quantum number l
•largely independent of quantum number n
They indicate the level of screening of the s,p,d,f etc
electrons by electrons of the inner shells
Another application of the idea of screening:
Electron configuration for the ground state of K (Z=19):
2
2
6
2
6
1s 2s 2 p 3s 3 p ...
Is the last electron in 3d or 4s ?
On average, the 3d electron is closer
to the nucleus than 4s, hence the energy
is lower for 3d in hydrogen.
r2 R
However, due to this bump,
a 4s electron can get closer to the nucleus
than a 3d, i.e. in many-electron atoms
the 4s better penetrates the closed shell and
its energy decreases. So it turns out that the last
electron in potassium is 4s.
2
GROTRIAN DIAGRAM
Transitions only occur
between adjacent columns
∆l = ±1 selection rule