# Lecture 21

advertisement ```Physics 249 Lecture 21, Oct 24th 2012
Reading: Chapter 7.6, 7.7, 7.8, then 7.4(magnetic moment) and 7.5
Homework due Friday Oct 26th. Available on web site.
1) Spin
In order to explain the way the orbitals are filled in the ground state of multi-electron
atoms Pauli suggested two ideas.
A forth (fifth) quantum number: spin.
It was observed that multiple transition energy lines could be seen from something as
simple as a lowest s wave orbit in the presence of an extra field. Different angular
momentum states can have different energies in the presence of an extra field. This
observation indicated that the electron might have an additional quantum number
associated with angular momentum.
This idea is analogous to the idea the gravitationally orbiting objects can have orbital and
spin (rotational) angular momentum.
The explanation that fits the data is that the electron has an additional intrinsic angular
momentum called spin. Note that the electron has no detectable internal structure so
there is nothing that should in principle be able to literally spin. We now understand that
“spin” is a result of solving the relativistic version of the Schrodinger equation known as
the Dirac equation. The average absolute value of the electron momentum is fairly high
indicating that a relativistic treatment is needed. Just as we discovered above that the
wave equation solutions are Eigenstates of angular momentum they are also Eigenstates
of spin.
𝑆 2 → ℏ2 𝑠(𝑠 + 1)
𝑆𝑧 → ℏ𝑚𝑠
s=½
ms= -½, ½
2) Pauli exclusion principle.
Consider a wave function solution for an atom with 2 electrons. As an illustrative
example the 1D particle in a box is simplest. If the potential energy between them is
small compared to the energy levels it can be ignored and the two particles move
independently making the wave equation separable..
𝜓𝑛1,𝑛2 (𝑥1 , 𝑥2 ) = 𝜓𝑛1 (𝑥1 )𝜓𝑛2 (𝑥2 )
𝑛1 𝜋𝑥1
𝑛2 𝜋𝑥2
𝜓𝑛1,𝑛2 (𝑥1 , 𝑥2 ) = 𝐶𝑠𝑖𝑛 (
) 𝑠𝑖𝑛 (
)
𝐿
𝐿
the observable quantity is the probability density. If the particles are identical (as
electrons are) then:
2
|𝜓𝑛1,𝑛2 (𝑥1 , 𝑥2 )| = |𝜓𝑛1,𝑛2 (𝑥2 , 𝑥1 )|
2
This is true if
The wave equation is symmetric
𝜓𝑛1,𝑛2 (𝑥1 , 𝑥2 ) = 𝜓𝑛1,𝑛2 (𝑥2 , 𝑥1 )
or anti-symmetric
𝜓𝑛1,𝑛2 (𝑥1 , 𝑥2 ) = −𝜓𝑛1,𝑛2 (𝑥2 , 𝑥1 )
our original equation does not obey these conditions. Two solutions are:
symmetric
𝜓𝑠 = 𝐶 (𝜓𝑛1,𝑛2 (𝑥1 , 𝑥2 ) + 𝜓𝑛1,𝑛2 (𝑥2 , 𝑥1 ) = 𝜓𝑛1 (𝑥1 )𝜓𝑛2 (𝑥2 ) + 𝜓𝑛1 (𝑥2 )𝜓𝑛2 (𝑥1 ))
anti-symmetric
𝜓𝐴 = 𝐶 (𝜓𝑛1,𝑛2 (𝑥1 , 𝑥2 ) + 𝜓𝑛1,𝑛2 (𝑥2 , 𝑥1 ) = 𝜓𝑛1 (𝑥1 )𝜓𝑛2 (𝑥2 ) − 𝜓𝑛1 (𝑥2 )𝜓𝑛2 (𝑥1 ))
If you put two identical particles in the same energy state for the anti-symmetric version
you get zero and thus zero probability for them to exist in that state.
The interpretation if these ideas is that there are two unique types of particles. Fermions,
such as electrons, that have anti-symetric wave functions and bosons, such as photons,
which have symmetric wave functions. Again this is proven in a more rigorous way
when we consider the solution to the relativistic wave equation.
The practical consequence is that no two electrons can exist in exactly the same quantum
state in a system. For multi-electron atoms this means that the orbitals fill up progressing
through the energy levels and angular momentum states including spin.
3) Multi-electron atoms
In general as you progress to multi-electron atoms the electrons will fill orbitals
progressing through ms, m, l, n quantum numbers.
Using spectroscopic notation, which indicates n, l type shell (s l=0, p l=1, d l=2, f, l=3, g
l=4…), and number of m and ms states.
nl#(m+ms)
Then
H 1: 1s
(superscript 1 is implied)
He 2: 1s2
Li 3: 1s22s
Be 4: 1s22s2
B 5: 1s22s2p
C 6: 1s22s2p2
…
Note that 2s fills before 2p. The 2s orbit has two radial maxima, one at very low radius.
The 2s orbital is less shielded from the nucleas by the 1s orbital. This means it sees a
higher effective proton charge Zeff and has a higher negative energy (i.e. lower energy)
or in terms of ionization potential a higher value. It is interesting to use the energy level
to determine the effective amount of unshielded charge a given orbit sees.
2
𝑍𝑒𝑓𝑓
𝐸1
𝐸𝑛 = −
2
𝑛
more on this below
There are subtleties to the filling patterns.
Spin causes the equivalent of a small circulating current or magnetic dipole. When filling
orbitals with multiple ms states the ms states will fill with spins aligned first filling one
direction of spin and then the other. This is due to the force that aligns magnetic dipoles
in a magnetic field. For a p orbital
m = -1, ms = -1/2
m = -1, ms = -1/2 + m = 0, ms = -1/2
m = -1, ms = -1/2 + m = 0, ms = -1/2 + m = 1, ms = -1/2
m = -1, ms = -1/2 + m = 0, ms = -1/2 + m = 1, ms = -1/2 + m = -1, ms = ½
…
Note, which spin fills first or the direction of m filling is in principle arbitrary unless an
additional field causes one or the other state to be lower energy.
More in next lecture
For reference the radial wave function
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