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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 βspinβ 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. π 2 β β2 π (π + 1) ππ§ β βππ 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.. ππ1,π2 (π₯1 , π₯2 ) = ππ1 (π₯1 )ππ2 (π₯2 ) π1 ππ₯1 π2 ππ₯2 ππ1,π2 (π₯1 , π₯2 ) = πΆπ ππ ( ) π ππ ( ) πΏ πΏ the observable quantity is the probability density. If the particles are identical (as electrons are) then: 2 |ππ1,π2 (π₯1 , π₯2 )| = |ππ1,π2 (π₯2 , π₯1 )| 2 This is true if The wave equation is symmetric ππ1,π2 (π₯1 , π₯2 ) = ππ1,π2 (π₯2 , π₯1 ) or anti-symmetric ππ1,π2 (π₯1 , π₯2 ) = βππ1,π2 (π₯2 , π₯1 ) our original equation does not obey these conditions. Two solutions are: symmetric ππ = πΆ (ππ1,π2 (π₯1 , π₯2 ) + ππ1,π2 (π₯2 , π₯1 ) = ππ1 (π₯1 )ππ2 (π₯2 ) + ππ1 (π₯2 )ππ2 (π₯1 )) anti-symmetric ππ΄ = πΆ (ππ1,π2 (π₯1 , π₯2 ) + ππ1,π2 (π₯2 , π₯1 ) = ππ1 (π₯1 )ππ2 (π₯2 ) β ππ1 (π₯2 )ππ2 (π₯1 )) 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=4β¦), 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 β¦ 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 ππππ πΈ1 πΈπ = β 2 π 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 β¦ 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