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Atomic Term Symbols

Atomic Term Symbols. The Stationary States of the Many-Electron Atom are determined by the overall properties of the Atom, not those of the individual electrons that exist in the atom. What are these overall properties? The total orbital angular momentum of the electrons, L

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Atomic Term Symbols

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  1. Atomic Term Symbols • The Stationary States of the Many-Electron Atom are determined by the overall properties of the Atom, not those of the individual electrons that exist in the atom. • What are these overall properties? • The total orbital angular momentum of the electrons, L • The total spin angular momentum of the electrons, S • The total angular momentum of the atom, J • Thus, the eigenstates of the Hamiltonian for the atom are labeled by these properties. These labels are called • Atomic Term Symbols.

  2. Atomic Term Symbols • The 'definition' of the term symbol: • 2S+1LJ • Note: In the symbol L must be replaced with its alphabetic 'code': • L=0 is S, • L=1 is P, • L=2 is D, • L=3 is F, • L=4 is G, • L=5 is H... • J is the vector sum of L and S: • J = (L+S), (L+S-1), (L+S-2), ....|L-S|

  3. Assigning Term Symbols: The ground state of hydrogen atom is one electron in the lowest energy atomic orbital: the 1s. Therefore the total orbital angular momentum of all (one) electrons is L=l=0, and the total electron spin is S=s=1/2. Applying the 'definition' of the term symbol results in a 2S1/2 atomic term for the ground state of the H atom. This symbol is ‘read’: “doublet Ess one half” Exciting the single hydrogenic electron to higher orbitals results in different atomic states or 'terms' of the atom. Note that an H atom with the electron in a 3d and 10d orbital both result in 2D3/2 and 2D5/2 terms, but at different energies.

  4. The lowest electron configuration of He is 1s2. The ground state of the neutral helium atom is therefore 1S0 since the total orbital and spin angular momentum of this configuration is zero. In fact, any electron configuration (orbital population) that consists of any combination of closed shells or subshells will result in this (totally symmetric 1S0 term. Therefore, in the designation of atomic terms, the contribution from closed subshell electrons may be neglected. What can’t be neglected? The resultant angular momenta of the open-shell electrons in the atom!

  5. Because the angular momenta of each electron adds to the rest of the electrons vectorially, there will, in general, be more than one possible result of this addition. It is our job to account for each and every possible way in which the properties of the electrons will sum up and sort those possibilities into those corresponding to distinct Terms (total energies; atomic eigenstates). We will seek to categorize each the ways in which the electrons add up by a complete description of the l, ml, ms (s is the same for all e-’s ) of each electron: This is called a microstate. A given microstate is not necessarily a good description of the state of the entire atom; sometimes many microstates are needed to describe a given Term.

  6. To determine the states (Terms) of a given Atom or Ion: • Write down the electronic configuration (ignore closed subshell electrons) • Determine the number of distinct microstates that can represent that configuration. If you have e electrons in a single open subshell of 2l+1 orbitals, this value is #microstates(single open subshell) = (2(2l+1))!/e!(2(2l+1)-e)! • Tabulate the number of microstates that have a given ML and MS • Decompose your table into terms by elimination • Test the total degeneracy of the resultant terms to account for all the microstates counted in parts 2 and 3 • Determine the lowest term for the configuration by Hunds Rules.

  7. Summary Examples H (1s1)ground state term symbol is 2S1/2 He (1s2)Ground state term symbol is 1S0

  8. He (1s12s1) An Excited State Configuration Terms: 1S0  , 3S1 { There is neither a   3S0     nor a     3S-1    Term }

  9. B (1s22s22p1) {ground state} Term symbol: 2P1/2, 2P3/2 The spin orbit splitting is regular so2P1/2is the ground state.

  10. C (1s22s22p2) {lowest electron configuration} Calculate the number of possible electron arrangements in the given configuration: There are 6!/4!2! = 15 microstates expected Write down all these possibilities :

  11. Tabulate the total numbers by ML and MS

  12. Decompose this table into terms

  13. Check with reality:

  14. N (1s2 2s2 2p3) {lowest energy configuration} This configuration has 6!/3!3! = 20 microstates Draw all the possibilities

  15. Tabulate the totals

  16. Assign Terms Assign Terms

  17. Check with reality

  18. What are the atomic state term symbols resulting from the lowest energy configuration of O (1s2 2s2 2p4) ? Aha! We don't haver to do this because the terms arising from p2 and p4 are exactly the same! Note, however, the ordering of the J levels is now inverted

  19. Reality:

  20. If you think you have mastered the process of determining the states arising from a given electronic configuration, you should try Ti(1s22s22p63s23p64s23d2)Here is the answer depicted graphically

  21. Some results of this procedure for your amusement:

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