Multielectron Atoms and Periodic Table

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1 GRE Question

2 Multielectron Atoms and Periodic Table Helium Atom 2 2m e ( ) + 2ke 2 2ke 2 + ke2 r 1 r 2 r 2 r 1 Electron-electron repulsion term destroys spherical symmetry. No analytic solution discovered (to date). If you ignore the e-e repulsion term (and ignore indistinguishability of electrons), = E ( r 1, r 2 ) n 1,l 1,m 1 ( r 1 ) n2,l 2,m 2 ( r 2 ) ~ 2 2m e r 2 (~r i ) 2ke 2 r i = E i (~r i ) Hydrogen- like wave func4on

3 Wave functions are hydrogen-like with Z=2 Energy of individual electrons: E n = 13.6 ev Z2 n 2 Ground-state energy: both electrons in n=1 shell E = (2)(8)(13.6) ev = 109 ev The actual ground-state energy of helium is -79 ev, so e-e repulsion accounts for roughly +30 ev. Bad Approximation to ignore e-e interaction! So how can we get a better approximation, accounting for the electron-electron interaction?

4 The Central Field Approximation (Hartree Method) Assume each electron moves independently in a spherically symmetric potential energy distribution due to both the nucleus and the negatively charged cloud of the other electrons. Charge density of other electrons: (~r )= (~r i ) 2 e X i (the sum over i represents summing over all electrons other than the electron in which we are determine the poten4al energy that it sees) The potential energy is then determined using standard electrostatic theory: V (r) = Zke2 r Z k (~r 0 ) e ~r ~r 0 d3 r Insert this into the Schrodinger equation and solve for new wave function

5 Hartree Method This central field approximation requires an iteration scheme: 1. Guess wave functions for each of the electrons 2. Use wave functions to determine the charge distribution each electron sees. 3. Solve the TISE for each electron using the potential energy associated with the charge distribution. 4. If the solutions to the TISE give rise to significantly different wave functions than those used in step #2, go back to step #2 and repeat. After convergence, the output of this procedure yields selfconsistent wave functions (and associate eigenenergies) for each electron in the atom. More sophisticated models account for spin and force antisymmetric multiparticle quantum states. (only important for valance electrons)

6 Features of the Hartree Method Due to spherical symmetry, potential energy is still a function of r. Thus the angular portion of the wave functions are still spherical harmonics: nlm l = R nl (r)y m l l (, ) Same rules for allowed values of l and m l apply as for hydrogen. The radial functions, determined by numerically solving the radial portion of the TISE, can differ significantly from those of the hydrogen atom (this is due to the e-e interactions). The eigenenergies now depend on both n and l (for hydrogen they only depend on n) Energies are degenerate only if V (r) / 1/r This dependence on l can be understood in terms of screening.

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8 The Dependence of Energy on l Screening Screening can be understood in terms of Gauss law. The outer electron sees an enclosed charge to be less than that of the nucleus. For an outer electron, the electric field due to the nucleus is partially screened by the inner electrons. If the outer electron s radial probability distribution is more spread out, it spends more time penetrating through the screening effect of the inner electrons, and it will be more strongly attached to the nucleus. There is less screening. For a given n, values of l having wave functions with a higher spread in r will have lower energies.

9 Clicker Question For a given value of n, for what values of l will an outer electron be less effectively screened by the inner electrons (and thus have a lower energy state)? 1. High values of l 2. Low values of l 3. Huh, I m so lost!

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11 The Dependence of Energy on l For a given value of n, states with low values of l have a higher dispersion in r, so there is a higher probability the outer electron is found closer to the nucleus. States with low values of l have lower energies than states with a higher value of l. Crudely, the ordering of energy levels can be determined by the ordering of the sum n+l. Ties generally go to lower n states.

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13 Orbital energies are lower in multielectron atoms than in the hydrogen atom

14 Structure of Periodic Table Structure can be understood using ordering of energy states and degeneracy of subshells.

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17 GRE QUESTIONS

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21 Ionization Energies What is the ionization energy of hydrogen? I 1 = 13.6 ev What is the ionization energy of helium, ignoring screening? Ignoring screening, it acts hydrogen-like: E 1 = Z n 2 ev = 54 ev Actual ionization energy is E 1 =24.6 ev. If you use above equation to define Z eff, for He, then Z eff =1.4 This is ionization energy is high! Nobel gas.

22 Clicker Question Lithium (Z=3): How do you expect the ionization energy of lithium to compare to hydrogen and helium? 1. higher than hydrogen, less than helium 2. higher than both hydrogen and helium 3. less than both hydrogen and helium 4. higher than hydrogen, less than helium

23 E 1 = Z2 eff 13.6 n 2 ev Ionization Energies For lithium, valence electron is in the 2s state, so n=2 We expect screening to be quite effective since the 2s radial probability distribution is further out than the 1s distribution. If we guess that Z eff is roughly 1, then we guess E ev. Actual ionization energy is 5.4 ev. (Z eff = 1.26). Why is Z eff > 1? This is a general property. Atoms with lone valence electrons in outer subshell will have low ionization energies. Ionization energy increases as you move to right within a subshell (due to higher nuclear charge). Nobel gases have very high ionization energies.

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29 Spectroscopic Notation Uhlenback and Goudsmit

30 Clicker Question What is the spectroscopic notation for the ground state of helium? S S S S S 0

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32 Residual Coulomb Interaction An electric interaction that compensates for the fact that the Hartree net potential V(r) acting on each valence electron describes only the average affect of the Coulomb interactions between that electron and the other valence electrons. Refinements are needed to account for Exchange force (symmetric vs anti-symmetric wavefunctions) Non-Spherically symmetric charge distribution (due to other valence electrons).

33 Clicker Question For the first excited state of helium, possible states are 2 1 S 0 (singlet) and 2 3 S 1 (triplet). Which one has a lower energy? 1. The singlet 2. The triplet 3. They should have the same energy!

34 JITT Responses Due to electron repulsion the states will have different energies. In the triplet state when the particles are at the same location the spatial wave function is zero, however in the singlet state the spatial wave function is nonzero.the closer the electrons get to each other, the more they repel therefore the singlet state will have a higher energy. I believe the singlet state (antisymmetric)would have lower energy because antisymmetric states minimize electronelectron interactions. the configuration with an anti-symmetric spacial state will result in lower energy, due to the fact that the electrons will tend to be farther apart and therefore have less coulomb repulsion energy.this corresponds to a symmetric spin state since the overall wavefunction has to be anti-symmetric, so the triplet has a lower energy.

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36 Hunds Rules The effects of the residual coulomb interaction give rise to Hunds Rules For a given subshell, states with a higher value of S will have lower energies. (Exchange Force) For a given subshell and given value of S, states with a higher value of L will have a lower energy. Sometimes Hund s Rules include the effects of LS coupling: For atoms with less than half-filled shells, the level with the lowest value of J lies lowest in energy.

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38 Ground State of Carbon What is the ground state of carbon (Z=6)? What is the spectroscopic notation of ground state? Maximize S: Thus S=1 (symmetric spin state) Need an antisymmetric spatial state, so the two 2p electrons need to have different values of m L. Values of m L indicate L=1 J=2,1,0; Thus lowest energy is 3 P 0.

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40 Summary: For lowest energies: minimize n + l (as much as possible) maximize S (exchange force) maximize L (residual coulomb) minimize J (for half-filled or less shell) (LS coupling) minimize m J (Zeeman effect if external field)

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