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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