Multi-Electron Atoms PDF

Multi-Electron Atoms  Many Electron Atoms: Central Field Approximation In first approximation, we can think that in a many-electron atom each of the Z atomic electrons moves in the same potential V(r) 2 Z...

Multi-Electron Atoms  Many Electron Atoms: Central Field Approximation In first approximation, we can think that in a many-electron atom each of the Z atomic electrons moves in the same potential V(r) 2 Ze V(r) = − 4πϵ0r i.e. each electron sees the nucleus. We assume that the combined potential energy function of an electron in an atom due to the electrostatic force of the nucleons and other electrons is replaced by a single central potential function V(r), which depends solely on the radial coordinate of the specific electron. This is known as the central field approximation.  However, there are some effects due to presence of other electrons which will modify the simple Coulomb potential. Most importantly, we must understand how electrons are distributed in an atom. Wolfgang Pauli Shell Structure (Magic Numbers) Full shells Ionization Energy more bound highest ionisation energy smallest atomic radius Closed shells define a set of magic numbers 2, 10, 18, 36, 54, 86,… Full shell + 1 least bound largest atomic radius Size (distance outermost e)  Many Electron Atom in a Central Field Sub-shells are filled in order of increasing energy following the Pauli principle. Energy of atom = sum of energies of each electron (specified by sub-shell). ‘Capacity’ of sub-shell = 2(2l + 1). Pauli principle ms = ±½ allowed values of m 2 electrons per ‘orbital’ (given by ℓ=0[s], ℓ=1[p], ℓ=2[d], ℓ=3[f]) Many Electron Atom in a Central Field Each electron is assigned to a single wave function specified by a unique combination of quantum numbers The quantum numbers that specify each electron wave functions are the ones we know: n ℓ m s=½ ms = ±½ n Principal quantum number. ℓ Orbital angular momentum quantum number. m Magnetic (spatial) quantisation quantum number. s = ½ Spin angular momentum quantum number. ms = ± ½ Spin (spatial) quantisation quantum number. The Pauli Exclusion Principle No two electrons in a single atom can have the same set of quantum numbers (n, l, ml, ms). Wolfgang Pauli (1900 - 1958) The Pauli Exclusion Principle What if … there was no Exclusion Principle? If there was no Exclusion Principle all electrons would settle in the lowest energy 1s state. The chemical and physical properties of the elements di ering by one electron would vary smoothly. However, the properties of the elements can vary dramatically from one electronic con guration to the next. fi ff Neon (Z = 10) is an unreactive gaseous element. Ne Fluorine (Z = 9) and sodium (Z = 11) are only one electron away are among the most reactive chemical elements in the periodic F table! Na  Copper (Z = 29) is not magnetic Nickel (Z = 28) is magnetic and and has excellent electrical has poor electrical conductivity. conductivity.  Electron Configurations Sub-shell filling Active atoms: unfilled sub-shells Valence particle or valence hole = 1 valence ‘particle’  Spectroscopic Notation l 0 1 2 3 4 5 s p d f g h Alphabetical order from here  Spectroscopic Notation n 1 2 3 4 5 Shell K L M N O 5s 3s 4s 5p 2s 3p 4p Subshell 1s 5d 2p 3d 4d 5f 4f 5g Degeneracy 2 8 18 32 50 6p 6 The Aufbau Principle 5d 10 5 4f 6s 14 2 B 2 2 1 1s 2s 2p 4 2 2 5p 6 Be 1s 2s 4d 5s 2 10 3 2 1 Li 1s 2s 4p 6 2 2 3d 4s 2 10 He 1s 1 1 3p 6 H 1s 3s 2 2p 6 (2,1,0, + 1/2) 2s 2 (2,0,0, + 1/2) (2,0,0, − 1/2) 1s 2 (1,0,0, + 1/2) (1,0,0, − 1/2)  Shells & Subshells All levels that have the same value of n lie (roughly) at the same average distance from the nucleus. The l = 0 subshells are penetrating orbits, i.e. have signi cant probabilities of being close to the nucleus. Reproduced from Modern Physics 4th Edition by K.S. Krane, published by Wiley. fi Energy Levels in Many-Electron Atoms If each electron of a multi-electron atom was only subject to a Coulomb potential, energy levels would be equivalent to those of the Hydrogen atom multiplied by the Z atomic number (squared) That is not the case! Observations show: 13.6 eV 2 Enl = − ZEff n 2 The force ‘felt’ by each electron on the outermost shell is lower than in the case of one-electron-only because Many electrons change the central of the screening (or shielding) of the closed-shells field V(r) and the energy levels. electrons.  Screening effects: Carbon, 6C ground state : 1s2 2s2 2p2 (capacity 2 2 6) Energy levels assuming a simple Coulomb potential Real Energy levels for the screened potential E=0 2s 2p 2s 2p E = - 9 ER 1s Level of energy numerically calculated using models that 1s consider screening effects E = -13.6 eV Z2/ n2 = - 36 ER where ER = 13.6 eV Screening The Li+ ion Many electron atoms are complicated! −e Earlier in the Bohr treatment of single electron atoms, we expect the electron energies to 2 scale as Z. +3e However, the ionisation energy of neutral Lithium is 5.39 eV (much smaller than the 13.6 eV for the H atom)! −2e Neutral Lithium can be considered as a a lled 2 1 1s shell with an outer 2s electron. fi There is no simple equation that allows us to predict the radii of electron orbits in multi-electron atoms. However, we expect that the n = 2 electron would be found at a greater distance from the centre of the nucleus than the n = 1 electrons. 1 The force on the outer 2s electron can be estimated using Gauss’ Law - (you will study this next semester in +3e Electromagnetism). In Gauss’ Law, we consider the net charge enclosed within 2 −2e a surface. If the nucleus and the 1s electrons lie within the 1 surface and the 2s electron outside, then the outer electron will experience a force due to the enclosed charge +3e − 2e = e. Gaussian Surface To a rst approximation, the neutral Lithium atom looks like a one-electron atom with an electron in an n = 2 orbit about a nucleus with an e ective charge Zeff = + e. This is known as electron screening: the outer electron is shielded from the nuclear charge by electrons from the inner shells. The energies of the excited states could be estimated using the equation 2 Zeff En = − 13.6 eV n 2 This model yields an ionisation energy for neutral Li of 3.40 eV, which is much closer to the measured value of 5.39 eV. The discrepancy arises from the probability density function for the 2s electron wavefunction. There is a signi cant probability that the 2s electron spends a signi cant fraction of time within the orbit of the n = 1 electrons and so feels the full force of the +3e nuclear charge. This would lead to an enhanced binding energy for this electron. fi fi ff fi  The probability density argument is further supported by considering the non-penetrating 2p and 3d electrons that occupy orbitals at much greater radii than the n = 1 electrons and experience an e ective nuclear charge of Zeff = + e due to the ‘screening’ by the two negative 1s electron charges. The agreement of the measured 2p and 3d electron energies (-3.50 eV and -1.51eV, respectively) is in much better agreement with the simple model. 2 Zeff 1 E2 = − 13.6 eV 2 = − 13.6 eV = − 3.54 eV 2 4 2 Zeff 1 E3 = − 13.6 eV 2 = − 13.6 eV = − 1.51 eV 3 9 ff  Test yourself 2 The ground state of Helium has the con guration 1s. Use the electron screening model to predict the energies of the following excited states of Helium. 1 1 (a) The 1s 2s state - (the measured value is -4.0 eV) 1 1 (b) The 1s 2p state - (the measured value is -3.4 eV) 1 1 (c) The 1s 3d state - (the measured value is -1.5 eV). Comment on the agreement your calculated value with the measured value. Solution For the outer electron in Helium , the nuclear charge (+2e) is screened by the 1s electron so the e ective charge seen by the outer electron is +e. ff fi Solution 2 Zeff 1 E2 = − 13.6 eV 2 = − 13.6 eV = − 3.4 eV 2 4 The measured value is -4.0 eV, which indicates that the 2s electron has a small penetration through the 1s distribution and experiences a tighter binding than predicted by this simple model. 1 1 The same energy value is obtained for the 1s 2p state since the energy depends solely on the principal quantum number n. The agreement with the experimental value is almost exact indicating that the 2p state has a lower penetration than the 2s electron. Solution The energy of the 3d state is predicted to be 2 Zeff 1 E3 = − 13.6 eV 2 = − 13.6 eV = − 1.5 eV 3 9 The agreement with the experimental value is exact indicating that the 3d electron has a lower penetration than the 2s electron. Angular Momentum Coupling in Multi-electron Atoms? Each electron will have an orbital angular momentum Li ⃗ and a spin Si ⃗ (with index i referring to the single electron) each associated with a quantum numbers l and s = 1/2, respectively. The total orbital angular momentum L ⃗ of electrons in an atom is given by the vector sum of all electrons: L⃗ = ∑ Li i The total spin S ⃗ of an atomic configuration is given by the vector sum of all optically active electrons: S⃗ = ∑ Si i The total angular momentum J ⃗ of an atomic configuration is given by the vector sum: J ⃗ = L ⃗ + S⃗ Case 1: L and S for electrons in closed (full) shells A full (n,ℓ) sub-shell of 2(2ℓ + 1) electrons: contributes to the energy of the atom has total Lz = 0 given by total sum of all mℓ has total Sz = 0 given by total sum of all ms has total Jz = 0 given by total sum of all mj For example: If 2 electrons are in the 1s level: L=0, S=0 (because of P.E.P.) If there are 6 electrons in 2p level: each electron has ℓ=1, but because of the P.E.P., they will have different m or different spin orientation Total mℓ = +1+0-1=0 Total ms= ½ - ½ =0 Case 1: L and S for electrons in closed (Full) shells A full (n,ℓ) sub-shell of 2(2ℓ + 1) electrons: contributes to the energy of the atom has total Lz = 0 given by total sum of all mℓ has total Sz = 0 given by total sum of all ms has total Jz = 0 given by total sum of all mj Hence it follows that a full sub-shell contributes zero to Jz. In fact, it can be shown that J = 0, S = 0 and L = 0 for full shells There is no contribution to the magnetic dipole moment from electrons in full shells Consequently: Only electrons in outermost shell are relevant. These are referred to as active electrons.  Case 2: Single electron in unfilled sub-shell If there is only one electron in the outermost sub-shell, the total angular momenta and energy levels are straightforward For instance, in Sodium, Z = 11. Active electron 2 2 6 1 1s , 2s , 2p , 3s. s state has l = 0. ∴L=0 Filled sub-shells J = |L + S| L = 0, S = 0 J = | 0 + 1/2 | J = 1/2 Case 3: Multi-electrons in unfilled sub-shell If two or more electrons are in an un lled sub-shell, we must evaluate the sum of the orbital and spin angular momenta. Two angular momenta with quantum numbers j1 and j2 can be combined to give | j1 − j2 | ≤ J ≤ j1 + j2 in integer steps. Convention: In atomic physics we use lower case letters for quantum numbers associated with a single electron, capital letters for the angular momenta of two or more electrons. fi Case 3: Multi-electrons in unfilled sub-shell Consider the quantum numbers l and s = 1/2 for each electron. 1. Find the vector sum all the orbital angular momentum of the active electrons ∑ L= li. i ∑ 2. Find the vector sum all the spin angular momentum of the active electrons S = si. i 3. Determine the total angular momentum J = | L + S | with values between | L − S | and | L + S |. Worked Example Consider a system of two valence electrons whose orbital quantum numbers are l1 =2 and l = 4, respectively. (a) Find the vector sum all the orbital angular momentum of the active electrons ∑ L= li. i ∑ (b) Find the vector sum all the spin angular momentum of the active electrons S = si. i (c) Determine the total angular momentum J = |L + S|.  ∑ (a) Find the vector sum all the orbital angular momentum of the active electrons L = li. i LMin = | l1 − l2 | LMax = l1 + l2 =4−2 =2+4 =2 =6 Possible values lie at integer steps in the range between these limits. L = 2, 3, 4, 5, 6  ∑ (b) Find the vector sum all the spin angular momentum of the active electrons S = si. i SMin = | s1 − s2 | SMax = s1 + s2 = 1/2 − 1/2 = 1/2 + 1/2 =0 =1 The possible values are S = 0 and S = 1. (c) Determine the total angular momentum J = |L + S|. The total angular momentum can take values in the range |L − S| ≤ J ≤ L + S JMin = | LMin − SMax | JMax = LMax + SMax =2−1 =6+1 =1 =7 The total angular momentum can take values J = 1, 2, 3, 4, 5, 6, 7. Spectroscopic notation Rather than the electronic notation (suitable for single-electron atom), another notation is used to define atom states: the spectroscopic notation Also referred to as Russell-Saunders notation S, L and J are the total spin, total orbital angular momentum and total angular momentum of electrons in the outermost unfilled sub-shell. 2S+1 is the spin multiplicity, denoting the number of possible values for the z- component of the spin. Example An atom with S = 1, L = 1, and J = 2, can be expressed as 3 P2 Spectroscopic notation If there is only one electron in outermost shell, it is easy: Sodium: 1 Ground state in electronic notation: 3s (1s22s22p6) S=1/2 because s-state means L=0; electron spin is ½, J = |0 + ½| = ½ L=0 2S 1/2 J =1/2 In spectroscopic notation, ground state is written as: Carbon (which I will show in detail as example in next slides): ground state is (1s2 2s2) 2p2 There are many more possibilities for J, as for L and S: S = |s1+ s2| which means it can be 0 or 1 L = |ℓ1 + ℓ2| which means it can be 0, 1 or 2 J = |L+ S| many combinations … What is the ground state then? And are all combinations possible?  Carbon Ground State 6C ground state : 1s2 2s2 2p2 (capacity 2 2 6) There are two valence electrons in the outermost shell SPIN: S = |s1+ s2| Each electron has s = ½. There are two possibilities because → | 1 + 2|, …| 1 − 2| in integer steps S = 0 (singlet) requires electrons in different ms states If I interchange the electrons, I do not get the same system  they are anti-symmetric S = 1 (triplet) can have electrons in same ms state If I interchange the electrons, I get the same system  they are symmetric 𝑆 𝑠 𝑠 𝑠 𝑠 Carbon Ground State: Spin 2 2 2 The ground state of carbon is 1s 2s 2p Consider S The capacity of a p state is 6. s1 = s2 = 1/2 The p state is un lled. S = 0, 1. There are 2 active electrons. S=0 S=1 Antisymmetric under Symmetric under electron exchange. electron exchange. fi Carbon Ground State: Orbital 2 2 2 The ground state of carbon is 1s 2s 2p The capacity of a p state is 6. The p state is un lled. There are 2 active electrons. l1 = l2 = 1 | l1 − l2 | ≤ L ≤ l1 + l2 L = 0, 1, 2. fi Carbon Ground State: Orbital L = 0, 1, 2. L = 0, 2. L=1 1 ml 2 ml mL 1 ml 2 ml mL 1 1 2 1 0 1 −1 −1 −2 1 −1 0 0 0 0 −0 −1 −1 Symmetric under Antisymmetric under electron exchange. electron exchange. What is the ground state of Carbon? The Pauli Exclusion Principle tells us that the electrons cannot have the same set of quantum numbers. Since n and l have the same value for both electrons … … ml and/or ms must be di erent. Which combinations are possible? S = 0, L = 0. (ms di erent, ml same) S = 0, L = 2. (ms di erent, ml same) S = 1, L = 1. (ms same, ml di erent) ff ff ff ff What is the ground state of Carbon? S = 0, L = 0. S = 0, L = 2. S = 1, L = 1. L antisymmetric, S antisymmetric L antisymmetric, S symmetric L symmetric, S antisymmetric J=0 J=2 J = 0,1,2 1 1 3 S0 D2 P0, 3P1, 3P2 How many states? What is the ground state of Carbon? To quantify the number of states, we need to count all possible orientations of S, L, and J, i.e mL and mJ This can be achieved by calculating (2S + 1)(2L + 1). S = 0, L = 0. 1×1=1 There is 1 1 S state with J = 0 S = 0, L = 2. 1×5=5 There are 5 1 D states with J = 2 S = 1, L = 1. 3×3=9 There are 9 1 P states with J = 0, 1, 2. There are 15 possible states. Do they all have the same energy? NO! The Pauli Exclusion Principle In a system of identical fermions, the total wavefunction describing the system must be antisymmetric with respect to the exchange of any two particles. Acceptable quantum states are described by wavefunctions with antisymmetric exchange symmetry. Wolfgang Pauli ΨOrbital(r1, r2)ΨSpin(r1, r2) = − ΨOrbital(r1, r2)ΨSpin(r1, r2) (1900 - 1958)  Corrections to Central Field Approximation The CFA is only an approximation. It does not account for Residual Coulomb interactions between outer electrons (screening). Spin-orbital interactions due to the angular momentum of active electrons. Formally, the Hamiltonian operator for the multi-electron atom has the form n 2 n ̂ ̂ 1 e ̂ ̂ ∑ 4πϵ0 | ri − rj | ∑ H = H0 + + ξi(ri)Li. Si i