Atomic Structure and the Fine structure constant α Niels Bohr Erwin Schrödinger Wolfgang Pauli Paul Dirac Lecture Notes Fundamental Constants 2015; W. Ubachs The Old Bohr Model An electron is held in orbit by the Coulomb force: (equals centripetal force) FCentripetal = FCoulomb mv2 Ze2 = 2 rn 4πε0rn Bohrs postulate: Quantization of angular momentum Ze2 r n22 v2r 2 = = h 4πε m m2 L = mvr = n = n 0 2π n2h2ε n2 h2ε r = 0 = r r = 0 = 0.529×10−10 m n πmZe2 Z 1 1 πme2 The size of the orbit is quantized, and we know the size of an atom ! Lecture Notes Fundamental Constants 2015; W. Ubachs The Old Bohr Model: Energy Quantisation 1 Ze2 Z 2 Quantisation of energy = 2 − = − En mv 2 R∞ 2 4πε0rn n 2 e2 m R = e The Rydberg constant is the scale unit ∞ 2 of energies in the atom 4πε0 2 2 2 Z Z Energies in the atom in atomic units E = − R∞ ⇒ − 1 Hartree = 2 Rydberg n n2 2n2 2 2 2 Z Z 2 2 e = − = − α with α = En 2 R∞ 2 mc n 2n 4πε0c dimensionless energy Lecture Notes Fundamental Constants 2015; W. Ubachs The Old Bohr Model; velocity of the electron Velocity in Bohr orbit vn=1 = Zαc e2 α = 4πε 0c Limit on the number of elements ? Classical argument Lecture Notes Fundamental Constants 2015; W. Ubachs Schrodinger Equation; Radial part: special case l=0 2 2 2 d dR 2 Ze − 2 + + + = 2 r V (r) 2 ( 1)R ER Prefactor for : − = 0 2mr dr dr 2mr 1/r ma 4πε0 = Find a solution for 0 2 Solution for the 4πε0 2 2 = 2 Ze a 2 − R"+ R' − R = ER length scale paramater Ze m 2µ r 4πε 0r 2 1 4πε0 Physical intuition; no density for r → ∞ a = a0 with a = Bohr radius Z 0 2 e me − trial: R(r) = Ae r / a A −r / a R R'= − e = − Solutions for the energy a a A R 2 R"= e−r / a = 2 e2 m 2 2 = − = − 2 e a a E Z 2 2ma 4πε0 2 2 1 2 Ze2 − − − = 2 E πε 2 Ground state in the 2m a ar 4 0r E = −Z R ∞ Bohr model (n=1) must hold for all values of r Lecture Notes Fundamental Constants 2015; W. Ubachs Quantum mechanics: same result The effect of the proton-mass in the atom Velocity vectors: Centripetal force M v1 = v m v2 m v2 µv2 m + M F = 1 1 = 2 2 = M r1 r2 r v = − v 2 m + M Quantisation of angular momentum: h Relative velocity L = µvr = n = n Relative coordinates: π dr 2 v = dt r = r1 − r2 Centre of Mass Kinetic energy 1 2 1 2 1 2 + = K = m1v1 + m2v2 = µv mr1 Mr2 0 2 2 2 Problem is similar, but Position vectors: Angular momentum L = m v r + m v r = µvr m µ M 1 1 1 2 2 2 r1 = r m + M With reduced mass r relative coordinate m mM r = − r µ = 2 m + M m + M Lecture Notes Fundamental Constants 2015; W. Ubachs Reduced mass in the old Bohr model isotope shifts Results Quantisation of radius in orbit: 2 2 2 n 4πε0 n me 1. Isotope shift rn = = a0 Z 2µ Z µ on an atomic transition e Energy levels in the Bohr model: 2. Effect of proton/electron mass ratio on the energy levels Z 2 µ E = − R∞ n 2 m µ mM M M / m µ n e red = / m = = = me m + M m + M 1+ M / m 1+ µ Rydberg constant: µ R = R∞ H m Conclusion: the atoms are not a good e probe to detect a variation of µ Lecture Notes Fundamental Constants 2015; W. Ubachs General conclusions on atoms and atomic structure 2 2 Note units (different units in this equation): Z Z 2 2 = − = − α EI 7 −1 En 2 R∞ 2 mc R∞ = − =1.0973731568549(83)×10 m n 2n hc dimensionless energy Conclusion 1: All atoms have the Rydberg as a scale for energy; they cannot be used to detect a variation of α µ M / m µ red = = m 1+ M / m 1+ µ Conclusion 2: the atoms are not a good probe to detect a variation of µ Lecture Notes Fundamental Constants 2015; W. Ubachs Relativistic effects in atoms Electron spin No classical analogue for this phenomenon Origin of the spin-concept -Stern-Gerlach experiment; 1 space quantization s = 2 Pauli: There is an additional “two-valuedness” in the spectra of atoms, behaving like an angular momentum Goudsmit and Uhlenbeck This may be interpreted/represented -Theory: the periodic system requires as an angular momentum an additional two-valuedness Lecture Notes Fundamental Constants 2015; W. Ubachs Electron spin as an angular momentum operator In analogy with the orbital angular momentum of the electron 1 L s = µ = −g µ g =1 2 L L B L A spin (intrinsic) angular momentum can be Spin is an angular momentum, so it defined: S µ = −g µ should satisfy S S B S 2 s,m = 2s(s +1) s,m s s a) in relativistic Dirac theory Sz s,ms = ms s,ms gS = 2 1 1 s = ,m = ± b) in quantum electrodynamics 2 s 2 gS = 2.00232... Note: the spin of the electron cannot be explained from a classically “spinning” electronic charge 2 Electron radius 2 e 2 v 1 m c = Angular momentum = ω = 2 = from EM-energy: e Le I mere 4πε0re from spin 5 re 2 Lecture Notes Fundamental Constants 2015; W. Ubachs Spin-orbit interaction Frame of nucleus: Spin of electron is a magnet with dipole v Frame of electron: µ µ = −g B S S e -e -e The dipole orients in the B-field with energy +Ze +Ze Ze2 V = −µ ⋅ B = S ⋅ L − v LS S 2 2 3 4πε0me c r The moving charged nucleus induces a magnetic field at the location of the A fully relativistic derivation electron, via Biot-Savart’s law = ζ ⋅ (Thomas Precession) yields VLS (r)S L µ Ze(− v)× r with B = 0 4π 3 Ze2 1 r ζ (r) = 1 2 2 3 Use L = mr ×v ; µ0ε0 = 8πε0me c r nl c2 Use: Ze L 1 2 Then B = = = int πε 2 3 r3 a3n3( +1/ 2)( +1) 4 0 mec r 3 Zαmc 2 3 Lecture Notes Fundamental Constants 2015; W. Ubachs n n ( +1/ 2)( +1) Fine structure in spectra due to Spin-orbit interaction In first order correction to energy Then the full interaction energy is: for state lsjm j 2 4 j( j +1)− ( +1)− s(s +1) E = lsjm V lsjm ESO = α Z hcR SO j SL j 2n3( +1/ 2)( +1) = lsjm j ζ nl L ⋅ S lsjm j S-states = 0, j = s ESO = 0 Evaluate the dot-product 2 J 2 = L + S = L2 + S 2 + 2L ⋅ S P-states =1, j = ±1/ 2 Then α 2Z 4hcR E = 1 2 2 2 SO 3 L ⋅ S sjm j = (J − L − S )sjm j 2n 2 1 = 2{j( j +1)− ( +1)− s(s +1)}sjm 2 j Show that the “centre-of-gravity” does not shift Lecture Notes Fundamental Constants 2015; W. Ubachs Kinetic Relativistic effects in atomic hydrogen Relativistic kinetic energy p4 Krel = Ψnjm − Ψnjm = rel 2 2 2 4 2 8m3c3 Ekin = p c + m c − mc = e 2 2 2 2 2 Z 4α 2 1 3 mc 1+ p / m c − mc = − (hc)R − 2n3 2 +1 8n p2 p4 mc2 1+ − + 2 2 4 4 2m c 8m c First relativistic correction term p4 K = − rel 3 2 8mec To be used in perturbation analysis: p = − ∇ operator does not i change wave function Lecture Notes Fundamental Constants 2015; W. Ubachs Relativistic effects in atomic hydrogen: SO + Kinetic Relativistic energy levels: Z 4α 2 2 3 Enj = En − (hc)R − 2n3 2 j +1 4n Fine structure splitting ~ Z4α2 j=1/2 levels degenerate Also the outcome of the Dirac equation ∂ψ (cα ⋅ p + βmc2 )ψ = ih ∂t P.A.M. Dirac Lecture Notes Fundamental Constants 2015; W. Ubachs Hyperfine structure in atomic hydrogen: 21 cm Nucleus has a spin as well, and therefore a magnetic moment e I µ = µI = gI µN ; N 2M p Interaction with electron spin, that may have density at the site of the nucleus (Fermi contact term) 1 I ⋅ S = I ⋅ J = (F 2 − J 2 − I 2 ) 2 Splitting : F=1 ↔ F=0 1.42 GHz F=1 or λ = 21 cm F=0 2 Scaling: g pα / µ Magnetic dipole transition Lecture Notes Fundamental Constants 2015; W. Ubachs Alkali Doublets Ze2 S ⋅ L V = SL 2 2 3 with 4πε0 2m c r 1 S ⋅ L = (J 2 − L2 − S 2 ) Na doublet 2 Selection rules: ∆ = ±1 ∆j = 0,±1 ∆s = 0 2P np 3/2 α 2Z 4hcR E = 2 SO 3 P1/2 2n ns 2 S1/2 Lecture Notes Fundamental Constants 2015; W. Ubachs The Alkali Doublet Method Lecture Notes Fundamental Constants 2015; W. Ubachs The Many Multiplet Method 1. 2. 3. 1. Strong transitions 2. Weak, narrow transitions 3. Hyperfine transitions Lecture Notes Fundamental Constants 2015; W. Ubachs The Many Multiplet Method Z 2 E = − R∞ n n2 2 e2 m R = e ∞ 2 4πε0 2 Lecture Notes Fundamental Constants 2015; W. Ubachs Relativistic corrections in the Many Multiplet Method Relativistic correction to energy level Further include Many body effects me4Z 2 (Zα )2 2 3 (Zα )2 1 ∆ = − − ∆ ≅ − ( ) n 2 3 n En C Z, j,l 2 n 2 j +1 4n ν j +1/ 2 (note: atomic units different) (Zα )2 ∆ ≅ E n n ν ( j +1/ 2) with: En is the Rydberg energy scaling ν is effective quantum number In many cases: C(Z, j,l) ≅ 0.6 These effects separate light atoms (low Z) from heavy atoms (high Z) Lecture Notes Fundamental Constants 2015; W.