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Atomic Spectra in Astrophysics Potsdam University : Wi 2016-17 : Dr. Lidia Oskinova [email protected] 01 Ioniztion of hydrogen and Saha equaiton Ionization stages: Saha law for hydrogen ne n p (2πme kT )3/2 − hνkT1c = e nH h3 Lets define the ionization fraction as ne x = n p +nH Using n p = ne and n p + nH = n one can re-write the Saha equation x2 1−x 1 3/2 − 13.6eV kT T e n p +nH ∝ Ionization fraction is determined by two parameters: temperature and density Temperature of the Universe T(z)=2.728(1+z) Density (n p +n H )(z)=1.6(1+z) 3 m -3 T=4000K x=0.6 T=3800K x=0.29 T=3000K x=0.00017 Hydrogen Atom Direct observation of H electron orbital (Stodolna et al. 2013, Phys. Rev. Lett. 110, 213001) The Schrödinger Equation for a H-like atom The Hamiltonian operator of H-like system Ĥ = Atomic Units −~2 2 ∇ 2µ Ze2 4πǫ0 r − Electron mass, m e =1.66 10 -27 kg Electron charge, e=1.6 10 -19 C Bohr radius, a 0 = 4πǫ0 ~2 me2 = 5.29 10 -11 m Dirack constant, h/2π = 1 a.u. Ĥ = 1 2 − 2µ ∇ − Z r For a system with energy E and wavefunction ψ: For H-like atom h 1 2 − 2µ ∇ − Z r i Ĥψ = Eψ − E ψ(~r) = 0 03 04 Wavefunctions and separating the variables m1 m2 m1 +m2 µ= r = (r, θ, φ) Coordinates ~ Reduced mass: For H-like atoms one can solve Schrödinger eq. analytically by separating the variables. Radial solutions Laguerre polinomials Angular solutions spherical harmonics ψ(r, θ, φ) = Rnl (r)Ylm (θ, φ) R(r) solutions exists only if main quantum number n=1,2,..., ∞ Y(τ,ψ): orbital quantum number l = 0,1,2,..., n-1 and magnetic quantum number m l =-l, -l+1,..l-1, l (2l+1 values) and spin quantum number m s =+1/2,-1/2 05 Quantum numbers n determines the energy of the atom l describes the electron angular momentum: ~[l(l + 1)]1/2 0 1 2 3 4 5 6 7 ... s p d f g h i k l ... m l the magnetic quntum number: determines level splitting in the presence of magnetic field. m~ is the projection of angular momentum on the z-axis ~[s(s + 1)]1/2 . √ 3/2~ for H-like atoms angular momentum s spin. The electron angular momentum is Electron spin is 1/2 s z projection of spin angular momentum. It can have s, -s+1,...,s1, s values. For one electron system only -1/2, +1/2. 06 Energy levels and Rydberg constant For bound states the solution of Schrödinger equation En = Z2 −R n2 The constant R takes different values for different H- like ions RH = µ R me ∞ The rydberg constant for a system of infinite nuclear mass R∞ = 109737.31 cm-1 For hydrogen measured experimentaly RH = 109677.581 cm-1 07 Angular momentum coupling in the Hydrogen atom Sources of angular momentum: electron orbital angular momentum l electron spin angular momentum s nuclear spin angular momentum i - determined by spin coupling of the various nucleons: protons and neutrons have spin 1/2. Only the total angular momentum is conserved. Individual momenta must be combined. This is done using a coupling scheme. Different schemes are possible, but ususally the strongest couplings are considered first. l + ~s = ~j Hydrogen the usual coupling scheme: ~ Next, combine electron and nuclear spin: ~j + ~i = Rules for vector addition: f~ |a − b| ≤ c ≤ a + b In quantum mechanics the result is quantized with a unity step. l1 = 2, l2 = 3, ~L = l~1 + l~2 → L = 1, 2, 3, 4, 5 08 The fine structure of hydrogen Fine structure in the energy split according to the value of j. For Hydrogen s=1/2. Lets take l=0,1,2,3. (2S+1) Configuration l s ns 0 1/2 LJ j 1/2 H atom ns1/2 Term n2 S Level n2 S1/2 np 1 1/2 1/2, 3/2 np1/2 , np3/2 n2 Po n2 Po1/2 , n2 Po3/2 nd 2 1/2 3/2, 5/2 nd3/2 , nd5/2 n2 D n2 D3/2 , n2 D5/2 nf 3 1/2 5/2, 7/2 nf5/2 , nf7/2 n2 Fo n2 Fo5/2 , n2 Fo7/2 E.g. n=2 give rise to three fine strucutre levels Which line is formed by spontaneus transitions from this level? 22 S 1/2 , 22 Po1/2 , 22 Po3/2 09 Selectrion Rules Selection rules are derived rigorously using quintum mechanics. Strong transitions are driven by electric dipoles. Weaker transitions could also be driven by quadrupoles. For electric dipole transitions in hydrogen ∆n any ∆l = +-1 ∆s = 0 (for H is always satisfied as s=1/2) ∆j = 0, +-1 ∆m j = 0, +-1 Lets consider Hα: n=2-3, ∆l = +-1 2s-3p, 2p-3s, 2p-3d are allowed, 2s-3s, 2p-3p, 2s-3d are not allowed. Considering fine structure, further constrains are put by ∆j = 0, +-1 rule Complex Atoms 11 Non-relativistic Schrödinger Equation hP 2 N −~ i=1 2 − ∇ 2me i Ze2 4πǫ0 ri + PN−1 PN i=1 e2 j=1+1 4πǫ0 |~ ri −~ r j| i − E × ψ(~ r1 , .., r~N ) = 0 The first sum is a kinetic energy operator for the motion of each electron and the Coulomb attraction between this electron and the nucleus. The second term: electron-electron repulsion term. Because of this term the equation cannot be solved analytically even for N=2. Therefore, to understand such systems it is necessary to introduce approximations 12 Central Field model The easiest approximation reduce the problem to a single particle situation. I.e. a potential does not depend on the angular position of electrons around the nucleus. Central field potential the force acting on each electron only depends on its distance from the nucleus. The Schrödinger eq. can be written for ith electron where h 2 i −~ ( 2me ∇2i + Vi (ri ) φi (ri ) = E i φi (ri ) Vi (ri ) is the angle-independent cental potential of each electron. The total energy of the system is then The solutions, E= P i Ei φi (ri ) , are called orbitals. Importnant to remember that this is an approximation. 13 By analogy with H, each atomic orbital can be written as the product of a radial and an angular function. φi (~ri ) = Rni li (ri )Yli mi (θi , φi ) The angular part does is independent of the other electrons. To specify central potential Vi (ri ) , one usually considers a potion of an electron in the central field of all the other electrons. Vi (ri ) = Ze2 4πǫ0 ri + P j,i D e2 4πǫ0 |~ ri −~ r j| E where <> represent average. Hence the radial part, Rni li (ri ) depends on all other electrons and cannot be obtained analytically. Usually it is done iteratively - the self-consistent field (SCF) problem 14 It is standard to use the H-atom orbital labels: n, l, m. The angular behaviour, given by and m is the same. However, radial part is not the same as in case of H. The total wavefunction of the system ψ(~ r1 , ..., r~N ) = φ1 (~ r1 ), ..., φN (r~N ) Lets consider spin and generalize orbital to spin- orbital Φi ( j) = φi (~ r j , σ j ) , where Φi ( j) means that spin-orbital i is a function of the space-spin coordinates of electron j. The space- spin wavefunction for the total system then Ψ(1, ..., N) = Φ(1), ..., Φ(N) Indistinguishable Particles Consider a system with two identical particles. If the wavefunction of these particles is Ψ(1, 2) , and the particles are indistinguishable, what property their wavefunction must have? |Ψ|2 . Then, if particles are indistinguishable |Ψ(1, 2)|2 = |Ψ(2, 1)|2 . Physically observable is Symmetric solution Ψ(1, 2)| = +Ψ(2, 1) , antisymmetric Ψ(1, 2)| = −Ψ(2, 1) . Pauli Principle: Wavefunctions are antisymmetric with respect to interchange of identical Fermions Fermions - any particles with half-integer spin (electrons, protons, neutrons). A two-electron wavefunction which obeys the Pauli Principle can be written as Ψ(1, 2) = √1 2 [Φa (1)Φb (2) − Φa (2)Φb (1)] = −Ψ(2, 1) 1 where √ is to preserve normalization. 2 15 The Pauli exclusion principle Consider a two- electron wavefunction. If the two spin- orbitals, Φa = Φb , then Ψ(1, 2) = 0 . This solution is not allowed. Hence solutions which have the two particle occupying the same spin-orbital are ecluded. The Pauli exclusion principle: No two electrons can occupy the same spin-orbital This exclusion is the key to atomic structure. It also provides the degeneracy pressure which holds up the gravitational collaps of white dwarfs and neutron stars. 16 Electron Configurations For a H-like atom, the energy is determined only by principle quntum number: E(1s)<E(2s)=E(2p)<E(3s)=E(3s)=E(3p)=E(3d)<E(4s)... For complex atoms the degeneracy on l is lifted. 17 18 Following the Pauli exclusion principle, each orbital labeled nl actually consists of orbitals with l+1 different m value, each with two possible values s. Thus the nl orbital can hold a maximum 2(2l+1) electrons. Write possible combinations of magnetic and spin quantum numbers for p electron 19 Following the Pauli exclusion principle, each orbital labeled nl actually consists of orbitals with l+1 different m value, each with two possible values s. Thus the nl orbital can hold a maximum 2(2l+1) electrons. An atomic configuration is given by distributing electrons among orbitals. First fill the atomic orbitals in energy order from the lowest energy orbitals upwards. Write the ground state configuration for carbon. Write the ground state configuration for neon. 20 Following the Pauli exclusion principle, each orbital labeled nl actually consists of orbitals with l+1 different m value, each with two possible values s. Thus the nl orbital can hold a maximum 2(2l+1) electrons. An atomic configuration is given by distributing electrons among orbitals. First fill the atomic orbitals in energy order from the lowest energy orbitals upwards. The ground state configuration for carbon 1s 2 2s 2 2p 2 The 2p orbital is not full. The ground state configuration for neon 1s 2 2s 2 2p 6 The 2p orbital has closed shell configuration. A closed shell or subshell makes no conribution to the total orbital or spin momentum, L or S. Atomic ions which have the same number of electrons form isoelectronic series. E.g. neon-like iron is 21 Following the Pauli exclusion principle, each orbital labeled nl actually consists of orbitals with l+1 different m value, each with two possible values s. Thus the nl orbital can hold a maximum 2(2l+1) electrons. An atomic configuration is given by distributing electrons among orbitals. First fill the atomic orbitals in energy order from the lowest energy orbitals upwards. The ground state configuration for carbon 1s 2 2s 2 2p 2 The 2p orbital is not full. The ground state configuration for neon 1s 2 2s 2 2p 6 The 2p orbital has closed shell configuration. A closed shell or subshell makes no conribution to the total orbital or spin momentum, L or S. Atomic ions which have the same number of electrons form isoelectronic series. E.g. neon-like iron is Fe XVII (Fe 16+ ) 22 The Periodic Table Dmitry Mendeleyev (1834-1907) The structure of peridic tableelectron configurations of elements Closed shells: n=1 orbitals Kshells Closed shells: n=2 orbitals Lshells Closed shells: n=3 orbitals M-shells etc. Atoms with the same electron configuration outside a closed shell share similar chemical and optical properties. Li, Na, K - a single electron outside a closed shell. Alkali metals. Excited states - when one of the outermost electrons jumps to higher orbital. He: 1s2s, 1s2p, 1s3s. Angular Momentum in Complex Atoms Complex atoms have more than one electron, hence several sources of angular momentum.Ignoring nuclear spin, the total conserved angular momentum, J is the sum of spin plus orbital momenta of all electrons. There are two coupling schemes of summing the individual electron angular momenta. L-S or Russell-Saunders coupling P~ ~ The total orbital angular momentum L = i li and the total electron angular P ~ si momentum S = i ~ These are then added to give Pauli Principle J~ = ~L + S~ closed shells and sub-shells (e.g. 1s 2 or 2p 6 ), have both L=0 and S=0. Hence, it is necessary to consider only electrons in open or partlyfilled shells. 23 2 2 24 Consider OIII with the configuration 1s 2s 2p3d Which electrons contribute to angular momentum? What are their l and s? Find L and S. Find J. Write possible terms. (2S+1) LJ 2 2 Consider OIII with the configuration 1s 2s 2p3d For the 2p electron: l 1 =1and s 1 =1/2. For the 3d electron: l 2 =2and s 2 =1/2. ~ = l~1 + l~2 → L = 1, 2, 3; L S~ = s~1 + s~2 → S = 0, 1; L S 1 0 (2S+1) J 1 LJ 1 1 0, 1, 2 J~ = ~L + S~ 2 0 2 2 1 1, 2, 3 3 0 3 3 1 2, 3, 4 Level 1 o P1 3 Po0 ,3 Po1 ,3 Po2 1 3 Do2 Do1 ,3 Do2 ,3 Do3 1 3 F 3o F 2o ,3 F 3o ,3 F 4o Twelve levels arise from 1s 2 2s 2 2p3d configuration of OIII. 25 j-j coupling P li + ~si → J~ = i ~ji Alternative scheme: 1) ~ji = ~ Consider again OIII and show that the result is the same. Why to schemes are needed? We considered non-relativistic Schrödinger equation. In this case levels with the same L and S have the same energy. In practice, relativistic effects split this degeneracy (these effects are introduced as a perturbation of the non-relativistic treatment). For light ions (lighter than Fe), relativisitc effects are weak. L-S coupling scheme is the most appropriate. For heavy ions (heavier than Fe), relativisitc effects are stron. j-j coupling scheme is the most appropriate. 26 27 Specroscopic Notations The standard notation is called spectroscopic notation and works with L-S coupling A term is a state of a configuration with a specific value of L and S: (2S+1) L (o) A state with S=0 is a singlet (2S+1=1); a state with S=1/2 is a doublet; a state with S=1 is a triplet. o means odd parity. When parity is even, no superscript. A level is a term with a specific value of J: (2S+1) L (o) J Each term can be split on 2J+1 sub-leveles called states which are designated by the total magnetic quantum number M J = -J, -J+1, ..., J-1, J. These states are degenerate in the absence of an eternal field. The splitting of levels into states in a magnetic field Zeeman effect. Parity of the Wavefunction 28 The parity of the wavefunction is dermined by how the wavefunction behaves upon inversion. Inversion, for an atom is is replacing vector r with -r. Probability distribution, i.e., square of the wavefunction must be unchanged, hence ~ 1 , ..., −r ~ N ) . Even parity states are given by + ψ , odd ψ(~ r1 , ..., r~N ) = ±ψ(−r arity states are given by − ψ . In practice, the parity is given by (−1)l1 +l2 +...+lN . As closed shells and sub-shells have an even number of electrons, it is necessary to consider only active electrons. For our example OIII with the configuration 1s 2 2s 2 2p3d we must consider only 2p and 3d electrons. What is the parity? The parity is importnant since it leads to a rigorous dipole slection rule, the Laporte rule: All electric dipole transitions connect states of opposite parity. Strong transitions can only link configurations with even to those with odd parity and vice versa