Atomic Spectra in Astrophysics

Atomic Spectra in Astrophysics Potsdam University : Wi 2016-17 : Dr. Lidia Oskinova lida@astro.physik.uni-potsdam.de

Complex Atoms

Non-relativistic Schrödinger Equation 02 [ N i=1 ( ) 2 2m e 2 i Ze2 4πǫ 0 r i + N 1 i=1 N j=1+1 e 2 4πǫ 0 r i r j E ] ψ( r 1,.., 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

Central Field model 03 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 [ ] ( 2 2 i+ V 2m e i (r i ) φ i (r i )=E i φ i (r i ) where V i (r i ) is the angle-independent cental potential of each electron. The total energy of the system is then E= i E i The solutions, φ i (r i ), are called orbitals. Importnant to remember that this is an approximation.

The Pauli exclusion principle 04 Consider a two- electron wavefunction. If the two spin- orbitals, Φ a =Φ b, 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.

05 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.

The Periodic Table Dmitry 06 Mendeleyev (1834-1907) The structure of peridic tableelectron configurations of elements Closed shells: n=1 orbitals K- shells Closed shells: n=2 orbitals L- shells 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 07 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 The total orbital angular momentum L= i l i and the total electron angular momentum S= i s i These are then added to give J= L+ S Pauli Principle 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.

Consider OIII with the configuration 1s 2 2s 2 2p3d 08 Which electrons contribute to angular momentum? What are their l and s? Find L and S. Find J. Write possible terms. (2S+1) L J

Consider OIII with the configuration 1s 2 2s 2 2p3d 09 For the 2p electron: l 1 =1and s 1 =1/2. For the 3d electron: l 2 =2and s 2 =1/2. L= l 1 + l 2 L=1, 2, 3; S= s 1 + s 2 S= 0, 1; L S J Level 1 0 1 1 P o 1 (2S+1) L J 1 1 0, 1, 2 3 P o 0,3 P o 1,3 P o 2 J= L+ S 2 0 2 1 D o 2 2 1 1, 2, 3 3 D o 1,3 D o 2,3 D o 3 3 0 3 1 F o 3 3 1 2, 3, 4 3 F o 2,3 F o 3,3 F o 4 Twelve levels arise from 1s 2 2s 2 2p3d configuration of OIII.

Specroscopic Notations 10 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 11 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 ψ( r 1,..., r N )=±ψ( r 1,..., r N ). Even parity states are given by +ψ, odd arity states are given by ψ. In practice, the parity is given by ( 1) l 1+l 2 +...+l N. 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

Example 1. The helium atom 12 The ground state 1s 2. This is a closed shell, with L=0, S=0, hence it gives rise to a single, even parity term 1 S, and level 1 S 0. The first ecited configuration is 1s2s. Write levels.

Example 1. The helium atom 13 1) The ground state 1s 2. This is a closed shell, with L=0, S=0, hence it gives rise to a single, even parity term 1 S, and level 1 S 0. 2) The first excited configuration is 1s2s. This has l 1 =l 2 =0 and hence L=0. But S=0, 1 giving rise to singlets and triplets. Hund s rule: For a given configuration, the state with the maimum spin multiplicity is lowest in energy. So the 3 S term ( 3 S 1 level) is lower in energy that the 1 S term ( 3 S 0 level). The spliting beween these terms is 0.8 ev. 3) The next excited configuration 1s2p. What is its parity?

Example 1. The helium atom 14 1) The ground state 1s 2. This is a closed shell, with L=0, S=0, hence it gives rise to a single, even parity term 1 S, and level 1 S 0. 2) The first excited configuration is 1s2s. This has l 1 =l 2 =0 and hence L=0. But S=0, 1 giving rise to singlets and triplets. Hund s rule: For a given configuration, the state with the maimum spin multiplicity is lowest in energy. So the 3 S term ( 3 S 1 level) is lower in energy that the 1 S term ( 3 S 0 level). The spliting beween these terms is 0.8 ev. 3) The next excited configuration 1s2p. It has odd parity. L=1, S=0, 1. The 3 P o term is lower than the 1 P o term, in this case by 0.25 ev. The 3 P o term splits into three levels: 3 P o 0, 3 P o 1, 3 P o 2

Example 2. The carbon atom 15 1) Consider the excited state configuration 1s 2 2s 2 2p3p, It is only necessary to consider 2p3p electrons. These give L=0, 1, 2 and S=0, 1, which give rise to the following terms all with even parity: 1 S, 3 S, 1 P, 3 P, 1 D, 3 D. 2) Now lets consider the ground state configuration: 1s 2 2s 2 2p 2. This configuration has l 1 = 1, s 1 = 1/2, l 2 = 1, s 2 = 1/2. Pauli principle restricts which terms are allowed. Which configuration is not allowed? (consider 2p 2 electrons). There are general methods of determining which terms are allowed for a configuration with a multiply occupied open shell. However, there are also some rules of thumb For a system with equivalent electrons (the same n and l), then the sum L+S must be even for the Pauli Principle to be satisfied. 3) The ground state configuration of carbon, CI, gives terms 1 S, 3 P, 1 D). The 3 P term has the highest spin this is ground state term. Other two terms have the same spin - which is lower in energy?

The second Hund s rule: For a given configuration and spin multiplicity, the state with the maximum orbital angular momentum is the lowest in energy. 16 The 1 D state of carbon is 1.42 ev lower than the 1 S state, but 1.26 ev higher than 3 P state. 4) In the examples above we considered only terms and neglected splitting by J.. However, this splitting increasingly importnant for hevier atoms. The 3 P ground state of carbon has L=1 and S=1, which gives J=0, 1, 2. Allowed leveles are 3 P 0, 3 P 1, 3 P 2. The energy order is given by the third Hund s rule. The lowest energy is obtained for lowest value of J in the normal case and for highest J value in the inverted case. The normal case is a shell which is less than half filled, e.g. 2p 2 inverted case is a shell which is more than half full, e.g. 2p 4 oxygen. in carbon. The found in atomic For carbon the energy order: 3 P 0 < 3 P 1 < 3 P 2, whereas for oxygen 3 P 2 < 3 P 1 < 3 P 0

Hund s rules 17 (1) For a given configuration, the term with maximum spin multiplicity has lowest energy. (2) For a given configuration and spin multiplicity, the term with the largest value of L lies lowest in energy. (3) For atoms with less than half-filled shells, the levels with the lowest value of J lies lowest in energy. (4) For atoms with more than half-filled shells, the levels with the highest value of J lies lowest in energy. Hund s rules are only applicable within L-S coupling. Furthermore, they are rigorous only for ground states.

Problem 1. 18 What is the ground state configuration, term, and level of the berilium atom, Be?

Problem 1. 19 What is the ground state configuration, term, and level of the berilium atom, Be? Be has 4 e, ground state 1s 2 2s 2. Both orbitals are fully occupied, giving a term 1 S and a level 1 S 0 One of the outer electrons is promoted to the 3d orbital. What terms and levels can this configuration have?

Problem 1. 20 What is the ground state configuration, term, and level of the berilium atom, Be? Be has 4 e, ground state 1s 2 2s 2. Both orbitals are fully occupied, giving a term 1 S and a level 1 S 0 One of the outer electrons is promoted to the 3d orbital. What terms and levels can this configuration have? The excited state has configuration 1s 2 2s 1 3d 1 giving l(2s)=0, s(2s)=1/2, l(3d)=2, s(3d)=1/2. Adding these gives L=2 and S=0, 1. These corresponds to terms 1 D, 3 D, and levels 1 D 2, 3 D 1, 3 D 2, 3 D 3.

Problem 2. 21 An excited helium atom has the configuration 3d 2. What values of the total orbital angular momentum quantum number L, and the total spin angular momentum qauntum number S are allowed?

Problem 2. 22 An excited helium atom has the configuration 3d 2. What values of the total orbital angular momentum quantum number L, and the total spin angular momentum qauntum number S are allowed? 4d 2 means l 1 =2, s 1 =1/2 and l 2 =2, s 2 =1/2. Adding these L=l 1 +l 2 =0, 1, 2, 3, 4 and S=s 1 +s 2 =0, 1 Without the Pauli Principle the terms could be formally writen 1 S, 1 P, 1 D, 1 F, 1 G, 3 S, 3 P, 3 D, 3 F, 3 G Pauli Principle the sum L+S must be even. Hence only 1 S, 3 P, 1 D, 3 F, 1 G are possible Use Hund s rules to find the energy order of these terms. For each term deduce the allowed values of the total angular momentum J. Give the full spectroscopic notations, including parity. Which level has the lowest energy.

Problem 2. 23 Use Hund s rules to find the energy order of these terms. For each term deduce the allowed values of the total angular momentum J. Give the full spectroscopic notations, including parity. Which level has the lowest energy? If Hund s rules are obeyed then energy increses for 3 F, 3 P, 1 G, 1 D, 1 S L S J= L S,.., L+S Level 0 0 0 1 S 0 (2S+1) L J J= L+ S 1 1 0, 1, 2 3 P 0, 3 P 1, 3 P 2 2 0 2 1 D 2 3 1 2, 3, 4 3 F 2, 3 F 2, 3 F 4 4 0 4 1 G 4 All terms have even parity since ( 1) l 1+l 2 = ( 1) 2+2 = 1