Determining the Absorption Coefficient

1. Determining the Absorption Coefficient It would be Good to know how to determine the coefficients of the Equation of Transfer!

2. The Absorption Coefficient • Occupation Numbers • N = Total Number of an Element • Ni,I = Number in the proper state (excitation and ionization) • Thermal Properties of the Gas (State Variables: Pg, T, Ne) • Atomic/Molecular Parameters: Excitation and Ionization Energies, Transition Probabilities, Broadening Parameters Absorption Coefficient

3. Bohr Theory The One Electron Case • e = Electron Charge (ESU) • Z = Atomic number of the nucleus • m = reduced mass of the nucleus • mnucleus*me/(mnucleus+me) ~ me • N = principal quantum number • rn = effective orbital radius • vn = speed in orbit Absorption Coefficient

4. f-f 1 2 3 4 Bohr Model b-b b-f ∞ • Total Energy (TE) = Kinetic Energy(KE) + Potential Energy(PE) • PE is Negative • For a bound state TE< 0 • Therefore: |KE| < |PE| n = Excitation “stage” i = Ionization Stage: I = neutral, II=first ionized Absorption Coefficient

5. Continuing ... Combine the Energy and Radial Force Equations Energy: Force: Substituting mv2 Absorption Coefficient

6. Bohr Postulates Angular Momentum is Quantitized Now put v into the force equation: Now put r into the energy equation: Absorption Coefficient

7. Hydrogen Atomic Parameters • Z = 1 • m = 1836.1 (1) / 1837.1 = 0.99946 me = 9.1033(10-28) gm • e = electron charge = 4.803(10-10) esu • h = 6.6252(10-27) erg s • k = 1.380622(10-16) erg K • Ei,n = -2.178617(10-11) / n2 ergs = -13.5985 / n2 eV • 1 eV = 1.6021(10-12) erg • For n = 1: Ei,1 = -13.5985 eV = -109678 cm-1 • 1 eV = 8065.465 cm-1 • Ei,n = -109678 / n2 cm-1 Absorption Coefficient

8. Continuing • Note that all the energies are negative and increasing • n =  ==> Ei, = 0 • The lowest energy state has the largest negative energy • For hydrogen-like atoms the energies can be deduced by multiplying by Z2 and the proper m. • Li++ (Li III): 32(-13.60) = -122.4 eV • hν = 122.4 eV = 122.4(1.6021(10-12) ergs) = 1.96(10-10) ergs • ν = 2.9598(1016) Hz ==> λ = 101 Å • What is the temperature? • hν = kT ==> T = 1.4(106) K Absorption Coefficient

9. Specifying Energies Energies are usually quoted WRT the ground state (n=1) • Excitation Potential: χi,n≡ Ei,n - Ei,1 • For an arbitrary n it is the absolute energy difference between the nth state and the ground state (n=1). • This means χi,1 = 0 and that χi, = -Ei,1 • χi, ≡ Ionization Energy (Potential) = Energy needed to lift an electron from the ground state into the continuum (minimum condition) • χi, = Ei, - Ei,1 Absorption Coefficient

10. Ionization Energy Requirement: Em = χi, + ε • ε goes into the thermal pool of the particles (absorption process) • One can ionize from any bound state • Energy required is: Ei,n = χi, - χi,n • Energy of the electron will be: hν - Ei,n Absorption Coefficient

11. A Free “Orbit” • Consider the quantum number n • which yields a positive energy • n = a positive value The solution for the orbit yields a hyperbola Absorption Coefficient

12. Bound Transitions • Consider a transition between two bound states (a,b) => (na,nb) [n’s are the quantum numbers] m and Z for H Absorption Coefficient

13. Hydrogen Lines Absorption Coefficient

14. Balmer Series of H The lower level is n = 2  na = 2 • First line (Hα) is the transition (absorption): 2 → 3. • For absorption: lower → upper na→ nb • For emission: upper → lower nb→ na • The “observed” wavelength is 6562.3 • The Rydberg formula gives vacuum wavelengths • We need to account for the index of refraction of air: Edlen’s Formula Absorption Coefficient

15. Edlen’s Formula Computes the Index of Refraction of Air • νλair = c/n or λVacuum = nλAir • Where σ = vacuum wavenumber in inverse microns. • For Hα: λVac = 0.6564038 microns • σ = 1.52345 • n = 1.000276237 • λAir = 6562.225 (calculated) Absorption Coefficient

16. Series Limits λ = -911.672 (1/2 - 1/na2)-1 • For na = 2 one gets λ = 3646.668 Å • Any photon which has λ < 3646.668 Å can move an electron from n = 2 to the continuum • Series Limits are defined as n =  to nLower Absorption Coefficient

17. Excitation Move an Electron Between Bound States • Ni,x = Number of ions (atoms) that have been ionized to state i and excited to level x • Ni,x depends upon the number of ways an electron can enter level x • Most levels have sublevels E from the “main” energy level • If there are gx of these levels then there are gx ways of entering the level • gx is called the statistical weight of the level (state) • Each of the sublevels has a complete set of quantum numbers • (n,l,ml,ms) or (n,l,j,mj) Absorption Coefficient

18. Quantum Numbers • n: Principal Quantum Number • : orbital angular momentum quantum number • L = (h/2π) 0    (n – 1) • m : -  m < +  • ms: ±½ • j: j = l + s => j = l ± s s = ½ • mj: -j  mj +j Absorption Coefficient

19. Hydrogen Quantum Numbers • n=1: (n = 1, l = 0, ml = 0, ms = ±½) • n=2: (n = 2, l = 0, ml = 0, ms = ±½) • (n = 2, l = 1, ml = -1, ms = ±½) • (n = 2, l = 1, ml = 0, ms = ±½) • (n = 2, l = 1, ml = 1, ms = ±½) • Therefore for n=1 g=2 and for n=2 g = 8 • For any H-like ion: gn = 2n2 • For any atom having level J (alternate set quantum number) with sublevels -J,...0,...+J the statistical weight is gJ = 2J + 1 • J is not necessarily an integer Absorption Coefficient

20. Notation 2S+1LJ • 2S+1: Multiplicity (Spin) S = ∑si • L: Azimuthal Quantum Number L = ∑i • J: L+S = total angular momentum vector Absorption Coefficient

21. Boltzmann Equation We shall start with the state-ratio form of the equation The E’s are with respect to the continuum. χ is the excitation potential of the state pi is the probability that the state exists Absorption Coefficient

22. The Partition Function • Note that pi depends on electron pressure and essentially quantifies perturbations of the electronic states. We define the partition function U as • These numbers are tabulated for each element as a function of ionization stage, temperature, and “pressure.” The latter is expressed as a lowering of the ionization potential. Absorption Coefficient

23. Rewrite the Boltzman Equation Number in state X WRT ground state Sum Ni,x Substitute Absorption Coefficient

24. Example What is the fraction of H that can undergo a Balmer transition at 5700K? • We need g2, UI(5700), and χ2 • g2 = 8 (n=2) • UI(T) = 2 • χ2 = 82259.0 cm-1 = 10.20 eV = 1.634(10-11) ergs • N2/NI = 8/2 e-(χ/kT) = 4 e-20.76 = 3.842(10-9) Absorption Coefficient

25. Ionization The Saha Equation • This is the Saha equation. It gives the ratio of ground state populations for adjoining ionization stages. What one really wants is the number in the entire stage rather than just the ground state. Remember that χi,0 = 0. Absorption Coefficient

26. Rewrite The Saha Equation Note that m = me Absorption Coefficient

27. Computing Stage Populations • Note that the Saha Equation does not tell you the populations but: • NT = NI + NII + NIII + NIV +... • Divide by NI • NT/NI = 1 + NII/NI + (NIII/NII)(NII/NI) + (NIV/NIII)(NIII/NII)(NII/NI) + ... • Each of the ratios is computable and NT is known. Absorption Coefficient

28. Example What is the degree of ionization of H at τ = 2/3 in the Sun? • T = 6035K and Pe = 26.5 dynes/cm2 • Pe NII/NI = 0.3334 T5/2 (2UII/UI) e-13.598*CF/kT • UII = 1: Partition function for a bare nucleus = 1 • UI = 2 • Pe NII/NI = 9.4332(108) e-26.15 = 4.162(10-3) • NII/NI = 1.57(10-4) Absorption Coefficient

29. Another Example The Same for Neutral and First Ionized Fe. • T = 6035K and Pe = 26.5 dynes/cm2 • Pe NII/NI = 0.3334 T5/2 (2UII/UI) e-7.87*CF/kT • UII = 47.4 • UI = 31.7 • Pe NII/NI = 2.82(109) 2.68(10-7) • NII/NI = 28.5 Absorption Coefficient

30. Data Sources • Atomic Energy Levels: NBS Monograph 35 and subsequent updates (NIST) • Line Lists • Compute from AEL • MIT Wavelength Tables • RMT - Charlotte Moore • Kurucz Line Lists • Literature for Specific Elements • Partition Functions • Drawin and Felenbok • Bolton in ApJ • Kurucz ATLAS Code • Literature (Physica Scripta) Absorption Coefficient