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Determining the Absorption Coefficient It would be Good to know how to determine the coefficients Determining the Absorption Coefficient It would be Good to know how to determine the coefficients

Determining the Absorption Coefficient It would be Good to know how to determine the coefficients - PowerPoint Presentation

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Determining the Absorption Coefficient It would be Good to know how to determine the coefficients - PPT Presentation

Determining the Absorption Coefficient It would be Good to know how to determine the coefficients of the Equation of Transfer Absorption Coefficient 2 The Absorption Coefficient Occupation Numbers N Total Number of an Element ID: 761956

coefficient absorption energy state absorption coefficient state energy number ionization quantum equation electron ground numbers excitation nii level potential

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

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

Absorption Coefficient 3 Bohr Theory e = Electron Charge (ESU) Z = Atomic number of the nucleus m ′ = reduced mass of the nucleus m nucleus*me/(mnucleus+me) ~ meN = principal quantum numberrn = effective orbital radiusvn = speed in orbit The One Electron Case

Absorption Coefficient 4 Bohr Model 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 3 2 4 1 f-f ∞ b-b b-f

Absorption Coefficient 5 Continuing ... Combine the Energy and Radial Force Equations Energy: Force: Substituting mv 2

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 Z = 1 m ′ = 1836.1 (1) / 1837.1 = 0.99946 m e = 9.1033(10 -28) gme = electron charge = 4.803(10-10) esuh = 6.6252(10-27) erg sk = 1.380622(10-16) erg KEi,n = -2.178617(10 -11 ) / n 2 ergs = -13.5985 / n 2 eV 1 eV = 1.6021(10-12 ) ergFor n = 1: Ei,1 = -13.5985 eV = -109678 cm-11 eV = 8065.54429 cm-1Ei,n = -109678 / n2 cm-1 Atomic Parameters

Absorption Coefficient 8 Continuing Note that all the energies are negative and increasing n = ∞ ==> E i ,∞ = 0The lowest energy state has the largest negative energyFor hydrogen-like atoms the energies can be deduced by multiplying by Z2 and the proper m′.Li++ (Li III): 3 2 (-13.60) = -122.4 eV h ν = 122.4 eV = 122.4(1.6021(10-12) ergs) = 1.96(10-10) ergsν = 2.9598(1016) Hz ==> λ = 101 ÅWhat is the temperature? hν = kT ==> T = 1.4(106) K

Absorption Coefficient 9 Specifying Energies Excitation Potential: χ i,n ≡ E i,n - Ei,1For 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, ∞ = -E i,1 χ i , ∞ ≡ Ionization Energy (Potential) = Energy needed to lift an electron from the ground state into the continuum (minimum condition)χi,∞ = E i,∞ - Ei,1 Energies are usually quoted WRT the ground state (n=1)

Absorption Coefficient 10 Ionization ε goes into the thermal pool of the particles (absorption process) One can ionize from any bound state Energy required is: E i,n = χi,∞ - χi,nEnergy of the electron will be: hν - E i,n Energy Requirement: E m = χi,∞ + ε

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) => (n a ,n b ) [n’s are the quantum numbers] m and Z for H

Absorption Coefficient 13 Hydrogen Lines

Absorption Coefficient 14 Balmer Series of H First line (H α ) is the transition (absorption): 2 → 3. For absorption: lower → upper na → nbFor 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 The lower level is n = 2  n a = 2

Absorption Coefficient 15 Edlen’s Formula νλ a ir = c/n or λ Vacuum = nλAirWhere σ = vacuum wavenumber in inverse microns.For Hα: λVac = 0.6564038 microns σ = 1.52345 n = 1.000276237 λ Air = 6562.225 (calculated) Computes the Index of Refraction of Air

Absorption Coefficient 16 Series Limits For n a = 2 one gets λ = 3646.668 ÅAny photon which has λ < 3646.668 Å can move an electron from n = 2 to the continuumSeries Limits are defined as n = ∞ to nLower λ = -911.672 (1/ ∞ 2 - 1/n a 2)-1 n Limit Name 1 912 Lyman 2 3646 Balmer 3 8204 Paschen 4 14590 Brackett 5 22790 Pfund

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

Absorption Coefficient 18 Quantum Numbers n: Principal Quantum Number l : orbital angular momentum quantum number L = l (h/2 π) 0 ≤ l ≤ (n – 1)ml : - l ≤ m ≤ + l m s : ±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, m l = 0, m s = ±½) n=2: (n = 2, l = 0, m l = 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, m s = ±½) Therefore for n=1 g=2 and for n=2 g = 8 For any H-like ion: g n = 2n 2For any atom having level J (alternate set quantum number) with sublevels -J,...0,...+J the statistical weight is gJ = 2J + 1J is not necessarily an integer

Absorption Coefficient 20 Notation 2S+1: Multiplicity (Spin) S = ∑ s i L: Azimuthal Quantum Number L = ∑ liJ: L+S = total angular momentum vector 2S+1 L J

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 p i is the probability that the state exists

Absorption Coefficient 22 The Partition Function Note that p i 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 N i,x Substitute

Absorption Coefficient 24 Example We need g 2 , U I (5700), and χ 2g2 = 8 (n=2)UI(T) = 2χ2 = 82259.0 cm-1 = 10.20 eV = 1.634(10-11) ergsN2/NI = 8/2 e -( χ /kT) = 4 e -20.76 = 3.842(10-9) What is the fraction of H that can undergo a Balmer transition at 5700K?

Absorption Coefficient 25 Ionization 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 χ i,0 = 0. The Saha Equation

Absorption Coefficient 26 Rewrite The Saha Equation Note that m = m e

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

Absorption Coefficient 28 Example T = 6035K and P e = 26.5 dynes/cm 2 P e NII/NI = 0.3334 T5/2 (2UII/UI) e-13.598*CF/kTUII = 1: Partition function for a bare nucleus = 1UI = 2Pe N II /N I = 9.4332(10 8 ) e -26.15 = 4.162(10-3) NII/NI = 1.57(10-4) What is the degree of ionization of H at τ = 2/3 in the Sun?

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

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 ListsLiterature for Specific ElementsPartition FunctionsDrawin and FelenbokBolton in ApJ Kurucz ATLAS Code Literature ( Physica Scripta )