Download Determining the Absorption Coefficient

Determining the Absorption
Coefficient
It would be Good to know how to
determine the coefficients of the
Equation of Transfer!
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
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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
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Bohr Model
b-b
1
b-f
2
3 4
f-f
∞
• Total Energy (TE) = Kinetic Energy(KE) +
Potential Energy(PE)
• PE is Negative
• For a bound state TE< 0
• Therefore: |KE| < |PE|
Ei,n
2
1
Ze
 mvn2 
2
rn
Absorption Coefficient
n = Excitation “stage”
i = Ionization Stage:
I = neutral, II=first ionized
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Continuing ...
Combine the Energy and Radial Force Equations
Energy:
Force:
Substituting mv2
Ei,n
1
Ze
2
 mvn 
2
rn
2
vn2
Ze2
Fn  m   2
rn
rn
Ei,n
Absorption Coefficient
Ze2

2rn
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Bohr Postulates
Angular Momentum is Quantitized
h
n
2
nh
vn 
2 mrn
mvn rn  n
Now put v into the force equation:
2
nh  3
Ze 2

 m 
rn   2

rn
 2 m 
n 2h 2
2

Ze
4 2mrn
n 2h 2
rn 
4 2 mZe 2
Now put r into the energy equation:
Ze 2
2 2mZ 2e 4
Ei ,n  

2 2
2
r
n
h
n
Absorption Coefficient
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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
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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
==>
Absorption Coefficient
T = 1.4(106) K
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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
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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
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A Free “Orbit”
Consider the quantum number n
n  1n which yields a positive energy
n = a positive value
Ei ,n  h  Ei ,n
4 2
2
4 2


2 m e Z
2 m e Z
 h 
2 2
h n
h 2n 2
2
The solution for the orbit yields a hyperbola
Absorption Coefficient
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Bound Transitions
Consider a transition between two bound states (a,b)
=> (na,nb) [n’s are the quantum numbers]
h  Ei ,nb  Ei ,na
2 2me 4 Z 2 2 2me 4 Z 2


2 2
h nb
h 2na2
2 2me 4 Z 2

h3
1
1

 n2 n2 
a 
 b
1
1
 3.2883797(10 )  2  2 
 nb na 
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1
c
1
   911.672  2  2 

 nb na 
Absorption Coefficient
1
m and Z for H
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Hydrogen Lines
Absorption Coefficient
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Balmer Series of H
The lower level is n = 2  na = 2
First line (Hα) is the transition (absorption): 2 → 3.
– For absorption:
– For emission:
lower → upper na → nb
upper → lower nb → na
1
1 1
   911.672  2  2   6564.038A

3 2 
c
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
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Edlen’s Formula
Computes the Index of Refraction of Air
νλair = c/n or λVacuum = nλAir
2949810 25540
(n  1)10  6432.8 

2
146  
41   2
8
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
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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
n
Limit
Name
1
912
Lyman
2
3646
Balmer
3
8204
Paschen
4
14590
Brackett
5
22790
Pfund
Absorption Coefficient
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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
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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
mj: -j  mj  +j
Absorption Coefficient
s=½
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Hydrogen Quantum Numbers
n=1:
n=2:
(n = 1, l = 0, ml = 0, ms = ±½)
(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
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Notation
2S+1L
J
• 2S+1: Multiplicity (Spin) S = ∑si
• L:
Azimuthal Quantum Number L = ∑i
• J:
L+S = total angular momentum vector
Absorption Coefficient
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Boltzmann Equation
We shall start with the state-ratio form of the equation
 Ei , x
Ni,x
gi , x e kT

 Ei , x 
N i , x
gi , xe kT
The E’s are with respect to the continuum.
x  Ei , x  )
gi , x  ( Ei ,kT

e
gi , x
i , x
N i , x gi , x kT

e
N i ,0 gi ,0
Ni   Ni,x
x
χ is the excitation potential of the state
 i , x
N i ,0

pi gi , x e kT

gi ,0
Absorption Coefficient
pi is the probability that
the state exists
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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
U (T )   pi gi , x e
 i , x
kT
x
• 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
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Rewrite the Boltzman Equation
i , x
N i , x gi , x kT

e
N i ,0 gi ,0
Number in state X WRT ground state
N i ,0
Ni 
U (T )
gi ,0
N i , x gi , x

e
Ni
gi ,0
 i , x
kT
Sum Ni,x
gi ,0
U (T )
Substitute
i , x
gi , x kT

e
U (T )
Absorption Coefficient
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Example
What is the fraction of H that can undergo a Balmer transition at 5700K?
N I ,2
g I ,2

e
NI
U (T )
  I ,2
kT
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
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Ionization
The Saha Equation

N i 1,0
2(2 mkT )3/ 2 gi 1,0 kTi ,
Ne 
e
3
N i ,0
h
gi ,0
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.
gi ,0 kTi ,0
gi ,0
N i ,0  N i
e
 Ni
U (T )
U (T )
N i 1,0
gi 1,0  kTi 1,0
gi 1,0
 N i 1
e
 N i 1
U (T )
U (T )
Absorption Coefficient
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Rewrite The Saha Equation
N i 1,0
2(2 mkT )
Ne 
N i ,0
h3
3/ 2
gi 1,0
e
gi ,0
 i ,
kT
Note that m = me

N i 1
2(2 mkT )3/ 2 U i 1 gi 1,0 gi ,0 kTi ,
Ne 
e
3
Ni
h
U i gi ,0 gi 1,0
N i 1
(2 mk )
Ne 
Ni
h3
3/ 2
2U i 1 3/ 2
T e
Ui
 i , 
kT
Pe  N e kT
N i 1
5 / 2 2U i 1
Pe  0.3334T
e
Ni
Ui
 i , 
kT
Absorption Coefficient
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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
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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
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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
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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
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