Spectral Line Intensities - Boltzmann, Saha Eqs.
Absorption in a line depends on:
- number of absorbers along the line-of-sight, and
-their cross section(s).
Absorp. ≈ naσl,
where na is the number of absorbers per unit vol.,
σ is the cross section, and
l is the (scattering) path length.
σl is also the cylinder volume.
The number density of absorbers na depends on,
- the number density of atoms of that element
- the fraction in the appropriate level or state.
The latter depends on a balance between the rate in vs. out of that state.
In processes (see sketch):
collisional (de)excitation
photoabsorption
spontaneous emission from above
Out:
analogous terms.
Thermal Equilibrium
Definition: Transitions into any (energy) level = transitions out of the level.
Note: if collisions are important, the population of the upper levels depends
on the r.m.s. relative speed, which depends on temperature - mv2/2 = 3kT/2
(equipartition).
But what are the relative numbers of particles in different levels? Or what is
the probability of an electron being found in a given level?
....assume that atoms have velocities given by the Maxwellian velocity
distribution (will not derive). In 1-d:
 m 
f ( vx ) = 
 2πkT 
1/ 2
e
−mv2x
2kT
is the probability of an atom having an x velocity in the range (vx, vx+dvx).
Maxwellian distribution
f(vx)
vx
Of more interest is the distribution of speeds v = (vx2 + vy2 +vz2)1/2.
Consider velocity space.... F = f(vx) f(vy) f(vz) is the local density of
velocity points in v-space.
Vz
dVz
dVx
dVy
Vx
Vy
F dvxdvydvz is the number of points in the little volume.
 m 

F=
 2πkT 
−m ( v2x +v 2y +v z2 )
3/2
2kT
e
 m 

=
 2πkT 
3/2
e
−mv 2
2 kT
which depends only on radius in v-space.
So the number of points in a thin shell is -
4πv2dv F = g(v)dv,
where the equality defines the speed distribution g.
We can define the kinetic energy distribution similarly -
F(E) dE =g(v)
dv. But,
1
v dE  2E 

E = mv 2 , dE = mv dv, v 2 dv =
=

2
m
m
1/ 2
dE
,
m
so,
2
3/2 −mv 
1/ 2
3/ 2 −E

m
2E
dE
m





2
 e 2kT dv = 4π 


 e kT
F( E )dE = 4πv 
 2πkT 
 m
m  2πkT 


Note: E1/2dE is essentially the energy space volume.
For particles with discrete energy levels i we have (again no derivation):
Fi = Cgi e
− Ei
kT
which leads to.....
,
Boltzmann's Law for the relative population of energy levels A, B in atom:
F B gB
=
e
F A gA
( EA −E B )
kT
,
where gB, gA are the statistical weights.
Formally, g = 2J + 1, where J is the total angular momentum quantum
number. Informally, g is the number of sublevels of equal energy. E.g. the
ground state of hydrogen has 2 sublevels (e- spin up or down), so g1 = 2.
The first excited state has a zero angular momentum part, and 3 l=1 states
(2p3), i.e. 4 states with e- spin up or down in any one of them, so g 2 = 4 x 2 =
8.
Note: atomic energy levels are measured in eV, so a useful conversion factor is :
1eV/k = 11,600 K.
––––––––––––––––––––––––––––––––
Ex. The relative population of the ground and first excited state in H:
ΔE = 13.6 (1/12 - 1/22) = 10.2 eV, with g2/g1 = 4.
T
N2/N1 = 4 exp(-118000/T)
103
10-51
104
10-5
105
1.2
106
2.7, but mostly ionized.
Related question - suppose we want to know the number in the ground state
compared to number in all states?
Let I ≥ 1,
F1 g1 (E i −E1 )/ kT
= e
,
Fi gi
F1
F1
−( E i −E1 )/ kT
Ftot = ∑ i Fi = ∑ i gi e
= U(T).
g1
g1
The sum U(T) is called the partition function. (It’s tough to compute.)
F1/Ftot = g1-1 U(T),
€
which answers the question above.
Ionization
If collisions are energetic enough, they can ionize the atom. The formula for
equilibrium ionization is similar to Boltzmann, but we not derive it. The
relative numbers of ions and neutrals of a given element is,
N + A(kT) 3 / 2 − χ o / kT
=
e
No
Ne
Saha eq.
where Ne is the number density of free electrons,
χ is the ionization energy, and
€o
A is a constant made up of several physical and atomic constants.
Note: the exponential term in the equation usually dominates.
This eq. can be directly generalized to higher ionization states.
The generalized Saha equation is -
N j +1N e  2πmkT  3 / 2 2U j +1(T) − χ / kT
=
e
,

2
 h 
Nj
U j (T)
where Nj refers to the population of the jth ionization state,
€
Uj is the partition function of the jth ion,
χ is the ionization energy of the jth ion.
E.g., Hydrogen
χ = 13.6 eV,
χ/k = 158,000 K
Partition functions:
Ho
H+
T = 5040
1.35
1.0
T=10,080
1.35
1.0
So,
N H + Ne
NHo
E.g.,

 3 / 2 −158000 /T
T
= 1.48
e
−11 
 5.55 ×10 
T = 10,000
N H + Ne
€
NHo
€
= 3.58 ×10 21 e−15.8 ≈ 4.9 ×1014
For
T = 1000,
N H + Ne
NHo
= 1.13 ×10 20 e−158 ≈ 2.7 ×10−49.
This abrupt change with T allows us to understand HII regions...
€
HII Regions
Are bubbles of ionized hydrogen surrounding hot stars. (Need the hot stars to
provide the high energy photons.)
Sharpless 212 in Hydrogen and Sulfur
Credit & Copyright: Lise Deharveng (Universite de Provence) et al., CNRS, OHP
O stars have large HII regions (≈ 200 pc)
B stars have smaller HII regions (≈ 10 pc)
UV photons ionize the gas ––> recombination produces hydrogen emission lines.
The two processes balance to give a very low fraction of neutral atoms in the HII
region.
Classical HII regions have sharp edges, but do not include the effects of stellar
winds or pressure induced motions.
In addition to H lines, HII region spectra often show lines of:
He, O+, O++, N+, Ne++.
Fich and Blitz, 1984, ApJ, 279, 125
Photodissociation regions
Surround HII regions in dense molecular clouds.
Ho
H+
H2