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. n a σl, where n a 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 n a 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 - mv 2 /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: f ( v x ) = m 2πkT 1/ 2 2 mv x 2kT e is the probability of an atom having an x velocity in the range (v x, v x +dv x ).
Maxwellian distribution f(v x ) v x
Of more interest is the distribution of speeds v = (v x 2 + v y 2 +v z 2) 1/2. Consider velocity space... F = f(v x ) f(v y ) f(v z ) is the local density of velocity points in v-space. V z dv z dv x dv y V x V y
F dv x dv y dv z is the number of points in the little volume. F = m 2πkT 3/2 e 2 m( v x2 +v y2 +v z ) 2kT = which depends only on radius in v-space. So the number of points in a thin shell is - m 2πkT where the equality defines the speed distribution g. 3/2 e mv 2 2 kt 4πv 2 dv F = g(v)dv, We can define the kinetic energy distribution similarly - dv. But, E = 1 2 mv2, de = mv dv, v 2 dv = v de m = 2E m so, F( E)dE = 4πv 2 3/2 m 2πkT e mv 2 2kT dv = 4π 2E m 1/ 2 de m Note: E 1/2 de is essentially the energy space volume. F(E) de =g(v) 1/ 2 de m, m 2πkT 3/ 2 E kt e
For particles with discrete energy levels i we have (again no derivation): F i = Cg i e E i kt, which leads to... Boltzmann's Law for the relative population of energy levels A, B in atom: F B F A = g B g A e where g B, g A are the statistical weights. ( E A E B ) kt, 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 g 1 = 2. The first excited state has a zero angular momentum part, and 3 l=1 states (2p 3 ), 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 (1/1 2-1/2 2 ) = 10.2 ev, with g 2 /g 1 = 4. T N 2 /N 1 = 4 exp(-118000/t) 10 3 10-51 10 4 10-5 10 5 1.2 10 6 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, F 1 F i = g 1 g i e (E i E 1 )/ kt, F tot = F i = F 1 i g i i e (E i E 1 )/ kt = F 1U(T). g 1 g 1 The sum U(T) is called the partition function. (It s tough to compute.) F 1 /F tot = g -1 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 + N o = A(kT)3 / 2 N e e χ o / kt Saha eq. where N e is the number density of free electrons, χ o is the ionization energy, and 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 +1 N e N j = 2πmkT 3 / 2 2U j +1 (T) e χ / kt, h 2 U j (T) where N j refers to the population of the j th ionization state, U j is the partition function of the j th ion, χ is the ionization energy of the j th ion.
E.g., Hydrogen χ = 13.6 ev, χ/k = 158,000 K Partition functions: H o H + T = 5040 1.35 1.0 T=10,080 1.35 1.0 So, N H + N e N H o T =1.48 5.55 10 11 3 / 2 158000 /T e E.g., T = 10,000 N H + N e N H o = 3.58 10 21 e 15.8 4.9 10 14
For T = 1000, N H + N e N H o =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. H o H + H 2