IONIZATION, SAHA EQUATION

Let the energies of two states, A and B, be EA and EB, and their statistical weights gA and gB, respectively. In LTE (Local Thermodynamic Equilibrium) the number of particles in the two states, NA and NB, satisfies Boltzman equation:

NA gA = exp[− (EA − EB ) /kT ] . (i.1) NB gB

Now, we shall consider two ions, “i” and “i+1”, of the same element. The ionization potential, i.e. the energy needed to ionize “i” from the ground state is χ, and the statistical weights of the −3 ground states of the two ions are gi and gi+1, respectively. The number densities, [ cm ], of the two types of ions and free electrons are ni, ni+1, and ne, respectively. We shall use the Boltzman equation (i.1) to estimate the number ratio ni+1/ni. The statistical weight of an ion in the lower ionization state to be used in the equation (i.1) is just gi. The statistical weight of an ion in the upper ionization state is gi+1 multiplied by the number of possible states in which a free electron may be put. As we know, in every cell of a phase space with a volume h3 there are two possible states for an electron, because there are two possible orientations of its spin. h =6.63 × 10−27 erg s is the Planck constant. The energy of a free electron with a momentum p with respect to the ground state of an ion in a lower ionization state is E = χ + p2/2m. The number of cells available for free 2 3 electrons with a momentum between p and p + dp is Ve4πp dp/h , where Ve =1/ne is the volume in ordinary space available per electron, and ne is the free electron number density. Now, we shall integrate over all available cells, taking the Boltzman factor into account

∞ 2 ni+1 gi+1 2Ve −(χ+p /2m)/kT 2 = 3 e 4πp dp = (i.2) ni gi h Z 0 ∞ gi+1 1 2 3/2 −χ/kT −x 1/2 3 (2mkT ) e 2π e x dx = gi ne h Z 0

gi+1 1 2 3/2 −χ/kT 3 (2πmkT ) e , gi ne h where we set x = p2/2mkT , and the value of the last integral was π1/2/2. The last equation is usually written as the Saha equation

1.5 ni+1ne (2πmkT ) 2gi+1 −χ/kT = 3 e . (i.3) ni h gi

It is customary to express the ionization energy in electron volts, 1eV = 1.602 × 10−12 erg. Ionization of the most abundant elements, hydrogen and helium, is important for the equation of state. For these elements we have:

ionization of hydrogen: χ = 13.54 eV , 2gi+1/gi = 1, first ionization of helium: χ = 24.48 eV , 2gi+1/gi = 4, second ionization of helium: χ = 54.17 eV , 2gi+1/gi = 1.

Consider now pure, partly ionized hydrogen. Let nH , nHI , and nHII be number density of all hydrogen, neutral hydrogen atoms, and hydrogen ions, respectively. We have nH = nHI + nHII , ne = nHII , and x ≡ nHII /nH is the degree of ionization. The density of gas is ρ = HnH , where H is the mass of a hydrogen atom. We may write the Saha equation as

i—1 2 1.5 nHII ne ρ x (2πmkT ) 2gi+1 −χ/kT = = 3 e . (i.4) nHI H 1 − x h gi The constants in cgs units are: H = 1.673 × 10−24, m = 9.11 × 10−28, k = 1.381 × 10−16, h = 6.63 × 10−27. With these constants we have

x2 log ρ + log = (i.5) 1 − x

5040 1 5 − 68240 =1.5log T − χ + log H (2πmk) . h 3 =1.5log T − − 8.394. T h i T The pressure of partially ionized hydrogen gas is simply k P = (n + n ) kT = (1+ x) ρT. (i.6) g H e H The internal energy should now include not only kinetic, but also ionization energy: χ U =1.5 (n + n ) kT + n χ =1.5P + x ρ. (i.7) g H e e H In these equations the degree of ionization should be treated as a function of density and tem- perature, x(ρ,T ). Differentiating equation (i.4) we obtain

∂x x (1 − x) χ = 1.5+ , (i.8a) − ∂lnT ρ (2 x)  kT  ∂x x (1 − x) = − . (i.8b) − ∂ ln ρT (2 x) As an example of the thermodynamic properties of partially ionized hydrogen we shall calculate specific heat of hydrogen gas at constant volume:

∂ (Ug/ρ) 1.5 ∂Pg χ ∂x cV,g ≡ = + = (i.9a)  ∂T ρ ρ  ∂T ρ H ∂T ρ

k 2 x (1 − x) = 1.5(1+ x)+(1.5+ χ/kT ) . H  (2 − x)  When the hydrogen is 50% ionized, i.e. x =0.5, then

k 1 2 c = 2.25+ (1.5+ χ/kT ) . (i.9b) V,g H  6  Typically, we have χ/kT ≫ 1, and therefore the specific heat for partly ionized gas is much higher than for neutral or fully ionized gas. Let us consider now a mixture of partially ionized hydrogen with radiation. The equation of state is k a P = (1+ x) ρT, P = T 4, P = P + P , β = P /P, (i.10a) g H r 3 g r g

k a P =(1+ x) ρT + T 4, (i.10b) H 3

k χ U =1.5(1+ x) ρT + aT 4 + x ρ, (i.10c) H H The equations in a differential form are: β d ln P = (4 − 3β) d ln T + βd ln ρ + dx (equation of state), (i.11a) 1+ x

i—2 dU 3 χ 3 χ β = (12 − 10.5β) d ln T + + βd ln ρ + + dx, (i.11b) P 2 kT  2 kT  1+ x

(2 − x) 3 χ dx = + d ln T − d ln ρ (Saha equation). (i.11c) x (1 − x) 2 kT 

From the Saha equation we obtain equations (i.8a) and (i.8b) . Those may be combined with the equations (i.11a) and (i.11b) to get

∂ ln P β ∂x =4 − 3β + , (i.12a)  ∂ ln T ρ (1 + x) ∂ ln T ρ ∂ ln P β ∂x = β + , (i.12b)  ∂ ln ρ T (1 + x) ∂ ln ρT T ∂U β χ ∂x = 12 − 10.5β + 1.5+ . (i.12c) P  ∂T ρ (1 + x)  kT  ∂ ln T ρ

Now we have all the derivatives that are needed to calculate the adiabatic relations: γ, ∇ad, and (∂ ln T/∂ ln ρ)S, as well as the specific heats, cV and cP , for the mixture of partially ionized hydrogen and radiation.

i—3