Revisiting Stellar Spectral Line Formation & Transitions (1)

 We start then with the Boltzmann equation, which we  can express:

N₂ / N₁   =     [g₂ / g₁ ]   exp (- E2 – E1) / kT

That is, for the atoms of a given element in a specified state of ionization, the ratio of the number of atoms N₂  with energy E2, to the number of atoms N₁  with energy E1, in different states of ionization is given by the above formula. The same form of the equation can also be used to find the ratio of probabilities, i.e. that the system will be found in any of the  g₂ degenerate states with energy E2 to the probability that the system is in any of the g₁ degenerate states E1, viz.

P(E2) / P(E1)   =     [g₂ / g₁ ]   exp (- E2 – E1) / kT

Thus, the Boltzmann equation can be posed in two forms.  In statistical mechanics we also have the partition function:

Z  =   å _j   exp ( - e _j )/ t

Which is just the summation over the Boltzmann factor (exp ( - e _j )/ t ) for all states j for which the number of particles (N) is constant. We will find it useful to rewrite it:

Z = g₁ +  å^¥ _{j = 2}   g _j  exp (- E _j – E1) / kT

Of interest now are the relative numbers of atoms in ionization stage i, which is written:

N _e N _{i + 1} / N _i  =

 2 Z _{i + 1} / Z _i (2 p m _e kT/ h ²) ^1.5  e ^{- c i/ kT}

    This is the Saha equation, named after the Indian astrophysicist who first derived it.  Here,  N _e  is the number of free electrons per unit volume and c _i  is the ionization potential of the ith ionization stage. Thus, the equation relates the number of atoms in two successive  ionization stages to the quantities that are relevant. As per our introduction to quantum mechanics, the factor ‘2’ in the equation refers to the two possible spins of the free hydrogen election with spin quantum number: m _s =  +½.

Recall that for thermodynamic equilibrium, the rate of ionization cannot exceed the rate of recombination[1].  In other words, the rate at which atoms in the ith stage are ionized (i.e. to the i +1st stage) must equal the rate at wich ions in that i +1st stage are recombining with free electrons to form ions in the ith stage. The latter depends on N _e N _{i + 1}   and the former on N _i . Hence, Saha’s equation simply expresses the fact these two processes must occur at the same rate.

 One can also rewrite the equation in a more manageable logarithmic form if one substitutes the numerical constants:

log(N _e N _{i + 1} /N _i)  = 

15.38 + log (2 Z _{i + 1} / Z _i ) + 1.5 log T – 5040 c _i/ T

The units here are important to note and are consistent with the ionization potential being measured in electron volts (eV). Therefore N _e  must be in particles per cubic centimeter.

    Yet another way to express the Saha equation is to introduce the electron pressure, P _e . This acknowledges that each separate species of particle makes its own contribution to the total gas pressure. The free electrons in a gas therefore produce a pressure given by: P _e = N _e kT.

Then we may write another log form of the Saha equation:

log(P _e N _{i + 1} /N _i)  = 

-0.48  + log (2 Z _{i + 1} / Z _i ) + 2.5 log T – 5040 c _i/ T 

Example Problem:

For a  hydrogen plasma find:

(P _e N _{i + 1} /N _i)  =  P _e N _HII /N _HI   

at a temperature of 5040 K, given the hydrogen partition functions are:

Z _{i + 1}  =  Z ₂    =   1   and:

Z _i =    Z ₂    =   2      with  c _i =   13.6 eV 

Solution:

By the Saha equation:   log(P _e N _{i + 1} /N _i)  = 

-0.48  + log (2 Z _{i + 1} / Z _i ) + 2.5 log T – 5040 c _i/ T

=   -0.48  +  log (2) + 2.5 log (5040)  –  5040 (13.6)/ 5040

=   -0.48 + 0.30  +   9.25 – 13.6  =  -4.83

Antilog (-4.53) =   2.95 x 10 ^-5

 A useful table that will come in handy for spectral line and ionization energy computations is presented below. 

Careful inspection of the table  shows two quantum states with the energy (-13.6 eV) and eight with (-3.40 eV).   Thus, two states are degenerate for the n=1 level and eight states are degenerate for the n=2 level.  Since g _n = 2n², then:

At the n=1 level the statistical weight is:  g ₁ = 2(1)² = 2

At the n=2 level the statistical weight is: g ₂ = 2(2)² = 8  

Suggested Problem:

Consider a gas of neutral hydrogen (H). Using the Boltzmann equation and the information in the table above, compute the temperature at which one will expect equal numbers of atoms in the ground state and the first excited state.

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[1] This must hold if the excitation and ionization equations are assumed valid, hence the numbers of atoms in a given level must not change with time.