Saha*Boltzmann Problem
You got your Ni I line and know the upper and lower level energies.
Now we want to figure out what fraction of all nickel N(Ni) is in your lower level --ionization stage r, excitation s-- N(r,s) at some temperature and electron pressure --we want N(r,s)/N.
The assignment involves working out this fraction at
T=Your line wavelength in angstroms (Kelvin),
Pe=2.5 [dyne/cm2].
So the temprature used for a 4294.092 angstrom line would be 4294.092 K. Hey, &lambda is independent of T, this is a trick so each student will have a different temperature! (note that Q=5040/T)
I assume you have familiarised yourself with the Saha Boltzmann notes...
***Nomenclature (new!)****
N stands for total number density of atoms, in this case oxygen.
Nr stands for the number density of atoms in the r-th stage of ionization,
Nr,s stands for the number density of atoms in the r-th stage of ionization and s-th stage of excitation.
Nr=sum of all Nr,s
N=sum of all Nr
You want Nr,s/N.
Note that astronomers use cgs units (still!) so the Saha factor is
Nr/N = Nr/(Nr-1+Nr+Nr+1) = 1/(Nr-1/Nr+1+1+Nr+1/Nr)
where
Nr+1/Nr = (Ur+1/Ur) 109.08-Q Ir/Q5/2Pe
The partition functions and ionization potentials for nickel are
Nickel atomic data:

The IP are 7.6398 18.16884 35.19 eV

Partition functions (Allen, Astrophysical Quantities)
   T      U(1)     U(2)    U(3)    U(4)
  low    21       10       21       25
 5040    29.51    10.47    --       --
10080    39.81    19.05    --       --

Ni partition function polynomial fits, T=temperature/1000 (Use these!)
   U1 = 20.22+2.129*T
   U2 = 6.97+(.0427+.146*T)*T
   U3 = 21.
   U4 = 25.

Now for the Boltzmann factor:
Nr,s/Nr = (gr,s/Ur) 10-Q cr,s
Where cr,s is the excitation potential of the lower level [eV]. If your level energy is in cm-1 (they usually are) divide by 8066 to convert to eV.
gr,s=2J+1 is the statistical weight of the lower level where J is the angular momentum quantum number, often called the J-value. It is just to the left of the level energy and is probably a half-integer between 1/2 and 11/2 for Fe II.
(So if J=7/2 then g=8.)

There you are! Multiply the Saha by the Boltzmann for the answer.
The Saha factor is between 0 and 1, the Boltzmann factor is probably small -- if ep=5 ev and T=10080 then the exponential factor is like ten to the minus 5/2... unless you got a resonance line!


SAHA BOLTZMANN worksheet
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Saha factor N(r)/N=N(2)/N:
 My wavelength w=
Temperature T=2w=
    Theta=5040/T=
Partition functions
              U1=
              U2=
              U3=28
              U4=6
Saha ratios
           N2/N1=
           N3/N2=
           N4/N3<1.E-07

           N(2)/N=
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Boltzmann factor
 g(r,s)=g(2,ep)=
 U(r,s)=U(2,T) =
 lower ep (eV) =
          theta=
   N(2,ep)/N(2)=
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Answer
   N(2,ep)/N(2) * N(2)/N =
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