mirror of
https://asciireactor.com/otho/as-500.git
synced 2024-11-24 03:15:05 +00:00
86 lines
2.6 KiB
Plaintext
86 lines
2.6 KiB
Plaintext
Homework 19 (marked homework 21 on the pdf)
|
||
|
||
1. Solve the NLTE balance equation (equation 3.24 in AGN3, also given on page 6, side 2, of the lecture notes) for the population of the ¹D level of O III, n(¹D), cm⁻³. That equation gives the dimensionless population ratio n(¹D) / n(3P) we want the population. Assume that the total density of O²⁺ is 10⁻⁴ of the hydrogen density, and that the electron density is equal to the hydrogen density. Vary the hydrogen density n(H) between 10⁻² to 10¹⁰ cm⁻³ in 1 dex steps. Plot log n(¹D) as the x-axis and log n(H) as the y-axis.
|
||
|
||
NTLE balance is given by
|
||
|
||
n₂/n₁ = nₑ q₁₂/A₂₁ (1 + nₑ q₂₁/A₂₁)⁻¹.
|
||
|
||
I model a two level O-III atom.
|
||
|
||
I assume density n[O²⁺]/n[H] = 10⁻⁴.
|
||
|
||
I assume hydrogen is completely ionized, i.e.,
|
||
|
||
nₑ = n[H].
|
||
|
||
I will vary n[H], so I need only find the coefficients q and A, which can be pulled from AGN3,
|
||
|
||
From combining the 3P levels,
|
||
A₂₁(1D→3P) = [6.8e-3 + 2.0e-2 + 1.7e-6] s⁻¹
|
||
= 0.0268 [s⁻¹].
|
||
|
||
The collision strengths are also given in AGN3,
|
||
γ₁₂ = 2.29
|
||
|
||
q₂₁ = 8.629e-6 γ₁₂ Tₑ^(-1/2) g₂⁻¹ [cm³/s]
|
||
= 1.98e-5 Tₑ^(-1/2) g₂⁻¹ [cm³/s].
|
||
|
||
Assuming electron temperature
|
||
Tₑ = 1e4 K.
|
||
|
||
g₁ = 9,
|
||
combining all the states of the 3P levels.
|
||
|
||
g₂ = 2*J + 1,
|
||
with J = 2 for 1D, so
|
||
g₂ = 5,
|
||
|
||
q₂₁ = 1.98e-5 * 1e-2 K^(-1/2) 1/5
|
||
= 3.96e-08 [cm³/s].
|
||
|
||
q₁₂ = q₂₁ g₂/g₁ exp(-Eₜₒₜ/kT).
|
||
|
||
Eₜₒₜ = 26169 K.
|
||
|
||
q₁₂ = 2.20e-8 exp(-26169/1e4) [cm³/s].
|
||
= 2.20e-8 * 0.073 [cm³/s]
|
||
= 1.61e-9 [cm³/s].
|
||
|
||
So, the population ratio is
|
||
|
||
n₂/n₁ = nₑ q₁₂/A₂₁ (1 + nₑ q₂₁/A₂₁)⁻¹
|
||
|
||
= nₑ (1.61e-9 [cm³/s])/(0.0268 [s⁻¹])
|
||
× (1 + nₑ (3.96e-08 [cm³/s])/(0.0268 [s⁻¹]))⁻¹
|
||
|
||
= nₑ (1.61e-9/0.0268 [cm³])
|
||
× (1 + nₑ (3.96e-08/0.0268) [cm³])⁻¹
|
||
|
||
= nₑ (6.01e-8 [cm³]) (1 + nₑ (1.48e-6) [cm³])⁻¹.
|
||
|
||
This can now be computed across the n[H] density spectrum.
|
||
|
||
Plot attached.
|
||
|
||
|
||
2. Solve for this population using the Boltzmann equation, assuming U=9. Add this to the plot.
|
||
|
||
From the boltzmann equation
|
||
|
||
n₂/n₁ = g₂/g₁ exp(-Eₜₒₜ/kt)
|
||
= 5/9 * 0.073
|
||
= 0.041.
|
||
|
||
|
||
|
||
3. Find the critical density and indicate that on the plot. How does this compare with the density where the ¹D level has come into LTE?
|
||
|
||
Critical density can be predicted as
|
||
|
||
n[crit] = A₂₁/q₂₁ = 6.78e5 [cm⁻³].
|
||
|
||
This line is drawn on the plot, and shows where the level populations predicted by the NLTE solution agrees with LTE predictions.
|
||
|
||
|