as-500/hw26/notes.mth

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Old approach:
4πj(u,l) = nᵤ (Aᵤₗ + qᵤₗ nₑ) hν.
Let's ignore collisional excitement, making
4πj(u,l) = n₂ Aᵤₗ hν.
From Allen, Table 4.11,
A[Hβ] = 8.419 × 10⁶ s⁻¹
If H was totally ionized this would be 0, right? Thought I could maybe find a recombination balance to determine a better ratio, but assuming H is not ionized seems to be the approach other students are taking.
From the Boltzman factors computed for hydrogen in hw14, I just reused my spreadsheet to find n₂ = (Z₂/∑Zᵢ * nₜₒₜ) at T=1e4 K:
n₂ = 0.275 [cm⁻³].
hν = 4.09 × 10⁻¹² [ergs].
4πj(u,l) = n₂ Aᵤₗ hν
= 0.275 [cm⁻³] 8.419 × 10⁶ [s⁻¹] 4.09 × 10⁻¹² [ergs]
*** = 9.47 × 10⁻⁶ [ergs cm⁻³ s⁻¹].