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24 lines
786 B
Plaintext
24 lines
786 B
Plaintext
Old approach:
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4πj(u,l) = nᵤ (Aᵤₗ + qᵤₗ nₑ) hν.
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Let's ignore collisional excitement, making
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4πj(u,l) = n₂ Aᵤₗ hν.
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From Allen, Table 4.11,
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A[Hβ] = 8.419 × 10⁶ s⁻¹
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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.
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From the Boltzman factors computed for hydrogen in hw14, I just reused my spreadsheet to find n₂ = (Z₂/∑Zᵢ * nₜₒₜ) at T=1e4 K:
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n₂ = 0.275 [cm⁻³].
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hν = 4.09 × 10⁻¹² [ergs].
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4πj(u,l) = n₂ Aᵤₗ hν
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= 0.275 [cm⁻³] 8.419 × 10⁶ [s⁻¹] 4.09 × 10⁻¹² [ergs]
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*** = 9.47 × 10⁻⁶ [ergs cm⁻³ s⁻¹]. |