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61 lines
1.6 KiB
Plaintext
61 lines
1.6 KiB
Plaintext
1.
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∫[ν₁,∞] ϕ[ν] a[ν] dν = Γ, the photoionization rate, in s⁻¹ under the units used below.
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ϕ[ν] = 10¹³ (ν/ν₀)⁻³ [photons cm⁻² s⁻¹ Ryd⁻¹].
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a[ν] = A₀/Z² (ν₁/ν)⁴ exp(4 - (4 tan⁻¹(ε)/ε)) (1 - exp(-2π/ε))⁻¹ [cm²],
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with A₀ = 6.30 × 10¹⁸ [cm²] and ε = √(ν/ν₁ - 1) and Z = 1,
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but, above the ionization threshold, this is approximately a power law, so
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a[ν] ≈ 6.30 × 10¹⁸ (ν/ν₀)⁻³ [cm²],
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with ν₀ = 3.288 × 10¹⁵ [s⁻¹].
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ϕ[ν] a[ν] dν
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= 10¹³ (ν/ν₀)⁻³ [cm⁻² s⁻¹ Ryd⁻¹] ×
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6.30 × 10¹⁸ (ν/ν₀)⁻³ [cm²]
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= 6.30 ×10³¹ (ν/ν₀)⁻⁶ [s⁻¹ Ryd⁻¹].
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∫[ν₀,∞] ϕ[ν] a[ν] dν
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= ∫[ν₀,∞] dν (6.30 ×10³¹ (ν/ν₀)⁻⁶ [s⁻¹ Ryd⁻¹])
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= 6.30 ×10³¹ (1/ν₀)⁻⁶ [s⁻¹ Ryd⁻¹] ∫[ν₀,∞] ν⁻⁶ dν
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= 6.30 ×10³¹ (3.288 × 10¹⁵)⁶ [s⁻⁶ Ryd⁻¹] [∞⁻⁵/(-5) - ((3.288× 10¹⁵ Ryd s⁻¹)⁻⁵/(-5)]
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= (1/5) 6.30 × 10³¹ (3.288 × 10¹⁵)⁶ [s⁻⁶ Ryd⁻¹] (3.288× 10¹⁵)⁻⁵ [Ryd s⁵].
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*** = 4.14288 × 10⁴⁶ s⁻¹.
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Seems really high... maybe I goofed something up, because 10¹³ hydrogen ionizing flux density seems sort of middle of the road. When we did AGN simulations, we went to around 10²⁴. Moving on for now, though...
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─────────────
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2.
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L = ϕ / (nₑ nₚ αᵦ(T))
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αᵦ(T) = 2.59 ×10⁻¹³
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n[H] = 10³ cm⁻³ = nₑ nₚ.
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ϕ = 10¹⁴ cm⁻² s⁻¹
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L = 10¹⁴ / (10³)² / 2.59 ×10⁻¹³ [cm]
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*** L = 3.861 × 10²⁰ cm.
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