as-500/hw22/text.mth

1.

∫[ν₁,∞] ϕ[ν] a[ν] dν = Γ, the photoionization rate, in s⁻¹ under the units used below.

ϕ[ν] = 10¹³ (ν/ν₀)⁻³ [photons cm⁻² s⁻¹ Ryd⁻¹].

a[ν] = A₀/Z² (ν₁/ν)⁴ exp(4 - (4 tan⁻¹(ε)/ε)) (1 - exp(-2π/ε))⁻¹ [cm²],

    with A₀ = 6.30 × 10¹⁸ [cm²] and ε = √(ν/ν₁ - 1) and Z = 1,

    but, above the ionization threshold, this is approximately a power law, so

a[ν] ≈ 6.30 × 10¹⁸ (ν/ν₀)⁻³ [cm²],
    
    with ν₀ = 3.288 × 10¹⁵ [s⁻¹].




    ϕ[ν] a[ν] dν 

        = 10¹³ (ν/ν₀)⁻³ [cm⁻² s⁻¹ Ryd⁻¹] ×
          6.30 × 10¹⁸ (ν/ν₀)⁻³ [cm²]

        = 6.30 ×10³¹ (ν/ν₀)⁻⁶ [s⁻¹ Ryd⁻¹].

∫[ν₀,∞] ϕ[ν] a[ν] dν 

    = ∫[ν₀,∞] dν (6.30 ×10³¹ (ν/ν₀)⁻⁶ [s⁻¹ Ryd⁻¹])

    = 6.30 ×10³¹ (1/ν₀)⁻⁶ [s⁻¹ Ryd⁻¹] ∫[ν₀,∞] ν⁻⁶ dν

    = 6.30 ×10³¹ (3.288 × 10¹⁵)⁶ [s⁻⁶ Ryd⁻¹] [∞⁻⁵/(-5) - ((3.288× 10¹⁵ Ryd s⁻¹)⁻⁵/(-5)]

    = (1/5) 6.30 × 10³¹ (3.288 × 10¹⁵)⁶ [s⁻⁶ Ryd⁻¹] (3.288× 10¹⁵)⁻⁵ [Ryd s⁵].

*** = 4.14288 × 10⁴⁶ s⁻¹.

    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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2.

L = ϕ / (nₑ nₚ αᵦ(T))

αᵦ(T) = 2.59 ×10⁻¹³

n[H] = 10³ cm⁻³ = nₑ nₚ.

ϕ = 10¹⁴ cm⁻² s⁻¹ 

L = 10¹⁴ / (10³)² / 2.59 ×10⁻¹³ [cm] 

*** L = 3.861 × 10²⁰ cm.