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94 lines
3.2 KiB
Plaintext
94 lines
3.2 KiB
Plaintext
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Homework 26, day 29, due Dec 7
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Let’s compute the radio brems to Balmer beta emissivity ratio for the Orion Nebula! As mentioned in class this is a great way to measure the interstellar reddening, since dust extinguishes visible light but not the radio. Let’s assume T=10⁴ K, n(H) = 10⁴ cm⁻³ , only consider H, and assume that it is fully ionized.
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1. What is the brems emissivity 4πνj[ν], erg cm⁻³ s⁻¹, at 21 cm?
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Conditions:
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T = 10⁴ K
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n(H) = 10⁴ cm⁻³
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H fully ionized.
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From Brandt, the emissivity j[ν] at is given by (note: n in m⁻³.)
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j[ν](ν) dν = (6.8 × 10⁻⁵¹ [J m³ K^(1/2)]) g(ν,T,Z) Z² nₑ nᵢ
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× exp(-hν/kT) T^(-1/2).
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The gaunt correction factor is approximated from the contour for
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log(T) = 4 [log(K)] and log(λ) = 5.32 [log(μm)].
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g[ν,T,Z] ≈ 5.
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nₑ = nᵢ = nₕ = 10⁴ [cm⁻³] = 10⁴ [cm⁻³] (100 [cm/m])³ = 10¹⁰ [m⁻³].
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For 21 cm,
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hν/k = 0.0685 [K].
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T = 10⁴ [K].
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j[ν](ν) dν = (6.8 × 10⁻⁵¹ [J m³ K^(1/2)])
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× g(ν,T,Z) Z² nₑ nᵢ exp(-hν/kT) T^(-1/2).
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= (6.8 × 10⁻⁵¹ [J m³ K^(1/2)])
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× (5) (10¹⁰ [m⁻³])²
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× exp(-0.0685 [K]/(10⁴ [K])) (10⁴ [K])^(-1/2)
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= (6.8e-51) * (5) * (1e20)
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* exp(-0.0685 [K]/(1e4 [K])) * (1e4)^(-1/2)
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[J m³ m⁻³ m⁻³]
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= (6.8e-51) * (5) * (1e20)
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* math.exp(-0.0685/1e4) * math.pow((1e4),(-1/2))
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[J m⁻³]
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= 3.4e-32 [J m⁻³].
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ν[21 cm] = 1.48 × 10⁹ s⁻¹.
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j[ν](ν) ν = 3.4e-32 [J m⁻³] * 1.48e9 [s⁻¹]
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= 3.4e-32 * 1.48e9 [J m⁻³ s⁻¹]
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= 5.03e-23 [J m⁻³ s⁻¹].
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j[ν](ν) ν = 5.03e-23 [J m⁻³ s⁻¹] × [10⁷ ergs/J] × [1/100 m/cm]³.
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*** = 5.02e-22 [ergs cm⁻³ s⁻¹].
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─────────────
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2. What is the emissivity in the Balmer beta line, the n=4-2 transition at 4861Å? The units of 4pj, are also erg cm -3 s -1 .
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The Balmer emission here is from recombination. From the table 4.4,
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4π j[Hβ] = 1.24 × 10⁻²⁵ [ergs cm³ s⁻¹] nₑ nᵢ
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= 1.24e-25 [ergs cm³ s⁻¹] 1e8 [cm⁻⁶]
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= 1.24e-17 [ergs cm⁻³ s⁻¹].
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─────────────
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3. What is the dimensionless ratio of the Balmer beta to radio brems emissivities? This is very nearly equal to the ratio of emission from the Orion Nebula. (I get roughly 2e4 but did this quickly).
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(1.24e-17 [ergs cm⁻³ s⁻¹])/(5.02e-22 [ergs cm⁻³ s⁻¹])
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= 1.24e-17 / 5.02e-22
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*** = 2.47e4.
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─────────────
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4. If the ratio measured with Space Telescope and the VLA is 2e3, find the optical depth due to dust at Balmer beta.
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Assuming dust is the only source of absorption and scattering,
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τ = ∫ (κ) dl ⟶ I = I₀ exp(-τ).
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exp(-τ) = I/I₀ = 2e3/2.47e4 * I₀﹐ᵦᵣ/Iᵦᵣ
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From ISM opacity,
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τᵦᵣ = ∫ 1e-24 [cm⁻¹] dl
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= 1e-24 [cm⁻¹] * 1344 [ly]
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= 1e-24 [cm⁻¹] * 1.27e21 [cm]
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= 0.0013.
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I₀﹐ᵦᵣ/Iᵦᵣ = exp(-0.0013) ≈ 1.
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-τ = ln(2e3/2.47e4)
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= -2.51.
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*** τ = 2.51.
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