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77 lines
2.6 KiB
Plaintext
77 lines
2.6 KiB
Plaintext
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Homework 15, due Oct 17
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1) Find the ratio of emission in the [O III] 4363 to 5007 lines as a function of temperature assuming LTE. Use the energies and TPs from NIST and assume LTE. Give this as an equation as a function of temperature.
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A₁﹐ᵦ
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j₁/j₂ = g₁/g₂ A₁/A₂ hν₁/hν₂ exp(-E₁₂/kT).
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j₁/j₂ = g₁/g₂ A₁/A₂ λ₂/λ₁ exp(-E₁₂/kT).
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g₁/g₂ = 1/3.
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A₁/A₂ = 94.292.
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λ₂/λ₁ = 1.1476.
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E₁₂ = 2.84 eV.
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exp(-E₁₂/kT) = exp(-2.84 eV/k × T) = exp(-32976 K⁻¹ × T).
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j₁/j₂ = 1/3 × 94.292 × 1.1476 × exp(-32976 K⁻¹ × T).
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j₁/j₂ = 36.070 × exp(-32976 K⁻¹ × T).
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─────────────
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2) Plot the line ratio as a function of temperature between 5000 K and 50 000 K. Make the y-axis a log.
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Attached.
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─────────────
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3) The line ratio is observed to be I(4363)/I(5007) = 0.0048 in Orion. What would be the LTE temperature? (Orion is not in LTE).
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*
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0.0048 = 36.070 × exp(-32976 K⁻¹ × T).
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0.0048/36.070 = exp(-32976 K⁻¹ × T).
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ln(0.0048/36.070) = -32976 K⁻¹ × T.
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(ln(0.0048/36.070)/-32976) K = T.
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T = 0.0027064 K.
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Way too small! Something wrong.
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─────────────
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4) What is the coefficient
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the Saha equation, in CGS units?
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⎛2 π mₑ k T ⎞3/2
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⎜---------- ⎟ = 2.41468×10¹⁵ g/(K erg s²) T^(3/2).
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⎝ h² ⎠
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─────────────
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5) Assume a hydrogen density of 1e15 cm-3. What is the hydrogen ionization fraction H+/H0 at 6000 K, 10 000 K, and 20 000 K?
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n[ion]/n[atom] = 1/nₑ
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× U[ion]/U[atom]
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× gₑ
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× (coefficient)
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× T^(3/2)
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× exp(-E[ion]/kT).
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nₑ = 10¹⁵ cm⁻³.
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U[ion] = 1.
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U[atom] = 10^0.3 ≈ 2.
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gₑ = 2.
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coefficient = 2.41468×10¹⁵ g/(K erg s²).
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Temperature Ion Fraction
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6000 K 4.196×10⁻⁶
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10000 K 3.365×10⁻¹
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20000 K 2.552×10³
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─────────────
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6) Use the Boltzmann equation to find the n=2 population at these three temperatures and give the Lya emissivity.
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Just following the process from hw14.
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4 π j = n₂ Aᵤₗ hν.
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hν = 1.63403 × 10⁻¹¹ ergs.
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n₂ differs by temperature: ~1.1×10⁷, ~3.9×10⁷, ~6.3×10⁷.
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Aᵤₗ = 6.27×10⁸ s⁻¹.
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Temperature Emissivity
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6000 K 1.136×10⁵ ergs
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10000 K 4.024×10⁵ ergs
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20000 K 6.525×10⁵ ergs
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