Homework 15, due Oct 17

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.
A₁﹐ᵦ

    j₁/j₂ = g₁/g₂ A₁/A₂ hν₁/hν₂ exp(-E₁₂/kT).
    j₁/j₂ = g₁/g₂ A₁/A₂ λ₂/λ₁ exp(-E₁₂/kT).

    g₁/g₂ = 1/3.
    A₁/A₂ = 94.292.
    λ₂/λ₁ = 1.1476.
    E₁₂ = 2.84 eV.
    exp(-E₁₂/kT) = exp(-2.84 eV/k × T) = exp(-32976 K⁻¹ × T).

    j₁/j₂ = 1/3 × 94.292 × 1.1476 × exp(-32976 K⁻¹ × T).
    j₁/j₂ = 36.070 × exp(-32976 K⁻¹ × T).

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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.

    Attached.

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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).
*

    0.0048 = 36.070 × exp(-32976 K⁻¹ × T).
    0.0048/36.070 = exp(-32976 K⁻¹ × T).
    ln(0.0048/36.070) = -32976 K⁻¹ × T.
    (ln(0.0048/36.070)/-32976) K = T.
    T = 0.0027064 K.

    Way too small! Something wrong.

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4) What is the coefficient
the Saha equation, in CGS units?

    ⎛2 π mₑ k T ⎞3/2
    ⎜---------- ⎟    = 2.41468×10¹⁵ g/(K erg s²) T^(3/2).
    ⎝      h²   ⎠

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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?

    n[ion]/n[atom] = 1/nₑ
                   × U[ion]/U[atom] 
                   × gₑ 
                   × (coefficient) 
                   × T^(3/2) 
                   × exp(-E[ion]/kT).

    nₑ = 10¹⁵ cm⁻³.
    U[ion] = 1.
    U[atom] = 10^0.3 ≈ 2.
    gₑ = 2.
    coefficient = 2.41468×10¹⁵ g/(K erg s²).

    Temperature     Ion Fraction
    6000 K          4.196×10⁻⁶
    10000 K         3.365×10⁻¹
    20000 K         2.552×10³

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6) Use the Boltzmann equation to find the n=2 population at these three temperatures and give the Lya emissivity.

    Just following the process from hw14.

    4 π j = n₂ Aᵤₗ hν.
    hν = 1.63403 × 10⁻¹¹ ergs.
    n₂ differs by temperature: ~1.1×10⁷, ~3.9×10⁷, ~6.3×10⁷.
    Aᵤₗ = 6.27×10⁸ s⁻¹.

    Temperature     Emissivity
    6000 K          1.136×10⁵ ergs
    10000 K         4.024×10⁵ ergs
    20000 K         6.525×10⁵ ergs