as-500/hw23/text

Homework 23, day 25, due Nov 28

Form a number by taking your birth month as a string and concatenating your birth day after it. For instance, my birth month is May, “05”, and birth day is 10, so I get 0510. Next multiply this by 5.5. In my case I get 2805. Finally, look through the Kaler catalogue to find a nebula with an NGC number as close as possible to your number. (Do not worry about precision – this is just so that we pick different nebulae). 

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    1) Which one did you pick?

        0530 * 5.5 = 2915.

        NGC 2899

    
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    2) What type of nebula is it? H II region, planetary nebula, supernova remnant, other??

        NGC 2899 is a Planetary Nebula.

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    3) Find your nebula on the web and post its image.

        Posted to discussion board. That's what this means, right?

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    4) Measure the gas kinetic temperature in your nebula by choosing the [O III] lines discussed in class, forming the ratio, and looking up the temperature on the plot in AGN3.

        Looking at the ratio from:

                [O III]
            ───────────  ⟵ 1 S₀
                │
                │
                │ λ = 4363 Å
                │
                │
                ↓
            ─────────── ⟵ 1 D₂
                  │
                  │
                  │ λ = 5007 Å, λ = 4959 Å  
                  │
                  │
                  ↓
            ─────────── ⟵ 3 P (3 combined levels)

        Where j is the emissivity at a wavelength.

            line ratio = j(5007 + 4959)/j(4363).


        For NGC 2899, 
            j(5007) = 94.77,
            j(4959) = 33.76, and
            j(4363) = 2.86.
    
        I didn't check what units the table is using. They will cancel. Oh, you know what, this appears to be a relative intensity to H-β, at any rate. 
    
        Line ratio = (94.77 + 33.76)/2.86 = 44.94.

        Log₁₀(line ratio) = 1.65.

        Extrcting the value from the plot, gives a temperature 

***         T ≈ 18500K.

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    5) Measure the electron density in your nebula by finding either the [O II] or [S II] lines and using the plot in AGN3.

        I will use S II, since it's readily available in front of me in the table.

        Line ratio = j(6731)/j(6717) = 6.55/6.57 = 1.00.

        Extracting from the table. this gives an electron density

***         nₑ ≈ 550 cm⁻³.

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    6) What is the gas pressure in dynes/cm 2 ? How does this compare with the pressure in the Earth’s atmosphere?

        Treating this as an ideal gas (of course these are charged particles, so not ideal), P = n R[O III] T, with R the specific gas constant.

    I have 
        T = 18500K, and 
        nₑ = n[H⁺] = 550 cm⁻³.

    I can find n[O III]/n[H II], assuming case B, where

        j(5006)/j(H-β) 
            = n[O III]/(nₚ nₑ) × (atomic physics factor = α)
            = α/(550 cm⁻³) n[O III]/n[H II].

    j(5006) = 94.77, and
    j(4861) = 10.00, so
    j(5006)/j(H-β) = 9.477.

    n[O III]/n[H II] = 9.477 (550 cm⁻³)/α = 5212.35 α⁻¹

    I'm not sure what atomic physics factor to use, here, and neither are my colleagues. We can check this out later, but Gary's door is closed at the moment. For now, I will leave it... should have asked on the discussions, but I am too late. If it's a similar factor from the earlier assignments, at least we know it had units cm⁻³, which is consistent with the following expression.

    n[O III] = 5212.35 n[H II] α⁻¹ 
             = (5212.35) (550 cm⁻³) α⁻¹ 
             = 2.867 × 10⁶ α⁻¹.

    P = n R[O III] T 
      = (2.867 × 10⁶ α⁻¹) (259.8 Joules/(kg K)) (18500 K)
      = (2.867 × 10⁶) (259.8 Joules/(kg K)) (18500 K) α⁻¹
      = 1.378 × 10¹³ α⁻¹ Joules/kg.

    I'm having trouble converting this... should be obvious. I'm out of time, though, so I'll hav eto come back to this. Thanks.