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			110 lines
		
	
	
		
			4.2 KiB
		
	
	
	
		
			Plaintext
		
	
	
	
	
	
			
		
		
	
	
			110 lines
		
	
	
		
			4.2 KiB
		
	
	
	
		
			Plaintext
		
	
	
	
	
	
| Homework 23, day 25, due Nov 28
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| 
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| 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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| 
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|     ─────────────
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|     1) Which one did you pick?
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| 
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|         0530 * 5.5 = 2915.
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| 
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|         NGC 2899
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| 
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|     
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|     ─────────────
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|     2) What type of nebula is it? H II region, planetary nebula, supernova remnant, other??
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| 
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|         NGC 2899 is a Planetary Nebula.
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| 
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|     ─────────────
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|     3) Find your nebula on the web and post its image.
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| 
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|         Posted to discussion board. That's what this means, right?
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| 
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|     ─────────────
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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.
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| 
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|         Looking at the ratio from:
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| 
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|                 [O III]
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|             ───────────  ⟵ 1 S₀
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|                 │
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|                 │
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|                 │ λ = 4363 Å
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|                 │
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|                 │
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|                 ↓
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|             ─────────── ⟵ 1 D₂
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|                   │
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|                   │
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|                   │ λ = 5007 Å, λ = 4959 Å  
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|                   │
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|                   │
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|                   ↓
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|             ─────────── ⟵ 3 P (3 combined levels)
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| 
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|         Where j is the emissivity at a wavelength.
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| 
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|             line ratio = j(5007 + 4959)/j(4363).
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| 
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| 
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|         For NGC 2899, 
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|             j(5007) = 94.77,
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|             j(4959) = 33.76, and
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|             j(4363) = 2.86.
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|     
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|         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. 
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|     
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|         Line ratio = (94.77 + 33.76)/2.86 = 44.94.
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| 
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|         Log₁₀(line ratio) = 1.65.
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| 
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|         Extrcting the value from the plot, gives a temperature 
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| 
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| ***         T ≈ 18500K.
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| 
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|     ─────────────
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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.
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| 
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|         I will use S II, since it's readily available in front of me in the table.
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| 
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|         Line ratio = j(6731)/j(6717) = 6.55/6.57 = 1.00.
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| 
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|         Extracting from the table. this gives an electron density
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| 
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| ***         nₑ ≈ 550 cm⁻³.
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| 
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|     ─────────────
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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?
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| 
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|         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.
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| 
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|     I have 
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|         T = 18500K, and 
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|         nₑ = n[H⁺] = 550 cm⁻³.
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| 
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|     I can find n[O III]/n[H II], assuming case B, where
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| 
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|         j(5006)/j(H-β) 
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|             = n[O III]/(nₚ nₑ) × (atomic physics factor = α)
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|             = α/(550 cm⁻³) n[O III]/n[H II].
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| 
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|     j(5006) = 94.77, and
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|     j(4861) = 10.00, so
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|     j(5006)/j(H-β) = 9.477.
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| 
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|     n[O III]/n[H II] = 9.477 (550 cm⁻³)/α = 5212.35 α⁻¹
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| 
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|     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.
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| 
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|     n[O III] = 5212.35 n[H II] α⁻¹ 
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|              = (5212.35) (550 cm⁻³) α⁻¹ 
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|              = 2.867 × 10⁶ α⁻¹.
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| 
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|     P = n R[O III] T 
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|       = (2.867 × 10⁶ α⁻¹) (259.8 Joules/(kg K)) (18500 K)
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|       = (2.867 × 10⁶) (259.8 Joules/(kg K)) (18500 K) α⁻¹
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|       = 1.378 × 10¹³ α⁻¹ Joules/kg.
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| 
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|     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.
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|       
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