mirror of
https://asciireactor.com/otho/as-500.git
synced 2024-11-22 19:15:06 +00:00
77 lines
2.6 KiB
Plaintext
77 lines
2.6 KiB
Plaintext
|
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).
|
|||
|
|
|
|||
|
|
─────────────
|
|||
|
|
2) Plot the line ratio as a function of temperature between 5000 K and 50 000 K. Make the y-axis a log.
|
|||
|
|
|
|||
|
|
Attached.
|
|||
|
|
|
|||
|
|
─────────────
|
|||
|
|
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.
|
|||
|
|
|
|||
|
|
─────────────
|
|||
|
|
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 ⎠
|
|||
|
|
|
|||
|
|
─────────────
|
|||
|
|
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
|
|||
|
|
|
|||
|
|
─────────────
|
|||
|
|
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
|