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Unformatted text preview: es more important at lower temperatures, so roughly €
αA∝T−0.7. Note that we have appended the subscript A. In general, a hydrogen atom can recombine either directly to the ground state, H + + e− → H 0 (1s) + γ , or indirectly to an excited state (n≥2). In the latter case the hydrogen atom reaches the ground state by emitting spectral lines (and possibly 2s1s continuum). The €
1 The repeated factor of 4π is not a typo: the 4πJ comes from the definition of J being per ν
ν
steradian of photon propagation direction, whereas the 4π on the right
hand side is associated with the surface area of a sphere of radius r surrounding the star. [ recombination coefficient αA includes both of these reaction channels. However, the reaction where the H atom goes directly to the ground state emits a photon with hν>13.6 eV. This photon can therefore ionize an H atom. In most nebulae, therefore, recombinations directly to the ground state do not result in a net recombination. Thus it may be a better approximation to exclude them. If we do so, we arrive at a different recombination coefficient, α B (H 0 ,10 4 K) = 2.6 × 10−13 cm3 s−1 . The two situations – where photons above 13.6 eV are re
absorbed or not – are called Case A recombination and Case B recombination respectively, and their reaction rates are €
called Case A (or Case B) recombination coefficients. Photoionization: Photoionization is simply the inverse process of (radiative) recombination. Therefore, it can be computed using the principle of detailed balance, which states that in thermal equilibrium, the forward and reverse rate of every reaction cancel. If we imagine a gas in thermal equilibrium at temperature T (with kT<<13.6 eV), the rate of recombinations directly to the ground state (in cm−3 s−1) is α1s(T)nenp. By comparison, the rate of photoionizations out of the ground state is: ∞
4 πJν ( r)
n (H 0 ) ∫ν ( H 0 )
aν dν . i
hν
The radiation density for a blackbody is however given by Wien’s law, €
2 hν 3
Jν = 2 e− hν / kT , c
so the principle of detailed balance states: €
∞
2ν 2
α1s (T ) n e n p = n (H 0 ) ∫ν ( H 0 ) 2 e− hν / kT aν dν . i
c
This can be compared to the Saha equation, €
n e n p ȹ 2πm e kT ȹ 3 / 2 − hν i ( H 0 ) / kT
= ȹ , ȹ e
n (H 0 ) ȹ h 2 Ⱥ
to give: €
ȹ h 2 ȹ 3 / 2 ∞ 2ν 2 − h [ν −ν ( H 0 )] / kT
i
α1s (T ) = ȹ aν dν . ȹ ∫ν ( H 0 ) 2 e
i
c
ȹ 2πm e kT Ⱥ
€ This allows us to obtain aν from the recombin...
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 Winter '08
 Sargent,A

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