Stromgren Sphere
A) Set up an equation of the size of a sphere that is ionized around a star assuming uniform photon flux in all directions, and uniform density in terms of the total ionization rate \( \eta\), the density \(n\) and the recombination cross section \(\alpha\). Assume that all recombinations result in photons that cannot ionize anything further. At steady state the total photoionization rate \(\eta\) is balance by the total recombination rate.
Given the steady state assumption, we can begin by solving for \(\eta\) the recombination rate, by simple unit conversion. We know from the previous post that the recombination rate, r, is \[r = n_e * n_p * \alpha \space [\frac{1}{s \cdot cm^3}] \]We know that the number of electrons and protons must be equal if the sphere is uniformly atomic hydrogen. \[n_e = n_p \rightarrow r = n^2 \alpha\]We want \(\eta\) in terms of \(\frac{1}{s}\) meaning we need to multiply by the volume of the total sphere, given that the ionization (and recombination) occurs equally in all directions. \[\eta = n^2 \alpha V = \frac{4}{3} \pi r_{ss}^3 n^2 \alpha \]Rearranging to solve for \(r_{ss}\), the radius of the Stromgren Sphere, we find \[r_{ss} = (\frac{3 \eta}{4 n^2 \alpha \pi})^{\frac{1}{3}}\]
B) Calculate the total number of ionizing photons emitted per second by the kind of star identified in the previous problem assuming its entire luminosity is due to photons with exactly the energy required for ionizing hydrogen atoms.
This problem asks us to solve \(\eta\) which is a rate in units \(\frac{photons}{second} \). To do this we relate the luminosity of the star from the previous problem to \(\eta\). We know that the luminosity is the amount of energy emitted by the surface of a star every second. \[L = 4 \pi r^2 \sigma T^4 \space [\frac{erg}{s}] \] From the given table we find that for an O type star (temperature >30,000K). \[r_{star} \approx 10*R_{\odot} = 6.96*10^{11} cm\]With this information we solve the luminosity \[L = 4 \pi r^2 \sigma T^4 \space [\frac{erg}{s}] = 3.92*10^{38} \frac{erg}{s}\]To convert this into the rate we need, we simply divide by the energy of a single ionizing photon, \(2.17*10^{-11} erg\). \[\eta = \frac{L}{2.17*10^{-11} erg} = 1.8*10^{49} \frac{photons}{second} \]
C) With information from (A) and (B), what is the size of a Stromgren Sphere around this kind of star assuming an initial hydrogen atom density of \(1 cm^{-3}\). The recombination rate \(\alpha = 3*10^{-9} cm^{-3} s^{-1}\). How does your answer compare to the HII region of Orion, which is ionized by a few massive stars and is 8pc across?
Plugging in the \(\eta\) from (B) into the equation for \(r_{ss}\) from (A) we can solve for the radius of the Stromgren Sphere. \[r_{ss} = (\frac{3 \eta}{4 n^2 \alpha \pi})^{\frac{1}{3}} = \frac{3 * 1.8 * 10^{49}}{4* 1^2 * 3*10^{-9} * \pi} = 1.57 * 10^{57} cm \approx 5pc \]Our answer is expectedly smaller than the Orion system which has a few massive stars ionizing the HII region.
D) Make a drawing of the Stromgren sphere surrounded by the neutral HI region. What is the size of the transition region at the edge of the Stromgren Sphere where atomic hydrogen and protons co-exist? How does the size of the transition region compare with the radius of the sphere? Does it make sense to think about the HI and HII regions as distinct?
From the drawing and the information given in the problem it becomes evident that to find the size of the transition area we must calculate the mean-free-path to find the width of the area in which hydrogen protons will be ionized by photons. If this path length is very large, the two areas could be considered distinct, whereas if it is small, much of the HI region will continue to be ionized by the photons that cross the transition region. We calculate mean-free-path \[l = \frac{1}{n * \alpha} \]Where n is the particle density and \(\alpha \) is the ionization cross-section as defined in the earlier section. \[l = \frac{1}{1*3*10^{-9}} \approx 3*10^8 cm\]This transition zone is significantly smaller than the the Stromgren radius of 5pc, thus we consider the two regions as non-distinct and expect interaction along the edge of the radius.
Nice work! Watch out for typos - 10^57cm is certainly much more than 5pc. :-)
ReplyDelete-Jamila (TF)