Electron at its finest

We are bounded in a nutshell of Infinite Space: Week 9: Worksheet #15: Problem #1: Electron at its finest 

1. Hydrogen Ionization: The interstellar medium is ionized by UV photons from stars. The goal of this worksheet is to explore the ionization process and how it regulates much of the observed structure of the ISM close to massive stars, and the diffuse ISM everywhere.

(a) Hydrogen energy levels: Outside of molecular clouds, the most abundant species in the ISM is atomic hydrogen (in molecular clouds it is molecular hydrogen). Whether the ISM is fully ionized or not will therefore depend on how easily atomic hydrogen is ionized. The ground electronic state of a hydrogen atom corresponds to an atom with the smallest (and hence most tightly bound) electron orbit around the nuclear proton that is consistent with a stationary electronic wave function, a standing wave. The electronic energy levels permitted by quantum mechanics are characterized by their quantum numbers n=1 2 3, where n=1 corresponds to the ground state. Make a drawing of the electronic energy levels of atomic hydrogen. Mark out the energy needed to excite an atom in its ground state to a free proton and electron. Illustrate what happens in case of photoionization.

(b) Ionizing stars: Remember that stars are blackbodies. Which kind of stars emit a majority of their photons with energies high enough to photo ionize (excite an electron into freedom) ground state hydrogen. Give your answer in both stellar surface temperature, and letter classification.

(c) Excitation state of hydrogen: But why do we only care about excitation from the ground state to free protons and electrons? After all if hydrogen is in an excited state (e.g. n=2) you could use many more of the available stellar photons to ionize the ISM. The lifetime of an excited state is \(\sim 10^{-9} s\). Let’s calculate the time scale of ionization right next to the star to test whether it is reasonable to assume that all hydrogen are in their ground state. First, set up an equation for the ionization rate for a single hydrogen atom in terms of the photon flux and the ionization cross section \(\sigma\). The ionization cross section is \(10^{-17} cm^2\) . Calculate the photon flux assuming that you are sitting right next to the star from (b) and that the star is emitting all its energy in the form of photons with the exact energy required to ionize atomic hydrogen.

How do the two time scales compare? Is it reasonable to assume that all hydrogen is in the ground state?

(d) Recombination: The inverse of photoionization is recombination. In a recombination event an electron and proton collide and become bound while emitting a photon. Illustrate a recombination event. Set up an equation for the recombination rate in terms of the number densities of protons, electrons and the rate coefficient \(\alpha\), which describes the efficiency at which a recombination occurs when an electron and proton collides. Note that a recombination can happen to any hydrogen energy level (n=1,2,3 etc). If the recombination takes the hydrogen immediately to the ground state you will produce a new ionizing photon. If the recombination take the hydrogen into any other level the emitted photon will not be able to ionize another hydrogen atom.

a.

b. Starting from the basic equations we have learned for blackbodies, we use Wein’s Law: \[ \lambda_{peak} = \frac{0.3 cm\cdot K}{ T}\] and the equation describing the energy of a photon: \[ E = h\nu\] and the relationship of wavelength and frequency: \[ c = \lambda \nu\] \[\nu = \frac{c}{\lambda},\] and now we can retroactively plug in the equations to describe how a specific photon energy is emitted by a star at a specific temperature: \[E = \frac{hc}{\lambda}\] \[ \lambda = \frac{hc}{E},\] \[ \frac{hc}{E} = \frac{0.3 cm\cdot K}{ T}\] \[T = \frac{0.3cm\cdot K \cdot E }{hc},\] Now that we have placed the equation in terms of the temperature, we can place the constants, including Planck’s Constant in eV, and values for energy of ionization:  \[ T = \frac{0.3cm\cdot K \cdot (13.6 eV) }{\left(3\times 10^{10} \frac{cm}{s} \right)(4.14 \times 10^{-15} eV \cdot s)},\]  This solves to the temperature at which stars will produce energy sufficient to cause the ionization of hydrogen. \[T = 3.29 \times 10^4 K \to ~ an~O/B~ star\] 

c. Next, we will use the temperature we have just found to describe the Ionization rate the star creates. Knowing the base equation for the rate: \[ I = j \sigma,\] we solve for the photon flux by first determining the flux at the surface of the star:  \[F = \sigma T^4 ,\]  and plugging the value for the temperature we had just found and the Stephan Boltzmann Constant: \[ T = \left(5.7 \times 10^{-5} \frac{1}{K^4} \right)( 3.3 \times 10^4 K )^4 ,\] which yields: \[ F = 6.75 \times 10^{13} \frac{ergs}{cm^2 s}.\] However, we now have to convert this into photon flux, meaning we apply dimensional analysis and find that: \[ F = 6.75 \times 10^{13} \frac{ergs}{cm^2 s} \times \frac{1 eV}{ 1.6 \times 10^{-12} ergs} \times \frac{1 ~photon}{13.6 eV},\] which in turn tells us the photon flux  \[ j = 3.1 \times 10^{24} \frac{photons}{cm^2 s},\] now going back to the original equation, we see that: \[I = j\sigma,\] and just plugging in the constant given in the problem and the value we just derived: \[I =  \left(3.1 \times 10^{24} \frac{photons}{cm^2 s}\right) (10^{-17} cm^2),\] the ionization rate is thus: \[ I = 3.1 \times 10^7 \frac{photons}{s}.\] Next we have to understand the timescale for atoms to get ionized: \[\frac{1}{I} = Timescale,\] such that: \[Timescale = \frac{1}{3.1 \times 10^7 \frac{photons}{s}} = 3.2 \times 10^{-8} \frac{s}{photon}\] Therefore, the timescale is 30 times that of the excited state duration, meaning the decay is much quicker than the time necessary to ionize the atom.

d. Recombination is when a proton and electron combine to form a stable atom once again after being ionized, which occurs at a rate of \[ Volumetric ~Recombination ~Rate = \frac{\# ~Recombinations}{t \cdot Volume},\] which is directly proportional to the density of protons and electrons found in the volume:  \[r_v \propto n_e n_p \] \[r_v = \alpha n_e n_p\]

As seen in the illustration above, the collision produces excess energy which is represented by \[  E_\gamma = E_{tot} - 13.6 eV, \] in the case of collision which produced another ionizing photon.