Week 9: WS15 #1

1. a) Hydrogen energy levels: Outside of molecular clouds, the most abundant species in the interstellar medium (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 the case of photoionization. 

Ideally a hydrogen atom will not resemble an avocado as much as this one does. 

The energy required to ionize a hydrogen atom is 13.6eV. If a photon of this energy is absorbed, the electron will jump from the ground state, n=1, to n=\(\infty\) and be freed from the proton.


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

The ionization energy of a ground-state electron in a hydrogen atom is 13.6eV, which is equivalent to 2.18x10^-11erg. We can use this to calculate the wavelength of the ionizing photon.
\(E=hc/\lambda\)
\(\lambda=hc/E\)
\(\lambda=\frac{(6.626\times10^{-27}erg.s)(3.0\times10{10}cm/s)}{2.18\times10^{-11}erg}=9.11\times10^{-6}cm\)
We can now calculate the temperature of the star that releases photons of this wavelength.
\(\lambda_{max}=b/T\)
\(T=b/\lambda_{max}\)
\(T=\frac{0.290cmK}{9.11\times10^{-6}cm}\approx32,000K\)
This corresponds to O-type stars, the hottest stars.

c) Excitation state of hydrogen: But why do we only care about excitation from the ground state to free protons and electron? After all, if hydrogen is in an excited state (eg. n=2) you could use many more of the available stellar photons to ionize the ISM. The lifetime of an excited state is ~ 10⁹s. 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 cm². 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 the atomic hydrogen. 
How do the two time scales compare? Is it reasonable to assume that all hydrogen is in the ground state? 

We can calculate the photon flux by dividing the total flux by the energy per photon:
\(F_\gamma=\frac{\sigma_{SB}T^4}{2.18\times10^{-11}erg}=\frac{(5.67\times10^{-5}erg/cm^2s)(32000K)^4}{2.18\times10^{-11}erg}=2.73\times10^7/cm^2s\)
The dimensions of photon flux are photons per time per area, so to get the rate, we multiply by the cross section: \(F_\gamma \sigma=(2.73\times10^7/cm^2s)(10^{-17}cm^2)=2.73\times10^7/s\).
By inverting this, we can calculate the ionization timescale: \(3.67\times10^{-8}s\). This is longer than the timescale of the excited electron by about one order of magnitude. This means that it is reasonable to assume that all hydrogen is in the ground state--it will return from the excited state to the ground state much faster than it will be re-ionized.

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 collide. Note that a recombination can happen to any hydrogen energy level. If the recombination takes the hydrogen immediately to the ground state you will produce a new ionizing photon. If the recombination takes the hydrogen into any other level, the emitted photon will not be able to ionize another hydrogen atom. 

The photon will be emitted as the proton and electron combine. This rate is given by \(r=\alpha n_pn_e\) where \(n_p\) is the number density of protons and \(n_e\) is the number density of electrons.