A) The most abundant species in the ISM outside of molecular clouds is atomic hydrogen. We need to determine how hydrogen is ionized in order to determine properties of the ISM. 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.
http://dev.physicslab.org/Document.aspx?doctype=3&filename=AtomicNuclear_BohrModelDerivation.xml
When the hydrogen atom receives enough energy from incoming light to jump up 13.6 eV in energy, it escapes it's atomic structure and ionizes into a proton and an electron.
B) Remember that stars are blackbodies. Which kind of stars emit a majority of their protons with energies high enough to photoionize (excited the electron to freedom) ground state hydrogen. Give your answer in both stellar temperature, and letter classification.
To solve for the temperature needed to create a 13.6 eV increase in energy we can simply use Planck's law and Wein's displacement law. First we know \[E = \frac{h c}{\lambda}\] which gives us \[13.6 eV = \frac{4.135*10^{-15} * 3*10^8}{\lambda} \rightarrow \lambda = 91.2 nm = 912 angstrom \]With this wavelength we can use the Wein displacement law to solve for temperature. \[\lambda T = 3 * 10^{-3} m \cdot K\] From here we can solve for temperature of our blackbody. \[T = \frac{3*10^{-3} m \cdot k}{91.2*10^{-9} m} = 32894 K \]This very high surface temperature gives the star an O Type classification on the Morgan-Keenan scale.
C) The ionization cross section is \(10^{-17} cm^2 \). Calculate the photon flux assuming the you are sitting right next to the star from (B) and that the star is emitting all of its energy in the form of photons with the exact energy required to ionize atomic hydrogen. How does this time scale compare to the excited time scale of a hydrogen atom, \(10^{-9} s \)? Is it reasonable to assume that all hydrogen is at the ground state?
We begin with the Stefan-Boltzmann law to relate the stellar flux to the temperature of the star. \[\sigma T^4 = F \rightarrow (5.67*10^{-5} \frac{erg}{s \cdot cm^2 \cdot K^4} * (32894 K)^4 = 6.64*10^{13} \frac{erg}{s \cdot cm^2} \] Using simple unit conversion we can find our timescale from this result by multiplying by an area (to cancel the cm^2) and dividing by an energy (to cancel the ergs).\[6.64*10^{13} \frac{erg}{s \cdot cm^2} * 10^{-17} cm^2 * \frac{1}{13.6eV = 2.18*10^{-11} erg} = 3.05*10^7 \frac{photons}{second} \]Converting this to a timescale through inversion we get \[3.05*10^7 \frac{photons}{second} = 3.28*10^{-8} \frac{seconds}{photon} \] This time scale is larger than the \(10^{-9}\) excited state lifetime meaning that the photon will almost always fall back to its ground state before the next photon has the chance to ionize the atom to its next energy level.
D) Draw a recombination event. Set up an equation for the recombination rate, r (which has units of cm^3 s^-1), in terms of the number densities of photons, electrons (np and ne) and the rate coefficient \(\alpha\), which describes the efficiency at which a recombination occurs when an electron and proton collide.
http://hendrix2.uoregon.edu/~imamura/123/lecture-6/recombination.jpg
To solve for the recombination rate, we simply need to match units. We know \[n_p = \frac{protons}{cm^3} \quad n_e = \frac{electrons}{cm^3} \quad \alpha = \frac{cm^3}{s}\] Thus if we want a rate in terms of \(\frac{1}{cm^3 \cdot s} \) we simply need to multiply the three terms. \[r = n_p * n_e * \alpha \]
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