Photo-ionisation

Interaction between a photon, from a nearby star say, and a molecule or atom, may lead to dissociation of the molecular bond, and/or ionisation of the atom. And so the process of molecule formation, which makes a cloud cool, get denser, and eventually form stars, can be completely reversed when the newly born star starts to ionise and heat the gas.

Recall from the first part of the course that massive stars, like O and B-type stars, are hot and hence emit lots of hydrogen ionising photons. These photons have a dramatic effect on the surrounding gas, and convert the hydrogen into it's HII form. And so massive bright stars are often surrounded by HII regions. These regions delineate spiral arms, because star formation occurs predominantly in spiral arms and massive stars don not live for very long.

HII regions are important, not only because they are beautiful to look at^4.2, but because their physics is reasonably well understood. And by observing the relative strengths of the emission lines that occur in their spectra, one can deduce the properties of the HII region, such as it's temperature, density, and the relative abundance of the elements.

There are two things I would like you to remember about this, one is to understand the physics that determines the properties of the spectrum that you get from such a nebula, and the second one to understand what sets their size (the so-called Strömgren radius

When an ion (for example HII, OIII, ...) recombines with an electron (and form HI, respectively OII), the electron does not necessarily have to fall directly to the lowest possible energy level, i.e. the ground state. Taking the example of HI, the ground state would correspond to the electronic $n=1$ state. Typically, the electron will cascade down to $n=1$, and the relative probability for the intermediate steps can be computed from quantum mechanics. However, this does not translate directly into the spectrum you observe. For example, suppose the electron makes a transition $n=2\rightarrow n=1$, and emits the corresponding photon. As this photon starts to move toward us, it may actually interact with another neutral hydrogen atom, and cause the electron of that atom to be excited from the $n=1$ to $n=2$ level. In which case the photon does not actually exit from the nebula! Suppose on the other hand, the photon makes an $n=3\rightarrow n=2$ transition. This photon has much more chance of leaving the nebula, since most of the neutral HI will be in the $n=1$ state, and not so much in the $n=2$ state. And so that photon will escape from the nebula. So curiously, it may be that lines with a low quantum mechanical probability, dominate a nebular spectrum, since photons from more likely transitions are unable to escape from the nebula. This is especially true for Planetary Nebulae spectra.^4.3

For an HII region, the dominant wavelength photon that escapes results from the $n=3\rightarrow n=2$ transition, denoted as H$\alpha$^4.4. This red line is the reason HII regions appear to fluoresce red. And by observing galaxies through a filter that only lets H$\alpha$ light through, one can easily find HII regions.

Suppose a source of ionising photons such as a hot star starts emitting ionising photons at a rate ${\dot N}$, in photons per second, and assume the source is surrounded by a homogeneous cloud of atomic hydrogen, with density $n$. The source will quickly ionise all hydrogen close to it. Let $R(t)$ be the radius of the ionisation front, within which most of the hydrogen is ionised, and outside of which the gas is mostly neutral. As $R$ increases between $R$ and $R+\Delta R$, the number $N$ of atoms that need to be ionised is

$$(4\pi/3)n[(R+\Delta R)^3-R^3]\approx 4\pi nR^2 \Delta R .$$ (4.1)

Since it takes the source a time $\Delta t=N/\dot N$ to produce this many photons, we find that the speed of the front is

$$v={\Delta R\over\Delta t}={\dot N\over 4\pi nR^2} .$$ (4.2)

Of course, this speed cannot be faster than the speed of light, and you see that, as the HII regions grows in size, the speed with which it grows decreases $\propto 1/R^2$. However, eventually some of the HII ions inside the ionisation front will start to recombine. Since extra photons are needed to re-ionise these, the speed of the front will slowly down even more. Eventually, an equilibrium is reached, in which the number of photo-ionisations within $R$ equals the number of recombinations. The radius of the HII region has grown to its maximal size: the Strömgren radius, $R_S$. To compute $R_S$, consider a small volume of the HII region. Since a recombination is an interaction between an electron and an HII ion, the rate at which HII ions recombine is proportional to the electron and the ion density:

$$\hbox{\rm recombination rate=} \alpha n_e N_{\rm HII} .$$ (4.3)

Since the recombination rate is in ions s$^{-1}$ volume$^{-1}$, we find that the recombination rate $\alpha$ must have dimensions of volume s$^{-1}$. If the gas is composed purely of hydrogen, and is very highly ionised, then $n_e\approx N_{\rm HII}\approx n_{\rm H}$, where $n_{\rm H}$ is the density of hydrogen, either HII or HI. The total number of recombinations within radius $R$ is then $(4/\pi/3) \alpha n_{\rm H}^2 R^3$. In equilibrium, this is the rate $\dot N$ at which the source produces ionising photons. Equating these, we find that

$$R_S = ({3\dot N\over 4\pi\alpha})^{1/3} n_{\rm H}^{-2/3} .$$ (4.4)

For example, assume $n_{\rm H}=5\times 10^3{\rm cm}^{-3}$ for the density of the cloud, and $\dot N=10^{49}{\rm s}^{-1}$ for the ionisation rate of the star. Then $R_S\approx 0.21\pc$, since $\alpha\approx 3.1\times 10^{-13}$ cm$^{3}$ s$^{-1}$ at a temperature of $T=8000K$ typical of HII regions.