ajb Posted May 13, 2007 Posted May 13, 2007 I have been "messing" Wien's law and the ionisation energies of Hydrogen and Helium to see of I can get a very naive guess at how hot stars are. Interestingly I get a temperature that is much higher than typical stars on the main sequence, but they are just within the range of the hottest stars observed. My approach was to assume that the temeperature of the star's surface is about the first ionisation energy of Hydrogen (13eV) and then Helium (25eV). This would be the "worse case scenario", atomic collisions would lower this temperature dramatically. I assume this as the photoshere of stars consists of plasma. This gives a surface temperature of 150,865 K and 290,125 K respectively. Stars at this temperature would radiate strongly in the high ultra-violet low x-rays. 150,865 K is just above the classification of O-type stars (30,000-100,000K). 290,125 K is just above record for a white dwarf, something like 250,000K. So it looks like my "stupid guess" gives an approximate upperbound on temperatures based on the composition of the star's surface. Anyone here know more about astrophysics like to comment?
John Cuthber Posted May 14, 2007 Posted May 14, 2007 "My approach was to assume that the temeperature of the star's surface is about the first ionisation energy of Hydrogen (13eV)" Why?
ajb Posted May 14, 2007 Author Posted May 14, 2007 "My approach was to assume that the temeperature of the star's surface is about the first ionisation energy of Hydrogen (13eV)" Why? I assume that the outer layer of the star is made of ionised hydrogen, hence i pick 13 eV. This is of course a gross simplification as we have a high photon density which would lower this energy. What supprised me was how this very stupid guess gives a very good approximate upper limit on the stars temperature.
insane_alien Posted May 14, 2007 Posted May 14, 2007 he is assuming that the gas on the surface is a plasma. that is the energy required for hydrogen to be a plasma. a better guess can be achieved by looking at the peak spectral emission of the sun(in the green part) and working from there.
ajb Posted May 14, 2007 Author Posted May 14, 2007 he is assuming that the gas on the surface is a plasma. that is the energy required for hydrogen to be a plasma. a better guess can be achieved by looking at the peak spectral emission of the sun(in the green part) and working from there. Exactly, I could get a much better guess at the sun's surface temperature by using Wein's law and the spectral peak as you say. Or maybe by using the Maxwell-Boltzmann distrubution and linking that to the sun's luminosity. But that is astrophysics. All I did was take a very wild stab in the dark.
swansont Posted May 14, 2007 Posted May 14, 2007 Wouldn't the ionization energy represent the equivalent of the latent heat for the phase change to plasma? As such, you'd expect a plateau at that temperature. But that shouldn't necessarily represent the surface temperature; to make a weak analogy, the temperature of water vapor just above a water surface below the boiling point (assuming "temperature" makes sense) should be higher than the surface before the vapor equilibrates with the surrounding material, because high-energy molecules preferentially evaporate (which is why evaporation cools the bulk material). So just because there is a plasma I don't think this means that the non-plasma will be at the same temperature. One needs to know why the outer layer is a plasma. And indeed the corona of our sun is much, much hotter than the surface temp.
John Cuthber Posted May 14, 2007 Posted May 14, 2007 It's not a phase change. It's possible to find a temperature where half the hydrogen is ionised. There isn't a temperature wher half of some water is boiled. I don't think it's really a barrier in the same way. If there's enough enery to ionise 1 hydrogen in ten then 1 in ten will be ionised.
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now