Blackbody Radiation: Temperature of Accreted Gas

AI Thread Summary
Gas accreting onto burned-out stars emits X-rays around 10 keV, suggesting a high temperature for the gas. To find this temperature, Wien's displacement law can be used, which relates the peak wavelength of emitted radiation to temperature. The conversion from energy to wavelength can be done using the formula E = hc/λ, ensuring unit consistency. While Wien's law provides a reasonable approximation, the actual temperature may vary due to the non-uniform nature of accretion disks. Ultimately, the discussion emphasizes the importance of understanding the underlying physics and formulas for accurate calculations.
tony873004
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This is actually a homework question, but its probably better suited for this forum.

Gas accreting onto burned-out stars like white dwarfs, neutron stars, and black holes can often be seen in the X-ray part of the EM spectrum, at energies of about 10 keV. What temperature does this suggest that the accreted gas is heated to?

I can't make sense of my class notes, and the textbook does not cover this well. I can't find a formula that relates wavelength to energy to give temperature. Formulas using Watts is the closest I can come, but Watts is Joules / s, and there is no time in this problem. What did I scribble down in my notes that I can't read??
 
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I think what you're looking for is Wien's displacement law:

\lambda_{max}T=2.898x10^{-3} m \cdot K

You can convert the energy of the X-Rays to wavelength using

E=\frac{hc}{\lambda}

Just be careful to watch your units. There are several conversions to be done, and the final answer should come out in Kelvin.
 
Grogs said:
I think what you're looking for is Wien's displacement law:

\lambda_{max}T=2.898x10^{-3} m \cdot K

You can convert the energy of the X-Rays to wavelength using

E=\frac{hc}{\lambda}

Just be careful to watch your units. There are several conversions to be done, and the final answer should come out in Kelvin.

Thank you very much! It took me a while, but I came up with something similar (the same, just written a little different)

t=\frac{2900 \mu m K}{\lambda_{peak}}
and I got 2*10^{7}K if I converted my units properly.
 
In the interest of understanding what you're doing with all those formulae, notice that the spectrum of a blackbody has a peak in the center, and that peak moves with temperature. Wien's law simply relates the temperature and the wavelength at the peak, allowing you to find one if you know the other.

- Warren
 
tony873004 said:
and I got 2*10^{7}K if I converted my units properly.

That's what I got too. You might find http://qonos.princeton.edu/nbond/unicalc/calculator.php useful. I have to do that kind of calculation a lot and Google calculator was really tedious, so I wrote my own to save time. It's mostly for astrophysics calculations at the moment.
 
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I think you should be using this formula:<u>~2.7kT.Incidentally,it gives the same order of magnitude as Wien's law,but there's no reason why anyone would choose Wien's law in this case...

Daniel.
 
dextercioby said:
I think you should be using this formula:<u>~2.7kT.Incidentally,it gives the same order of magnitude as Wien's law,but there's no reason why anyone would choose Wien's law in this case...

Well, if it were a pure blackbody, then Wein's law would be fine. In reality, accretion disks are not at a uniform temperature, even in the simplest model. I get the impression, from the wording, that the question is just looking for an order of magnitude. The true astronomer thing to do would be to divide by Boltzmann's constant and be done with it. :biggrin:
 
Since the problem doesn't specify the model one would have to use,i think the BB rad.is the only viable option.As for order of magnitude,indeed,incorrect formulas may provide good answers...

...Since the errors are very large.

Daniel.
 
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