Estimating Width of 21cm Line from Atomic Hydrogen Cloud

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SUMMARY

The discussion centers on estimating the width of a 21cm spectral line from an interstellar cloud of atomic hydrogen at a temperature of 300K. The participant initially calculated a width of 2.59x10^-4 cm using the full width at half maximum (FWHM) formula, but the expected answer is 2.0x10^-4 cm. The discrepancy arises from the omission of the factor of ln(2) in the formula used, which is crucial for accurate calculations. Additionally, alternative methods such as using exponential width or different velocity distributions may yield varying results.

PREREQUISITES
  • Understanding of Doppler broadening and its implications in astrophysics
  • Familiarity with the Boltzmann constant and its role in thermal physics
  • Knowledge of the full width at half maximum (FWHM) concept
  • Basic principles of spectral line formation in atomic hydrogen
NEXT STEPS
  • Study the derivation of the FWHM formula for spectral lines in astrophysics
  • Learn about the impact of temperature on spectral line broadening
  • Research the differences between Gaussian and non-Gaussian velocity distributions
  • Explore the concept of exponential width in spectral analysis
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Astrophysics students, researchers in spectral analysis, and anyone interested in understanding the properties of atomic hydrogen clouds and their spectral lines.

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Homework Statement



Estimate the width of a 21cm spectral line produced by an interstellar cloud of atomic hydrogen with a temperature of 300K.


Homework Equations



From wiki (link below), I found an equation for full width at half maximum, relating it to Boltzmann constant, temperature T, mass m, and speed of light c.

http://en.wikipedia.org/wiki/Doppler_broadening

I used the mass of one hydrogen atom for m.


The Attempt at a Solution



I got 2.59x10^-4 cm but the answer should be 2.0x10^-4 cm.

Any idea where I went wrong? Is the "width" mentioned in the question equal to the full width at half maximum?

Thanks.
 
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Many astrophysics texts use a less complex approximation for the FWHM line width. Commonly, the derived formula will simply be

\Delta\lambda_{FWHM}\approx\frac{2\lambda}{c}\sqrt{\frac{2kT}{m}}

If a more in-depth analysis is performed, taking into account the atoms motions in various directions relative to each other and the observers line of sight, you get the more precise formula containing the factor of \ln(2) which is missing from the above equation, but present in the one you used.

Your answer is technically more correct, just not the one you were expected to give.
 
But omitting ln2 gives a larger value of 3.11x10^-6m. The given answer should be 2x10^-6m.
 
Ah, yes, how stupid of me.

I've had a look through a few textbooks, and I'm not sure what they did to arrive at that solution. The only other thing I could think of would be that rather than using the FWHM line width, they were using the exponential width, i.e. the peak width where intensity drops to 1/e what it is at the peak, which would give a slightly smaller value. Unfortunately I don't know the formula for finding that offhand. I suppose they could also be assuming some other non-Gaussian velocity distribution (like Maxwellian or something), but that would be unusual for such a question.
 

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