Estimating Width of 21cm Line from Atomic Hydrogen Cloud

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Homework Help Overview

The problem involves estimating the width of a 21cm spectral line produced by an interstellar cloud of atomic hydrogen at a temperature of 300K, focusing on the interpretation of the width in terms of spectral line broadening phenomena.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the use of different formulas for calculating the full width at half maximum (FWHM) and question whether the width refers to FWHM or another measure. They explore the implications of using various approximations and constants in their calculations.

Discussion Status

The discussion is ongoing with participants providing insights into different approaches to the problem. Some have suggested that the original poster's calculation may be correct but not aligned with the expected answer, indicating a potential misunderstanding of the width definition. Others are exploring the possibility of alternative methods or assumptions that could lead to the expected result.

Contextual Notes

There is mention of differing interpretations of the width measurement, including the potential use of exponential width versus FWHM, and considerations of the velocity distribution of atoms in the cloud.

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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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