2
$\begingroup$

This is the question I'm trying to answer, and the given solution is below it.

Question

Solution

The bit I don't understand is why they used the equation in the highlighted section.

The only equation I've found that looks similar is:

Equation linking Doppler wavelength shift to rest wavelength and radial velocity

So my best guess is that this is the equation used (but with frequency instead) and since the broadening width is given in the question and not the shift, the '0.5' in the equation was used to find the shift (from the rest frequency to either end of the broadening). Does anyone know whether this is along the right lines or am I way off?

Sorry if the explanation isn't the best!

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

If the width of a spectral line is due to symmetric velocity shifts towards and away from an observer, then half the width of the line corresponds to the magnitude of that velocity.

All that remains then is to recognise that for non-relativisic speeds, the ratio of the frequency shift to the "rest frequency" is approximately equal to the ratio of the velocity shift to the speed of light.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thank you so much! $\endgroup$
    – user374355
    Commented Aug 7, 2023 at 22:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.