0
$\begingroup$

How is it possible to observe Fraunhofer lines in the emission spectrum of any star, since the elements absorbing the radiation couldn't possibly absorb all the radiation corresponding to a particular wavelength. For example, to be able to see the Na 'D' line in Sun's spectrum, a small amount of Na present in the Sun should absorb all the radiation corresponding to the D line wavelength. Even for highly abundant elements such as H and He, why is it not possible for at least some amount of radiation in their absorption spectra to escape and reach the observer? Furthermore, even if we grant that all of the original radiation is absorbed by excitation of electrons, is it not possible for them to eventually de-excite and emit the same radiation (to be precise, do no de-excitations occur at high temperatures)? If any of these things happen, then we shouldn't be able to see dark lines in the spectrum of stars. Most sources ascribe the presence of the lines to the presence of the elements but don't go into the details of how this really occurs. Please let me know. Thank you!

$\endgroup$
3
  • 1
    $\begingroup$ Isn't it just that while the absorption is frequency-specific, the emission of the absorbed energy occurs via normal thermal blackbody radiation, thus at all sorts of other frequencies? $\endgroup$
    – PcMan
    Mar 2 at 6:57
  • $\begingroup$ Yes technically that's how black body radiation should occur. My doubt is why do we see the Fraunhofer lines then? $\endgroup$ Mar 2 at 7:46
  • $\begingroup$ Blackbody radiation is not a mechanism. $\endgroup$
    – ProfRob
    Mar 2 at 7:59
2
$\begingroup$

Absorption lines are not black. You are correct that some radiation still emerges in our direction, even in the central wavelengths of the sodium D lines.

The presence of absorption lines is because of the temperature gradient in the Sun and that the blackbody radiation field is $\propto T^4$.

As we move outwards in the solar photosphere, at some point, the integrated material above that point is insufficient to stop most photons escaping. This is a definition of where the photosphere is.

However, the depth of this point is wavelength dependent. If we are looking at a "continuum" wavelength, then the opacity is relatively low (dominated by H- recombination) and the radiation escaping comes from a relatively deep layer. If we are looking at the centre of a Na D-line then the opacity is high and the emission comes from a region higher up in the photosphere. Because of the temperature gradient, the former layer is hotter than the latter. And because of the $T^4$ dependence, the former is bright while the latter is comparatively dark.

$\endgroup$
1
  • $\begingroup$ Thank you for the answer! $\endgroup$ Mar 3 at 9:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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