3
$\begingroup$

I have learned so many concepts under astrophysics and unfortunately, I have muddled everything together... Let me try to illustrate my problem:

  1. When a star is in main sequence, it fuses hydrogen to produce helium and energy that is mostly given off as light (This is what I learned from watching a few youtube videos)

  2. Also, a star is a black body and has a spectrum that looks like this: enter image description here

  3. Furthermore, if I split the light that comes from the sun, I get an emission spectrum of a few colors which correspond to the colors absorbed by the elements on the sun's surface

To wrap up what I am confused about...

Does the process go like this...

  1. During the fusion of hydrogen, energy is given off as EM waves

  2. The amount of the different sorts of EM waves is shown by the curve above

  3. As the EM waves move up the surface, it is absorbed by the elements on the surface and then re-radiated... this is the emission spectrum

I am having trouble linking up the different concepts together and articles on the net are complicating things further... could someone please tell me which part of my concepts is wrong?

$\endgroup$
4
$\begingroup$

The light that we see coming from the Sun is mainly due to black body radiation at its surface. The spectrum of black body radiation is statistical in origin, and as long as there are enough processes contributing to it the black body spectrum is independant of the microscopic details and depends only on the temperature. There is a discussion of this in the answers to the question What are the various physical mechanisms for energy transfer to the photon during blackbody emission?.

It's true that the fusion reactions in the Sun's core give off photons, but this is only part of the energy output. Energy is also produced as kinetic energy of neutrinos and the helium nuclei. However the energy is given off, it rapidly thermalises with the plasma and ends up as kinetic energy of the components of the plasma (mostly protons and helium nuclei). Energy reaches the surface by a mixture of convection and radiation and ends up heating the surface to about 6,000K. The surface emits black body radiation mainly due to transient dipole formation.

The different spectra you give in your question are simply due to different surface temperatures. The spectra from the surface is unconnected with excatly what's going on in the core.

$\endgroup$
  • $\begingroup$ So, the hydrogen fusion at the core produces energy which reaches the surface of the sun in the form of heat. Does this energy heat up the surface and raise temperature which in turn causes the surface to give off black body radiation... and the color we see is due to the peak wavelength? I read the linked answer but it was a bit too advanced for me. $\endgroup$ – Eliza Apr 16 '14 at 11:14
  • 1
    $\begingroup$ @Eliza: yes, your description is basically correct. $\endgroup$ – John Rennie Apr 16 '14 at 11:22
  • 1
    $\begingroup$ @Eliza: yes, exactly! If you heat your iron to 5,700K then it will produce light with the same spectrum as the Sun (well, it wouldn't because iron would have vaporised at that temperature, but in principle the spectrum would be the same). Black body radiation is a universal phenomenon. $\endgroup$ – John Rennie Apr 16 '14 at 11:36
  • 1
    $\begingroup$ so any object can be a black body and the radiation it gives off is dependent on the temperature? $\endgroup$ – Eliza Apr 16 '14 at 11:47
  • 1
    $\begingroup$ @Eliza: yes. For example you and I are both emitting black body radiation with a spectrum corresponding to a temperature of 36.8°C (or whatever skin temperature is). The spectrum is related to the temperature by Planck's law. $\endgroup$ – John Rennie Apr 16 '14 at 11:54
-1
$\begingroup$

Production of Emission Lines of Iron in the Sun

EMISSION SPECTRUM OF IRON

λ 4,000A 5,000A 6,000A 7,000A

T 7,250K 5,800K 4,800K 4,150K

PLASMA (100 %) PLASMA (10%)+ VAPOUR(90%) PLASMA (1%)+ VAPOUR(99%) VAPOUR (100%)

The visible Emission Spectrum of Iron contains around two hundred emission lines from violet to red which represent the large number of Electron energy losses when Electrons drop from a higher Quantum Energy Level to a lower Quantum Energy Level

A Neutral 56Fe atom has an Electron Configuration of 2, 8, 14, 2

Within the Sun's Photosphere Iron Nuclei above 7,250K start collecting Electrons at the lowest energy level (2) and in the process Photon Energy at Wavelengths less than 4,000A are emitted as Ultra-Violet Light

As the Iron Nuclei cool further, the Electrons progressively drop into the second (8)and third (14) energy levels progressively producing Photons at (5,800K 5,000A Green), (4,800K 6,000A Orange) and (4,150K 7,000A Red)

The final two valence Electrons produce Infra-Red radiation at temperatures below 4,150K

Iron changes from Vapour to Liquid at 3000K

The Energy density of the Blue, Indigo, Violet emission lines is much larger than the density of the Red, Orange, Yellow emission lines which is due to the higher Temperature at the Blue, Indigo, and Violet emission lines. The Energy density of the Green emission lines lies between those of Violet and Red

Joel Savory

$\endgroup$
  • $\begingroup$ Not very clear what point you are trying to make. $\endgroup$ – paisanco Sep 30 '16 at 3:31

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.