How Does Kirchoff's Law Relate to Spectrum Intensity and Wavelength?

[SebastianRM]

TL;DR

In the book Foundations of Astrophysics (R&P) chapter 5, section 5.2.
It states: "A solid, liquid or dense gas produces a continuous spectrum".

What is the author meaning by that, I literally just read in section 5.1, that the depending on the energy loss of the electrons that make up the substance, this energy will be released as photons. Since this is particular for each atom, it explained how we can tell which element is which.
So if the spectrum is Intensity vs Wavelength, how is it that the first statement is true.

 

[mathman]

In those media the photons undergo lots of scattering which reduces their energy and as a result, tends to smooth out the spectrum.

 

[stefan r]

A spectrum looks like a hump with a long tail. There will usually be little spikes. If the spikes are pointing up (more intense) they are emission. If the spikes are pointing down they are absorption. The hump itself is a continuous spectrum. The spikes tell you what elements or compounds are involved. Sometimes a computer will delete the background so it looks like a graph of just spikes.
Sometimes the signal from the spike is large enough that you cannot see any background or black body emission on the graph. It is still there if you zoom in.

 

[rishijohn90]

Your views resolve my doubt.
Thank you

 

[sophiecentaur]

Summary: In the book Foundations of Astrophysics (R&P) chapter 5, section 5.2.
It states: "A solid, liquid or dense gas produces a continuous spectrum".

So if the spectrum is Intensity vs Wavelength, how is it that the first statement is true.

That statement is oversimplified. Any source hot enough to be seen by virtue of its internally generated energy will be mostly black body but additional absorption and emission spectra from the outer parts will always be detectable.
I have a friend who does astrophotography and he has taken up spectroscopy. It's really not to hard for an amateur; you use a slit over the star of interest, followed by a diffraction grating. You get a pretty looking spectrum and the image data shows absorption lines and bands. All that's in his back garden.

 

[andrew s 1905]

That statement is oversimplified. Any source hot enough to be seen by virtue of its internally generated energy will be mostly black body but additional absorption and emission spectra from the outer parts will always be detectable.
I have a friend who does astrophotography and he has taken up spectroscopy. It's really not to hard for an amateur; you use a slit over the star of interest, followed by a diffraction grating. You get a pretty looking spectrum and the image data shows absorption lines and bands. All that's in his back garden.

They can do more than pretty spectra with amateur spectrocopists contributing to real science include studies of Be stars, classification of supernova, monitoring of symbiotic stars and much more all with back garden observatories .
Regards Andrew

 

[sophiecentaur]

They can do more than pretty spectra with amateur spectrocopists contributing to real science include studies of Be stars, classification of supernova, monitoring of symbiotic stars and much more all with back garden observatories .
Regards Andrew

Amateur astronomers have a vast potential for contributing to Space Science because there are so many different targets available, most of which cannot be observed by the expensive professional experiments.
Amateur naturalists have a similar potential for ground breaking observations.

 

[tech99]

In those media the photons undergo lots of scattering which reduces their energy and as a result, tends to smooth out the spectrum.

I cannot understand how the wavelength of an emission could alter as it is weakened. We do not see colour change with distance.

 

[sophiecentaur]

I cannot understand how the wavelength of an emission could alter as it is weakened. We do not see colour change with distance.

Inelastic scattering would take energy from or add it to photons (hot gas).