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Originally Posted by Michael Mozina
Let me explain how you are playing both sides of the street here, so maybe you will see where I'm coming from. On one hand you accept that we "see" various photons from a distance *AND* that we can identify them by their very unique wavelength. This is something I agree with actually. Each photon has a unique signature that is related to the valence shells and energy states and quantum discharges of from various atoms. We both agree that it is possible to identify the element by the wavelength.
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I'm not playing both sides, what you're missing (altough it has been pointed to you many times) is that light is emitted in more than just one way:
1. Light emitted by atom/molecule transition from one energy level to another; since those levels are quantized the emitted spectrum is discreet and unique for each chemical element, therefore this kind of emission is used in spectroscopy.
2. Blackbody radiation, emitted in a continuous spectrum and depending only on the temperature. (I must point out that both those mechanisms have the same origin: the energy is emitted in quanta, so you accept or reject them both.)
3. Synchrotron/cyclotron radiation: produced by an accelerated charged particle. It has a continuous spectrum.
There are other mechanisms that produce light, but already you may see that our sun give a spectra that is a superposition from (at least) three other spectra, two continuous and one discreet.
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Originally Posted by Michael Mozina
You then turn right around and hand me a fifty cent math formula that presumably renders the first argument mute. Do you see the problem here Baloo? We can't say that photon wavelength is related to the element and heat and then turn right around and suggest it does not matter.
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It does matter as long as you're aware of all the means by which the said photon could be generated. You're not so you think that all the photons are generated in the same way, which is false.
Why don't you ask dr. Manuel or dr. Bruce what do they think about this "50 cents" formula? You'll be surprised.
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Originally Posted by Michael Mozina
You are trying to suggest that ANY element can emit 171A, 195A, and 284A wavelengths is raised to the right temperature.
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A body having about 102000 °K will give a continuous spectrum having a peak intensity at exactly 284 A.
You can do your own calculations for what spectra will have a body with a specific temperature using the formula at the bottom of this page:
http://scienceworld.wolfram.com/phys...cementLaw.html
Of course a highly ionised Fe ion will give also a photon having exactly 284 A; fortunately for us the sun has only 6000 °K at the surface, therefore it's thermal emission at 284 A is very weak and is not masking the Fe ions emission.
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Originally Posted by Michael Mozina
Other than iron, show me one element that is capable of releasing all three of these wavelenghts using real nuclear chemical data and show me the materials used.
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Any element heated at 100000-160000 °K will produce a spectra having a peak intensity at a wavelength between 171-284 A.
And there is more; actually any body that surround you is emitting at 171 A: your computer screen, your chair, yourself. By using that "50 cents" formula you can compute how much energy you're releasing at this wavelength. (you'll need to know your temperature: it should be around 300 °K
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Originally Posted by Michael Mozina
I am not "pretending" anything. I could (wouldnt recommend it) look at the sun with my own two eyes and "see the light" any time I wish. I do see neon, and I see it in the penumbral filaments and not in the umbra.
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Let see what you've stated so far:
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Originally Posted by Michael Mozina
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Originally Posted by Baloo
What is, in your opinion, the temperature of the sun's photosphere?
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5800 - 6000K |
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Originally Posted by Michael Mozina
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Originally Posted by Baloo
what wavelength have the photons emitted by a surface having a 6000° Kelvin temperature?
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In this case I think most of the photons being released at 6000 degrees come from neon plasma
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Seems like a great question and it warrants study. |
So you've said that you don't know how a neon surface having 6000°K look like. Then how do you know that is a neon surface having 6000°K in those pictures?
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Originally Posted by Michael Mozina
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Originally Posted by Baloo
Neon gives off visible light? Could you tell me in which conditions the neon produce visible light?
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Electricity is typically required. The next time you're in a pub, take a look at the neon signs. You can get them in all colors, including white. |
Mister Mozina, do I need to explain you the difference between an incadescent and a fluorescent light bulb?
The neon inside a fluorescent bulb don't give off VISIBLE light; that's why the fluorescent bulbs are coated with phosphor. The phosphor is the one actually shining in a neon lamp... since the sun give visible light would you concede that is actually phosphor on it's surface and not neon?
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Originally Posted by Michael Mozina
Exactly Baloo. They have to be iron ions flowing through those arcs. You just explained it yourself perfectly. There can be no other light source here other than iron ion photons. That *IS* the light source and HAS TO BE the light source.
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Mister Mozina, nobody disagreed that there is iron in those arcs. But due to the used filters in those pictures all the other elements (that are emitting on other wavelengths) are not visible. Is like using sunglasses at night: you'll see the white cars, but the blue, brown, black, red cars will be invisible to you.