What explanation can be proposed for t he color bands that are superimposed on the fluorescent spectrum?
818868
I suspect this refers to how flourescent light bulbs generate a quasi-continuous spectrum as opposed to the discrete emission lines like other self respecting plasma discharge tubes.  \n\nHere is a website that has several pages on Flourescent Lamps:\n\nhttp://home.howstuffworks.com/fluorescent-lamp.htm\n\nThe first two or three are the ones useful to you.  Also there are several links on the last page under More Great Links that will help you in your researches.  Happy hunting!\n\nBTW: This reeks of an intro science class lab write-up question.  TAs at help desks or profs during office hours can help a great deal more effectively than something like yahoo answers, if you want to truly understand.
348134
Web_Maven got it right but only if you already know the right answer. Here is how it goes.\n \nThe first part you got, when you mix lights of color red, green and blue you can get other colors and white. But why doesn't this work with pigments? Because pigments do not emit light, what they do, is absorb all the other colors except the one(s) you see. If you throw white light on a red pigment, it will absorb all colors except red. If you mix the three pigments, the mix absorbs all colors.
920352
Did you measure the absorbance of a standard in addition to the unknown sample?\n\nIf so, you could plug the values into (As / Au) = (Cs / Cu)\nWhere\nAs = Absorbance of the Standard\nAu = Absorbance of the Unknown\nCs = Concentration of the Standard\nCu = Concentration of the Unknown?\n\nAlternatively, did you get a value for the molar absorptivity of PNP at the wavelength you used? \nIf so, you could plug it into\n\nA = (epsilon)(b)(c)\n\nWhere\nepsilon is the molar absorptivity,\nb is the length of the path through which the light passed,\nc is the concentration of the chromogenic substance.\n\nAnd you could rearrange that to solve for c.\n\nHope it helps
166583
That's light in the visible spectrum. I'd imagine a great deal of compounds could, especially since "great absorbance" means nothing, in scientific terms.
733136
All elements when heated give off a signature in the light spectrum. For example if you heated a piece of pure iron until it glowed then put a prism in front  of the light, you would see the spectrum of light -- obvious stuff. However because it is the metal iron there would be slots-- frequencies that would be missing. These signatures are all very different from each other depending on which element(s) are giving off light. When something is red shifted the signature is still there but it's place in the spectrum is in the wrong place. It has shifted to the red end of the spectrum-- it is in the longer or red end of it.\n\nThere is two ways to think of this, the other is that when heated elements will give off light in only so many frequencies- the rest of the frequencies are missing.
317548
You are confusing emissive mixing with subtractive mixing. When dealing with pigments, one customarily uses the colors Cyan, Magenta, and Yellow as primary colors. This is because you are actually dealing with a subtractive spectrum. The colors of the pigments are the part of the spectrum that is not absorbed by the material, so mixing them is not an additive procedure, but a subtractive procedure. Mixing all of the three colors above will not get you white, but (in a perfect world) will subtract every color from being reflected, giving you black, the absence of photons. In the real world, the pigments are imperfect and you actually get a murky dark greenish color, which is why an independent black ink is added to printing processes (CMYK).\nRed, Green and Blue are primary for the emissive spectrum, these are used for mixing colors of pure light (ie., CRTs and LCDs). Mixing all of these colors (use red, green, and blue filters in front of white lights) will get you white, as it should. See the article below for more detail.
963067
Emission lines of different wavelength depending upon what is being studied.
682859
If you could find a gas that had a semi-stable quasistate that when excited would release a photon with a wavelength corresponding to "black", then yes.  However, I don't know exactly what band that would be.  The closest you'd be able to get to black would be a dark ultraviolet, or if you could overlap in phase multiple color lasers which could possibly (but unlikely) add together to make the appearance of black.  Most likely, you're out of luck.
778920
different types of chlorophyls are composed of different pigments which absorb different wavelengths of light (different colors in the spectrum)
671037
The sodium vapor is absorbing the electromagnetic energy in that specific band.  It is actually two lines, close together, in the yellow-orange part of the visibile spectrum.\n\nThe energy of that yellow-orange light contains precisely the quanta of energy to excite electrons into higher energy states, so the sodium vapors abosrb very strongly at those frequencies.  The other frequencies are at the wrong energy level, and are "ignored".
501676
Given a bit of thought:  \nIf you are basing this decoding entirely on light, then there is a possibility that occurred to me.  Your mention of polarized ink gave me an idea.  \n\nFrom the basic idea, the color white is merely something that reflects all of the visible wavelengths back to you, not absorbing any colors at all.  This makes it seem very bright, because all of that energy reaches your eyes.  \n\nIf you can create a pure white ink (or same white as the paper) from pigments -- probably water-based so that they would not form texture -- then you may be able to create a white ink that is truly a mixture of the wavelengths.  Then if you have a pair of glasses that filter one color (like a camera), you would see the remaining colors and the writing would be visible.\n\nNow, I have never tried this and don't know that it would work, but given the energies of the color wavelengths, I think it would be easier to see if you filter the blue/violet end of the spectrum so that you see the red end.  This should show with more contrast, making it easier to read.  \n\nI do not know that it would be possible to so precisely filter one particular color.  If you create this white ink, it is possible that some mixing would mean your colors would not be perfect.  If you don't see anything right away, try all 7 combinations of the filters for a camera lens -- red blue and yellow each alone, two of each mixed, and all three together.  (And any others you can find.)  \n\nThis is just a springboard idea, and I do not know that it would work.  But if you do get it working let me know!  I would be very interested in knowing if my hypothesis has a practical application!
335985
Hi there,\n\nTemperature vs colour goes red, orange, yellow, green, blue, violet. It is simply the spectrum of light (think rainbow) because that is what it is based on.\n\nSo red light is actually the coolest, and as you go up the temperature scale, you run through the spectrum to violet.\n\nWhite light is a combination of ALL the different colours of light, so something which glows white hot is just releasing all different levels of energy, so it appears white. So in actual fact, blue (or more accurately violet) is hotter, because if something is blue hot, then it is releasing ALL really high levels of energy, but you rarely get anything which is completely blue hot - there is almost always some sort of mixture, and usually, the object glows white tinged with blue (which just means that the object is releasing all different levels of energy, but predominantly or mostly blue!)\n\nJust for your interest, the way this works is, when you heat something up, you are exposing it to invisible energy called "infra red" which is just heat. So when you touch something and it feels hot, you are simply sensing infra red energy!\nThis energy is "weaker" than light (so light is actually more energetic than heat), but if you let an object get hot enough, it stops being able to release the heat as just infra red and starts to release it as visible light. The hotter the object gets, the further along the spectrum is the light emitted.\nActually, if the object absorbs enough energy it will start to release Ultra Violet (UV) and X-Rays!
487877
The UV-VIS spectrum is due to the electronic transitions of the molecule. This is characteristic of a compound. Qualitative and quantitative estimations of compounds are possible by this non destructive technique. \n\nThe absorption (A) of a solution at a particular wavelength is given by Beer-Lambert's law ect = A where c is the concentration of the compound , t is the thickness of the cell and e is the molar extinction coefficient characteristic of the compound at a given wavelength. This principle is used for quantitative measurements. \n\nCary 5E is a high resolution spectrophotometer to study the electronic spectrum of solutions, single crystals and powder samples (in nujol mull) in the range of 3150-185 nm. This is a double beam instrument controlled by a microprocessor. Base line correction, repetitive scan, kinetics, derivative and concentration modes are possible. Diffuse reflectance is available for solid samples. VT accessory in the range 5-90 degrees Centigrade is available\n\nThis principle is used for quantitative measurements.
1004069
Ignore the above answers they are all horribly wrong. This is a chemistry question.\n\nColors in chemistry are due to two main phenomena. \n\n1. Light Absorption\n2. Charge Transfer\n\n\nFor the first case an electron can be promoted to a higher energy state by absorbing light energy. White light consists of wavelengths (about) 400-800 nanometers. Wavelength is inversly related to frequency which is related to the energy of light. The photonic energy of the light from certain wavelengths is absorbed to promote an electron. The rest of the light is transmitted. This transmitted light is what we see. But also keep in mind that the eye works strangely and we see mixed colors a little different.. I dont know why but just keep it in mind.\n\nThe second phenomena is charge transfer. This occurs when  an electron is transferred from one atom to to another. An example is KMnO4 (which has no d electrons) but is purple! This is because a electron is transferred from the oxygen to the manganese. \n\nAnd there are some other mechanisms for which colors are produced. One of the latest is now understood by a physics professor at ucla. Hes been studying how sound can interact with a bubble to produce light.
170333
What happens here is the reverse of dispersion that is the spectrum of colours obtaines which u see as the colours of rainbow or light emerging out of prism. When u spun the top all the colours get mixed to form their parent colour which is white. The same phenomenon is evident when wheel decorated with all the colours of rainbow is rotated fast.
283630
no i don't have any info. but i think we're able to detect all of the colors of the spectrum.
1139150
A good question. We perceive color because each of the cones in our eyes is receptive to only a specific range of frequencies, and so light of a given frequency will only stimulate some of these cells and not others. Any combination of light that stimulates the same cells by the same amount will be perceived as the same, even if spectroscopically it is quite different. Yellow light, such as that emitted by sodium vapor, falls well within the frequency range of both the "red" cones and the "green" cones, and so stimulates both, causing us to perceive yellow. However, if we were to provide both red and green light simultaneously, the red cones would be stimulated by red light and the green cones by green light, and we would also perceive yellow. Because we have but three types of cones, we can, both in principle and in practice, provide the perception of (almost) any color of light by varying luminosities of just the three primary colors, which is why we say that other colors are combinations of these three primary colors. However, monochromatic light of a frequency in between the frequencies of the primary colors will also appear to be a secondary color, even though it is obviously not a mixture of primary colors. Thus the grade school simplification of other colors as being literally composed of the three primary colors is, I'm afraid, quite misleading.