rhodamine 6g uv vis spectra?
414639
Click on the source link.  The site contains the absorption as well as fluorescence spectra.
860076
Sorry I can't help you, I didn't pay close enough attention during UV-Vis Spec. But I don't see why you would need the lamda (sp?) max for b. But really, I'm not sure. Good luck!
125296
use spectrophotometric method(uv-vis spectrometry).Few things u shud remember.....\n1) Use Labert-Beer's law\n2) u shud know the molar extingtion coefficient value of that particular species\n3) concentration of the species\n4) Calculate lambdamax. using LB law\nIt is an indirect method.\n\nUV-VIS. Spectroscopy will give u the lambdamax value directly...alongwith absorbance
995857
It's all about the color spectrum- ROYGBIV- And think more Blue, Indigo, Violet , more insects. More red, orange- less insects.\n\nMost insects see better in UV and are thus generally attracted to \nshort wave lengths toward that end of the spectrum. Insects do not see in the red end of the spectrum.
353617
Optical Spectrum Analyzer (OSA) is probably what you're looking for...nice for discovering chemical composition of light source and probably a zillion other different things.
875835
OD=E*d*C\n\nWhere OD=optical density\nE the molar extinction coefficient\nd the length of the path of the light through the solution (width of the cuvette)\nC the concentration in mole/L of the solute.\n\nYou need d and C as well to get E.\nThis is a fair approximation especially for student problems.\n\nIn reality if you wanted to find E, you would do a standard curve by measuring the OD of samples of known concentration.\nThen you would use linear regression to find the equation of the curve and from its slope you would calculate E (slope=E*d)\n\nA good standard curve and thus accurate calculation requires care to ensure the accuracy of the standards, several points (ideally 9 at least) and being within the linear are of the method.\nThe region of highest accuracy is 0.2-0.8. You can go even below or above this range (though not above 1.0) but this is the range where the photometric error is minimal.
681686
Na emits light at a primary wavelength of 589.0 nm (plus at a couple other wavelengths).  This light is a characteristic orangey-yellow, and is what makes street-lamps the colour they are, since many of them are sodium lamps.
963067
Emission lines of different wavelength depending upon what is being studied.
568363
Of course, sunlight is made up of many colors.  However, we can think about "original" (or "monochromatic", or more realistically nearly monochromatic) yellow lights from other sources (like a laser).\n\nWith the composed yellow, you can separate the green and the red back out.  One way to do this would be with a prism.  While the monochromatic yellow light would just stay yellow, the composed red+green would separate back out, with each color in its position in the rainbow.\n\nA more sophisticated way to do this would be with a spectrometer.  The monochromatic yellow would show just the one wavelength, in the yellow part of the spectrum.  The composed light would show two wavelengths: one each in the green and red parts of the spectrum.  It would show no light in the yellow part of the spectrum.\n\nSnawshia is right about the speed: in a vacuum, they'll have the same speed.  In other media (water, glass, etc.), each color may have a slightly different speed.
939890
well Kc first of all this is physics, not chem.\nsecondly the wavelength cannot be in Hz, frequency is measured in Hz so probably your ques should be \na certain violet light has a frequency of 6.0*10^14 Hz. What is its wavelength?\nif this is correct all you need is the equation\nVelocity  =  frequency X wavelength\n 3.0 * 10^8 = 6.0 * 10 ^14  X W\nW = ( 3.0 * 10 ^8 ) / ( 6.0 * 10 ^14 )\n     = 0.5 * 10 ^ -6 \n     = 5 * 10 ^ -7 meters\n\nI hope this 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.
759217
Burn it in a sprectrophotometer.  Look at the emitted light.
890060
This solution probably has fluorescence at UV (ultra violet) wave length. try looking at it in a room with red light, if there is no light using the lenses then it's confirmed. but you can't simply identify it this way. you have to know the exact wave length that it has fluorescence at. even then you might not be able to identify it. try using UV-visible spectroscopy and when you have its spectra, you can identify it. for more information see: principles of instrumental analysis by Skoog
725621
6Rb(s)+N2(g)--->2Rb3N(s)
573389
it's reactive to light
48684
C- longer wavelength.  Red is the long (low fequency) end of the spectrum, and violet is at the short end (high frequency).
909784
The visible series is also called the Balmer series. The energy transitions are from n>2 to n=2. The least energetic and hence longest wavelength line would be the transition from n(h)=3 to n(l)=2. The equation is: \n\n1/lambda = RH (1/n(l)^2 - 1/n(h)^2) \nFor RH use Rydberg constant of 10972130 m^-1 (or .010972 nm ^-1), n(l) is 2, n(h) is 3.\nSolving gives lambda = 656nm, this appears as red.