# Tag Info

0

The phosphors lining the glass tube of a fluorescent light do a pretty good job of smearing the atomic mercury line emission spectrum into something closer to black body radiation. There could be a phosphor mix that accomplishes what you want...

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When you scatter light off of a material there is a photon-phonon interaction which will shift the photon frequency depending on the phonon energy (Raman scattering, Brillouin scattering). The effect is quite small, however. How much broadening do you need? Rayleigh scattering through a warm, high density gas will probably go quite far in messing up the ...

1

This might not be feasible for your setup, but you could try rapidly rotating your light source, which would Doppler broaden your spectral lines. You're correct in that the speed would need to be a substantial with respect to the speed of light. As an example, if your frequency is 500nm let's say. If you'd like it to spead out on the order of a single ...

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I should say that the comments and answers so far have nothing to do with the effect that is shown on the plot. Note, in particular, how the intensity of broadband background depends on the wavelength of the laser used for Raman scattering. This indicates that the background is not related to the properties of the molecule itself, but rather to its ...

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The spectra are actually always continuous if the resolution in frequency of the spectral analyzer and sensitivity in intensity of the detector is high enough. The spectrum of radiation emitted by hot atomic gases and molecules is called discrete because it has very sharp spectral peaks and looks as if formed of lines. But they are actually peaks with ...

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Let's put it clear first: for Raman scattering there is no excited state at all, the light just bounces of a molecule. If the photo has the right energy, it can bring the molecule to an excited state. Different things can happen to a molecule in this state - in most of the cases the energy will be dissipated through collisions, but in a rare case the ...

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In theory, perhaps. It is possible, using multilayer dielectric coatings, to produce a surface which is reflective in very narrow bands (in this case, the Sun's dark lines)and transmissive (or absorptive) elsewhere. In practice, the spectral "blurring" caused by atmospheric transmission/absorption/re-emission effects would make this effect pretty much ...

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The NIST Atomic Spectra Database is a decent source for general-purpose lookup and identification of spectral transitions and levels. You will probably be more interested in their spectral line info.

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Unlike electrical or magnetic field, which acts differently on particles of different charge, for the gravitational field there is equivalence principle, which means that electrons and nuclei would experience the same acceleration due to gravity. There is, of course, the overall shift of energy level if the atom is observed from the place with the different ...

3

The sun's spectrum is very complex, and indeed there are a lot of "lines" both light and dark (emission and absorption) amidst a sea of what looks to be continuous frequencies. Note that the atoms you study in a textbook are idealizations. In a hot object such as the sun, some photons come to us by way of atomic emissions, but the speeds of the atoms that ...

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Certain wavelengths of electromagnetic spectrum is emitted when the electrons in an atom to move from a higher level to a lower. The wavelength that is emitted depends upon the number of shells the electrons move down. When an electron/electrons move into certain number shells, white light is emitted i.e. when two hydrogen atoms fuse into an Helium atom, ...

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$\sin \theta = (m \lambda)/d = (5 \times 1228^{-6})/(1/600) = 3.684$ $d = (1/600)\times 3 = 0.005 \lambda = (d \sin \theta) /m = (0.005 \times 3.684)/4 = 0.004605\: \mathrm{mm} = 4605\: \mathrm{nm}$

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Here you have a list of non-oxygen containing 4-element pnictides. It might not be a thorough list, but it might help.

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