# Tag Info

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Your $\vec{r}$ is just a position vector pointing from the origin. Thus, what you are calculating is the matrix elements of the dipole moment (omitting the charge), which characterize the intensity of coupling between the atom and the electromagnetic field. Where the dipole moment is zero, such a coupling is absent, whereas the non-zero values characterize ...

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Equate the function to half its peak value and solve. For eample, if $b\gg a$, then the envelope of the overall function is defined by the sinc function, which has a maximum of 1, so it reaches the FWHM points at $${\rm sinc}^2 \left( \frac{ax}{2} \right) = \frac{1}{2}$$ This must be solved numerically. The solution is $ax/2 \simeq \pm 1.392$. Thus the FWHM ...

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You would do just as the name implies. For example, the envelop has a height of about 1. So half max is 0.5. The width of the envelop at 0.5 appears to be 0.25 - -0.25 which equals 0.5. You could also use the function to determine where the maximum occured and how wide it is at half that max. Since you've plotted the data you should have all the info that ...

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In reality, the R, G, and B that your LED’s are emitting are not single, narrow lines, but are instead rather broadband. See this question. To the extent that your LEDs’ spectra overlap, you can indeed make a spectrophotometer using it as a light source since you’ll have a continuous overall spectrum. But this would not be recommended because you’ll have ...

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If I shine emulated orange light through a sample and onto a detector, will the effect be in any way similar to shining actual orange light... There is no "actual" orange light. There are only an infinite number of power spectra that appear orange to our eyes. If you want "orange" in a spectrophotometer, then you almost certainly want a ...

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If the source is very large, then an E.M. Wave from the more distant part of the source lags behind that from the near side. A very short single pulse can only come from a small source.

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If a star was moving toward Earth, its light waves will appear to be pushed together. This decreasing distance between Earth and the star in effect, shortens the wavelength of the light received by Earth. This is called the Doppler effect. The star's spectral lines move toward the blue end of the spectrum. Now consider the complete opposite, where a star is ...

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Yes, the relative motion will cause a doppler shift in the wavelengths observed. Also if your space is expanding, i.e the objects are not changing position in space, just the space between points is expanding, you also get a red shift in the light as the light traverses space which you can think of as space stretching the wave out if you like. Furthermore, ...

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There is a difference between thermal radiation, arising from material where the populations of excited states in atoms/ions and the distribution of particle speeds can be characterized by a temperature, and blackbody radiation, where all those things are true plus the radiation field is in equilibrium with the matter. Colour refers here to a perception. It ...

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My thoughts so far on explanations include the fact that the peaks aren't infinitely thin - i.e. they have some thickness to them. Is this from inherent quantum mechanical uncertainty in energy of the photons that need to be absorbed (maybe from the Heisenberg relation for $\Delta E$ and $\Delta t$)? Yes. Are there Doppler broadening effects occurring? ...

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Indeed the absorption line is not infinitely thin. If it were, the absorption would take infinitely long! There are several sources of line broadening. Lines are broadened by the effect of lifetime. The transition takes a finite time and therefore a finite number of wavelengths. This results in broadening. Also the molecule feels effects from its environment....

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The elements in the hot atmosphere surrounding the Sun absorb their preferred wavelengths of light being beamed out of the sun, but they then re-emit those absorbed photons in random directions- most of which are not aimed at us here on earth. So we see a solar spectrum with dark lines in it- representing those "missing" photons.

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Molecular spectrum is not just a superposition of the atomic spectra, it also has vibrational and rotational degrees of freedom corresponding to the oscillations of the inter-atomic bond lengths and the rotation of a molecule as a whole (or rotation of its parts in case of complex molecules). The frequencies of these degrees of freedom are usually much ...

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I've tried using a 650 nm laser as the source but the spectrum appears very smooth with the peak at the corresponding wavelength. This sounds like what you would expect from this source. Interference patterns are not magical features where you tap on the laser and it suddenly acquires interference $-$ they come from the superposition of two or more distinct ...

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Electron releases the work function energy to be bound to metals again.

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In that case the electron absorbs the energy of the incident photon (which will be greater than the amount of the work function). So the energy after this interaction would be contained in the scattered electron. So $$E_{\gamma} = K_e + \phi$$ where $E_{\gamma}$ is the energy of the incident photon, $K_e$ is the kinetic energy of the ejected electron and $\... 0 The gold standard for this is the NIST Atomic Spectra Database, which curates comprehensive lists of energy levels, together with their energies, angular momenta (i.e. term symbols) and Landé$g$factors, as well as indications of how well-described they are by Hartree-Fock single-determinant approximations; transitions between levels, including their ... 2 Dopant affects band gap of material. The band gap determines the wavelength of light that the material absorbs. The color of material depends on what light it absorbs. Maybe P/N dope affects color of material. If you draw the band gap of material using any method (Density functional Theory or etc), you can predict color of material some extent. In real world,... 2 A spectrometer normally uses a diffraction grating. The interference pattern from a grating for a single wavelength concentrates the light into a single angle at each order. (The better the grating, the tighter the angle.) Try reflecting a laser beam from an unused CD or DVD (in a manner that does not put an output into your eye). (I got a pocket sized ... 12 As you've already noted, we cannot detect cosmic antimatter from its spectrum. As the answers at How would we tell antimatter galaxies apart? indicate, there are two ways we could detect cosmic antimatter. Firstly, we would see the tell-tale 511 keV gamma ray signature of electron + positron annihilation reactions coming from the border of the antimatter ... 8 Antihydrogen atoms have been created in the lab and their basic spectral characteristics confirmed as identical to hydrogen. So we cannot tell by directly observing an object. But we infer it from the fact that no gross interactions between matter and antimatter have been observed. A full answer is given at How would we tell antimatter galaxies apart?, but ... 1 The data for levels would be on the NIST database, but I don't know where or how you could generate a chart of levels and transitions as you ask for (except by hand, of course). The data base does have details of multiply ionized states as well. To find$\mathrm{Nd}^{3+}$you would search for "Nd IV", as "ND I" (and "Nd$0$" ... 2 The two processes you've mentioned, finite lifetimes and Doppler effects, have different effects: The finite excited state lifetime results in an uncertainty in the transition energy due to the energy-time uncertainty principle - heuristically,$\Delta E \Delta t \ge \frac{\hbar}2$means that a short lifetime will increase$\Delta E$, so the emitted photons ... 0 Since$E = h\nu\$, broadening the spectral (frequency) width of photons necessarily broadens the energy spectrum. Doppler broadening means that atoms are moving at different directions and speed from thermal motion. When an atom runs upstream into a photon, the frequency is increased from the Doppler effect. Also the photon "hits harder", even ...

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