# Tag Info

1

If you want to do that yourself, you can look at the SDSS catalog. It provides freely available spectra over several years. The problem is, their resolution is right about the speed of the Earth, $30 km/s$ (vs $29.8 km/s$) (I wonder if they are related). There are higher resolution sources, but they are not as easy to grab (to my knowledge, but I am not a ...

2

See Vogel, H.C. "On the spectrographic method of determining the velocity of stars in the line of sight" Monthly Notices of the Royal Astronomical Society, Vol. 52, pp. 87-97 (1891). http://adsabs.harvard.edu/full/1891MNRAS..52...87V 47 nearby stars were studied and the Doppler shifts of these stars were calculated on various days of the year.

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Using the NIST Spectral Database, entering sodium (Na) with $\lambda$ between 588nm and 590nm gives wavelengths of 588.9950954nm and 589.5924237nm.

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The numbers are not measured, they are part of a model that explains why the hydrogen atom emits/absorbs only at those wavelengths. The first series of Hydrogen lines to be discovered was the Balmer Series, and nobody knew why they were discrete lines instead of a continuous spectrum. Johann Balmer discovered that the lines all had wavelengths equal to, ...

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This question is closely related to the question "If photon energies are continuous and atomic energy levels are discrete, how can atoms absorb photons?". While this technically isn't an exact duplicate of the link above, a similar conceptual explanation applies, namely that for complicated strongly-coupled quantum systems (like a hot chunk of metal), the ...

0

This is quite a natural confusion. You are correct, were the Solar spectrum purely due to the spectral output of the atoms composing it, we would not be able to get a continuous spectrum. However, the light emitted by the Sun is due to its temperature. All objects that are above $-273.15^{\circ}C$ (so, all objects) emit radiation at a continuous spectrum ...

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A magnetic dipole transition can be modelled as a time-dependent perturbation $V_{\text{md}}(t) = {e\over 2 m}(\vec{L} + 2\vec{S})\cdot \vec{B}e^{-i \omega t}$. Fermi's Golden Rule tells us that the transition rate for $b-X,1$ is proportional to the matrix element of the perturbation between the initial and final states, W \propto \langle \psi_b|{e\over ...

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