# Tag Info

## Hot answers tagged spectroscopy

21

To be a little pedantic, nobody has yet done precision spectroscopy of antihydrogen, though the recent success in trapping it at CERN (all over the news this week, paper here) is an early step toward that. It's possible that there are small differences in the spectrum of antihydrogen and hydrogen, though these differences can't be all that large, or they ...

17

The full quantum analysis of the hydrogen atom is a quantum two body problem, however, one of those bodies is extremely massive compared to the other, so that this problem, as a first approximation, is analysed by solving either the first quantised (i.e. for one quantum particle in a classical environment) Schrödinger or Dirac equations for inverse square ...

15

When astronomers started to get spectra of stars and began classifying them, the initial classification was based on the strength of the Balmer absorption lines in the spectra. The Balmer lines are created by electons in hydrogen atoms that are currently in the second energy level (N=2) absorbing energy and jumping up to higher levels. The stars with the ...

12

Almost all exoplanets observed are near F, G, and K stars. In part, this is because astronomers are looking for earth-like planets, so they look at stars similar to our Sun, but there are also some physical reasons. Sahu et al (2006) have provided some evidence that red dwarfs (class M) are more likely to have planets than other spectral types, though it is ...

12

Good quantum number are associated with operators that commute with the Hamiltonian. They correspond to conseved quantities. Overall angular momentum is conserved, but the portions of it due to orbital motion and due to spins are not themselves conserved. n is 'bad' in that there's no conserved physical quantity related to radial motion. A decently ...

10

Given what we know about planetary formation (Link 1, Link 2, Link 3 and Link 4), and the theories around it, it would probably be a safe bet to say that ALL stars end up having some left over material that might become planets. I think the bigger question is how many of those planetary orbits stay stable enough throughout the life of the star? All these ...

9

There's a few ways the temperature can be measured remotely. The easiest way is to measure the amount, and for bonus, spectrum, of the radiated heat. All objects greater than Absolute Zero radiate a certain amount of energy. The wavelength spectrum can be determined by Planck's Law, and the amount of energy by the Stefan-Boltzmann law, both of which are ...

8

You've asked a lot of questions there, and I'll try to answer them one by one. First, though, I want to ask what post you're reading about metallicity in the core vs. out here in the 'burbs because I don't think it is correct. Obviously, for example, we exist and we're ~26,000 light-years (half-way out) from the galactic center and we have a fair amount of ...

8

The major tool in investigating the composition of an astronomical body is spectroscopy. This technique makes use of the fact that a body composing certain elements/compounds will shine more brightly(or less) at particular wavelengths. The pattern of 'lines' taken by a spectrograph can then be used to infer the original composition. This technique is used ...

6

EM radiation, including light, is a spectrum of different wavelengths. Spectroscopy is the detailed analysis of a light signal by wavelength. Ordinary color images break up light into 3 channels (red, green, and blue), but spectroscopy is generally concerned with breaking up light into a higher number of bands (e.g. 10, 100, or more), and a spectrometer is ...

6

The "metallicity" of a star simply means how much elements other than hydrogen or helium it contains. In this case, a "metal" means anything that's not on the first row of the periodic table of elements. Thus, a "metal-rich star" is one that contains lots of (for example) carbon, oxygen, nitrogen, etc. The term does not refer to metals in the strict ...

6

A galaxy spectrum is a quite complex and complicated topic, and many entire careers are fully devoted to understanding them, so this can only be a simplified answer. It is still quite lengthy, though, so if you're impatient, I've summarised it at the bottom. A blend of starlight of different spectral types makes up the continuum. The light is emitted by ...

5

the physical reason why this is happening is that absorption of a medium is frequency dependent. Mathematical description The most prominent example of a natural description might be the Lambert-Beer law that states that the change of a quantity $q$, $dq/dx$ is related to its value at $x$ multiplied by a scalar factor $\lambda$, which might depend on some ...

5

Blackbody radiation for a white hot object emits the spectrum from infrared to ultraviolet. See: http://en.wikipedia.org/wiki/Black_body Graphite is a decent approximation of a blackbody radiator. So graphite heated to white hot will emit the full spectrum of visible light. Note however, that the spectrum will not be flat. There will be more energy on ...

5

There are continuous spectra, emission spectra, and absorption spectra. Dense materials, such as the deeper parts of our sun, have continuous spectra. Hot, low-density gases have emission spectra (a black spectrum with bright lines in it). The continuous spectrum generated deep in our sun passes through the cooler and lower-density outer regions of the sun, ...

5

Just to add to gigacyan's answer, the harmonic oscillator Hamiltonian may be written in terms of raising and lowering operators: \begin{eqnarray} \hat{H}\psi&=&-\frac{1}{2m}\frac{\partial^2\psi}{\partial x^2}+\frac{1}{2}m\omega^2x^2\psi\nonumber\\ &=&(a^{\dagger}a+\frac{1}{2})\omega\psi \end{eqnarray} where ...

5

A spectrometer does something similar to what a prism does: light goes in, and gets split up into a spectrum. If you shine white light through a prism, a rainbow comes out the other side. But not all things give off white light. In fact, each element, when excited gives off a unique set of wavelengths that act like its signature: these are called emission ...

5

Velocity and wavelength shifts are connected by the Doppler effect, with which you may already be familiar. Basically, objects that are approaching appear bluer and those receding appear redder. Now, suppose body of material emits some spectral line and that the particles of that material have line-of-sight velocities that range from $-10^4\text{km.s}^{-1}$ ...

4

The redshift due to cosmological expansion is identical to a Doppler shift in its effect on the spectrum of any source. To be specific, both phenomena "stretch" all wavelengths by the same factor. There's a very good reason for this: in a suitable coordinate system, the cosmological redshift is a Doppler shift. You'll find statements in some textbooks ...

4

Is this right? An electron could stay a little time in a unstable energy level? Short answer is -- yes, this is a correct standard description of the process. Let me elaborate. In quantum mechanics all observables, including energy, are described with hermitian operators. Eigenvectors of these operators are the states with definite value of the ...

4

Superfluid Helium-4 has a very well studied excitation structure -- at very low momenta, there is a low energy excitation, the phonon, that corresponds to a periodic density fluctuation in the superfluid with well defined wave-number and an energy $E = c \hbar k$ (c being the speed of sound in the superfluid). Though others might quibble with me over ...

4

The resolution would ultimately depend on the size of the laser beam where it crosses the molecular beam. The smallest size you'll be able to obtain with the beam is on the order of the optical wavelength (there are some numerical factors that depend on the exact geometry, but if you're just looking for a ballpark figure, that will give you a lower bound). ...

4

Dear Josh, the wave functions are only perfectly symmetric and/or perfectly antisymmetric - each of the factors is - in the case of two particles. As Fabian correctly says, for more than 2 particles, the wave function isn't perfectly symmetric and isn't perfectly antisymmetric with respect to particular transpositions of the two particles. The general ...

4

A good source of spectroscopic data is the NIST Chemistry WebBook. It compiles all possible information about polyatomic molecules including their UV/Vis and IR spectra. You will find references to corresponding papers and, as I see, the new version includes a Java applet that plots the UV spectrum. Here is a direct link to THT spectrum.

4

Just compare the resolution of the two: Prism depending on n, there is no good material n>1.7 (besides diamond) depending on base length if you use a equilateral triangle have to use more than one to overcome this prism absorb light, you have got scattering (stray light) too Now a grating: optimize it for your wavelength choose lines per milimeter ...

4

The spectral classes were originally developed by spectroscopists, before stellar astrophysics was nearly as developed as it is today. So they were based on the presence, absence, or relative strength of spectral lines, rather than the overall blackbody spectrum, which would make more sense. Unfortunately, since most lines first become more prominent and ...

4

Originally, the names were descriptions of series of spectral lines recorded on photographic film. Introduction to Modern Optics By Grant R. Fowles

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