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4

To follow the information in Chris White's answer - essentially, you would want a medium that allows you to see the spectra. There are several online resources that could help you in this experiment, in particular, the CD spectrometer, which can be constructed simply and on that website, it shows several examples of how everyday light sources can be ...


6

Lasers by definition only emit a single wavelength of light. You use one if you want that wavelength or if you want your photons to be in phase. You don't care about the photon phases, and you want to sample all wavelengths, so a laser is very much the wrong tool. If you just want collimation of the light, mirrors, lenses, or even just well-separated ...


0

One important advantage of mass spectrometry over optical methods that has not been mentioned yet: sensitivity. If you have just a handful of molecules you are interested in, there is only a tiny possibility that you will hit them by photons, in most cases they will go undetected. However, if you ionize molecules, you can move them with electrostatic lenses ...


1

would it not work better/more easily to heat a sample into a its gas phase, shine a very bright light through it, and apply normal spectroscopy? This is very similar to atomic absorption spectroscopy. Except that the light is a heated filament with the elements of interest incorporated so it emits the spectral lines for those elements. You measure the ...


3

The traditional spectroscopy you describe is still used; different techniques apply better in different circumstances. For example, there are chemical bonds whose energy is in the ultraviolet. UV spectroscopy is good if you are dealing with such molecules, perhaps things with lots of $\require{mhchem}\ce{C=C}$ bonds. The information you get can tell you ...


2

Optical spectroscopy will tell you about different kinds of bonds that are present in a molecule. However, it is very hard if not impossible to distinguish, say, $\require{mhchem}\ce{CH3CH2CH2CH2CH3}$ (pentane) from $\require{mhchem}\ce{CH3CH2CH2CH2CH2CH3}$ (hexane) with this method - you might see a difference in intensity of the different bond types, but ...


3

Almost but not quite. Qualitatively the spectrum is the same with the $1/n^2$ spacing, but the scale of the spectrum is set by the reduced mass $\mu$, $$\mu = \frac{1}{\frac{1}{m_l}+ \frac{1}{m_p}}$$ where $m_p$ is the proton mass and $m_l$ is the lepton (muon or electron) mass. Since $m_p \approx 2000 m_e$, it is not a large error to take $\mu = m_e$ for an ...


2

Rather than write something unintelligible, I'll quote from a page on cesium clocks. According to quantum theory, atoms can only exist in certain discrete ("quantized") energy states depending on what orbits about their nuclei are occupied by their electrons. Different transitions are possible; those in question refer to a change in the electron and ...


2

As you point out in your ray diagram, the point source no longer looks like a point source after the diffraction grating. You could get around this by collimating the input beam with another lens (which you did), with an aperture, or by moving the point source far away from the grating (effectively using the grating as an aperture). So, I think the answer ...


2

If you had a single hydrogen atom, and you watched for a single transition, then yes, you would only see emission at a single frequency. There would be one line in your spectrograph so to speak. But rarely do you have just one atom. And quite often an atom undergoes multiple transitions while you are watching it. When looking at a large ensemble, whether it ...


1

There is no luck involved. Quite the opposite. Firstly, as many of the comments say, spectal lines arise in a very wide range of frequencies well outside the visible range. Secondly, and IMO more importantly for this answer, evolution is NOT random. And we see in the visible range because: "Most" of the Sun's radiant energy is in this band, in the sense ...


2

The Balmer series for hydrogen is an example with some visible lines, and some lines outside the visible spectrum. On the other hand, the Lyman series for hydrogen is completely outside the visible spectrum. Since every element has an infinite number of spectral lines, it would instead be very unlucky if they all somehow fell outside the visible spectrum.



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