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Wikipedia:Reference desk/Archives/Science/2018 March 18

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March 17

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Continuous, emission and absorption spectrums

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Does anyone know where I can find find the continuous, emission and absorption images for the elements? I am creating a periodic table of the elements and really want to have these added. I found the visible spectrum images on wikimedia, which is fine, and am content on using them. I have looked and searched and found nothing really, except for hydrogen mostly. The kalzium program on Linux distros has the emission and absorption images, but they are small (and the program hasn't been updated for 12 or more years). 68.68.64.65 (talk) —Preceding undated comment added 22:25, 17 March 2018 (UTC)[reply]

Yes, Abductive, the images. Here are the ones I have been using so far: https://en.wikipedia.org/wiki/Spectral_line 68.68.64.65 (talk) 03:14, 19 March 2018 (UTC) — Preceding unsigned comment added by 69.63.88.54 (talk) 22:30, 18 March 2018 (UTC)[reply]

They seem to be a high quality and complete set of emission spectra in the visible range. Other than that you also want absorption spectra, how do those compare to what you want (spectral range, spectral resolution, graphical size, etc.)? DMacks (talk) 03:29, 19 March 2018 (UTC)[reply]

DMacks, I am trying to follow the makers of kalzium (in a way). I just wanted to add the 3 mentioned spectrums (like they had the emission and absorption in kalzium). I am satisfied with what I have now, I just thought that having the 3 would make it look "better". But I never realized looking for visible spectra without lines, just emission lines and just absorption lines spectra would be so difficult. Very grateful for what I found on Wikipedia. I couldn't find them anywhere else. Thanks folks for all your replies and help. 68.68.64.65 (talk) 14:28, 19 March 2018 (UTC)[reply]

March 18

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Does light knock at least some electrons off from any material?

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When light hits a material (any material), can we assume that at least some electrons will be displaced? If we put the material in the dark and illuminate it unevenly, how could we analyze the surface to find spots where light hit? --Doroletho (talk) 14:05, 18 March 2018 (UTC)[reply]

First of all, not all light photons will have any effect. They need to be of a high enough energy to do so, which means a high enough frequency - see photoelectric effect and Einstein's first, and Nobel-winning, 1905 paper. (light travels at a constant speed, so its energy depends on its frequency, not like the speed and kinetic energy of a massive particle.)
Secondly, metals and conductors will allow electrons to move freely over their surface. So even if they were displaced, they'd rearrange almost immediately. If the material is a semiconductor though, the effect of the light can be to render that spot conductive. As an insulator, the pattern of electrical charge is fixed in place; first as an even distribution, then as the remainder which wasn't exposed to the light. This is the basis of xerographic photocopying. Andy Dingley (talk) 15:40, 18 March 2018 (UTC)[reply]
  • That would depend a bit on the context, and how fine a spatial resolution you want.
Classically, the instrument was the electroscope, mostly the gold leaf electroscope. It's also possible to make an electroscope with a fine wire probe, which can be scanned across an area to 'read' the charge over it. Like most sensitive charge-reading instruments, the reading process is both destructive (it removes the charge that was there) and also needs resetting after each measurement where charge was found. An electronic electrometer is a more modern version of this, and more convenient.
To see the spatial distribution of the charge, xerography can be used. With a semiconductor-coated metal drum (selenium or a semiconductor doped organic polymer) place a constant charge over the whole surface. Then write on it with light (either laser spot or a reflection of the photocopying target), then dust with a fine pigment powder. Light makes the semiconductor conductive, dissipating the charge through the metal drum. The pigment is attracted to the remaining charge, but only in the unexposed areas. Pressing a sheet of paper over this transfers the pigment, giving an image of the remaining charge, and the pigment can be fixed in place by heat fusing it to the paper. Andy Dingley (talk) 17:26, 18 March 2018 (UTC)[reply]
Lots of equipment is used to study a material's surface - there's even a name for the entire field of study: surface physics!
Wikipedia has an article, Surface metrology. I have seen commercial equipment for these purposes, called by any of various generic names: "surface metrology station," "laser metrology machine," "profiling machine," "surface roughness meter," ... and so on. For example, you can purchase a "Panasonic Advanced Metrology System Solution" from your local ... place ... that sells semiconductor fabrication test-equipment. Just don't ask how much it costs. As the promotional literature reminds you, "...what does poor quality really cost?"
Other equipment that can be used to study material surface physics of course includes the conventional optical microscope; the electron microscope in all its forms; the atomic force microscope; the Raman spectrometer; the four point probe; and many other unique and specialized types of equipment. Nimur (talk) 17:59, 18 March 2018 (UTC)[reply]
Semi-important nitpick: light only always travels at the same speed in a vacuum. It travels more slowly through a medium, and this is what gives rise to things such as refraction. I'm noting this because it appears to me that "light always travels at the same speed" is a common misconception. More than once I've seen people introduced to things like Cherenkov radiation express confusion because they believe this. --47.146.60.177 (talk) 03:01, 20 March 2018 (UTC)[reply]
See Ionizing radiation for the cut-off for appreciable effects of photon energies on electrons, broadly speaking. Acroterion (talk) 18:02, 18 March 2018 (UTC)[reply]
Well, if you use that term, proceed with caution: while it is true that "ionizing radiation" knocks electrons out of their atomic orbit, even "non-ionizing" radiation can add energy to an electron: as Andy linked above, that is called the photoelectric effect. In some materials, like metals and semiconductors, an energized electron has greater electron mobility and may migrate, even if its "parent atom" is not "ionized." If we're not extremely careful with terminology, we can lead to great confusion: when mobile electrons flow in a crystal lattice, we do not usually say that the individual atoms are "ionized." In detailed study of solid-state crystal lattices, we often use the term electron gas or "free electrons" to describe sufficiently-mobile electrons that are not specifically associated with individual nuclei. Importantly: these electrons and their weakly-associated nuclei are not ionized: the total amount of energy is too low to separate the electrons completely. In specific: there is not enough energy to move the electrons to infinite distance from the material lattice. Broadly speaking, ionizing radiation provides exactly enough energy to move the electrons to an infinite distance from their atomic nuclei: this exact quantity of energy is called the ionization potential. Nimur (talk) 18:17, 18 March 2018 (UTC)[reply]
Agreed, that's why I qualified my answer. Since the question was about displacement, as opposed to raising an electron's energy level, it seems relevant if the OP isn't familiar with the concept. Acroterion (talk) 18:20, 18 March 2018 (UTC)[reply]
Yes, thank you for the clarification. The key distinction is "how far" the electron can be displaced. Nimur (talk) 18:24, 18 March 2018 (UTC)[reply]
Some substances have a very high ionization energy. For example helium and neon. These need vacuum ultraviolet to shift an electron. So normal light does nothing too them and they are transparent. Many other transparent materials will need electromagnetic radiation in the ultraviolet to knock off electrons. Graeme Bartlett (talk) 22:37, 18 March 2018 (UTC)[reply]
  • Note that we're operating in the realm of quantum mechanics, where particles have a nonzero probability of escaping from an energy well, even with no boost at all. A boost too small to kick the particle out of the energy well by itself will nevertheless increase this probability. So the basic answer is that when we're dealing with Avogadro's number of particles, pretty much any incident light, regardless of how low its energy is, will increase the number of escaping electrons to some degree. Looie496 (talk) 01:02, 20 March 2018 (UTC)[reply]