So, I'm wondering what makes things look green, right down to the atom. Why do some chemicals and minerals reflect green light while others don't? What is at the heart of this? Wrad 04:13, 15 September 2007 (UTC)[reply]

A small note: some objects will give off green light – it's not just a matter of reflection. Vranak 04:55, 15 September 2007 (UTC)[reply]

How so? Wrad 04:56, 15 September 2007 (UTC)[reply]

In order to really understand this you'll likely need to read about light. There is a visible spectrum of light that we can sense with our eyes. So when we see something, light is reflecting off of it (life from the Sun or a light bulb). The color we see has to do with what wavelengths of light reflect off of that object. So, if you see something that is green, than the object off of which that light was reflected absorbed all but the green wavelength of light. If you see something yellow, than it absorbed all but the yellow wavelength of light. Black objects absorb almost all light. White objects reflect it all back.
Mrdeath5493 05:06, 15 September 2007 (UTC)[reply]

I'm aware of all this. I want to go deeper. Wrad 06:51, 15 September 2007 (UTC)[reply]

A little bit deeper into why things would absorb certain colors. The absorbance of the photons of a specific wavelength (and energy) depends on the structure of the molecules in the object, and the available electronic transitions in the molecule. For example, if a material absorbs yellow, it is because there is a transition corresponding to the energy of the "yellow" photons. If it doesn't absorb green, there is no transition the the corresponding energy of the "green" photons. Also see energy level, although I'm afraid the articles aren't written too well, and someone without physics/chemistry background, especially quantum, may not get all of it. There are probably better articles out there. --Bennybp 06:08, 15 September 2007 (UTC)[reply]

I'll have a look. Wrad 06:51, 15 September 2007 (UTC)[reply]

Our article on chromophores is fairly easy to understand (compared with those given above) - a chromophore is an area of a molecule such an azo dye (used to dye fabrics) which contains groups such as benzene rings and double bonds. As a result, electrons can move from one area to another with ease, which means that energy levels are close together, and therefore photons of visible light can be absorbed or reflected (many chemicals are white or transparent because these energy levels are too far apart, and therefore correspond with high-energy light instead such as ultraviolet). Laïka 07:13, 15 September 2007 (UTC)[reply]

I don't know if you re-worded the original question or if I was just to tired to completely miss the fact that the original question wanted specifics... :P
Mrdeath5493 03:52, 18 September 2007 (UTC)[reply]

scientist say that when photon hits an electron revolving in the orbit of an atom ; then the electron absorb the photon and raise to upper orbit for a small time; after that time the electron release the same energy that it absorbed from the photon and get back to it's previous orbit . now i want to know :

   1.how the electron can raise to upper orbit while the upper orbit is filled by electron already ?
   2.what is the exact time that the excitedelectron stay in upper orbit ? 

is this time constant in all case ?

   3.when the excited electron return to it's normal state ; it radiate the absorbed energy . 

what is the direction of the radiated energy ?

   4.on which factor this direction depends on? 
   5.could this direction could be changed in any process ?
   4.is the energy of the excited electron exactly same before and after the process ?
 

—Preceding unsigned comment added by Shamiul (talk • contribs) 07:29, 15 September 2007 (UTC)[reply]

Question1 - the electron is raised to an empty orbit.

Question2 - no idea I'd like an answer to this too.

Question3,4,5 - It's usually assumed that the energy is radiated randomly in any direction.. However I can't guaranteee this is always true.

Question6 - Yes.87.102.47.225 08:38, 15 September 2007 (UTC)[reply]

In Question 1 you are probably thinking of the Pauli exclusion principle, which prevents two electrons being in the same quantum state. Note that an electron's state is not determined by its orbital alone; it is determined by a combination of four different quantum numbers - see electron configuration - so several electrons can occupy the same orbital. However, it is generally correct that only the outermost, highest energy electrons (called valence electrons) are excited by incoming photons, as they are the ones that have nearby empty states to move into.

For Question 3, our article on spontaneous emission says the direction and phase of a spontaneously emitted photon are both random. However, for Questions 4 and 5, see our article on stimulated emission, a different emission process in which the phase and direction of the emitted photon are not random. Gandalf61 08:46, 15 September 2007 (UTC)[reply]

The answer to question 2 boils down to 'it depends'. In general, excited states that are further from their ground state decay more rapidly—the more excess energy in the system, the fast that energy gets shed. Beyond that rule of thumb, you need to start getting into some hardcore quantum mechanics. The lifetime of an excited state depends strongly on the molecules involved, their environment, and whether or not a transition is 'forbidden'. Since excited state lifetimes are sensitive to their surroundings, they are often used in research as a sensitive probe to study the interactions of molecules (see, for example, Fluorescence lifetime imaging).

To answer the second part of the question, the relaxation of a molecule from its excited state to a lower energy state is a stochastic process—like radioactive decay, one can make statements about the probability of a molecule returning to a lower-energy state, but one can't pin a specific time on a particular molecule.

The answer to questions 3, 4, and 5, are, again, 'it depends'. If the excited state has a long lifetime, then the molecule will be able to move about and tumble around between the time that it absorbs a photon and the time that it emits one. If a lot of tumbling occurs, it will appear that there is no relationship between the incoming and outgoing photons. However, if the excited state lifetime is short, or a molecule is bound so that it isn't free to rotate, the incoming photon and outgoing photon will tend to be polarized in the same plane. Measurements of fluorescence anisotropy – that is, measurements of how much of the original polarization is preserved – are used by chemists to determine how fast molecules are tumbling. TenOfAllTrades(talk) 15:31, 15 September 2007 (UTC)[reply]

Damn, you guys/gals are GOOD! Edison 00:10, 16 September 2007 (UTC)[reply]

In today's featured picture, Image:Deathvalleysky nps big.jpg, Why does the milky way appear to have a dark stripe? I thought it was comprised of lots of stars and should thus be a band of brightness. -- SGBailey 08:23, 15 September 2007 (UTC)[reply]

The Milky Way not only contains lots of stars and gas, but also a considerable amount of dust. This tends to obscure light in the human visual range, though telescopes sensitive to other wavelengths can often see through the dust clouds. --Pekaje 12:49, 15 September 2007 (UTC)[reply]

Much of the dust is in the form of carbon particles, similar to smoke, as well as other more rock like particles. Infrared can pass through dust more easily. Graeme Bartlett 22:21, 15 September 2007 (UTC)[reply]

Have a look at this by now classic picture. Observe how the center is obscured by dust in the visual-light (optical) image but we can see through to it in the infrared range. Because infrared light is blocked by the atmosphere such pictures have to be taken from space (here done with the COBE and the IRAS spacecraft. —Preceding unsigned comment added by Sanders muc (talk • contribs) 22:54, 15 September 2007 (UTC)[reply]

Thanks -- SGBailey 09:27, 17 September 2007 (UTC)[reply]

please can you anybody explain the meaning of organing cloths —Preceding unsigned comment added by 59.94.97.201 (talk) 10:06, 15 September 2007 (UTC)[reply]

Organic or organing? In what context? Do you mean organic clothes such as those made from cotton and hemp?--Shantavira|^{feed me} 11:28, 15 September 2007 (UTC)[reply]

I'm going to take a stab at this one. Assuming you're buying clothing and you see one specifically labelled organic cloth. It means it's made from cotton, hemp, silk, or other natural materials (like Shantavira said), and not materials like nylon or polyester. --Wirbelwindヴィルヴェルヴィント (talk) 17:19, 15 September 2007 (UTC)[reply]

I would take it to mean that the cotton or other natural fibre was produced using organic farming techniques, such as no pesticide, GMOs or artificial fertilizers. Graeme Bartlett 21:15, 15 September 2007 (UTC)[reply]

I'd also expect them to use natural dyes (if any), such as beet juice. StuRat 04:41, 18 September 2007 (UTC)[reply]

What are the commercially available 'non-botanically categorisable' deodorising Neptune Plants? Are they seaweeds, hydrozoan colonies, bryozoans, algae or artificial? 81.140.42.10 10:34, 15 September 2007 (UTC)Alan Lewis[reply]

Welcome to Wikipedia. You can easily look up this topic yourself. Please see neptune plant. For future questions, try using the search box at the top left of the screen. It's much quicker, and you will probably find a clearer answer. If you still don't understand, add a further question below by clicking the "edit" button to the right of your question title. --Shantavira|^{feed me} 14:14, 15 September 2007 (UTC)[reply]

1. If red wine vinegar were distilled, what would result? Would it become a sort of horrible brandy? If white wine vinegar were distilled? If rice vinegar were distilled?
2. Could alcohol be distilled using absorption by a sponge rather than condensation? Imagine a quantity of wine in a vessel, the top of which is blocked by a sponge. Heating the vessel to the boiling point of ethanol, would the ethanol vapors tend to be absorbed by the sponge, or would they simply evaporate through the sponge into the surrounding air?
3. What results from the distillation of beer? Anything worthwhile?
Any insights or comments would be appreciated, if possible in layman's terms.
-Kent, 129.174.54.141 14:53, 15 September 2007 (UTC)[reply]

1. In general, vinegars – including wine vinegars – contain little to no alcohol. The wine vinegars are produced in two fermentation steps. The first converts grape sugars to ethanol (ethyl alcohol) in the wine; the second fermentation then converts the ethanol to ethanoic acid (acetic acid). You could increase the concentration of acetic acid through distillation, but you wouldn't be able to recover much (if any) alcohol that way.

2. There are a number of different techniques that can be used to separate or enrich the amount of alcohol in solution. In principle, it would be possible to use a 'sponge' with very tiny pores that would allow water molecules through but not allow ethanol molecules to pass. This sort of idea is used in so-called molecular sieves.

3. It depends on the type of beer. Most beers don't do well with conventional heat distillation, as heat or bright light can cause chemical reactions that unfavourably change their flavour (see lightstruck beer). Other types of distillation can be used, however. Freeze distillation is regularly used to make ice beer, for instance. TenOfAllTrades(talk) 15:14, 15 September 2007 (UTC)[reply]

To expand on 3, the answer is "a spirit". Whether it's a worthwhile drinking spirit, or something you would use only for cleaning engine parts, depends on how you handle, distill, condition, and adulterate it. Remember that all {?) drinking spirits are made first by preparing a beer-like fermented drink, and then distilling that. That basic beer ("wash") could be filtered and clarified and you could drink it (I've drunk unfiltered, yeasty, wash at a distillery tour, and it tasted like a very rough dark beer). So if you distill proper beer you'll end up with something you could call vodka; if you distill it and condition it differently you could call the result whiskey. I'm not sure the molecules described the lightstruck article TenOfAllTrades pointed out will make it to the final spirit - the rigors of being boiled several times will destroy lots of molecules that would otherwise flavor the final spirit. Despite the hocus-pocus that fancy spirit producers would have you believe (whiskey distillery tours are full of claims of uniqueness that border on witchcraft) the output of the still is largely industrial-grade ethanol, flavored only by those few molecules hardy enough to survive all that boiling (the raw spirit from the Lagavulin Single Malt still takes like ethanol and smoke). All of the color, and most of the flavor, of spirit drinks like whiskey comes from the conditioning (that is, from sitting around in a sherry cask for a decade or so). —Preceding unsigned comment added by 86.131.223.148 (talk) 20:16, 15 September 2007 (UTC)[reply]

i am trying to write a book about this subject but cant find a city near the pipeline can soem1 help me?76.22.186.240 16:58, 15 September 2007 (UTC)[reply]

Rampart, Alaska looks to be the closest town, but perhaps is not to be counted as a city. Fairbanks is a city though. Did you look at Trans-Alaska Pipeline System Graeme Bartlett 21:55, 15 September 2007 (UTC)[reply]

When using a regular ruler with marked millimetres, I should be able to get two decimals places in respect to Significant Figures (1.22cm), this works out to 3 SigFigs. Assuming I am measuring two objects with the same ruler, one of which is 1.22 cm long and the other is 11.44 cm long. How should I write the one that is 11.44cm long? "11.44" has 4 Sigfigs, contrary to "1.22"'s 3 SigFigs. However, I am still able to get the same accurate reading in respect to the number of decimals. Should "11.44" be rounded to "11.4"? Thanks. Acceptable 20:29, 15 September 2007 (UTC)[reply]

Do you have a requirement to only use 3 figures? I would suggest that you do not round off your more accurate measurement unless you have a good reason to. The accuracy varies by a factor of 10 over the 3 significant figures range. 1.00 and 9.99 vary from 1% to 0.1% accuracy. And if you measured something smaller, say 0.96 cm you are suddenly down to 2 figures, but your measurement is still almost as accurate as your 1.22 cm measurement. Graeme Bartlett 21:28, 15 September 2007 (UTC)[reply]

I don't think significant figures really matter much standing by themselves (where there should be a ± standard deviation). Rather, I believe significant figures are most useful when doing arithmetic, where the question of "how many decimal places do I keep" always seems to come up. It's not really a measure of accuracy, but a systematic way to keep accuracy in check when doing math (ie measure two things with the ruler and divide them, and suddenly you have numbers down to the micrometer scale and beyond). See Significant Figures#Importance --Bennybp 15:47, 16 September 2007 (UTC)[reply]

If the ruler is marked in millimeters, the usual rule is that you can only measure the accuracy to the nearest millimeter, with an accuracy of ± 0.5 mm. Or in centimeters, that means to the nearest 0.1 cm, with an accuracy of ± 0.05 cm. You seem to think you can measure to 10X that accuracy. Even if you can eyeball the blank spots between the marks with that accuracy, there's no guarantee that the marks on the ruler are placed precisely enough for that.

Also, the number of significant figures on a measurement frequently is higher for larger measurements. StuRat 04:21, 18 September 2007 (UTC)[reply]

There are two sources of error in using your ruler. One is that you can only compare the length of your object to the marks on the ruler to a precision of 1mm. This limits the number of decimal places you can keep - but because the size of the error is not dependent on the size of the thing being measured, it doesn't mean you have to limit the number of significant digits. However, the length of your ruler depends on the temperature (since the ruler expands and contracts) - which introduces another error that is a percentage of the length of the thing you are measuring - so large objects would be measured with more millimeters of error than small objects. That kind of error limits the number of significant digits. Different causes of error result in different kinds of error. SteveBaker 14:26, 18 September 2007 (UTC)[reply]

Does reductio ad absurdum apply when discussing quantum mechanics?--88.110.108.191 21:17, 15 September 2007 (UTC)[reply]

It could be, reductio ad absurdum is a logic technique, so if you can make a statement about quantum mechanics, and then find it implies something untrue, or contradictory, then you know the original statement is false. However just because a statement sounds crazy, weird, or absurd, it is not the same as being false. The field of quantum theory has many strange conclusions, so you will have to to more careful not to let your common sense interfere! None of this would I count as scientific method though. Graeme Bartlett 21:47, 15 September 2007 (UTC)[reply]

I will add that reductio ad absurdum is always true, it may or may not be used, but if you claim it is untrue, all of maths and most science theory will be unavailable for your use! Graeme Bartlett 22:01, 15 September 2007 (UTC)[reply]

Depends on what flavor of QM you subscribe to. The famous Schrödinger's cat thought problem is a classic example of this; Schrödinger proposed it as an example of how ridiculous the Copenhagen interpretation of QM's approach to the world is, but it was taken up as a great example of how to view the world through a QM lens, and is now famous as a way of thinking about QM rather than an argument against its way of thinking.

So anyway. According to Schrödinger's point of view, yes, it applies. According to Bohr's, no, it doesn't, because you expect the quantum world to look absurd from a macroscopical logical point of view, and showing it to be so doesn't disprove a thing. --24.147.86.187 21:44, 15 September 2007 (UTC)[reply]

Reductio ad absurdum depends on the law of excluded middle - put simply, each statement is either true or false. In most moden interpetations of QM, this law does not always apply. In the Schrödinger's cat thought experiment, the statement "the cat is dead" is neither true nor false before we observe the cat. It is not just that we do not know whether or not the cat is dead - the cat really, truly, physically is in an indeterminate state - it is neither alive nor dead. In the double-slit experiment the statement "photon X passed through the right-hand slit" is neither true nor false unless we place a detector at one of the slits - in which case we destroy the interference pattern. See quantum logic for an attempt to develop a logic that is consistent with QM. Gandalf61 09:18, 16 September 2007 (UTC)[reply]

It is possible to use this argument. If you found some theoretical aspect of quantum mechanics implied that the colour of everything in the universe had to be pink - then you'd be able to say that that piece of theory must be false because the universe isn't entirely pink. The tricky part is that perhaps that bit is correct - but some other part of our scientific laws turns out to be wrong. So logical principles do not perhaps entirely hold. What logic could tell you would be something like "If all of our other laws are right and if this law is also right then the universe would be pink. Since the universe isn't pink and we still assume that all of other laws are correct - then this new law must be wrong." SteveBaker 14:21, 18 September 2007 (UTC)[reply]

I know that Glycogen, starch, and dextrin are polymer carbohydrates and glucose is an example of a monomer etc, but I have another question. Are monomer carbohydrates effectively monosaccharides (divided into pentoses and hextoses?) and polymer carbohydrates polysaccharides, but then where do disaccharides fall? Are there any fundamental differences in the function and actions of these different classes of carbohydrates?

Phew, lots of questions. Thanks, RHB - Talk 21:18, 15 September 2007 (UTC)[reply]

The monomers ( which are mono saccharides) dissolve easily in water. However the polymers (which are poly saccharides) are much less soluble. This means that the polymers are trapped inside a living cell, whereas the monomers, can travel in blood or sap and move through cell walls. Disaccharides are made from a pair of monosaccharides. They are soluble in water, but may have to be broken down to move through a cell membrane. Note that some oligosaccharides (containing a few nonosaccharides entities) are used to decorate some proteins glycoprotein and appear on the ouside of the cell membrane. These are important for recognising cells by antibodies etc. Graeme Bartlett 21:35, 15 September 2007 (UTC)[reply]

disaccharides are neither monosaccharides nor polysaccharides, they're just that: dimers, disaccharides. Bendž|Ť 08:30, 16 September 2007 (UTC)[reply]

Any ideas for generating a circular timebase synchronized to the input signal? I think I need a quadrature o/p sinewave vco, but having difficulty getting a design for it. —Preceding unsigned comment added by 88.109.30.31 (talk) 21:50, 15 September 2007 (UTC)[reply]

Normally you would be talking about a fixed frequency for a circular scan. All that you would do is trigger from the signal, or possibly use a phase locked loop to synchronise to the input frequency. You are right, you would need a sinewave with a 90° shifted wave (cosine wave!) to operate the X and Y inputs. Do you then plan to use your input signal to modulate the brightness on the display, or do you want to control the distance from the central point on the screen to generate a polar plot? Graeme Bartlett 22:08, 15 September 2007 (UTC)[reply]

Some displays will demodulate an input signal into I and Q components and then use these to drive the X and Y deflections on the oscilloscope. Graeme Bartlett 22:11, 15 September 2007 (UTC)[reply]

Yeah you're right. But I need a quad o/p osc whose frequency I can vary somehow to give me timebase expansion (like on a linear sweep scope). Yes the time base needs to be phase locked to the input signal-- so I need a VCO quad osc- right? I planned to have a pseudo polar display where the i/p signal just modulates the Y axis. This is far simpler (and looks better) than having the i/p signal modulate the trace amplitude radially.--88.109.30.31 22:27, 15 September 2007 (UTC)[reply]

One option is the Costas loop. It uses a VCO with a 90° shifter. It should be possible to approximate this with an analogue circuit across your frequency of interest. Another method is to use a frequency squarer (multiply the input signal by itself), this will double the frequency, and you can lock onto this with a 90 degree phase sift easily extracted. Graeme Bartlett 22:39, 15 September 2007 (UTC)[reply]

Use a two-channel arbitrary waveform generator programmed with a sine wave on one channel and a cosine wave on the other. You can then easily vary the frequency of the waveform that is output by the AWG.

Atlant 14:33, 17 September 2007 (UTC)[reply]

I dont have access to an AWG. What i think I need is a quadrature o/p sine wave VCO. I need sine waves (up to 4 MHz) to make the trace circular. The previous suggestion from Graeme only gives you a square wave o/p. I seem to remember a quad o/p sine oscillator made by Feedback Instruments but cant find the details now. Anyone know how those worked? —Preceding unsigned comment added by 88.111.135.162 (talk) 15:38, 17 September 2007 (UTC)[reply]

Okay, wouldn't the output of an integrator, by definition, be shifted 90 degrees from a sinusoidal input?

Atlant 18:22, 17 September 2007 (UTC)[reply]

You are correct that an integrator (or differentiator) would indeed produce an exact 90 deg phase shift for an input sinewave, but I would need to compensate for the 6 dB fall (rise) per octave somehow. This is not a trivial problem.(unless you or someone else knows different)--88.111.135.162 20:37, 17 September 2007 (UTC)[reply]

Well, designs exist for "levelling amps" (automatic gain control circuits). With modern ICs (such as DC-controlled "volume controls"), it shouldn't be too difficult if you're not too picky about the exact circularity of your trace.

Atlant 00:09, 18 September 2007 (UTC)[reply]

OK but my timebase range of 40 Hz to 4 MHz is about 16 octaves, giving a 96 dB range I have to level. I suppose I could change the gain of the integrator by hand. Also, if the integrator were part of a vco circuit, this may work. State variable oscillator? —Preceding unsigned comment added by 88.111.135.162 (talk) 03:20, 18 September 2007 (UTC)[reply]

That would be true if you were doing it "all as one range". But if your sweep is divided into (say) decades or the even-more-common 1, 2, 5 sequence, you can range-switch the values in the integrator as well, avoiding the need to handle the full 96 dB amplitude shift.

Meanwhile, I was wondering: you never really answered our question about how you intend to display the signal/data. Is it going to be a brightness modulation of the circular trace? Or are you going to try to use the signal data to deflect along the "radius" of the circular sweep? That is, create a deflection that is always at right angles to the instantaneous direction of the sweep? If the later, you have a pretty complex problem to create THAT deflection signal as well; you'll need some tricky analog math to keep the signal deflecting the beam in the right, constantly-changing direction.

If you're really trying to accomplish this (radial deflection from the circular trace), maybe analog techniques aren't your best bet. Maybe you should just let a computer digitize the input signal and do all the mathematical transforms in the digital domain, eventually displaying the result on its raster scanned screen? It might not be as "retro-cool" as a vector graphics display on a phosphor screen, but it would be LOT easier, and you could do things like "thermal scale" variable persistence (if appropriate for your application).

Atlant 12:31, 18 September 2007 (UTC)[reply]

Point one: agreed. 1,2,5 range would be practical.

Point 2:Yes I did answer it.

I planned to have a pseudo polar display where the i/p signal just modulates the Y axis. This is far simpler (and looks better) than having the i/p signal modulate the trace amplitude radially.

point 3:agreed radial modulation is difficult. see reply to point 2 above.

Thanks for your comments, but I think you'll agree its a fairly difficult problem if you want a display BW of even 10MHz. I'll persue the analog approach as its going to give me potentially faster display than digitising.--88.111.135.162 14:09, 18 September 2007 (UTC)[reply]

The kinds of circuit I suggested do produce sine waves. They are techniques to lock on a sinewave to a input signal, which for the suggestions I made would be a PSK or QAM modulated signal. If you have more arbitary inputs you may have to do other things. The VCO in the COSTAS loop produces a sine wave. The designs don't say how to build a 90° phase shifter, which is hard to do accross a broad frequency range Hilbert transform. The frequency squarer does not produce a square wave, it multiplied the input by itself, a sine times a sine is still a sine wave at twice the frequency. However if your input is a square wave, it will make a square wave. Graeme Bartlett 01:28, 22 September 2007 (UTC)[reply]

What is the cause of heart palpitations and missing beats? —Preceding unsigned comment added by 88.109.30.31 (talk) 23:14, 15 September 2007 (UTC)[reply]

Our article on palpitation has a great deal of information. TenOfAllTrades(talk) 23:34, 15 September 2007 (UTC)[reply]