How does the length of a visible light wave compare with the typical spacing of atoms in a solid?
442820
The wavelength of visible light is approximately 400 to 800 nm.  In a crystal, such as table salt (NaCl) the spacing between the Cl atoms is 0.563 nm, with the Na atoms half way in between.  The numbers are similar for other solids.  So, atoms are spaced about 500 times the wavelength of light, but that is a rough number.
43048
Mostly, it's a matter of the form of bonds.  Large electronic orbitals and conjugated orbitals are more likely to absorb light, while short, stiff bonds are more likely to let it pass.  Heavy metals have big orbitals, and characteristically absorb light, while shrot bonds, like C-H or O-H or N-C or whatever are more likely to let light pass.  But as stated above, even with this there can be issues where opacity occurs due to light being scattered by refraction at many pahse changes (like in ice that's filled with small bubbles).
745572
Wow, I have a test on this tomorrow =D\n\nWell, ground state is basically saying that it is a nuetral atom.  Are you stating for the location of the electron, so for Calcium -\nStarting from the beginning...\n\n1s^2\n2s^22p^2p^2p^2\n3s^23p^2p^2p^2\n4s^2\nOr the Abbreviated [Ar] 4s^2\n\nEdit -  I guess I should clarify what each symbol means...\nThe first quantum number is the energy level, going from 1-7\nThe second quantum number, or in this case, letter, states the type of orbit that the electron follows, the letters are s p d f\nThe third quantum number is an extension of number 2, basically, there is always s (1), ppp (3) , ddddd (5), ffffffffffffff (14), and refers to the different orientations of that orbit, so for f, there are 14 different ways it can be presented around the nucleus,\nThe Fourth quantum number states how many electrons are present in the orbit. So the 4th is the 2 in 1s^2, saying the level is full, for there can only be two electrons occupying each energy orbit.\n\nFor absorption, an electron enters the excited state, where it jumps an energy level, when it recieves energy.  This is absorption. For emission, it occures after absorption, in which an electron leaves the excited state, through emitting a photon, a kind of energy, thus why it is emission.  It then drops an energy level, eventually to the ground state.\n\nFor the light spectrum, the black lines show the photon signature that the electron releases.  When it releases the energy in the form of a photon, it can be seen as light on the spectrum, and if the wave length of the emission is correct, it can be visable. The black lines just indicate the wave length of the emitted photon. For example, when electron in potassium leave the excited state while burning, they give off a purplish pinkish light. That's their signature.\n\nFor the excited state, that's explained in emission and absorption.\n\nThe flaw in Bohr's model? Well, electron do not travel in set paths.  Bohr's model illustrated exact paths according to energy levels. the flaw was that it was only possible in the Hydrogen atom. In other atoms, the electrons' negative charges oppose each other and throw the other out.  It only works in hydrogen because there only is one electron.  In truth, electron travel a more random path within their energy levels, more sporadic, but in a basic area, or orbit.  This pattern is called the electron cloud, because of the uncertainty on where an electron will be at any given time.\n\nWoo, thanks for the warm-up on my test tomorrow. Good luck!
64215
almost certainly it would be in the visible region of the spectrum. after all it does have that pretty color!\nvisible region is roughly between 400-700 nanometers wavelength.
617450
Some molecules absorb visible light, e.g. due to conjugation and pi to pi antibonding transitions, with the highest intensity at a particular wavelength.  One can prepare standard solutions, with accurately devised known concentrations, the measured intensities of these solutions are correlated with these standard solutions, covering a particular dynamic range.  Assuming Beer's Law, the plot will resemble a linear curve, and a least fit regression line and its equation can be obtained; also by correcting for the blank if this is has not bee accounted for by the spectrometer.  The unknown pure solution, or a matrix derived alternative method if the solution is not pure that there are significant impurities in the unknown solution, will also be measured at this wavelength and it's intensity, often the y value, will be plugged into the regression equation to estimate the concentration.
219596
visible light has wavelengths from 400 to 700 nm. Frequency is inversely proportional to wavelength. The longer the wavelength (lower frequency), up to 700nm, the closer the color will be to red. As wavelength decreases, the color will shift through orange, yellow, green, blue, to violet at 400 nm. See; the color spectrum (more info in linked wikipedia article)
1090689
The empty space is not so empty.\nBillions of atoms and much more photons;\n1000 dm3 and 1,000,000 cm3.
258719
The speed of light in a vacuum is the same for all colors, and this includes all electromagnetic waves outside the human visible spectrum, such as ultraviolet, microwaves, infrared, gamma rays, x-rays, and so on. \n\nIn materials, however, the speed of light is not only slower than in a vacuum, it's also different for each color.  This is why we have dispersion in transparent materials which is responsible for phenomena like chromatic abberation in lenses, rainbows formed by droplets of rain, and splitting of light into a spectrum by a prism.  If the speed of light was slower in a material than in a vacuum, but the same for all frequencies (colors) of light, we'd have refraction, but no dispersion.   \n\nAnd, while nothing can have a velocity higher than that of light in a vacuum, things can go faster than light in a transparent medium.  This causes the electromagnetic equivalent of a "sonic boom", which is called Cherenkov radiation.  \n\nAs for a light year being a "perfect unit of measure", well, it has a rigorous definition developed from basic units of nature on which everyone can agree.  But, so does the meter, or the second, and almost everything other than the kilogram.  \n\nHowever, if you were to send a beam of light into space, you'd find that depending on the frequency it may travel somewhat less than a standard "light year" in one year's time.  This is because space isn't actually a perfect vacuum, and so it does actually have an index of refraction. For many wavelengths, it will be a full light year distance traveled, but some frequencies will travel slightly slower due to the tiny amount of matter, averaging one hydrogen atom per cubic meter, that is present even in deep space.
83782
as u know atoms are microspical, normally the distance btween bonded ions are microscopical too!! infact the space between 2 ions are sometimes lesser than the diamiter of the atom itself!\n\nhere's a demonstration for u to understand this phinomina more clearly; take about 10 pieces of paper and pin tiny holes randomly on each paper. now keep each piece of paper on top of the other; eventually u will not see the otherside of the paper through it (or the visibility is very low than a single holy piece of paper), even though each single paper had singnificant nmber of holes to see through!\n\nA thing is formed of many million layers of this ions. so even though it has microbial 'holes' the loght cant go through & since these holes are microbial, the thing still feels solid because there are more dencity of ion than space!\n\nhope u understood wut i try to mean LOL
166583
That's light in the visible spectrum. I'd imagine a great deal of compounds could, especially since "great absorbance" means nothing, in scientific terms.
1113461
It's 0.00002777778 inches, or approximately 705.55 nanometers.  It's an inch divided into 36000 parts, so it is pretty small.\n\nI believe the wavelength of red light falls in a range of 630 to 760 nanometers or approximately 1/36000 of an inch.
752656
The size of an object you can resolve with an optical system has a theoretical limit given by the wavelenght of the light you're using divided by the diameter of the optics multiplied by 1.22. If you plug the numbers in you find for a 1 foot diameter optical system at visible light (~400 nm) at an altitude of 100km, the smallest thing you could see is on the order of a foot, give or take (rough calculation). You could definitely not read a postage stamp or a wrist watch. Maybe a license plate with some bigger optics, but this is ignoring all sorts of atmospheric scintillation which screws things up.
457019
Also depends on your scale. At microscopic scales, a strings length is fractal, and thereby infinite.\n\nJust think of measuring a coastline length at a microscopic scale.
753625
the internuclear distance between the centre of an atom to another
52947
The narrower the slit, the wider the length between peaks in the diffraction pattern.
727800
Those are states of matter (but there's also plasma).  Light isn't matter, it's a wave, like sound.
960070
Two dimensions should be explanation enough. Assume the de Broglie wavelength of an electron is some distance in nanometers. If we draw the nucleus as a dot and draw the electron as a continuous wave around the nucleus, the wave has to be the length of some multiple of its wavelength. If the wave were not the length of a discrete multiple of its wavelength it would not be continuous because a full circle's rotation would cause the wave to jump. If we constrain the atomic model to behave like this, then the answer is simple. \n\nConsider the electron wave that is four wavelengths long. This wave is a function which describes electron density. The next energy level has to correspond to a wave that is five wavelengths long. In this way, the wave cannot increase gradually, but rather the levels are split into discrete shells, one for each number of wavelengths in the wave.