iN WHAT Portion of th electromagnetic spectrum would you expect to find these transitions? Explain please. If you do i will vote u best answer.
207481
they are within the visble range;\nThe series of lines in the hydrogen spectrum associated with transitions to or from the first energy level or ground state. It lies in the ultraviolet (Lyman-alpha at 1215.67 Å; Lyman-beta at 1026 Å) with a series limit is at 912 Å. The He II Lyman lines have almost exactly one-quarter the wavelength of their hydrogen equivalent (He II Lyman-beta, 303.78 Å; He II Lyman limit, 227 Å.)
406118
All kinds of electromagnetic radiation reflect as light.\nSo do microwaves.\n\nTh
442169
You can use the Rydberg formula, which relates the reciprocal of the wavelength of the photon absorbed or emitted to the squares of the reciprocals of the principal quantum numbers.\n\n1/wavelength = R(1/n_l^2 - 1/n_u^2)\n\nSo, for the listed values of n, \n\nA.  1/w = R(1/ 2^2 - 1/4^2) = 0.1875R\nB.  1/w = -0.75R\nC.  1/w = 0.21R\nD.  1/w = -0.04861R\n\nSince energy and frequency are proportional to 1/wavelength,\nlooks like the order is D, A, C, B.
360131
1. light quantum\n\n2. the visible spectrum is very small in comparison to the rest of the spectrum (see first link)\n\n3. visible light.
304528
Maybe this explanation can help? http://en.wikipedia.org/wiki/Spectral_density
371019
Sorry, you're going to have to be a bit more specific than that. Electromagnetics is a rather broad subject.\n\n~X~
533960
The visible light region is from 0.7 µm to 0.4µm of the spectrum. The color of an object that we see is the color or wavelength of light reflected from the visible spectrum. All other colors or wavelength are absorbed. the following chemicals are different in terms of wavelength because of their different nature's regarding reflection of particular wavelength or the absorption of other.
375970
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!
119843
Not alot actually. Photons bounce around like crazy. When white light strikes a surface some of it is absorbed by the molecules in it. The rest bounces off. When that light hits your eye you see the light whithout the wavelength the surface absorbed giving you the perception of colour. If all light was stopped by matter it would be pitchblack all around you. \n\nNo surface reflects 100% of the infalling light however. Some is always absorbed by atoms and molecules (that become excited ot more energetic) no mather how reflective the surface or transparent the substance. Water may appear completely transparent but only a few hundred meters beneath the surface of the ocean it is quite dark because the water has absorbed the sunlight.\n\nLight ends when a photon is unable to move. A photon is composed of an electric wave and a magnetic wave. Pure energy that is. The higher the frequency of the photon the less likely it is to be stopped by an atom (atoms are composed mainly of empty space). So gamma rays (the photons with the highest energy) are able to pass through matter. But most photons will be stopped by an atom eventually and when that happens the energy of the atom is taken up by the electron in it. The electron moves into a higher orbit: the atom has become excited. Usually, when the electron moves back into its ordinary orbit, the energy is rereleased as heat (low energy infrared photons). Sometimes the energy is rereleased as a photon of a different colour (frequency). The substance glows.
618411
Shorter wavelength=higher frequency.  I know radio waves have  longest wavelength so must be lowest frequency.  and I know from longest to shortest wavelength it is radio...IR...ROYGBlueIV..Ultra so I would guess ultraviolet has highest frequency.\nSince all electrmagnetic radiation on the spectrum moves (through a vacuum) at the speed of light, if the wavelength is longer, fewer of those wavelengths are going to pass a certain point, thus a lower frequency (or Hertz I believe is the term).  The shorter the wavelength, the more wavelengths are going to pass a certain point in one second so their freq is higher.
290858
gamma rays\nvisible light\ninfrared light\nmicrowaves
487084
The correct answer is 3.\nThe 1st spectral line is caused by the electron falling from state 2 to state1.  The 2nd line is caused by a shift from 3 to 2, and the third, a shift from 3 to 1.
36741
I didn't give it a thumbs down, and I think the answer is fine as far as it goes.  However, one problem that I noticed is that you have a circular definition.  You said electromagnetic waves are oscillating electric and magnetic waves.  I assume you meant to say electromagnetic waves are oscillating electric and magnetic *fields*, which is correct.
514856
Gamma rays have the smallest wavelengths and the most energy on the electromagnetic spectrum. There isn't much physical difference between gamma rays and x-rays besides the fact that the gamma rays vibrate faster. They measure 10 to the negative power of 11 all the way to 10 to the negative power of 14 (meters). I hope I've answered your question. My knowledge of the electromagnetic spectrum only extends so far.
382103
No they cant be seen .. just the visible radiations can be seen .. as the figure in this link http://en.wikipedia.org/wiki/Image:Electromagnetic-Spectrum.png shows the visible area and that the radio waves are out of that area ..
265213
All natural light sources generate a "spectrum" of frequencies that approaches the "black body" radiation curve. The peak of that curve denotes the temperature of the emitting object. Certain frequencies can be affected in transit via interaction with materials as in emission and reflection nebula (ie. various places within the Orion constellation). The sun emits UV radiation commensurate with its black body ideal.\n\nUNLESS someone can show that the moon absorbes ALL UV frequencies on reflection then by definition:\n\nMOONLIGHT DOES INDEED CONTAIN UV FREQUENCIES\n\n===============================================\nSome of you people amaze me, you answer questions that you know NOTHING about. Answers like "No because it is reflected light" are a case in point. Is your blue shirt invisible in the mirror because the light coming from your shirt is lost as it reflects off of the mirror. I THINK NOT! Reflected light IS LIGHT. Some of it is absorbed, reflected, refracted but it is light. YES; the UV portion in moonlight is very weak, hey guess what -ALL THE LIGHT frequencies in moonlight are weak; THATS WHY MOONLIGHT IS DIM.\n==============================\nUV is NOT produced only from objects above 24000 degrees. It is produced at ALL temperatures, but the lower the temperature the lower the amount of UV that is produced. PLEASE Google "black body radiation"
274275
Yes.\nReality, as we accept, exists within a perceptible spectrum/frequency range. We can measure many other wavelengths that go unnoticed by humans. In a way, it's like insects that see in the UV or IR wavelengths. Their "reality", in this sense, is much unlike our own.\nScience has only identified X% of spectrums and, considering the numerous anomalies that occur on a regular basis, there is no telling what happens or exists in these undiscovered frequencies.\nThen again, I'm just a computer scientist, not a physicist.