The excitation of atoms

shiningknight · Feb 2, 2005

Hello people of Earth... I have but a simple question that nobody seems to be able to answer...and I'm getting a tad upset about it. Here's my question(and hopefully this is the right place to post it!): I understand that if you apply a lot of energy to an atom, it will leave the ground state energy level and will then go to an excited level. I also understand that the level of excitation depends on the amount of energy that is applied to the atom via heat, light, or electricity. Therefore my question is, could one say that the greater the amount of energy introduced/received by an atom then, the faster it will enter into the excited level? Also....When an atom becomes excited, it's electron "jumps" into a higher-energy orbital. And when it makes the transition to a lower energy orbital from that of a higher one it emits a light photon(emission). If the time it takes for an atom to become excited is decreased, due to the larger amount of energy being introduced to it, wouldn't that mean that the emission speed of that atom's electron would increase in proportion to the increased amount of energy applied? So thats my question(s). Hopefully I made enough sense for someone to give me a nice and detailed answer. Thanks. Untill then... -ShiningKnight-

newtonian · Feb 2, 2005

ShiningKnight - Not my field but I will research it and get back to you. I do know that if an atom receives considerable energy it can lose an electron or gain one and thus become ionized. And that most of the IGM (=intergalactic medium, the thin density space between galaxies) gas clouds are ionized. This has to do with the emission of photons (also caused by energy interactions as per your post) - the spectral analysis of the wavelengths of these emissions prove that these clouds are less than one part in one million neutral! However, your question has to do with how much time it takes for an atom to increase its energy state so as to cause its electrons (I assume you want per electron or per shell) to jump to a higher energy shell? I don't even know at what speed any electron changes location from one shell to another, let alone the change in speed due to higher energy as a cause. So, for me, that will take some research time. Perhaps someone else knows - otherwise, please be patient.

newtonian · Feb 2, 2005

ShiningKnight- Still researching, but I have a clue. Certain metals carry an electron current better than other materials because they are better conductors of mobile electron flow, called electric current. The reason for this is that these metals do not bind the electrons in their outermost shell very strongly. That is, they become free fairly easily and therefore can easily be induced to become mobile electrons. This is a clue that the answer to your question will be different depending on which atom you are referring to, as the energy required to remove an outer electron from a conducting metal like copper or aluminum will be less than the energy required to remove an outer electron from a non-conducting atom or compound. Still researching.....

nacnud · Feb 2, 2005

I would explain but I think that wikipedia has better descriptions of these sorts of events. Atomic orbitals Fluorescence

newtonian · Feb 2, 2005

Another clue to complexity: electrons are not in a specific position, but have a certain probability of being in a specific region or volume. When I was in high school (many years ago) physics and chemistry were being revised on this issue - from specific planet-like orbits to regions. The older version produced equations formulated by Danish physicist Niels Bohr in 1913 involving quantized atom structure. He defined electron shells in the following way mathematically: Bohr postulated that the variant angular momentums of electrons in different orbital shells "could only take those values equal to Planck's constant multiplied by an integer (a whole number) and divided by another constant (2 times pi, the familiar geometrical constant)."- "The World of Science," Volume 13 - "Physics in Everyday Life," 1991, page 90. Of course, the angular momentum would depend on the radius of the electron from the nucleus, and hence which orbital shell it revolved in. Bohr's predicted values worked well for the simple hydrogen atom which has only one electron. It does not work well with more complex atoms. To understand more complex atoms, one needs to come to grips with the wave characteristics of sub-atomic particles - with the primary point being the difference in behavior of a wave and a particle. Er - I see your question could have a very complex answer. Still researching.....

newtonian · Feb 2, 2005

nacnud - Thank you for the informative links. My time is limited on this. Shining Knight - Note nacnud's links which add detail to my following posted research: So I will simplify and use the averages, or mean values. The Orbital shells have been worked out using wave theory, specifically wave equations by Erwin Shrodinger. Physicists (chemists?) simplify matters using averages based on the simple hydrogen atom, and specifically hydrogen photon (light) emission spectrums, notably the following spectral lines in the hydrogen spectrum: Sharp (s) Principle (p) diffuse (d) fundamental (f) These spectral lines of hydrogen correspond to these 4 basic types of electron orbitals: s, p, d, f types. S type orbitals are the simplest, as they are spherical. Atoms can simply have a number of s orbitals with greater radius and energy levels, or atoms can be much more complex with the inclusion of these other types of orbitals. P type orbitals have twin lobes - as usual exact electron position is a matter of probability. There are 3 p orbitals for hydrogen atoms, mutually perpendicular to each other in 3 dimensions, lying along the x,y and z axis (axes? plural) as cartesian coordinates. D type orbitals are still more complex, there are 5 of them, and these consist of specific directional lobes around the nucleus of the atom. ?There are seven f orbitals with shapes that are very hard to represent graphically.? - Ibid, page 68. Now, to see how your question is very complex if you want a precise answer: ?Although the Shrodinger wave equation for the sole electron in a hydrogen atom can be solved fairly easily, it is less simple for atoms with more than a few electrons, even using the largest computers.? - Ibid, p. 68 Still researching..... Meanwhile, since we have much better computers than available in 1991 - can anyone post an updat

newtonian · Feb 2, 2005

ShiningKnight - A possible simple answer: If quantum theory is correct in these matters, and the resulting wave mechanics are correct (I am a doubting Thomas when it comes to popular theories): The SPEED of electrons from one orbital shell to another may be quantized, and the increased energy input may simply increase the RATE of electrons moving towards higher energy shells rather than increasing the VELOCITY of the electrons involvled. That is an educated guess. However, I am not sure - since the positions of electrons are not precise but a matter of probability, and also the state of any specific electron is not accurately represented as a particle, but rather as a wave function. In fact, increased energy of an electron would logically effect the frequency of the wave of an electron. Still, quantum theory would state that an electron must have one of a specific quanta of energy or angular momentum- and thus my above educated guess. It will be a long time before I can post again on this thread (perhaps a few days) - hopefully others will give better responses. NOtE: there is an additional complicating factor: spin and the resulting magnetic fields produced. Electrons also have a quantized spin, either 1/2 + or 1/2 -. This cause double emission lines in very precise spectrographs involving photon emission - i.e. spectrum of light. It is called "fine structure." The corresponding effect of spins of neutrons and protons, with the resulting magnetic fields from spinning protons, on spectra is called "hyperfine structure"- again dual lines corresponding to the dual possible spins, but 1,000 times smaller than the fine structure effect caused by electron spin. Obviously, magnetic fields of protons and electrons will effect their motion - I don't know how such variation can be forced into quantum numbers as respects speed of electrons from shell to she

nexium · Feb 2, 2005

What you typed is inline with what has been observed, except I don't think the emmission speed is changed by the input energy, nor the excitation speed. There are likely at least minor errors in current main stream beliefs about such matters. Neil

Saiph · Feb 2, 2005

1) More energy doesn't necessarily mean higher excitation levels, or faster excitation. It all depends on the wavelength of the individual photons. Transitions require a very specific wavelength of light to occur. Any more, or any less, and it doesn't happen. No matter how many photons you pack in there. Small lie actually, if you have 2 photons with half the energy, in there at the exact right moment, it'll happen, same with other ratio's and ammounts. But photons are small, this doesn't happen often enough under any reasonable conditions. 2) Now, assuming you have the right wavelength of light: More intensity does mean more frequent excitation events. This makes some sense. Think of a game of bumper cars. If you only have two cars in the ring, and they drive around randomly, they won't hit often (excitation will be rare). As you put more in, the impact rate goes up, (the rate of excitation increases). 3) De-excitation is a statistical occurance. Various states of excitation have various half-lives. The duration of the state is basically independent of the surroundings (unless the surroundings change the properties of the state, by lowering or raising the energy, then you've got a different level...). Having more photons around, isn't going to increase or decrease the rate of excitation. 4) The actual speed of excitation and de-excitation (the time for the electron to start and finisht the transition, not the number of times per minute it wants to transition) is also independent of the surroundings. 5) There is a way to "pre-maturely" de-excite an atom, and that's stimulated emission. If you get a photon of the appropriate wavelength (usually the same that would be emitted by the transition) to pass by, it sets up a harmonic with the electron, and cuases the electron to transistion, giving off a photon with the same properties, (phase, momentum, direction even). This is the property behind lasers btw. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

shiningknight · Feb 2, 2005

I thank all of you that have responded. From what I'm able to pick up, the energy applied and the amount of time leading to excitation of the atom are not in proportion of each other. Right? I mean, wouldn't it make sense that the more energy applied to the atom, the shorter the amount of time needed to excite it? Then again, Newtonian's educated guess makes just as much sense. Is there any possible way to decrease the amount of time it takes for any particular atom to become excited? Once again, I thank those of you who have responded to my question/thread. Untill then... -ShiningKnight-

newtonian · Feb 2, 2005

ShiningKnight - You are welcome. I will bow to more educated ones like Saiph for now as I will be very busy for at least two days. I will be thinking about it, though. I will follow this thread and post again later. My thoughts off the top of my head would be to look for influence specifically from the following sources: Magnetic influence. Spin influence, as from antimatter (opposite spin) particles. Increased temperature and atomic vibrations. The matching of nuclear vibrational resonances like the exact triple matching of Helium, Berrylium and carbon in stars that makes carbon synthesis in stars possible. The input of various forms of energy, notably with wavelengths similar to the wavelength of the electrons involved - or various particle wavelength equivalents. For the latter, note particle wave theories as well as actual observations in atomic accelerator experiments. Extreme shock waves such as from supernovae, again with particular wavelengths. Finding out the exact cause of the extreme heat in the IGM, as well as why it has increased in recent billions of years. Etc.

Saiph · Feb 3, 2005

I can't think of anything that makes the excitation go faster. I can get you excitations more often...but each excitation takes the same amount of time. Analogy time: Take a bullet, and place it and other identicle ones in a revolver. The firing mechanism you use gives you a certain rate of fire. Now, put the same bullets in a faster device, namely an assault rifle, or other automatice weapon (heck, even a machine gun). And in each case, you get a different rate of fire. However, the time required for the bullet to fire (the charge ignites, accelerate the bullet, etc) is not altered by the firing mechanism. The time involved in firing is a property of the bullet. All the firing mechanism can do, is initiate the firing process in at a more rapid rate. Excitation is the same thing. I can think of ways to atoms to become excited "more often", but I cannot think of any ways to shorten the time required to finish the excitation process. Now, de-excitation on the other hand, you can prematurely end, via stimulated emission. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"This is my Timey Wimey Detector. Goes "bing" when there's stuff. It also fries eggs at 30 paces, wether you want it to or not actually. I've learned to stay away from hens: It's not pretty when they blow" -- The Tenth Doctor, "Blink" </div>

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