the interaction of laser beam with the energy levels of U
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A two-color three-photon ionization technique, using two pulsed dye lasers, was employed to explore the high-lying energy levels of U I. A number of new U I levels in the 34 290--35 120-cm/sup -1/ region was observed.^Uncertainty in the assignment of J values to some levels known from earlier studies was also reduced or removed.\n\nHUH? Is that what you want?\n\nEDIT\nAbstract. Two-colour resonance ionization spectroscopy was used for identification of 240 new levels of 238U in the energy range of 38 500 to 41 200 cm-1. A precision of +or-0.03 cm-1 (1 SD) for the energies was obtained by using laser-induced fluorescence signals of I2 for spectral calibration. The angular momentum quantum number, J, was uniquely assigned for 134 of these levels by using angular momentum selection rules. \n\nThats a little more specific\nAnd some more\n\nAbstract. Visible and infrared absorption measurements on the  ion in tetragonal zircon-type matrix  are reported and analysed in terms of the standard parametrization scheme. The observed 17 main peaks and a number of less intense lines have been assigned and fitted to most of the 32 allowed electric dipole transitions with the root mean square error equal to . The free-ion parameters obtained for the model Hamiltonian, , ,  and , as well as the corresponding crystal-field parameters, , , ,  and , agree fairly well with the initial theoretical estimations. The results are discussed in relation to the previous spectroscopic study on the scheelite-type matrix .
45381
E=hf \nwhere,\nh=6.63e-34 (planck's constant)\nE=(6.63e-34)(2.56e3)\n =1.7e-30\nCalculate for photon\n(1000j/sec)x(1 photon/1.70x-30)\n=5.88e32 photons/sec
107
Materials that form an inverse energy distribution?  Not quite.\n\nThere are three basic components for getting a laser to work.  First, one needs an excitation source (pumping source) such as a flashlamp, an electrical spark, etc.  Second, one needs a material (the gain medium) which can exhibit what's known as population inversion - the capacity for a material to raise the electron energy level from the lower to the upper energy states.  Nearly all materials will exhibit this behavior with some more efficiently than others.  Finally, one needs to direct the stimulated energy by using a resonator.\n\nThere are many combinations of pump sources, lasing materials and optical configurations depending on the intended use.  There are solid state (ruby, Nd:YAG), gas (nitrogen, helium-neon) and semiconductor (gallium-arsenide) lasers that come in a variety of wavelengths (colors) to choose from, again, depending on what you want to do with it.\n\nJust a little bit of background: My degree is in Physics, and I've been working in the lasers and electro-optics field for the past 10+ years.  You can see what my graduate lab at SDSU looked like at: http://www.equationlab.com/mtp/quantumblog/about.html
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Whether or not this is homework, you deserve an answer that lets you compute the energy of any wavelength photon. The formula is based on Planck's constant (usually represented as h), and says the energy of a single photon is equal to Planck's constant times the photon's frequency nu, or e = h*nu. Planck's constant h = 6.626*10^-34 Joule-second = 4.136*10^-15 electron volt-second.\nIf inputing wavelength lambda, first compute frequency in Hertz nu as c/lambda.\nThus for lambda = 515*10-9 m, nu=3*10^8/515*10-9 = 5.825*10^14 Hz. h*nu = 3.860*10^-19 Joule = 2.409 eV.\n(See ref.)\nP.S. I don't know where answer #1 came from. It says that about 12,000 photons would give you one Joule. Not likely.
657067
Yes! in the Sub Atomic particle. Photons behave the same way.
477990
put the counter closer cos alpha raysonly travel a short distance...maybe that will work?... \n\nor get something that will alter with radiation... i think there a material that changes colour when hit with radiation....\n\nOOOOOOOOOrr... get something that emits a longer range radiation ray when the alpha partiles hit it...that way u can get a good approx of the count
416739
http://en.wikipedia.org/wiki/Rydberg_constant\n\nTake a look at the formula. That should be enough to get you started.
184938
It depends on the wavelength of the light and its energy.
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Its not just for a microscopic particle. Its for any particle. The effect is only large for very small particles though.\n\nThink of it this way.\n\nIf you want to know precisely where a particle is you will have to use a very short wavelength of light to measure its position - you cannot be more accurate than the wavelength of the light. But very short wavelength light carries a lot of momentum. Each photon has momentum h/wavelength (h is Planks constant) - the shorter the wavelength the greater the momentum.\n\nBut if you hit the particle with a photon with a lot of momentum, it will accelerate it. As a result you will not be able to get an accurate measurement for its momentum. To measure its momentum accurately you would have to use a very long wavelength photon.\n\nThese are competing requirements.\n\nThe Hesienberg relationship quantifies these competing requirements by saying that the error in position multiplied by the error in momentum cannot be lower than hbar (Planks constant over 2 pi).\n\nPlanks constant is very tiny - around 6.62 e-34 Js. So it only matter for tiny particles.
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http://www.orau.org/ptp/museumdirectory.htm\n\ngo to this link.  it has one of the finest museums on-line dealing with the detection of radiation...and other neat stuff
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Sanity check. Ultraviolet light is energetic light, which will break bonds, which is why it can be dangerous. UV light is shorter than about 400 nm. So after throwing around the big exponents, we want an answer less than 400 nm, but probably larger than 100 because then it's x-rays, which is usually overkill for breaking molecules apart. \n941 kJ / mol means 941,000 joules are sufficient to break \n6.02x10^23 molecular bonds. ( 1 mol is Avagadro's number)\ndivide it out and get how much to break just one N2 bond:\n941,000 J / 6.02x10^23 molecules = 1.56 x 10^-18 joules\nWhat wavelength of light corresponds to that amount of energy being packed into one photon? \nE = hc / lambda, where h = 6.63x10^-34 m2 kg/s (Planck), c = 2.228x10^8 m/s (speed of light), and lambda is wavelength in meters (staying in mks units, E will be in joules). Solve for lambda, where E is 1.56 x 10^-18. \nlambda = hc/E = 1.274 x 10^-7 meters\nmultiply by 10^9 to get nm, then lambda = 127.4 nm\nLooks good to me...  There might be a simpler cookbook equation to get this answer, but it's good to know how to do it from the basic constants of nature. \n\nNote: previous answer left out factor of Avagadro's number and did no sanity check...
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lithium-6 + neutron ------> helium-4 +  hydrogen-3\n\nC
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y dont u do ur homework urself .ok heres abrief intro.\n\n1.see the level of frequency changes when beam of ultra violet light rays enter to one medium from another medium(to glass from air).\n\n2.when light undergoes refraction it frequency changes its speed acc. to the medium such as(glass,air and water)\n\n3.yes there will be,as light enters to one medium from another medium.
364383
This is a very simplistic description of a complex process.\n\nWell because He, Ar ,Ne have more electrons, protons and so are different to H in that the interactions between the electrons themselves and between the electrons and their nuclei are different.\n\nClearly if there are more electrons then there are more  to be able to jump into higher energy levels and then return via a number of different ways (single or multiple steps). So the absorption/emission spectra would reflect the variations in these elements. Since the steps require different energy to hydrogen owing to differences in orbital sizes and electron numbers then the energy quanta required/emitted will also be different and so different lines on the spectrum.
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so you plot the proton number on the y-axis, and the different melting points of the elements across the x- axis.. actually are sure it has to be a bar graph.. i think a line/scatter graph would be more accurate maybe?
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You can ask atomic research agency. they are using this type of software.
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Light spectra is essentially due to the electronic levels, not the nucleus (the nucleus emission/absorption would be at much higher frequency, in the gamma range); and there are no notable difference between the electronic energy level due to the extra neutron deuterium has in its nucleus; therefore very little difference in the visible band between protium and deuterium.