What is the maximum # of emission lines possible for a hydrogen atom in the n=5 state?
919471
Three Pfund series transitions from n=5 state (to n=1, n=2 and n=3)\n1<--5 @ 95 nm\n2<--5 @ 434 nm\n3<--5 @ 1282 nm
199003
a proton has a weight of approximately 1836.15 electrons
362329
Consider that one electron will repel another electron (-ve does not attract -ve), each bond between a carbon and hydrogen is cased by a pair of electrons (2 electrons) so you have a total of eight electrons... this results in neutralising the charge on all the atoms.\n\nCarbon has 2s1, 2s2, 2p2 electrons which means it is theoretically 4 electrons short of its closest nobel gas (nobel gases are neutral).\nHydrogen has 1s1 electrons which means it is theoretically 1 electron short. Note that there is no p orbital in first row of periodic table.\n\nAs such each hydrogen shares its sole electron with the carbon, donating a total of 4 electrons, hence 4 hydrogens.\nThe Hydrogens get their electrons from the carbon, the carbon has 2s2 and 2p2 electrons, totalling 4 to share, 1 to each hydrogen... we do not count the s1 orbital because it is unlikely to operate in forming bonding.\n\nNow each pair repels eachother so they are furthest apart, if we do the mathematics...\nblah blah...\nWe end up with an optimum angle of 107.5 degrees between eah hydrogen, any more than that and a different hydrogen will repel it back, and likewise if less. this forms a 'tetrahedral' shape.\n\nIf you were to switch a hydrogen with another element... say chlorine, to produce CH3Cl (Chloro-methane) then the hydrogens (being the smaller atoms with less electrons) will be pushed further from the chlorine which will tend to determine the molecule's appearance. Ch3Cl is not a perfect tetrahedral as its angles will not equal 107.5, but we still call it that for simplicity's sake.\n\nI recommend that for further information do a search on the following phrases:\n\nVSEPR\nValence Shell Electron Pair Repulsion theory\nMolecular geometry\nMethane\nOrganic Chemistry
838195
Good question - it depends on which theory you want to use. Currently, it is estimated that there are 5 hydrogen atoms per cubic metre. But this does not necessarily mean that there is "nothing" between these atoms.\n\nAccording to quantum physics, even the space that we most often referred to a a vacuum, may be composed of various exotic unobservable (for now) components.\n\nIn quantum physics a vacuum is not the same thing as "nothing." One theory supposes that all space is filled with the Higgs Ocean composed of an undetected Higgs field.\n\nThe current idea in quantum physics is that the only place where there is "nothing" is between the strings making up all matter and energy. The concept of "nothingness" is a complex and extensive subject and very difficult for us to visualize.\n\nOf course there is much more to these concepts than a few words in a brief paragraph can relay.\n\nGood Luck
357280
There must be a maximum. As an object gets hotter and hotter the molecules vibrate faster and faster. They clearly cannot exceed the speed of light so there must be a maximum.
798748
Bohr's theory worked great to explain the spectrum of hydrogen, but it doesn't work for anything with more than a single electron.  The previous answer is correct, also...
13297
Atoms have a nucleus ( composed of protons and neutrons) and a cloud of electrons. The number of the different components varies between different elements.\n\nElectrons can only have specific energy levels in an atom - that is the energy is not continious but quantized. This is different from, say, the solar system, because a planet could theoretically be in any orbit. But we know from quantum mechanics (and many, many experiments) that electons can only occupy certain "orbits."\n\nSometimes an electron can be bumped into a higher orbit (the higher orbits have less energy, because they are not as tightly bound) or kicked out of the atom altogether. If a second electron moves into the spot it will release energy as it falls into that level.\n\nHydrogen is the easiest element to study. The nucleus is a single proton, and it can have only one electron. The different energy levels can be illustrated like this:\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c2\n(though in reality electrons don't travel on nice orbits)\n\nIf an electron in the second lowest level does a "quantum leap" to the very lowest, it will emit energy - light - at a very specific frequency. In this case 656 nanometers, which is red. If, on the other hand, an electron falls from the 3rd level down to the 1st level the emitted light is blue.\n\nA "spectrum" will show all of these distinct levels, with the brighter ones being the most common. This shows the spectrum of hydrogen with a diagram of the quantum leaps:\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c4\n\nIn other atoms it becomes more complex, not every jump is allowed, and in the precense of a magnetic field a "fine structure" is seen in the spectrum, as some of the main lines look double.\n\nThe result of all of this is that each different atom will give a different characteristic spectrum. So, you can use this to identify different elements in anything from crime scene samples to distant galaxies.
312552
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 .
774231
an octet configuration with eight valence electrons each (on each atom) ?
101595
it is a series of lines, due primarily to the presence of hydrogen and helium and to lesser extent carbon, nitrogen and a few others. the most important lines show up in the visible part of the EM spectrum.
979155
alright. Electron configurations. I had the exact same problem three weeks ago. I was SO confused. But i got it in the end.\n\nLet me explain.\nThere are two types of electron configurations. One is the simple electron shell configuration. Let's say 2-8-16 and the other is called spdf notation like 1s2 2s2(i can' t type the two after the s like an exponent so bear with me)\n\nNow the first one is simple. Shell notation. An atoms outer shells are divided into shells that each have a maximum limit for the number of electrons that they can contain. the first can hold 2, the second 8 and so on. They all follow a rule. It is always 2 time the shell number squared. lets say level 1. Square it. You get 1. Multiply it by 2. You get 2. Therefore 2 is the maximum # that level one can hold. You said 15. So fill each level so that each contains the max. It would be 2-8-5. (The last one is five so that the all ad up to 15). Got it?\n\nNext, we have spdf notation. This takes a lot of time to explain so i'll just give you a site to go to. \n\nhttp://www.fordhamprep.org/gcurran/sho/sho/lessons/lesson36.htm\n\nGood luck!
16762
The atomic number is the number of protons.
1090457
a the number of outer electron that will bond with other atoms to fill thei valence shells
1234
n=3 means that there are s, p, and d shells involved.  The s shell can accomidate 2 electrons, the p, 6 and the d, 10, so a total of 18 electrons can be accommodated, so that's 9 pairs of electrons.
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1312 kJ is for ONE MOLE of hydrogen atoms to be ionized.  You'd have to divide by Avogadro's number to find kJ per atom.\n\nIf you want an explanation instead of just an answer, here it is:\n\nThe energy of a Bohr orbit is given by the equation:\n\nE = (-hcRH)/n^2, where h=Planck's constant, c = speed of light, RH = Rydberg constant for H.  The term "-hcRH" is -2.18E-21 kJ.\n\nFor an atom of hydrogen to be ionized, its electron has to go from n=1 (the lowest energy orbit) to n=infinity (the boundary of the atom).  Plugging in n=1, the energy becomes -2.18E-21 kJ.  Plugging in n=infinity, the energy goes to zero.  So the energy needed for an electron to travel from n=1 to n=infinity is E(final) - E(initial), which is 0 - (-2.18E-21 kJ) = 2.18E-21 kJ.  This is the ionization energy for one atom.
360598
N has atomic number 7- so electronic configuration is 2,5 whish emans its has 5 valence electrons in the L shell
641972
Photons.\n\nThere are about 10^9 photons for every nucleon in the universe.