write the electron configuration for the first excited state of the Na atom?
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Considering the conflicting answers, perhaps I should add something:\n\na sodium *atom* carries 11 electrons and has a neutral charge, therefore having a configuration of 2:8:1 as one person wrote -- 2 electrons in the first (ground) level, 8 electrons at the second level, and 1 (valence electron) in the third level.  As a configuration written 1s2 2s2 2p6 3s1 -- 2 electrons in the first (ground) level, 8 electrons total at the second level (2 in s, 6 in p), and 1 electron in the third level.  These are the two possible ways of writing the electron configuration -- remembering of course the superscript that we don't have here!  \n\nThe other answers that I saw were describing sodium as having 10 electrons.  These with 10 electrons are not sodium atoms but sodium *ions.*  If sodium becomes ionized it loses an electron and becomes positively charged -- in that case losing the valence electron in the third level.
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Ground State - 1s2 2s2 2p6 3s1 : 2,8,1\n\nElectron moves from the 2p orbital into the 3s orbital.\nExcited State - 1s2 2s2 2p5 3s2 : 2,7,2
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You need to know more than just the electron configuration.  There is contraction from left to right on the periodic table, plus additional periodic trends that must be considered.   \n\nEven more importantly, you need to define whether you are looking for the largest element IN ITS NATURAL  STATE (i.e. Cl2, Hg metal, etc) or AS THE ION.   Atomic and ionic radii differ.\n\nMore clarity is needed here.
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Look at your periodic table. \n\nHydrogen - 1s1\nHelium - 1s2\nLithium - 1s2 2s1\nGo to Neon - 1s2 2s2 2p6\nArgon - 1s2 2s2 2p6 3s2 3p6\nKrypton - 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6\n\nSo, let's say you need Silver (Ag)\n[Kr] 5s2 4d9\n\nIt's perfectly acceptable to use a noble gas in brackets to represent the configuration.\n\nI hope this give you an idea how to write it all.
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You should consult a chemistry textbook, if possible, or if not, look online for an answer; I'm sure there is information or a guide describing how to do it better than I can in a few short sentences. That said, the basic way to draw a molecule (often abbreviated mlcl) is to use each element's symbol(ie sodium=Na, carbon=C, etc) to denote an atom, and to draw a single, solid line for each bond between elements. This line/bond represents 2 electrons, aka one orbital. Most elements like to have 8 electrons in their valence (outermost) shells in order to be stable, but this doesn't always happen; there are many exceptions, hence my suggestion to look at a textbook or online text/guide.\n\nIt gets more complicated when you're dealing with things like organic chemistry or complexes or chelating agents where sharing of bonds/electrons is pretty common and there are more types of bonds. For a simple example, if you had a 6-carbon atom ring, each carbon has 6 electrons in its valence; it wants 8, so what happens, does it get frustrated with its misfortune and self destruct, destroy our little universe? Of course not... there are a total of (6)(6)=36 electrons, and we need (6)(8)=48. How is this possible? Clearly, 36 is not equal to 48. We can't create more. So what happens? As it turns out, carbon, along with most other elements, passed kindergarten with flying colors- they SHARE them, and since it's a symmetrical hexagon shape of identical atoms, they are shared equally. There are all kinds of crazy combos in which the electrons are shared unequally between atoms because of something called electronegativity, but our rules for drawing those lopsided things aren't very different. However, that's for another 'lesson;' don't worry about it for now.\n\nSo, in order to make the carbon happy, we have to share 12, or 6 pairs, of electrons. ([48 needed]-[36 that we have] = 12) \nSince this carbon ring has multiple resonance structures (atom config the same but electron pairs on different atoms), we draw six carbons in a hexagon, with one solid line bond between each, and a dotted circle in the middle- this accounts for all of its resonance structures and provides us with more info, in addition to only needing one drawing to display that info, than using single bond lines only.\n\nSo in conclusion, I've said way more than I really had to... from the level of generality of your question I assume that you don't know too much about chemistry; if you are confused with what I said, then do as I said- look up the info online or get a book. It's not too easy to understand right away but with some patience you should be able to figure it out just fine; good luck.
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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 .
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We have used short laser pulses to examine the process by which two ground state sodium atoms cooled to 0.5 mK in a magnetooptic trap absorb two photons tuned to the atomic 3S &#92;rightarrow 3P transition to produce a molecular ion by associative ionization. The 10 ps pulses are split into two pulses with the second pulse delayed up to T = 12 ns relative to the first. For T < 4 ns we see an enhancement of the ion signal, while for T > 4 ns the signal decays with a time constant of roughly 10 ns. The first laser pulse excites the atom pair at long range to an attractive R^-3 potential from the 3S + 3P asymptote. The second pulse promotes the pair to an attractive R^-5 potential from the 3P + 3P asymptote, where associative ionization can occur at short internuclear distances. We have simulated the behavior with a conceptually simple yet accurate model that predicts an enhancement of the flux of atoms caused by longer times spent on the intermediate potential.
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i presume you have the energy values for the different levels?\n\nif so you use the equation E=hc/lamda to find the wavelength for each energy transition in the series.\n\nThe red line will be about 575nm and the purple line will be about 375nm.
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B.the electron subshell
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google it and type 'periodic table, first twenty elements'\nn u'll get sumthing, far quicker than lettin sum1 write them up 4 u.
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Paint the protons one color, the neutrons another, and the electrons another. Put the nucleus together with the correct number of protons & neutrons. Put the correct # of electrons on it so that they appear to have some distance between them and the nucleus in an evenly spread manner. pretty simple. Styrofoam balls and toothpicks work great.
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Every atom has two unique numbers: Atomic Number (Z) and mass number (A). Atomic number Z represents the number of protons in the nucleus of an atom. Mass number A represents the number of neutrons and protons in the nucleus of an atom. When atoms are expressed on a periodic table, they are written with a number on the top and a number on the bottom. Z is written on the top and A is written on the bottom. Also when elements are represented with the name and a number on the upper lefthand side corner. This number represents the electrical charge of the atom. This is the number that tells us if the atom gained or gave away electrons.\n\nTo calculate the number of protons: Simply just look at the atomic number and it is the number of protons. For example Number of protons in a sodium atom is 11.(Z=11)\n\nTo calculate the number of electrons: Subtract electrical charge of the atom from the atomic number. For example Na+1 tells us that sodium atom gave an electron away. So 11-1=10 electrons present.\n\nBut the number of electrons is equal to the number of protons when the atom is not charged.\n\nTo calculate the number of neutrons: Subtract the atomic number from the mass number. For example, to find the number of neutrons in the nucleus of a sodium atom 23-11=12 neutrons.
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the amount of electrons in an atom.\n\nif you're going to use noble gas configuration, you would use the noble gas and then continue the configurations.\n\nex. i want the configuration of... flourine (F)\nfor noble gas configuration i would say: (He) 2s 2 2p 6\n\nthe parenthasese is actually a bracket, and the numbers that are spaced should be in the upper righthand corner of the letter before it.
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Here is what I got\nhttp://img.photobucket.com/albums/v628/kerahna/lewis.jpg\n\nPretend the Xs are all dots. I just draw Xs to keep track of which electron belongs to which atom. They are all the same.\nI haven't checked the formal charges yet, I'm going to do that now.\n\nEDIT: The formal charges are good.
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The general valence shell electronic configuration is\nns2 np3.\nAnd the elements along the group are :-\n1.Nitrogen (N)\n2.Phosphorous (P)\n3.Arsenic (As)\n4.Antimony (Sb)\n5.Bismuth (Bi)\n6.Ununpentium (Uup)\n\nCheck it out in various books above 10th grade
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As in sodium and potassium - Group I and magnesium and calcium - Group II?\nIf you look at the arrangement of the electrons that an atom of the above elements has:\nNa: 11, arranged as 2,8,1\nK: 19, arranged as 2,8,8,1\nBoth Na and K have 1 electron in their most outer shell. Hence they are in the 1st column, or Group I.\nMg: 12, arranged as 2,8,2\nCa: 20, arranged as 2,8,8,2\nBoth Mg and Ca have 2 electrons in their most outer shell. Hence they are in the 2nd column, or Group II.\n\nHope this helps :)
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The full electronic configuration of mercury is:\n1s2\n2s2 2p6\n3s2 3p6 3d10\n4s2 4p6 4d10 4f14\n5s2 5p6 5d10\n6s2