The Periodic Table of the Elements PHYS-107 Fall 2004 - Dr. Philip Edward Kaldon1234567891011121314151617181 2H HeHydrogen Helium1.00794 4.002634 5678910Li Be B C N O F NeLithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon6.941 9.01218 10.811 12.011 14.0067 15.9994 18.9984 20.179711 12 13 14 15 16 17 18Na Mg Al Si P S Cl ArSodium Magnesium Aluminum Silicon Phosphorus Sulfur Chlorine Argon22.9898 24.305 26.9815 28.0855 30.9738 32.066 35.4527 39.94819 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br KrPotassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton39.0983 40.078 44.9559 47.867 50.9415 51.9961 54.9381 55.845 58.9332 58.6934 63.546 65.39 69.723 72.61 74.9216 78.96 79.904 83.837 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I XeRubidium Strontium Yttrium Zirconium NiobiumMolybdenumTechnetium Ruthenium Rhodium Palladium Silver Cadmium Indium Tin Antimony Tellurium Iodine Xenon85.4678 87.62 88.9059 91.224 92.9064 95.94 97.9072 101.07 102.906 106.42 107.868 112.411 114.818 118.71 121.76 127.6 126.904 131.2955 56 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86Cs Ba Lu Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At RnCesium Barium Lutetium Hafnium Tantalum Tungsten Rhenium Osmium Iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Astatine Radon132.9054 137.327 174.967 178.49 180.9479 183.84 186.207 190.23 192.217 195.08 196.9665 200.59 204.3833 207.2 208.9804 208.9824 209.9871 222.017687 88 103 104 105 106 107 108 109 110 111 112‡114 116 118Fr Ra LrRf †Db Sg Bh Hs Mt Uun Uuu Uud Uuq Uuh UuoFrancium Radium LawrenciumRutherfordiumDubnium Seaborgium Bohrium Hassium Meitnerium UnunniliumUnununoium-? Ununduoium-? Ununquadium UnunhexiumUnunoctium223.02 226.025 262.11 261.11 262.114 263.118 262.12 265 266 269 272 277 “289” 289 29357 58 59 60 61 62 63 64 65 66 67 68 69 70 71La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb LuLanthanum CeriumPraseodymiumNeodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium88.9059 140.115 140.908 144.24 144.913 150.36 151.965 157.25 158.925 162.5 164.93 167.26 168.934 173.04 174.96789 90 91 92 93 94 95 96 97 98 99 100 101 102 103Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No LrActinium ThoriumProtactiniumUranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium FermiumMendeleviumNobelium Lawrencium227.028 232.038 231.036 238.029 237.048 244.064 243.061 247.07 247.07 251.08 252.083 257.095 258.1 259.101 262.11Revised: 31 August 2002 ‡ New reports (1997-99) of elements 112, 114 (one event), 116 and 118 offer hope for an island of stability and trans-300. “Scientists discover 2 ‘superheavy’ elements.” CHE, June 18, 1999, p. A18. But… in 2002, there are now reports that some (not all) of these events could be the result of scientific fraud. “Atomic Lies.” CHE, August 16, 2002, p. A16. Investigations continue.† IUPAC’s revised slated of element names and abbreviations as of early 1997. “New compromise offered on heavy-element names.” C&EN, Feb. 24, 1997, p. 12. Names guessed for 111-118.How to find a radius of an electron:1.) Start with a nucleus with Z protons. A singleelectron in orbit around this nucleus undergoes UniformCircular Motion, with the Coulomb Force as thecentripetal force.FkZermvrE==2222.) Simplifying, we get:kZermv22=orvkZemr22=Note that r and v are now locked together.3.) deBroglie said that matter has waves:l ==hphmv4.) The Bohr atom quantizes the states – in this case, wecan think of the electron deBroglie matter wave as astanding wave:2plrn=5.) Solve (4) for r and substitute (3) for l, and we seethat the angular momentum (L = mvr) is quantized.mvr nhn==2ph6.) Square (5) and then substitute v2 from (2):mvr n222 22= hmkZermrn22222= hmkZe r n222= hrnmkZe=222h7.) Note that the mass here is the mass of theelectron, (me = 9.11 × 10-31 kg), so that the onlyvariables in this equation are n and Z.rnmkZenZmkenZanee==FHGIKJ=22222220hh8.) The stuff in the parentheses represents theradius of the n = 1 orbit of hydrogen (Z = 1):amkeJskg CmeNmC02234231 9 19 21010538 10911 10 9 10 1602 100528 1022===´×´´ ´=´--×--h.( . )( )( . ).ch9.) The Three-Body Problem says that we can’tdo this same kind of analysis with any atomsthat have more than one electron. But… wecan create a hydrogenic (means hydrogen-like)ion for all other elements by stripping off allbut one of the electrons, and putting in thecorrect Z for the nuclear charge.How to find the Energy of an electron:10.) The kinetic energy is ½ mv². We can findthis using (2) and (7):1222222222422222422292 1943134222182222 2 29 10 1602 10 9 11 102 10538 10218 10 13622mvkZerkZe m kZenkZemnZnkemZnCkgJsZnJZneVeneeeNmC== = =FHGIKJ=´´´´×FHGGIKJJ=´=×----hh h(). ....chchch(since 1 electron volt = 1 eV = 1.602 × 10-19 J).11.) The total energy of the electron is the sum of theP.E. plus K.E. terms. The P.E. for Coulomb’s Law is:PE W FdkZern=- =- =-2Or P.E. = -2K.E. This form has the advantage that theP.E. is zero at infinity, and is always negative. The totalenergy, En = K.E. + P.E. = K.E. - 2K.E. = - K.E. Thuswe can write the total energy as a function of n and Z as:EZnJZneVn=- ´ =--221822218 10 136..So for Hydrogen (Z=1), the first three levels are:nrnEn1 0.528 Å -13.6 eV2 2.11 Å -3.40 eV3 4.75 Å -1.51 eV(1 Å = 1 × 10-10 m)How to find a transition photon:12.) In hydrogen, for an n=3 electron to drop to then=2 state, this is:DE = 3.4 eV - 1.51 eV = 1.89 eV = 3.03 × 10-19 J = h f.fEhJJsHzcfHzmnmms==´´×=´==´´=´ =---303 106 626 10457 102 998 10457 10656 10 6561934148147......lThis is red light.NOTE: Don’t confuse the wavelength l here in (12)with the deBroglie wavelength l in