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Lecture 49/7/05 Homework concerns Atomic Structure
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Cathode Ray Tubes Glass tube with most of the air removed and 2 electrodes Cathode ray goes between electrodes with applied voltage Run in straight lines Cause gases to glow Can heat metal Can be deflected by a magnetic field Attracted toward positively charged plates Gives off light when they strike a fluorescent screen
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JJ Thomson (1897) Determined the charge to mass ratio for electrons 5.60 x 10 -9 g/Coulombs Used Cathode Ray Tube with both magnetic and electrical field
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Robert Millikan (1911-1913) Measured the charge on an electron Oil drop experiment Found each to have a charge as a multiple of 1.60 x 10 -19 C Current value: -1.602176 x 10 -19 C Used to find mass of an electron 9.109382 x 10 -28 g
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Fig. 2-2, p.42
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Electrons (cont.) Later showed that cathode rays (electrons) were the same as β particles
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Canal Rays Eugene Goldstein (1886) First evidence of fundamental positive particle Different gases gave different charge to mass ratios Ernest Rutherford called them ‘protons’
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Neutrons James Chadwick (student of Rutherford) Observed radiation released when particles from radioactive Polonium struck Beryllium 1932 No charge Slightly heavier than proton
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Plum-pudding model JJ Thomson Atom composed of a uniform sphere of positively charged matter with electrons embedded in the sphere
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Rutherford’s Gold foil experiment
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New model Most of the mass and all of the positive charge in the center
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ATOMIC COMPOSITION Protons positive electrical charge mass = 1.672623 x 10 -24 g relative mass = 1.007 atomic mass units (amu) Electrons negative electrical charge relative mass = 0.0005 amu Neutrons no electrical charge mass = 1.009 amu
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Atomic Number (z) Number of protons in nucleus Unique for every element
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Mass Number, A C atom with 6 protons and 6 neutrons is the mass standard = 12 atomic mass units Mass Number (A)= # protons + # neutrons A = 5 p + 5 n = 10 amu A = 5 p + 5 n = 10 amu
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ISOTOPES Atoms of the same element (same Z) but different mass number (A).
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Isotopes Because of the existence of isotopes, the mass of a collection of atoms has an average value. Average mass = ATOMIC WEIGHT Boron is 20% 10 B and 80% 11 B. That is, 11 B is 80 percent abundant on earth. For boron atomic weight = 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu
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Isotopes & Atomic Weight 6 Li = 7.5% abundant and 7 Li = 92.5% Atomic weight of Li = ______________ 28 Si = 92.23%, 29 Si = 4.67%, 30 Si = 3.10% Atomic weight of Si = ______________
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