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Electrons Cathode Ray Tube Experiments Millikan oil-drop experiment Davy, Faraday, Stoney, Thomson PowerPoint Presentation
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Electrons Cathode Ray Tube Experiments Millikan oil-drop experiment Davy, Faraday, Stoney, Thomson

Electrons Cathode Ray Tube Experiments Millikan oil-drop experiment Davy, Faraday, Stoney, Thomson

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Electrons Cathode Ray Tube Experiments Millikan oil-drop experiment Davy, Faraday, Stoney, Thomson

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  1. Fundamental Particles of Matter • Electrons • Cathode Ray Tube Experiments • Millikan oil-drop experiment • Davy, Faraday, Stoney, Thomson • Protons • Cathode Ray Tube Experiments • Rutheford scattering experiment • Neutrons • -Bombardment experiments • Chadwick

  2. Cathode rays • Travel in straight lines • Negatively charged • Have mass • Cathode rays = Electron beam • Thomson determined the e/m ratio by studying the deflection of the beam in electric and magnetic fields

  3. Millikan Drop analysis • Determined the charge (e) on an electron from the experiment • e/m = 1.75882*108 C/g from the cathode ray exp. • m = 1.75882*108 C/g * e = 9.10940*10-28 g

  4. Original “Plum Pudding” Model • Protons • Cathode Ray Tube Experiments • Atom  cation+ + e- • Canal Rays = Positive (cation) rays • H2 2H+ + 2e- • Determined the e/m of a proton • Rutheford Scattering Experiment • Nucleus New Model

  5. Atomic Number • Moseley • Bombarded Elements with high energy e- • Produced different wavelengths for different element • Each element differs from the preceding element by having one more proton • An atom is defined by the # of protons

  6. Isotopes - Atoms of the same element (have the same # of protons) that have different masses (have differing number of neutrons) • Mass number A = Z (#p) + #n Nuclide Symbol E A Z 11 5 10 5 #n=6 #n=5 B B

  7. Mass Spectrometry - measures the charge-to-mass (e/m) ratio of charged particles. If we put Neon in a mass spectrometer we could determine how many natural isotopes there are for neon and the percent abundance of each isotope 20Ne 90.48% 21Ne 0.27% 22Ne 9.25% atomic mass of Ne = 19.99244*0.9048 + 20.99384*0.0027 + 20.99384*0.925 atomic mass Ne = 20.1979 amu