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Lecture 12sf - Structure of the Atom Cathode ray tubes(late...

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Structure of the Atom Cathode ray tubes (late 19th century) had a low pressure gas and two electrodes under high voltage. A beam of light emanated from the negative electrode (cathode). This beam of light was deflected by both a magnetic and electric field, and the direction of the deflection showed that it consisted of negatively charged particles. These particles had the same properties regardless of the gas in the tube. J.J. Thomson made a quantitative measurement of the deflection of the beam when put through electric and magnetic fields. Using equations of electricity and magnetism, he was able to calculate the charge to mass ratio of these particles. = -1.758 x 10 8 Your textbook reports this as the reverse ratio of mass to charge e m = -5.686 x 10 -12 coulomb kg e represents the charge of the electron m represents the mass of the electron. About 10 years later, Robert Millikan performed the “oil drop” experiment which determined the charge of the electron. Millikan irradiated small sprayed oil drops with X-rays, which gave them a negative charge. These drops were then suspended between two charged plates. By varying the charge on the plates, he could just suspend the oil drop, meaning the electrical force up (the negatively charged drop is attracted to the upper positive plate) is equal to the gravitational force down.
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Knowing the voltage on the plates, and through certain other measurements, Millikan could calculate the charge (in coulombs) on each oil drop. Doing this experiment many times, he came up with different charges on each drop. Some typical numbers are: 3.2 x 10 -19 coulombs 6.4 x 10 -19 coulombs 4.8 x 10 -19 coulombs 9.6 x 10 -19 coulombs 8.0 x 10 -19 coulombs 11.2 x 10 -19 coulombs This was done literally hundreds of time. Analysis shows that every measurement is an integer multiple of 1.6 x 10 -19 coulombs. That is, each droplet of oil must have an integer number of excess electrons: 2 electrons, 4 electrons, 3 electrons, etc, and the charge of each electron is this common multiple: 1.6 x 10 -19 coulomb. Modern measurements show that e = 1.602 x 10 -19 coulombs Combining this with J.J. Thomson’s measurement: = -1.758 x 10 8 m = mass of electron = 9.11 x 10 -28 grams
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Rutherford Scattering Experiment This is a classic experiment described in almost all general chemistry textbooks. Radium, a source of alpha particles (helium nuclei of mass 4 units), is aimed at thin gold foil. The alpha particles are probes? What happens to them when they encounter gold atoms? The result was unexpected. Almost all the alpha particles went through undeflected, as if nothing was in their path. However, about 1 in 20,000 alpha particles were deflected at very large angles, sometimes coming back in the direction from which they were emitted.
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