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Hw2K-G1-BasicAtom - HW Set 2 KEY BASIC ATOMIC STRUCTURE CHM...

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HW Set 2 - KEY BASIC ATOMIC STRUCTURE CHM 2045 HW3K-G1.ATM Sub-Atomic Particles 1. The beams of electrons or cathode-rays from both tubes are identical despite the difference their cathode compositions. Since the same particles originate from different elements, the electrons could be assumed to part of the composition of all atoms. 2. a) Since the electron carries the smallest quantity of negative charge (by definition, unit negative charge or -1), the addition of an electron to a neutral body gives that body -1 charge. The addition of one electron to a neutral atom produces an ion with -1 charge: A + 1e - ! A -1 where A is any atom The addition of two electrons to a neutral atom yields an ion with a 2- charge, and so on. A + 2e - ! A 2- b) Loss of electron(s) by a neutral atom yields ions with unit positive charge(s). A ! A + + 1e - A ! A 2+ + 2e - 3. The droplets contain different charges because there may be 1, 2, 3 or more excess electrons on the droplet. The charge of one electron is likely to be the lowest common factor in all the observed charges (and the smallest difference in charge that may be calculated between any two droplets). Assuming this is so, we calculate the apparent electronic charge from each drop as follows: A 1.60 x 10 -19 coul/1e - = 1.60 x 10 -19 coul/e - B 3.15 x 10 -19 coul/2e - = 1.58 x 10 -19 coul/e - C 4.81 x 10 -19 coul/3e - = 1.60 x 10 -19 coul/e - D 6.31 x 10 -19 coul/4e - = 1.58 x 10 -19 coul/e - The average value from these four observations is 1.59 x 10 -19 coul/e - . 4. Mass of an electron: e = - 1.76 x 10 8 coul therefore: m g m = e = -1.60 x 10 -19 coul = 9.09 x 10 -28 g -1.76 x 10 8 coul -1.76 x 10 8 coul g g
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2 Mass of H + , e = +9.58 x 10 4 coul m g m = e = +1.60 x 10 -19 coul = 1.67 x 10 -24 g +9.58 x 10 4 coul +9.58 x 10 4 coul g g Electron mass as a percentage of H atom mass: % = mass e - x 100 = 9.09 x 10 -28 g x 100 = 0.0544% mass proton + mass e - 1.67 x 10 -24 g + 9.09 x 10 -28 g Note that this extremely small percentage would be much smaller for the other isotopes of hydrogen as well as atoms of other elements. All other elements have atomic nuclei with at least as many neutrons as protons. When computing the mass of an atom, ion, or molecule it will be practical to ignore the mass of the electrons. 5. Neutrons have no electrostatic charge and consequently would not generate the significant magnetic field observed for a charged particle in motion. The external magnetic field used to characterize moving charged particles such as electrons and protons would not have a major effect upon a path of a neutron beam. 6. a) Radius of a proton = d = 1.5 x 10 -13 cm = 0.75 x 10 -13 cm 2 2 Volume of a proton = 4 B r 3 = 4 (3.14)(0.75 x 10 -13 cm) 3 = 1.8 x 10 -39 cm 3 3 3 Density = mass = 1.67 x 10 -24 g = 9.3 x 10 14 g volume 1.8 x 10 -39 cm 3 cm 3 Stop and think about this for a moment. As noted, osmium is the densest element (even
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