MSJ2e_Ch02_ISM - Chapter 2 Atoms and Elements 23 Chapter 2...

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Chapter 2: Atoms and Elements 23 Chapter 2: Atoms and Elements Solutions for Chapter 2 Questions for Review and Thought Review Questions 1. The coulomb (C) is the fundamental unit of electrical charge. 2. Millikan devised an experiment to determine the charge of an electron. It consisted of a chamber with electrically charged plates on the top and bottom. (Figure 2.2). Tiny oil droplets were sprayed into the chamber. As the droplets settled slowly through the chamber, they were exposed to x-rays. This caused electrons from gas molecules in the air to be transferred to the oil droplets. Using a small telescope to observe the tiny droplets, he then adjusted the electrical charge on the plates so that the upward pull of the electrostatic charges just balanced the downward motion of the droplet due to gravity. From this, he was able to calculate the charge on the droplets. Different droplets had different charges, but these charges were all multiples of the same small number, which he decided must represent the smallest fundamental negative charge: the charge of one electron. With the mass-to-charge ratio determined by Thomson, and the charge of the electron, the mass of the electron could also be calculated. 3. (a) The proton is about 1800 times heavier than the electron. (b) The charge on the proton has the opposite sign of the charge of the electron. 4. (a) The surprising results in Rutherford’s gold foil experiment were that some of the alpha particles were deflected backwards. Rutherford compared this surprise to how you’d feel if you shot a cannon at a piece of paper and had the paper deflect the cannon ball back at you! (b) Rutherford calculated that the nucleus is about 10,000 times smaller than the atom. 5. In a neutral atom, the number of protons is equal to the number of electrons. Units and Unit Conversions 6 . Define the problem : If the nucleus were scaled to a diameter of 4 cm, determine the diameter of the atom. Develop a plan : Find the accepted relationship between the size of the nucleus and the size of the atom. Use size relationships to get the diameter of the “artificially large” atom. Execute the plan : The atom is about 10,000 times bigger than the nucleus. 10,000 × 4 cm = 40,000 cm Check your answer: A much larger nucleus means a much larger atom. This large atomic diameter result looks right. 7. Define the problem : A piece of paper is exactly 11 cm high. Use conversion factors to change the units to centimeters, millimeters, and meters. Develop a plan : Use the exact relationship between inches and centimeters to convert between inches and cm. Use metric relationship between cm and m, and m and mm to convert cm into mm. Use the metric relationship between cm and m to convert cm into m. Execute the plan : 11 in × 2.54 cm 1in = 27.94 cm
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Chapter 2: Atoms and Elements 24 27.94 cm × 1m 100 cm × 1000 mm = 279.4 mm 27.94 cm × 100 cm = 0.2794 m Check your answers: The number of centimeters should be larger than the number of inches, since an inch is larger than a centimeter. The number of millimeters should be larger than the number of centimeters, since a millimeter is smaller than a centimeter.
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MSJ2e_Ch02_ISM - Chapter 2 Atoms and Elements 23 Chapter 2...

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