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2 Atoms, Molecules, and Ions Solutions to Exercises 2.15 (a) Most of the volume of an atom is empty space in which electrons move. Most alpha particles passed through this space. The path of the massive alpha particle would not be significantly altered by interaction with a "puny" electron. (b) Most of the mass of an atom is contained in a very small, dense area called the nucleus. The few alpha particles that hit the massive, positively charged gold nuclei were strongly repelled and essentially deflected back in the direction they came from. (c) The Be nuclei have a much smaller volume and positive charge than the Au nuclei; the charge repulsion between the alpha particles and the Be nuclei will be less, and there will be fewer direct hits because the Be nuclei have an even smaller volume than the Au nuclei. Fewer alpha particles will be scattered in general and fewer will be strongly back scattered. 2.16 (a) The droplets carry different total charges because there may be 1, 2, 3, or more electrons on the droplet. (b) The electronic charge is likely to be the lowest common factor in all the observed charges. (c) Assuming this is so, we calculate the apparent electronic charge from each drop as follows: A: 1.60 10⁻¹⁹ / = 1.60 C B: 3.15 10⁻¹ / 2 = 1.58 C C: 4.81 10⁻¹ / 3 = 1.60 C D: 6.31 10⁻¹⁹ / 4 = 1.58 C The reported value is the average of these four values. Since each calculated charge has three significant figures, the average will also have three significant figures. (1.60 10⁻¹⁹ C + 1.58 10⁻¹⁹ C + 1.60 C + 1.58 X C) / 4 = 1.59 10⁻¹⁹ C Modern View of Atomic Structure; Atomic Weights (sections 2.3-2.4) 2.17 (a) 1.35 Å 1 10⁻¹⁰ 1 Å m X 1 1nm 10⁻⁹ m = 0.135 nm 1.35 Å 1 10⁻¹⁰ 1Å m 1 1 10⁻¹² pm m = 1.35 X 10² or 135 pm = 100 pm) (b) Aligned Au atoms have diameters touching. d = 2r = 2(1.35 Å) = 2.70 Å 1.0 mm 1000 1m mm 1 10⁻¹⁰ Å m 1 Au 2.70 atom Å = 3.70 10⁶ Au atoms (c) V = =4/3πr³. r = 1.35 Å 1 10⁻¹⁰ 1Å m 100 cm = 1.35 10⁻⁸ cm m V = cm³ = 1.03 10⁻²³ cm³ 21