A technician using a scanner to monitor the uptake of radioactive iodine in a patient s thyroid. P.48

Transcription

1 A technician using a scanner to monitor the uptake of radioactive iodine in a patient s thyroid. P.48

2 Rodioactivity In the late nineteenth century scientists discovered that certain elements produce high-energy radiation. Studies in the early twentieth century demonstrated three types of radioactive emission: gamma ( ) rays, beta ( ) particles, and alpha ( ) particles. A ray is high-energy light ; a particle is a highspeed electron; and an particle has a 2 charge, that is, a charge twice that of the electron and with the opposite sign. P.48

3 The Nuclear Atom In 1911 Ernest Rutherford (Fig. 2.11), who performed many of the pioneering experiments to explore radioactivity, carried out an experiment to test Thomson s plum pudding model. The experiment involved direction a particles at a thin sheet of metal foil, as illustrated in Fig P.48

4 Figure 2.11 P.48

5 Figure 2.11 Ernest Rutherford ( ) was born on a farm in New Zealand. In 1895 he placed second in a scholarship competition to attend Cambridge University but was awarded the scholarship when the winner decided to stay home and get married. As a scientist in England, Rutherford did much of the early work on characterizing radioactivity. He named the and particles and the ray and coined the term half-life to describe an important attribute of radioactive elements. His experiments on the behavior of particles striking thin metal foils led him to postulate the nuclear atom. He also invented the name proton for the nucleus of the hydrogen atom. He received the Nobel Prize in chemistry in P.48

6 Figure 2.12 Rutherford s experiment on -particle bombardment of metal foil. P.49

7 Rutherford reasoned that if Thomson s model were accurate, the massive a particles should crash through the thin foil like cannonballs through gauze, as shown in Fig. 2.13(a). Although most of the particles passed straight through, many of the particles were deflected at large angles, as shown in Fig. 3.13(b), and some were reflected, never hitting the detector. P.48

8 Figure 2.13 (a) The expected results of the metal foil experiment if Thomson s model were correct. (b) Actual results. P.49

9 In Rutherford s mind these results could be explained only in terms of a nuclear atom an atom with a dense center of positive charge (the nucleus) with electrons moving around the nucleus at a distance that is large relative to the nuclear radius. P.49

10 2.5 The Modern View of Atomic Structure: An Introduction The simplest view of the atom is that it consists of a tiny nucleus (with a diameter of about cm) and electrons that move about the nucleus at an average distance of about 10 8 cm from it (see Fig. 2.14). The nucleus is assumed to contain protons, which have a positive charge equal in magnitude to the electron s negative charge, and neutrons, which have virtually the same mass as a proton but no charge. The masses and charges of the electron, proton, and neutron are shown in Table 2.1. P.49

11 Figure 2.14 A nuclear atom viewed in cross section. Note that this drawing is not to scale. P.50

12 TABLE 2.1 The Mass and Charge of the Electron, Proton, and Neutron P.50

13 An important question to consider at this point is, If all atoms are composed of these same components, why do different atoms have different chemical properties? In Fig.2.15, these two atoms are isotopes, or atoms with the same number of protons but different numbers of neutrons. P.50

14 Figure 2.15 Two isotopes of sodium. Both have 11 protons and 11 electrons, but they differ in the number of neutrons in their nuclei. P.50

15 Note that the symbol for one particular type of sodium atom is written Mass number Atomic number Na Element symbol where the atomic number Z (number of protons) is written as a subscript, and the mass number A (the total number of protons and neutrons) is written as a superscript. (The particular atom represented here is called sodium twenty-three. It has 11 electrons, 11 protons, and 12 neutrons.) P.50

16 If the atomic nucleus were the size of this ball bearing, a typical atom would be the size of this stadium. P.50

17 Sample Exercise 2.2 Writing the Symbols for Atoms Write the symbol for the atom that has an atomic number of 9 and a mass number of 19. How many electrons and how many neutrons does this atom have? Solution The atomic number 9 means the atom has 9 protons. This element is called fluorine, symbolized by F. The atom is represented as P.52

18 Sample Exercise 2.2 and is called fluorine nineteen. Since the atom has 9 protons, it also must have 9 electrons to achieve electrical neutrality. The mass number gives the total number of protons and neutrons, which means that this atom has 10 neutrons. See Exercises 2.43 though 2.46 P.52

19 Law of Conversation of Mass *The total mass of substances does not change during a chemical reaction. Lavoisier stated the law of mass conservation on the basis of his combustion experiment. 2Hg + O 2 2 HgO (the mass before the reaction = the mass after the reaction) 180g glucose + 192g oxygen gas 264 g carbon dioxide + 108g water (372 g material before change 372 g material after change) Mass conservation means that, based on all chemical experience, matter can not created or destroyed.

20 Applying the Law of Conservation of Mass. A g sample of magnesium is allowed to burn in g of oxygen gas. The sole product is magnesium oxide. After the reaction, no magnesium remains and the mass of unreacted oxygen is g. What mass of magnesium oxide is produced? Mass before reaction = g Mg g O 2 = g Mass after reaction =? g MgO g O 2 = g? g MgO = g g = g Law of Conservation of Mass The total mass of substance present after a chemical reaction is the same as the total mass of a substance before the reaction.

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22 Law of Constant (Definite) Composition No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass. Percent by mass (mass percent, mass%) is the fraction by mass expressed as percentage. mass of element = mass of compound x part by mass of element one part by mass of compound Sample A Composition Sample B g g g H % H = g H g O % O = g O

23 Using the Law of Constant Composition. A g sample of magnesium, when combined with oxygen, yields g of magnesium oxide. A second magnesium sample with a mass g is also combined with oxygen. What mass of magnesium oxide is produced from this second sample? From the first experiment, we can establish the proportion of magnesium in magnesium oxide g magnesium g magnesium oxide This proportion must be the same in all samples g magnesium x = g magnesium oxide g magnesium oxide 0.1g magnesium

24 Law of Multiple Proportions If two elements form more than two compounds, the masses of one element combined with a fixed mass of the second are in the ratio of small whole numbers. Ex. CO and CO 2 In one oxide, 1.00 g of carbon is combined with g of oxygen, and the other, with g of oxygen =

25 A CO B CO 2 A H 2 O B H 2 O 2 A N 2 O B NO 2 A 量 12 : 16 = 1 : 4/3 2 : 16 = 1 : 8 28 : 16 = 1 : 4/7 B 量 12 : 32 = 1 : 8/3 2 : 32 = 1: : 32 = 1 : 16/7 BA 例 8/3 : 4/3 = 2 : 1 16 : 8 = 2 : 1 16/7 : 4/7 = 4 : 1

26 Please complete the following figure according to the law of constant composition (definite proportions). What is the formula for (1)~(4) and the mass for X, Y and Z. oxygen 8 g (2) nitrogen X g (4) Si Z g CO 2 (1) NH 3 SiC carbon Y g (3) hydrogen 1 g