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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website CHAPTER 23 Nuclear Chemistry INTRODUCTION NUCLEAR IN ATOMIC NUCLEI. THIS BRANCH OF CHEMISTRY BEGAN WITH THE DIS- COVERY OF NATURAL RADIOACTIVITY BY 23.2 NUCLEAR STABILITY ANTOINE BECQUEREL AND GREW AS A RESULT OF SUBSEQUENT INVESTIGATIONS BY CURIE 23.1 THE NATURE OF NUCLEAR REACTIONS CHEMISTRY IS THE STUDY OF REACTIONS INVOLVING CHANGES PIERRE AND 23.3 NATURAL RADIOACTIVITY MARIE 23.4 NUCLEAR TRANSMUTATION AND MANY OTHERS. NUCLEAR 23.5 NUCLEAR FISSION CHEMISTRY IS VERY MUCH IN THE NEWS TODAY. IN AD- 23.6 NUCLEAR FUSION DITION TO APPLICATIONS IN THE MANUFACTURE OF ATOMIC BOMBS, HY- 23.7 USES OF ISOTOPES DROGEN BOMBS, AND NEUTRON BOMBS, EVEN THE PEACEFUL USE OF 23.8 BIOLOGICAL EFFECTS OF RADIATION NUCLEAR ENERGY HAS BECOME CONTROVERSIAL, IN PART BECAUSE OF SAFETY CONCERNS ABOUT NUCLEAR POWER PLANTS AND ALSO BECAUSE OF PROBLEMS WITH DISPOSAL OF RADIOACTIVE WASTES. IN THIS CHAP- TER WE WILL STUDY NUCLEAR REACTIONS, THE STABILITY OF THE ATOMIC NUCLEUS, RADIOACTIVITY, AND THE EFFECTS OF RADIATION ON BIOLOGICAL SYSTEMS. 903 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 904 NUCLEAR CHEMISTRY 23.1 THE NATURE OF NUCLEAR REACTIONS With the exception of hydrogen (1 H), all nuclei contain two kinds of fundamental par1 ticles, called protons and neutrons. Some nuclei are unstable; they emit particles and/or electromagnetic radiation spontaneously (see Section 2.2). The name for this phenomenon is radioactivity. All elements having an atomic number greater than 83 are radioactive. For example, the isotope of polonium, polonium-210 (210 Po), decays spon84 taneously to 206 Pb by emitting an particle. 82 Another type of radioactivity, known as nuclear transmutation, results from the bombardment of nuclei by neutrons, protons, or other nuclei. An example of a nuclear transmutation is the conversion of atmospheric 14 N to 14 C and 1 H, which results when 7 6 1 the nitrogen isotope captures a neutron (from the sun). In some cases, heavier elements are synthesized from lighter elements. This type of transmutation occurs naturally in outer space, but it can also be achieved artificially, as we will see in Section 23.4. Radioactive decay and nuclear transmutation are nuclear reactions, which differ significantly from ordinary chemical reactions. Table 23.1 summarizes the differences. BALANCING NUCLEAR EQUATIONS In order to discuss nuclear reactions in any depth, we need to understand how to write and balance the equations. Writing a nuclear equation differs somewhat from writing equations for chemical reactions. In addition to writing the symbols for various chemical elements, we must also explicitly indicate protons, neutrons, and electrons. In fact, we must show the numbers of protons and neutrons present in every species in such an equation. The symbols for elementary particles are as follows: 1 1p or 1H 1 proton 1 0n neutron 0 1e or 0 1 electron 0 1e or 0 1 positron 4 2 He or 4 2 particle In accordance with the notation used in Section 2.3, the superscript in each case denotes the mass number (the total number of neutrons and protons present) and the subscript is the atomic number (the number of protons). Thus, the “atomic number” of a TABLE 23.1 Comparison of Chemical Reactions and Nuclear Reactions CHEMICAL REACTIONS 1. Atoms are rearranged by the breaking and forming of chemical bonds. 2. Only electrons in atomic orbitals are involved in the breaking and forming of bonds. 3. Reactions are accompanied by absorption or release of relatively small amounts of energy. 4. Rates of reaction are influenced by temperature, pressure, concentration, and catalysts. Back Forward NUCLEAR REACTIONS 1. Elements (or isotopes of the same elements) are converted from one to another. 2. Protons, neutrons, electrons, and other elementary particles may be involved. Main Menu TOC Study Guide TOC 3. Reactions are accompanied by absorption or release of tremendous amounts of energy. 4. Rates of reaction normally are not affected by temperature, pressure, and catalysts. Textbook Website MHHE Website 23.1 THE NATURE OF NUCLEAR REACTIONS 905 proton is 1, because there is one proton present, and the “mass number” is also 1, because there is one proton but no neutrons present. On the other hand, the “mass number” of a neutron is 1, but its “atomic number” is zero, because there are no protons present. For the electron, the “mass number” is zero (there are neither protons nor neutrons present), but the “atomic number” is 1, because the electron possesses a unit negative charge. The symbol 0e represents an electron in or from an atomic orbital. The symbol 1 0 represents an electron that, although physically identical to any other electron, 1 comes from a nucleus (in a decay process in which a neutron is converted to a proton and an electron) and not from an atomic orbital. The positron has the same mass as the electron, but bears a 1 charge. The particle has two protons and two neutrons, so its atomic number is 2 and its mass number is 4. In balancing any nuclear equation, we observe the following rules: The total number of protons plus neutrons in the products and in the reactants must be the same (conservation of mass number). • The total number of nuclear charges in the products and in the reactants must be the same (conservation of atomic number). • If we know the atomic numbers and mass numbers of all the species but one in a nuclear equation, we can identify the unknown species by applying these rules, as shown in the following example, which illustrates how to balance nuclear decay equations. EXAMPLE 23.1 Balance the following nuclear equations (that is, identify the product X): (a) 212 84 Po 88n 208 Pb 82 X (b) 137 55 Cs 137 56 Ba X 88n (a) The mass number and atomic number are 212 and 84, respectively, on the left-hand side and 208 and 82, respectively, on the right-hand side. Thus, X must have a mass number of 4 and an atomic number of 2, which means that it is an particle. The balanced equation is Answer 212 84 Po 88n 208 Pb 82 4 2 (b) In this case the mass number is the same on both sides of the equation, but the atomic number of the product is 1 more than that of the reactant. The only way this change can come about is to have a neutron in the Cs nucleus transformed into a 0 proton and an electron; that is, 1 n 88n 1 p 1 (note that this process does not 0 1 alter the mass number). Thus, the balanced equation is We use the 0 notation here be1 cause the electron came from the nucleus. Similar problems: 23.5, 23.6. 137 55 Cs 88n 137 Ba 56 0 1 Note that the equation in (a) and (b) are balanced for nuclear particles but not for electrical charges. To balance the charges, we would need to add two electrons on the right-hand side of (a) and express barium as a cation (Ba ) in (b). Comment PRACTICE EXERCISE Identify X in the following nuclear equation: 78 33 As Back Forward Main Menu TOC 88n Study Guide TOC 0 1 X Textbook Website MHHE Website 906 NUCLEAR CHEMISTRY 23.2 NUCLEAR STABILITY The nucleus occupies a very small portion of the total volume of an atom, but it contains most of the atom’s mass because both the protons and the neutrons reside there. In studying the stability of the atomic nucleus, it is helpful to know something about its density, because it tells us how tightly the particles are packed together. As a sample calculation, let us assume that a nucleus has a radius of 5 10 3 pm and a mass of 1 10 22 g. These figures correspond roughly to a nucleus containing 30 protons and 30 neutrons. Density is mass/volume, and we can calculate the volume from the known radius (the volume of a sphere is 4 r3, where r is the radius of the sphere). 3 First we convert the pm units to cm. Then we calculate the density in g/cm3: r density 5 mass volume 2 To dramatize the almost incomprehensibly high density, it has been suggested that it is equivalent to packing the mass of all the world’s automobiles into one thimble. 10 3 pm 1 1 10 12 m 1 pm 10 22 g 4 3 3r 4 3 100 cm 1m 1 (5 10 10 22 13 5 10 13 cm g cm)3 1014 g/cm3 This is an exceedingly high density. The highest density known for an element is 22.6 g/cm3, for osmium (Os). Thus the average atomic nucleus is roughly 9 1012 (or 9 trillion) times more dense than the densest element known! The enormously high density of the nucleus prompts us to wonder what holds the particles together so tightly. From Coulomb’s law we know that like charges repel and unlike charges attract one another. We would thus expect the protons to repel one another strongly, particularly when we consider how close they must be to each other. This indeed is so. However, in addition to the repulsion, there are also short-range attractions between proton and proton, proton and neutron, and neutron and neutron. The stability of any nucleus is determined by the difference between coulombic repulsion and the short-range attraction. If repulsion outweighs attraction, the nucleus disintegrates, emitting particles and/or radiation. If attractive forces prevail, the nucleus is stable. The principal factor that determines whether a nucleus is stable is the neutron-toproton ratio (n/p). For stable atoms of elements having low atomic number, the n/p value is close to 1. As the atomic number increases, the neutron-to-proton ratios of the stable nuclei become greater than 1. This deviation at higher atomic numbers arises because a larger number of neutrons is needed to counteract the strong repulsion among the protons and stabilize the nucleus. The following rules are useful in predicting nuclear stability: Nuclei that contain 2, 8, 20, 50, 82, or 126 protons or neutrons are generally more stable than nuclei that do not possess these numbers. For example, there are ten stable isotopes of tin (Sn) with the atomic number 50 and only two stable isotopes of antimony (Sb) with the atomic number 51. The numbers 2, 8, 20, 50, 82, and 126 are called magic numbers. The significance of these numbers for nuclear stability is similar to the numbers of electrons associated with the very stable noble gases (that is, 2, 10, 18, 36, 54, and 86 electrons). • Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of these particles (Table 23.2). • All isotopes of the elements with atomic numbers higher than 83 are radioactive. All isotopes of technetium (Tc, Z 43) and promethium (Pm, Z 61) are radioactive. • Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.2 NUCLEAR STABILITY 907 TABLE 23.2 Number of Stable Isotopes with Even and Odd Numbers of Protons and Neutrons PROTONS NEUTRONS Odd Odd Even Even Odd Even Odd Even NUMBER OF STABLE ISOTOPES 4 50 53 157 Figure 23.1 shows a plot of the number of neutrons versus the number of protons in various isotopes. The stable nuclei are located in an area of the graph known as the belt of stability. Most radioactive nuclei lie outside this belt. Above the stability belt, the nuclei have higher neutron-to-proton ratios than those within the belt (for the same number of protons). To lower this ratio (and hence move down toward the belt of stability), these nuclei undergo the following process, called -particle emission: 1 0n 0 1 88n 1 p 1 Beta-particle emission leads to an increase in the number of protons in the nucleus and a simultaneous decrease in the number of neutrons. Some examples are 14 14 6 C 88n 7 N 40 40 19 K 88n 20 Ca 97 97 40 Zr 88n 41 Nb FIGURE 23.1 Plot of neutrons versus protons for various stable isotopes, represented by dots. The straight line represents the points at which the neutron-toproton ratio equals 1. The shaded area represents the belt of stability. 0 1 0 1 0 1 120 100 Number of neutrons 80 Belt of stability 60 Neutrons/Protons = 1 40 20 0 Back Forward Main Menu 20 40 60 Numbers of protons TOC Study Guide TOC 80 Textbook Website MHHE Website 908 NUCLEAR CHEMISTRY Below the stability belt the nuclei have lower neutron-to-proton ratios than those in the belt (for the same number of protons). To increase this ratio (and hence move up toward the belt of stability), these nuclei either emit a positron 1 1p 88n 1 n 0 0 1 or undergo electron capture. An example of positron emission is 38 19 K We use 0e rather than 0 here 1 1 because the electron came from an atomic orbital and not from the nucleus. 0 1 88n 38 Ar 18 Electron capture is the capture of an electron — usually a 1s electron — by the nucleus. The captured electron combines with a proton to form a neutron so that the atomic number decreases by one while the mass number remains the same. This process has the same net effect as positron emission: 0 37 1 e 88n 17 Cl 0 55 1 e 88n 25 Mn 37 18 Ar 55 26 Fe NUCLEAR BINDING ENERGY A quantitative measure of nuclear stability is the nuclear binding energy, which is the energy required to break up a nucleus into its component protons and neutrons. This quantity represents the conversion of mass to energy that occurs during an exothermic nuclear reaction. The concept of nuclear binding energy evolved from studies of nuclear properties showing that the masses of nuclei are always less than the sum of the masses of the nucleons, which is a general term for the protons and neutrons in a nucleus. For example, the 19 F isotope has an atomic mass of 18.9984 amu. The nucleus has 9 protons 9 and 10 neutrons and therefore a total of 19 nucleons. Using the known masses of the 1 1 H atom (1.007825 amu) and the neutron (1.008665 amu), we can carry out the following analysis. The mass of 9 1 H atoms (that is, the mass of 9 protons and 9 elec1 trons) is 9 1.007825 amu 9.070425 amu and the mass of 10 neutrons is 10 1.008665 amu Therefore, the atomic mass of a trons, protons, and neutrons is 19 9F 9.070425 amu There is no change in the electron’s mass since it is not a nucleon. 10.08665 amu atom calculated from the known numbers of elec10.08665 amu 19.15708 amu which is larger than 18.9984 amu (the measured mass of 19 F) by 0.1587 amu. 9 The difference between the mass of an atom and the sum of the masses of its protons, neutrons, and electrons is called the mass defect. Relativity theory tells us that the loss in mass shows up as energy (heat) given off to the surroundings. Thus the formation of 19 F is exothermic. According to Einstein’s mass-energy equivalence rela9 tionship (E mc2, where E is energy, m is mass, and c is the velocity of light), we can calculate the amount of energy released. We start by writing E ( m)c2 (23.1) where E and m are defined as follows: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.2 E energy of product m mass of product NUCLEAR STABILITY 909 energy of reactants mass of reactants Thus we have for the change in mass m 18.9984 amu 19.15708 amu 0.1587 amu Because 19 F has a mass that is less than the mass calculated from the number of elec9 trons and nucleons present, m is a negative quantity. Consequently, E is also a negative quantity; that is, energy is released to the surroundings as a result of the formation of the fluorine-19 nucleus. So we calculate E as follows: E 108 m/s)2 ( 0.1587 amu)(3.00 16 1.43 10 22 amu m /s With the conversion factors 1 kg 6.022 1026 amu 22 1 kg m /s 1J we obtain E 1.43 2.37 1016 10 11 amu m2 s2 1.00 kg 6.022 1026 amu 1J 1 kg m2/s2 J This is the amount of energy released when one fluorine-19 nucleus is formed from 9 protons and 10 neutrons. The nuclear binding energy of the nucleus is 2.37 10 11 J, which is the amount of energy needed to decompose the nucleus into separate protons and neutrons. In the formation of 1 mole of fluorine nuclei, for instance, the energy released is E ( 2.37 10 11 J)(6.022 1.43 1013 J/mol 1.43 1023/mol) 1010 kJ/mol The nuclear binding energy, therefore, is 1.43 1010 kJ for 1 mole of fluorine-19 nuclei, which is a tremendously large quantity when we consider that the enthalpies of ordinary chemical reactions are of the order of only 200 kJ. The procedure we have followed can be used to calculate the nuclear binding energy of any nucleus. As we have noted, nuclear binding energy is an indication of the stability of a nucleus. However, in comparing the stability of any two nuclei we must account for the fact that they have different numbers of nucleons. For this reason it is more meaningful to use the nuclear binding energy per nucleon, defined as nuclear binding energy per nucleon nuclear binding energy number of nucleons For the fluorine-19 nucleus, nuclear binding energy per nucleon 2.37 10 11 J 19 nucleons 1.25 Back Forward Main Menu TOC Study Guide TOC 10 12 J/nucleon Textbook Website MHHE Website NUCLEAR CHEMISTRY FIGURE 23.2 Plot of nuclear binding energy per nucleon versus mass number. Nuclear binding energy per nucleon (J) 910 56Fe 4 1.5 × 10–12 He 238U 1.2 × 10–12 9 × 10–13 6 × 10–13 3 × 10–13 2H 0 20 40 60 80 100 120 140 160 180 200 220 240 260 Mass number The nuclear binding energy per nucleon allows us to compare the stability of all nuclei on a common basis. Figure 23.2 shows the variation of nuclear binding energy per nucleon plotted against mass number. As you can see, the curve rises rather steeply. The highest binding energies per nucleon belong to elements with intermediate mass numbers—between 40 and 100—and are greatest for elements in the iron, cobalt, and nickel region (the Group 8B elements) of the periodic table. This means that the net attractive forces among the particles (protons and neutrons) are greatest for the nuclei of these elements. Nuclear binding energy and nuclear binding energy per nucleon are calculated for an iodine nucleus in the following example. EXAMPLE 23.2 The atomic mass of 127 I is 126.9004 amu. Calculate the nuclear binding energy of 53 this nucleus and the corresponding nuclear binding energy per nucleon. Answer There are 53 protons and 74 neutrons in the nucleus. The mass of 53 1 H 1 atoms is 53 1.007825 amu 53.41473 amu and the mass of 74 neutrons is 74 1.008665 amu Therefore, the predicted mass for and the mass defect is m 127 53 I 74.64121 amu is 53.41473 126.9004 amu 74.64121 128.05594 amu, 128.05594 amu 1.1555 amu The energy released is E ( m)c2 ( 1.1555 amu)(3.00 1.04 Back Forward Main Menu TOC 108 m/s)2 1017 amu m2/s2 Study Guide TOC Textbook Website MHHE Website 23.3 E 1.04 1.73 The neutron-to-proton ratio is 1.4, which places iodine-127 in the belt of stability. 1017 10 10 amu m2 s2 1.00 kg 6.022 1026 amu 911 1J 1 kg m2/s2 J Thus the nuclear binding energy is 1.73 nucleon is obtained as follows: 1.73 10 10 J 127 nucleons Similar problems: 23.19, 23.20. NATURAL RADIOACTIVITY 10 10 1.36 10 J. The nuclear binding energy per 12 J/nucleon PRACTICE EXERCISE Calculate the nuclear binding energy (in J) and the binding energy per nucleon of 209 83 Bi (208.9804 amu). 23.3 NATURAL RADIOACTIVITY Nuclei outside the belt of stability, as well as nuclei with more than 83 protons, tend to be unstable. The spontaneous emission by unstable nuclei of particles or electromagnetic radiation, or both, is known as radioactivity. The main types of radiation are: particles (or doubly charged helium nuclei, He2 ); particles (or electrons); rays, which are very-short-wavelength (0.1 nm to 10 4 nm) electromagnetic waves; positron emission; and electron capture. The disintegration of a radioactive nucleus is often the beginning of a radioactive decay series, which is a sequence of nuclear reactions that ultimately result in the formation of a stable isotope. Table 23.3 shows the decay series of naturally occurring uranium-238, which involves 14 steps. This decay scheme, known as the uranium decay series, also shows the half-lives of all the products. It is important to be able to balance the nuclear reaction for each of the steps in a radioactive decay series. For example, the first step in the uranium decay series is the decay of uranium-238 to thorium-234, with the emission of an particle. Hence, the reaction is 238 92 U 88n 234 Th 90 4 2 The next step is represented by 234 90 Th 88n 234 Pa 91 0 1 and so on. In a discussion of radioactive decay steps, the beginning radioactive isotope is called the parent and the product, the daughter. KINETICS OF RADIOACTIVE DECAY All radioactive decays obey first-order kinetics. Therefore the rate of radioactive decay at any time t is given by rate of decay at time t N where is the first-order rate constant and N is the number of radioactive nuclei present at time t. (We use instead of k for rate constant in accord with the notation used by nuclear scientists.) According to Equation (13.3), the number of radioactive nuclei at time zero (N0) and time t (Nt) is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 912 NUCLEAR CHEMISTRY TABLE 23.3 The Uranium Decay Series 238 92 U 109 yr 4.51 234 90 Th 24.1 days 234 91 Pa 1.17 min 234 92 U 2.47 7.5 226 88 222 86 104 yr 1.60 230 90 105 yr 103 yr Th Ra Rn 3.82 days 218 84 Po 3.05 min 0.04% 218 85 214 82 At Pb 2s 26.8 min 214 83 Bi 99.96% 1.6 10 4 214 84 19.7 min 210 81 Po Tl s 1.32 min 210 82 Pb 20.4 yr 210 83 Bi 100% 210 84 5.01 days 206 81 Po Tl 4.20 min 138 days 206 82 Pb ln N0 Nt t and the corresponding half-life of the reaction is given by Equation (13.5): t1 2 We do not have to wait 4.51 109 yr to make a half-life measurement of uranium-238. Its value can be calculated from the rate constant using Equation (13.5). Back Forward 0.693 The half-lives (hence the rate constants) of radioactive isotopes vary greatly from nucleus to nucleus. For example, looking at Table 23.3, we find two extreme cases: Main Menu 238 92 U 214 84 Po TOC 88n 234 Th 90 88n 210 82 Pb Study Guide TOC 4 2 4 2 t1 2 4.51 t1 2 1.6 Textbook Website 109 yr 10 4 s MHHE Website 23.3 NATURAL RADIOACTIVITY 913 The ratio of these two rate constants after conversion to the same time unit is about 1 1021, an enormously large number. Furthermore, the rate constants are unaffected by changes in environmental conditions such as temperature and pressure. These highly unusual features are not seen in ordinary chemical reactions (see Table 23.1). DATING BASED ON RADIOACTIVE DECAY The half-lives of radioactive isotopes have been used as “atomic clocks” to determine the ages of certain objects. Some examples of dating by radioactive decay measurements will be described here. Radiocarbon Dating The carbon-14 isotope is produced when atmospheric nitrogen is bombarded by cosmic rays: 14 7N 1 0n 88n 14 C 6 1 1H The radioactive carbon-14 isotope decays according to the equation 14 6C 0 1 88n 14 N 7 This decay series is the basis of the radiocarbon dating technique described on p. 527. Dating Using Uranium-238 Isotopes We can think of the first step as the rate-determining step in the overall process. Because some of the intermediate products in the uranium series have very long halflives (see Table 23.3), this series is particularly suitable for estimating the age of rocks in the earth and of extraterrestrial objects. The half-life for the first step (238 U to 234 Th) 92 90 is 4.51 109 yr. This is about 20,000 times the second largest value (that is, 2.47 105 yr), which is the half-life for 234 U to 230 Th. Therefore, as a good approximation 92 90 we can assume that the half-life for the overall process (that is, from 238 U to 206 Pb) is 92 82 governed solely by the first step: 238 92 U 238U t1 _ 238U 206Pb 206 g/2 238 g/2 4.51 × 109 yr FIGURE 23.3 After one halflife, half of the original uranium238 is converted to lead-206. Forward Main Menu 8 4 2 6 0 1 t1 2 4.51 109 yr In naturally occurring uranium minerals we should and do find some lead-206 isotopes formed by radioactive decay. Assuming that no lead was present when the mineral was formed and that the mineral has not undergone chemical changes that would allow the lead-206 isotope to be separated from the parent uranium-238, it is possible to estimate the age of the rocks from the mass ratio of 206 Pb to 238 U. The 82 92 above equation tells us that for every mole, or 238 g, of uranium that undergoes complete decay, 1 mole, or 206 g, of lead is formed. If only half a mole of uranium-238 has undergone decay, the mass ratio Pb-206/U-238 becomes 2 Back 88n 206 Pb 82 0.866 and the process would have taken a half-life of 4.51 109 yr to complete (Figure 23.3). Ratios lower than 0.866 mean that the rocks are less than 4.51 109 yr old, and higher ratios suggest a greater age. Interestingly, studies based on the uranium series as well as other decay series put the age of the oldest rocks and, therefore, probably the age of Earth itself at 4.5 109, or 4.5 billion, years. TOC Study Guide TOC Textbook Website MHHE Website 914 NUCLEAR CHEMISTRY Dating Using Potassium-40 Isotopes This is one of the most important techniques in geochemistry. The radioactive potassium-40 isotope decays by several different modes, but the relevant one as far as dating is concerned is that of electron capture: 0 1e 40 19 K 88n 40 Ar 18 t1 2 1.2 109 yr The accumulation of gaseous argon-40 is used to gauge the age of a specimen. When a potassium-40 atom in a mineral decays, argon-40 is trapped in the lattice of the mineral and can escape only if the material is melted. Melting, therefore, is the procedure for analyzing a mineral sample in the laboratory. The amount of argon-40 present can be conveniently measured with a mass spectrometer (see p. 76). Knowing the ratio of argon-40 to potassium-40 in the mineral and the half-life of decay makes it possible to establish the ages of rocks ranging from millions to billions of years old. 23.4 NUCLEAR TRANSMUTATION The scope of nuclear chemistry would be rather narrow if study were limited to natural radioactive elements. An experiment performed by Rutherford in 1919, however, suggested the possibility of producing radioactivity artificially. When he bombarded a sample of nitrogen with particles, the following reaction took place: 14 7N 4 2 88n 17O 8 1 1p An oxygen-17 isotope was produced with the emission of a proton. This reaction demonstrated for the first time the feasibility of converting one element into another, by the process of nuclear transmutation. Nuclear transmutation differs from radioactive decay in that the former is brought about by the collision of two particles. 4 The above reaction can be abbreviated as 17N( ,p)17O. Note that in the parenthe8 ses the bombarding particle is written first, followed by the ejected particle. The following example illustrates the use of this notation to represent nuclear transmutations. EXAMPLE 23.3 Write the balanced equation for the nuclear reaction resents the deuterium nucleus (that is, 2H). 1 56 26Fe(d, )54Mn, where d rep25 The abbreviation tells us that when iron-56 is bombarded with a deuterium nucleus, it produces the manganese-54 nucleus plus an particle, 4He. Thus, the 2 equation for this reaction is Answer 56 26 Fe Similar problems: 23.33, 23.34. 2 1H 88n 4 2 54 25 Mn PRACTICE EXERCISE Write a balanced equation for 106 46 Pd( ,p)109 Ag. 47 Although light elements are generally not radioactive, they can be made so by bombarding their nuclei with appropriate particles. As we saw earlier, the radioactive carbon-14 isotope can be prepared by bombarding nitrogen-14 with neutrons. Tritium, 3 1H, is prepared according to the following bombardment: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.4 6 3 Li Alternating voltage ± ± Magnetic field Tritium decays with the emission of 3 1H Target 88n 3 H 1 4 2 particles: 88n 3 He 2 0 1 t1 2 12.5 yr Many synthetic isotopes are prepared by using neutrons as projectiles. This approach is particularly convenient because neutrons carry no charges and therefore are not repelled by the targets — the nuclei. In contrast, when the projectiles are positively charged particles (for example, protons or particles), they must have considerable kinetic energy in order to overcome the electrostatic repulsion between themselves and the target atoms. The synthesis of phosphorus from aluminum is one example: Dees FIGURE 23.4 Schematic diagram of a cyclotron particle accelerator. The particle (an ion) to be accelerated starts at the center and is forced to move in a spiral path through the influence of electric and magnetic fields until it emerges at a high velocity. The magnetic fields are perpendicular to the plane of the dees (socalled because of their shape), which are hollow and serve as electrodes. 1 0n 915 NUCLEAR TRANSMUTATION 27 13 Al 4 2 88n 30 P 15 1 0n A particle accelerator uses electric and magnetic fields to increase the kinetic energy of charged species so that a reaction will occur (Figure 23.4). Alternating the polarity (that is, and ) on specially constructed plates causes the particles to accelerate along a spiral path. When they have the energy necessary to initiate the desired nuclear reaction, they are guided out of the accelerator into a collision with a target substance. Various designs have been developed for particle accelerators, one of which accelerates particles along a linear path of about 3 km (Figure 23.5). It is now possible to accelerate particles to a speed well above 90 percent of the speed of light. (According to Einstein’s theory of relativity, it is impossible for a particle to move at the speed of light. The only exception is the photon, which has a zero rest mass.) The extremely energetic particles produced in accelerators are employed by physicists to smash atomic nuclei to fragments. Studying the debris from such disintegrations provides valuable information about nuclear structure and binding forces. FIGURE 23.5 A section of a linear particle accelerator. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 916 NUCLEAR CHEMISTRY TABLE 23.4 The Transuranium Elements ATOMIC NUMBER NAME 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Hahnium Seaborgium Nielsbohrium Hassium Meitnerium SYMBOL Np Pu Am Cm Bk Cf Es Fm Md No Lr Rf Ha Sg Ns Hs Mt PREPARATION 238 1 239 92U 0n 88n 93Np 0 239 239 1 93Np 88n 94Pu 239 1 240 94Pu 0n 88n 95Am 239 4 242 94Pu 2 88n 96Cm 241 4 243 95Am 2 88n 97Bk 242 4 245 96Cm 2 88n 98Cf 238 1 U 15 0n 88n 253Es 92 99 238 1 255 170n 88n 100Fm 92U 253 4 256 99Es 2 88n 101Md 246 12 254 96Cm 6C 88n 102No 252 10 257 98Cf 5B 88n 103Lr 249 Cf 12C 88n 257Rf 98 6 104 249 15 260 98Cf 7N 88n 105Ha 249 18 263 98Cf 8O 88n 106Sg 209 54 262 83Bi 24Cr 88n 107Ns 208 58 265 82Pb 26Fe 88n 108Hs 209 58 266 83Bi 26Fe 88n 109Mt 0 1 0 1 1 0n 2 1n 0 1 0n 70 1 80 1 1 0n 4 1n 0 1 5 0n 4 1n 0 4 1n 0 4 1n 0 1 0n 1 0n 1 0n THE TRANSURANIUM ELEMENTS Particle accelerators made it possible to synthesize the so-called transuranium elements, elements with atomic numbers greater than 92. Neptunium (Z 93) was first prepared in 1940. Since then, 20 other transuranium elements have been synthesized. All isotopes of these elements are radioactive. Table 23.4 lists the transuranium elements and the reactions through which they are formed. 23.5 NUCLEAR FISSION Nuclear fission is the process in which a heavy nucleus (mass number > 200) divides to form smaller nuclei of intermediate mass and one or more neutrons. Because the heavy nucleus is less stable than its products (see Figure 23.2), this process releases a large amount of energy. The first nuclear fission reaction to be studied was that of uranium-235 bombarded with slow neutrons, whose speed is comparable to that of air molecules at room temperature. Under these conditions, uranium-235 undergoes fission, as shown in Figure 23.6. Actually, this reaction is very complex: More than 30 different elements have 90 Sr 38 235 U 92 143 Xe 54 FIGURE 23.6 Nuclear fission of U-235. When a U-235 nucleus captures a neutron (red dot), it undergoes fission to yield two smaller nuclei. On the average, 2.4 neutrons are emitted for every U-235 nucleus that divides. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.5 NUCLEAR FISSION 917 Relative amounts of fission product been found among the fission products (Figure 23.7). A representative reaction is 235 92U 80 100 120 140 160 Mass number FIGURE 23.7 Relative yields of the products resulting from the fission of U-235, as a function of mass number. TABLE 23.5 Nuclear Binding Energies of 235U and Its Fission Products NUCLEAR BINDING ENERGY 235 U Sr 143 Xe 90 2.82 1.23 1.92 10 10 10 10 10 10 J J J 1 0n 88n 90 Sr 38 143 54 Xe 3 1n 0 Although many heavy nuclei can be made to undergo fission, only the fission of naturally occurring uranium-235 and of the artificial isotope plutonium-239 has any practical importance. Table 23.5 shows the nuclear binding energies of uranium-235 and its fission products. As the table shows, the binding energy per nucleon for uranium235 is less than the sum of the binding energies for strontium-90 and xenon-143. Therefore, when a uranium-235 nucleus is split into two smaller nuclei, a certain amount of energy is released. Let us estimate the magnitude of this energy. The difference between the binding energies of the reactants and products is (1.23 10 10 1.92 10 10) J (2.82 10 10) J, or 3.3 10 11 J per uranium-235 nucleus. For 1 mole of uranium-235, the energy released would be (3.3 10 11)(6.02 1023), or 2.0 1013 J. This is an extremely exothermic reaction, considering that the heat of combustion of 1 ton of coal is only about 8 107 J. The significant feature of uranium-235 fission is not just the enormous amount of energy released, but the fact that more neutrons are produced than are originally captured in the process. This property makes possible a nuclear chain reaction, which is a self-sustaining sequence of nuclear fission reactions. The neutrons generated during the initial stages of fission can induce fission in other uranium-235 nuclei, which in turn produce more neutrons, and so on. In less than a second, the reaction can become uncontrollable, liberating a tremendous amount of heat to the surroundings. Figure 23.8 shows two types of fission reactions. For a chain reaction to occur, enough uranium-235 must be present in the sample to capture the neutrons. Otherwise, many of the neutrons will escape from the sample and the chain reaction will not occur, as depicted in Figure 23.8(a). In this situation the mass of the sample is said to be subcritical. Figure 23.8(b) shows what happens when the amount of the fissionable material is equal to or greater than the critical mass, the minimum mass of fissionable material required to generate a self-sustaining nuclear chain reaction. In this case most of the neutrons will be captured by uranium-235 nuclei, and a chain reaction will occur. FIGURE 23.8 Two types of nuclear fission. (a) If the mass of U-235 is subcritical, no chain reaction will result. Many of the neutrons produced will escape to the surroundings. (b) If a critical mass is present, many of the neutrons emitted during the fission process will be captured by other U-235 nuclei and a chain reaction will occur. (a) Back Forward Main Menu TOC Study Guide TOC (b) Textbook Website MHHE Website 918 NUCLEAR CHEMISTRY TNT explosive THE ATOMIC BOMB The first application of nuclear fission was in the development of the atomic bomb. How is such a bomb made and detonated? The crucial factor in the bomb’s design is the determination of the critical mass for the bomb. A small atomic bomb is equivalent to 20,000 tons of TNT (trinitrotoluene). Since 1 ton of TNT releases about 4 109 J of energy, 20,000 tons would produce 8 1013 J. Earlier we saw that 1 mole, or 235 g, of uranium-235 liberates 2.0 1013 J of energy when it undergoes fission. Thus the mass of the isotope present in a small bomb must be at least 235 g Subcritical U-235 wedge FIGURE 23.9 Schematic cross section of an atomic bomb. The TNT explosives are set off first. The explosion forces the sections of fissionable material together to form an amount considerably larger than the critical mass. 8 1013 J 2.0 1013 J 1 kg For obvious reasons, an atomic bomb is never assembled with the critical mass already present. Instead, the critical mass is formed by using a conventional explosive, such as TNT, to force the fissionable sections together, as shown in Figure 23.9. Neutrons from a source at the center of the device trigger the nuclear chain reaction. Uranium235 was the fissionable material in the bomb dropped on Hiroshima, Japan, on August 6, 1945. Plutonium-239 was used in the bomb exploded over Nagasaki three days later. The fission reactions generated were similar in these two cases, as was the extent of the destruction. NUCLEAR REACTORS In Europe, nuclear reactors provide about 40 percent of the electrical energy consumed. A peaceful but controversial application of nuclear fission is the generation of electricity using heat from a controlled chain reaction in a nuclear reactor. Currently, nuclear reactors provide about 20 percent of the electrical energy in the United States. This is a small but by no means negligible contribution to the nation’s energy production. Several different types of nuclear reactors are in operation; we will briefly discuss the main features of three of them, along with their advantages and disadvantages. Light Water Reactors Most of the nuclear reactors in the United States are light water reactors. Figure 23.10 is a schematic diagram of such a reactor, and Figure 23.11 shows the refueling process in the core of a nuclear reactor. An important aspect of the fission process is the speed of the neutrons. Slow neutrons split uranium-235 nuclei more efficiently than do fast ones. Because fission reactions are highly exothermic, the neutrons produced usually move at high velocities. For greater efficiency they must be slowed down before they can be used to induce nuclear disintegration. To accomplish this goal, scientists use moderators, which are substances that can reduce the kinetic energy of neutrons. A good moderator must satisfy several requirements: It should be nontoxic and inexpensive (as very large quantities of it are necessary); and it should resist conversion into a radioactive substance by neutron bombardment. Furthermore, it is advantageous for the moderator to be a fluid so that it can also be used as a coolant. No substance fulfills all these requirements, although water comes closer than many others that have been considered. Nuclear reactors that use light water (H2O) as a moderator are called light water reactors because 1 H is the lightest isotope of the element hydrogen. 1 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.5 NUCLEAR FISSION 919 Shield Steam To steam turbine Shield Water Pump Control rod Uranium fuel FIGURE 23.10 Schematic diagram of a nuclear fission reactor. The fission process is controlled by cadmium or boron rods. The heat generated by the process is used to produce steam for the generation of electricity via a heat exchange system. The nuclear fuel consists of uranium, usually in the form of its oxide, U3O8 (Figure 23.12). Naturally occurring uranium contains about 0.7 percent of the uranium-235 isotope, which is too low a concentration to sustain a small-scale chain reaction. For effective operation of a light water reactor, uranium-235 must be enriched to a concentration of 3 or 4 percent. In principle, the main difference between an atomic bomb and a nuclear reactor is that the chain reaction that takes place in a nuclear reactor is kept under control at all times. The factor limiting the rate of the reaction is the number of neutrons present. This can be controlled by lowering cadmium or boron rods between the fuel elements. These rods capture neutrons according to the equations FIGURE 23.12 Uranium oxide, U3O8. FIGURE 23.11 Refueling the core of a nuclear reactor. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 920 NUCLEAR CHEMISTRY 113 48 Cd 10 5B 1 0n 1 0n 88n 114 Cd 48 88n 7 Li 3 4 2 where denotes gamma rays. Without the control rods the reactor core would melt from the heat generated and release radioactive materials into the environment. Nuclear reactors have rather elaborate cooling systems that absorb the heat given off by the nuclear reaction and transfer it outside the reactor core, where it is used to produce enough steam to drive an electric generator. In this respect, a nuclear power plant is similar to a conventional power plant that burns fossil fuel. In both cases, large quantities of cooling water are needed to condense steam for reuse. Thus, most nuclear power plants are built near a river or a lake. Unfortunately this method of cooling causes thermal pollution (see Section 12.4). Heavy Water Reactors Another type of nuclear reactor uses D2O, or heavy water, as the moderator, rather than H2O. Deuterium absorbs neutrons much less efficiently than does ordinary hydrogen. Since fewer neutrons are absorbed, the reactor is more efficient and does not require enriched uranium. The fact that deuterium is a less efficient moderator has a negative impact on the operation of the reactor, because more neutrons leak out of the reactor. However, this is not a serious disadvantage. The main advantage of a heavy water reactor is that it eliminates the need for building expensive uranium enrichment facilities. However, D2O must be prepared by either fractional distillation or electrolysis of ordinary water, which can be very expensive considering the amount of water used in a nuclear reactor. In countries where hydroelectric power is abundant, the cost of producing D2O by electrolysis can be reasonably low. At present, Canada is the only nation successfully using heavy water nuclear reactors. The fact that no enriched uranium is required in a heavy water reactor allows a country to enjoy the benefits of nuclear power without undertaking work that is closely associated with weapons technology. Breeder Reactors A breeder reactor uses uranium fuel, but unlike a conventional nuclear reactor, it produces more fissionable materials than it uses. We know that when uranium-238 is bombarded with fast neutrons, the following reactions take place: 238 92 U 1 0n 239 92 U 239 93 Np Plutonium-239 forms plutonium oxide, which can be readily separated from uranium. 88n 239 U 92 88n 239 Np 93 88n 239 94 Pu 0 1 0 1 t1 2 23.4 min t1 2 2.35 days In this manner the nonfissionable uranium-238 is transmuted into the fissionable isotope plutonium-239 (Figure 23.13). In a typical breeder reactor, nuclear fuel containing uranium-235 or plutonium239 is mixed with uranium-238 so that breeding takes place within the core. For every uranium-235 (or plutonium-239) nucleus undergoing fission, more than one neutron is captured by uranium-238 to generate plutonium-239. Thus, the stockpile of fissionable material can be steadily increased as the starting nuclear fuels are consumed. It takes Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.5 NUCLEAR FISSION 921 about 7 to 10 years to regenerate the sizable amount of material needed to refuel the original reactor and to fuel another reactor of comparable size. This interval is called the doubling time. Another fertile isotope is 232 Th. Upon capturing slow neutrons, thorium is trans90 muted to uranium-233, which, like uranium-235, is a fissionable isotope: 232 1 233 90 Th 0 n 88n 90 Th 233 233 90 Th 88n 91 Pa 233 233 91Pa 88n 92 U FIGURE 23.13 The red glow of the radioactive plutonium-239 isotope. The orange color is due to the presence of its oxide. 0 1 0 1 t1 2 22 min t1 2 27.4 days Uranium-233 is stable enough for long-term storage. Although the amounts of uranium-238 and thorium-232 in Earth’s crust are relatively plentiful (4 ppm and 12 ppm by mass, respectively), the development of breeder reactors has been very slow. To date, the United States does not have a single operating breeder reactor, and only a few have been built in other countries, such as France and Russia. One problem is economics; breeder reactors are more expensive to build than conventional reactors. There are also more technical difficulties associated with the construction of such reactors. As a result, the future of breeder reactors, in the United States at least, is rather uncertain. Hazards of Nuclear Energy Molten glass is poured over nuclear waste before burial. Back Forward Main Menu Many people, including environmentalists, regard nuclear fission as a highly undesirable method of energy production. Many fission products such as strontium-90 are dangerous radioactive isotopes with long half-lives. Plutonium-239, used as a nuclear fuel and produced in breeder reactors, is one of the most toxic substances known. It is an alpha emitter with a half-life of 24,400 yr. Accidents, too, present many dangers. An accident at the Three Mile Island reactor in Pennsylvania in 1979 first brought the potential hazards of nuclear plants to public attention. In this instance very little radiation escaped the reactor, but the plant remained closed for more than a decade while repairs were made and safety issues addressed. Only a few years later, on April 26, 1986, a reactor at the Chernobyl nuclear plant in Belarus surged out of control. The fire and explosion that followed released much radioactive material into the environment. People working near the plant died within weeks as a result of the exposure to the intense radiation. The long-term effect of the radioactive fallout from this incident has not yet been clearly assessed, although agriculture and dairy farming were affected by the fallout. The number of potential cancer deaths attributable to the radiation contamination is estimated to be between a few thousand and more than 100,000. In addition to the risk of accidents, the problem of radioactive waste disposal has not been satisfactorily resolved even for safely operated nuclear plants. Many suggestions have been made as to where to store or dispose of nuclear waste, including burial underground, burial beneath the ocean floor, and storage in deep geologic formations. But none of these sites has proved absolutely safe in the long run. Leakage of radioactive wastes into underground water, for example, can endanger nearby communities. The ideal disposal site would seem to be the sun, where a bit more radiation would make little difference, but this kind of operation requires 100 percent reliability in space technology. Because of the hazards, the future of nuclear reactors is clouded. What was once hailed as the ultimate solution to our energy needs in the twenty-first century is now TOC Study Guide TOC Textbook Website MHHE Website 922 NUCLEAR CHEMISTRY Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Nature’s Own Fission Reactor It all started with a routine analysis in May 1972 at the nuclear fuel processing plant in Pierrelatte, France. A staff member was checking the isotope ratio of U-235 to U-238 in a uranium ore and obtained a puzzling result. It had long been known that the relative natural occurrence of U-235 and U-238 is 0.7202 percent and 99.2798 percent, respectively. In this case, however, the amount of U-235 present was only 0.7171 percent. This may seem like a very small deviation, but the measurements were so precise that this difference was considered highly significant. The ore had come from the Oklo mine in the Gabon Republic, a small country on the west coast of Africa. Subsequent analyses of other samples showed that some contained even less U-235, in some cases as little as 0.44 percent. The logical explanation for the low percentages of U-235 was that a nuclear fission reaction at the mine must have consumed some of the U-235 isotopes. But how did this happen? There are several conditions under which such a nuclear fission reaction could take place. In the presence of heavy water, for example, a chain reaction is possible with unenriched uranium. Without heavy water, such a fission reaction could still occur if the uranium ore and the moderator were arranged according to some specific geometric constraints at the site of the reaction. Both of the possibilities seem rather farfetched. The most plausible explanation is that the uranium ore originally present in the mine was enriched with U-235 and that a nuclear fission reaction took place with light water, as in a conventional nuclear reactor. As mentioned earlier, the natural abundance of U-235 is 0.7202 percent, but it has not always been that low. The half-lives of U-235 and U-238 are 700 million and 4.51 billion years, respectively. This means that U-235 must have been more abundant in the past, because it has a shorter half-life. In fact, at the time Earth was formed, the natural abundance of U-235 was as high as 17 percent! Since the lowest concentration of U-235 required for the operation of a fission reactor is 1 percent, a nuclear chain reaction could have taken place as recently as 400 million years ago. By analyzing the amounts of radioactive Forward Main Menu TOC fission products left in the ore, scientists concluded that the Gabon “reactor” operated about 2 billion years ago. Having an enriched uranium sample is only one of the requirements for starting a controlled chain reaction. There must also have been a sufficient amount of the ore and an appropriate moderator present. It appears that as a result of a geological transformation, uranium ore was continually being washed into the Oklo region to yield concentrated deposits. The moderator needed for the fission process was largely water, present as water of crystallization in the sedimentary ore. Thus, in a series of extraordinary events, a natural nuclear fission reactor operated at the time when the first life forms appeared on Earth. As is often the case in scientific endeavors, humans are not necessarily the innovators but merely the imitators of nature. Photo showing the natural nuclear reactor site (lower righthand corner) at Oklo, Gabon Republic. Study Guide TOC Textbook Website MHHE Website 23.6 NUCLEAR FUSION 923 being debated and questioned by both the scientific community and laypeople. It seems likely that the controversy will continue for some time. 23.6 NUCLEAR FUSION In contrast to the nuclear fission process, nuclear fusion, the combining of small nuclei into larger ones, is largely exempt from the waste disposal problem. Figure 23.2 showed that for the lightest elements, nuclear stability increases with increasing mass number. This behavior suggests that if two light nuclei combine or fuse together to form a larger, more stable nucleus, an appreciable amount of energy will be released in the process. This is the basis for ongoing research into the harnessing of nuclear fusion for the production of energy. Nuclear fusion occurs constantly in the sun (Figure 23.14). The sun is made up mostly of hydrogen and helium. In its interior, where temperatures reach about 15 million degrees Celsius, the following fusion reactions are believed to take place: FIGURE 23.14 Nuclear fusion keeps the temperature in the interior of the sun at about 15 million °C. 1 1H 3 He 2 1 1H 2 1H 3 He 2 1 1H 88n 3He 2 88n 4He 2 21H 1 88n 2H 1 0 1 Because fusion reactions take place only at very high temperatures, they are often called thermonuclear reactions. FUSION REACTORS A major concern in choosing the proper nuclear fusion process for energy production is the temperature necessary to carry out the process. Some promising reactions are REACTION 2 1H 2 1H 6 3 Li 2 3 1 1H 88n 1H 1H 3 4 H 88n 2He 1 n 1 0 2 4 1H 88n 2 2He ENERGY RELEASED 6.3 2.8 3.6 10 10 10 13 J J 12 J 12 These reactions take place at extremely high temperatures, on the order of 100 million degrees Celsius, to overcome the repulsive forces between the nuclei. The first reaction is particularly attractive because the world’s supply of deuterium is virtually inexhaustible. The total volume of water on Earth is about 1.5 1021 L. Since the natural abundance of deuterium is 1.5 10 2 percent, the total amount of deuterium present is roughly 4.5 1021 g, or 5.0 1015 tons. The cost of preparing deuterium is minimal compared with the value of the energy released by the reaction. In contrast to the fission process, nuclear fusion looks like a very promising energy source, at least “on paper.” Although thermal pollution would be a problem, fusion has the following advantages: (1) The fuels are cheap and almost inexhaustible and (2) the process produces little radioactive waste. If a fusion machine were turned off, it would shut down completely and instantly, without any danger of a meltdown. If nuclear fusion is so great, why isn’t there even one fusion reactor producing energy? Although we command the scientific knowledge to design such a reactor, the technical difficulties have not yet been solved. The basic problem is finding a way to hold the nuclei together long enough, and at the appropriate temperature, for fusion to Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 924 NUCLEAR CHEMISTRY FIGURE 23.15 A magnetic plasma confinement design called tokamak. Plasma Magnet occur. At temperatures of about 100 million degrees Celsius, molecules cannot exist, and most or all of the atoms are stripped of their electrons. This state of matter, a gaseous mixture of positive ions and electrons, is called plasma. The problem of containing this plasma is a formidable one. What solid container can exist at such temperatures? None, unless the amount of plasma is small; but then the solid surface would immediately cool the sample and quench the fusion reaction. One approach to solving this problem is to use magnetic confinement. Since a plasma consists of charged particles moving at high speeds, a magnetic field will exert force on it. As Figure 23.15 shows, the plasma moves through a doughnut-shaped tunnel, confined by a complex magnetic field. Thus the plasma never comes in contact with the walls of the container. Another promising design employs high-power lasers to initiate the fusion reaction. In test runs a number of laser beams transfer energy to a small fuel pellet, heating it and causing it to implode, that is, to collapse inward from all sides and compress into a small volume (Figure 23.16). Consequently, fusion occurs. Like the magnetic confinement approach, laser fusion presents a number of technical difficulties that still need to be overcome before it can be put to practical use on a large scale. THE HYDROGEN BOMB The technical problems inherent in the design of a nuclear fusion reactor do not affect the production of a hydrogen bomb, also called a thermonuclear bomb. In this case the objective is all power and no control. Hydrogen bombs do not contain gaseous hydrogen or gaseous deuterium; they contain solid lithium deuteride (LiD), which can be packed very tightly. The detonation of a hydrogen bomb occurs in two stages—first a fission reaction and then a fusion reaction. The required temperature for fusion is achieved with an atomic bomb. Immediately after the atomic bomb explodes, the following fusion reactions occur, releasing vast amounts of energy (Figure 23.17): 6 3 Li 2 1H Back Forward Main Menu TOC 2 1H 2 1H 88n 2 4 2 88n 3 H 1 Study Guide TOC 1 1H Textbook Website MHHE Website 23.6 NUCLEAR FUSION 925 FIGURE 23.16 This small-scale fusion reaction was created at the Lawrence Livermore National Laboratory using the world’s most powerful laser, Nova. There is no critical mass in a fusion bomb, and the force of the explosion is limited only by the quantity of reactants present. Thermonuclear bombs are described as being “cleaner” than atomic bombs because the only radioactive isotopes they produce are tritium, which is a weak -particle emitter (t 1 12.5 yr), and the products of the 2 fission starter. Their damaging effects on the environment can be aggravated, however, by incorporating in the construction some nonfissionable material such as cobalt. Upon bombardment by neutrons, cobalt-59 is converted to cobalt-60, which is a very strong -ray emitter with a half-life of 5.2 yr. The presence of radioactive cobalt isotopes in the debris or fallout from a thermonuclear explosion would be fatal to those who survived the initial blast. FIGURE 23.17 Explosion of a thermonuclear bomb. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 926 NUCLEAR CHEMISTRY 23.7 USES OF ISOTOPES Radioactive and stable isotopes alike have many applications in science and medicine. We have previously described the use of isotopes in the study of reaction mechanisms (see Section 13.5) and in dating artifacts (p. 527 and Section 23.3). In this section we will discuss a few more examples. STRUCTURAL DETERMINATION The formula of the thiosulfate ion is S2O2 . For some years chemists were uncertain 3 as to whether the two sulfur atoms occupied equivalent positions in the ion. The thiosulfate ion is prepared by treatment of the sulfite ion with elemental sulfur: SO2 (aq) 3 S(s) 88n S2O2 (aq) 3 When thiosulfate is treated with dilute acid, the reaction is reversed. The sulfite ion is reformed and elemental sulfur precipitates: H S2O2 (aq) 88n SO2 (aq) 3 3 S(s) (23.2) If this sequence is started with elemental sulfur enriched with the radioactive sulfur35 isotope, the isotope acts as a “label” for S atoms. All the labels are found in the sulfur precipitate in Equation (23.2); none of them appears in the final sulfite ions. Clearly, then, the two atoms of sulfur in S2O2 are not structurally equivalent, as would be the 3 case if the structure were OSOSO SOOOOOOOOOS QQQQQ 2 Otherwise, the radioactive isotope would be present in both the elemental sulfur precipitate and the sulfite ion. Based on spectroscopic studies, we now know that the structure of the thiosulfate ion is S OS O A O SOO S OOS S Q Q A SQS O 2 STUDY OF PHOTOSYNTHESIS The study of photosynthesis is also rich with isotope applications. The overall photosynthesis reaction can be represented as 6CO2 6H2O 88n C6H12O6 6O2 18 In Section 13.5 we learned that the O isotope was used to determine the source of O2. The radioactive 14C isotope helped to determine the path of carbon in photosynthesis. Starting with 14CO2, it was possible to isolate the intermediate products during photosynthesis and measure the amount of radioactivity of each carbon-containing compound. In this manner the path from CO2 through various intermediate compounds to carbohydrate could be clearly charted. Isotopes, especially radioactive isotopes that are used to trace the path of the atoms of an element in a chemical or biological process, are called tracers. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.7 USES OF ISOTOPES 927 ISOTOPES IN MEDICINE Technetium was the first artificially prepared element. Tracers are used also for diagnosis in medicine. Sodium-24 (a emitter with a halflife of 14.8 h) injected into the bloodstream as a salt solution can be monitored to trace the flow of blood and detect possible constrictions or obstructions in the circulatory system. Iodine-131 (a emitter with a half-life of 8 days) has been used to test the activity of the thyroid gland. A malfunctioning thyroid can be detected by giving the patient a drink of a solution containing a known amount of Na131I and measuring the radioactivity just above the thyroid to see if the iodine is absorbed at the normal rate. Of course, the amounts of radioisotope used in the human body must always be kept small; otherwise, the patient might suffer permanent damage from the high-energy radiation. Another radioactive isotope of iodine, iodine-123 (a -ray emitter), is used to image the brain (Figure 23.18). Technetium is one of the most useful elements in nuclear medicine. Although technetium is a transition metal, all its isotopes are radioactive. Therefore, technetium does not occur naturally on Earth. In the laboratory it is prepared by the nuclear reactions 98 42 Mo 1 0n 88n 99 Mo 42 99 42 Mo 88n 99m Tc 43 0 1 where the superscript m denotes that the technetium-99 isotope is produced in its excited nuclear state. This isotope has a half-life of about 6 hours, decaying by radiation to technetium-99 in its nuclear ground state. Thus it is a valuable diagnostic tool. The patient either drinks or is injected with a solution containing 99mTc. By detecting the rays emitted by 99mTc, doctors can obtain images of organs such as the heart, liver, and lungs. A major advantage of using radioactive isotopes as tracers is that they are easy to detect. Their presence even in very small amounts can be detected by photographic techniques or by devices known as counters. Figure 23.19 is a diagram of a Geiger counter, an instrument widely used in scientific work and medical laboratories to detect radiation. FIGURE 23.18 A compound labeled with iodine-123 is used to image the brain. Left: A normal brain. Right: The brain of a patient with Alzheimer’s disease. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 928 NUCLEAR CHEMISTRY FIGURE 23.19 Schematic diagram of a Geiger counter. Radiation ( , , or rays) entering through the window ionized the argon gas to generate a small current flow between the electrodes. This current is amplified and is used to flash a light or operate a counter with a clicking sound. Cathode Anode Insulator Window Argon gas Amplifier and counter High voltage 23.8 BIOLOGICAL EFFECTS OF RADIATION In this section we will examine briefly the effects of radiation on biological systems. But first let us define quantitative measures of radiation. The fundamental unit of radioactivity is the curie (Ci); 1 Ci corresponds to exactly 3.70 1010 nuclear disintegrations per second. This decay rate is equivalent to that of 1 g of radium. A millicurie (mCi) is one-thousandth of a curie. Thus, 10 mCi of a carbon-14 sample is the quantity that undergoes (10 10 3)(3.70 1010) 3.70 108 disintegrations per second. The intensity of radiation depends on the number of disintegrations as well as on the energy and type of radiation emitted. One common unit for the absorbed dose of radiation is the rad (radiation absorbed dose), which is the amount of radiation that results in the absorption of 1 10 5 J per gram of irradiated material. The biological effect of radiation depends on the part of the body irradiated and the type of radiation. For this reason the rad is often multiplied by a factor called RBE (relative biological effectiveness). The product is called a rem (roentgen equivalent for man): 1 rem 1 rad 1 RBE Of the three types of nuclear radiation, particles usually have the least penetrating power. Beta particles are more penetrating than particles, but less so than rays. Gamma rays have very short wavelengths and high energies. Furthermore, since they carry no charge, they cannot be stopped by shielding materials as easily as and particles. However, if or emitters are ingested, their damaging effects are greatly aggravated because the organs will be constantly subject to damaging radiation at close range. For example, strontium-90, a emitter, can replace calcium in bones, where it does the greatest damage. Table 23.6 lists the average amounts of radiation an American receives every year. It should be pointed out that for short-term exposures to radiation, a dosage of 50 – 200 rem will cause a decrease in white blood cell counts and other complications, while Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 23.8 BIOLOGICAL EFFECTS OF RADIATION 929 TABLE 23.6 Average Yearly Radiation Doses for Americans SOURCE DOSE (mrem/yr)* Cosmic rays Ground and surroundings Human body† Medical and dental X rays Air travel Fallout from weapons tests Nuclear waste Total *1 mrem 1 millirem 1 10 3 20 – 50 25 26 50 – 75 5 5 2 133 – 188 rem. † The radioactivity in the body comes from food and air. a dosage of 500 rem or greater may result in death within weeks. Current safety standards permit nuclear workers to be exposed to no more than 5 rem per year and specify a maximum of 0.5 rem of human-made radiation per year for the general public. The chemical basis of radiation damage is that of ionizing radiation. Radiation of either particles or rays can remove electrons from atoms and molecules in its path, leading to the formation of ions and radicals. Radicals (also called free radicals) are molecular fragments having one or more unpaired electrons; they are usually shortlived and highly reactive. For example, when water is irradiated with rays, the following reactions take place: radiation H2O 88888n H2O H2O H2O 88n H3O e OH hydroxyl radical The electron (in the hydrated form) can subsequently react with water or with a hydrogen ion to form atomic hydrogen, and with oxygen to produce the superoxide ion, O2 (a radical): e Chromosomes are the parts of the cell that contain the genetic material (DNA). Back Forward Main Menu O2 88n O2 In the tissues the superoxide ions and other free radicals attack cell membranes and a host of organic compounds, such as enzymes and DNA molecules. Organic compounds can themselves be directly ionized and destroyed by high-energy radiation. It has long been known that exposure to high-energy radiation can induce cancer in humans and other animals. Cancer is characterized by uncontrolled cellular growth. On the other hand, it is also well established that cancer cells can be destroyed by proper radiation treatment. In radiation therapy, a compromise is sought. The radiation to which the patient is exposed must be sufficient to destroy cancer cells without killing too many normal cells and, it is hoped, without inducing another form of cancer. Radiation damage to living systems is generally classified as somatic or genetic. Somatic injuries are those that affect the organism during its own lifetime. Sunburn, skin rash, cancer, and cataracts are examples of somatic damage. Genetic damage means inheritable changes or gene mutations. For example, a person whose chromosomes have been damaged or altered by radiation may have deformed offspring. TOC Study Guide TOC Textbook Website MHHE Website 930 NUCLEAR CHEMISTRY Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Food Irradiation If you eat processed food, you have probably eaten ingredients exposed to radioactive rays. In the United States, up to 10 percent of herbs and spices are irradiated to control mold, zapped with X rays at a dose equal to 60 million chest X rays. Although food irradiation has been used in one way or another for more than 40 years, it faces an uncertain future in this country. Back in 1953 the U.S. Army started an experimental program of food irradiation so that deployed troops could have fresh food without refrigeration. The procedure is a simple one. Food is exposed to high levels of radiation to kill insects and harmful bacteria. It is then packaged in airtight containers, in which it can be stored for months without deterioration. The radiation sources for most food preservation are cobalt-60 and cesium-137, both of which are emitters, although X rays and electron beams can also be used to irradiate food. The benefits of food irradiation are obvious — it reduces energy demand by eliminating the need for refrigeration, and it prolongs the shelf life of various foods, which is of vital importance for poor countries. Yet there is considerable opposition to this procedure. First, there is a fear that irradiated food may itself become radioactive. No such evidence has been found. A more serious objection is that irradiation can de- Strawberries irradiated at 200 kilorads (right) are still fresh after 15 days’ storage at 4°C; those not irradiated are moldy. stroy the nutrients such as vitamins and amino acids. Furthermore, the ionizing radiation produces reactive species, such as the hydroxyl radical, which then react with the organic molecules to produce potentially harmful substances. Interestingly, the same effects are produced when food is cooked by heat. Food Irradiation Dosages and Their Effects† DOSAGE EFFECT Low dose (Up to 100 kilorad) Medium dose (100–1000 kilorads) High dose (1000 to 10,000 kilorads) Inhibits sprouting of potatoes, onions, garlics. Inactivates trichinae in pork. Kills or prevents insects from reproducing in grains, fruits, and vegetables after harvest. Delays spoilage of meat, poultry and fish by killing spoilage microorganism. Reduces salmonella and other food-borne pathogens in meat, fish, and poultry. Extends shelf life by delaying mold growth on strawberries and some other fruits. Sterilizes meat, poultry, fish, and some other foods. Kills microorganisms and insects in spices and seasoning. † Source: Chemical & Engineering News, May 5 (1986). Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 931 KEY EQUATION • SUMMARY OF FACTS AND CONCEPTS 1. For stable nuclei of low atomic number, the neutron-to-proton ratio is close to 1. For heavier stable nuclei, the ratio becomes greater than 1. All nuclei with 84 or more protons are unstable and radioactive. Nuclei with even atomic numbers tend to have a greater number of stable isotopes than those with odd atomic numbers. 2. Nuclear binding energy is a quantitative measure of nuclear stability. Nuclear binding energy can be calculated from a knowledge of the mass defect of the nucleus. 3. Radioactive nuclei emit particles, particles, positrons, or rays. The equation for a nuclear reaction includes the particles emitted, and both the mass numbers and the atomic numbers must balance. 4. Uranium-238 is the parent of a natural radioactive decay series that can be used to determine the ages of rocks. 5. Artificial radioactive elements are created by bombarding other elements with accelerated neutrons, protons, or particles. 6. Nuclear fission is the splitting of a large nucleus into two smaller nuclei and one or more neutrons. When the free neutrons are captured efficiently by other nuclei, a chain reaction can occur. 7. Nuclear reactors use the heat from a controlled nuclear fission reaction to produce power. The three important types of reactors are light water reactors, heavy water reactors, and breeder reactors. 8. Nuclear fusion, the type of reaction that occurs in the sun, is the combination of two light nuclei to form one heavy nucleus. Fusion takes place only at very high temperatures, so high that controlled large-scale nuclear fusion has so far not been achieved. 9. Radioactive isotopes are easy to detect and thus make excellent tracers in chemical reactions and in medical practice. 10. High-energy radiation damages living systems by causing ionization and the formation of free radicals. E ( m)c2 (23.1) Relation between mass defect and energy released. KEY WORDS Breeder reactor, p. 920 Critical mass, p. 917 Mass defect, p. 908 Moderators, p. 918 Nuclear Nuclear Nuclear Nuclear binding energy, p. 908 chain reaction, p. 917 fission, p. 916 fusion, p. 923 Nuclear transmutation, p. 904 Plasma, p. 924 Positron, p. 905 Radical, p. 929 Radioactive decay series, p. 911 Thermonuclear reaction, p. 923 Tracer, p. 926 Transuranium elements, p. 916 QUESTIONS AND PROBLEMS NUCLEAR REACTIONS Review Questions 23.1 How do nuclear reactions differ from ordinary chemical reactions? 23.2 What are the steps in balancing nuclear equations? 23.3 What is the difference between 0e and 0 ? 1 1 23.4 What is the difference between an electron and a positron? Problems 23.5 Complete the following nuclear equations and identify X in each case: Back Forward Main Menu TOC X (a) 26 Mg 1p 88n 4 12 1 2 (b) 59 Co 2H 88n 60 Co X 27 1 27 (c) 235 U 1 n 88n 94 Kr 139 Ba 3X 0 36 56 92 (d) 53Cr 4 88n 1 n X 24 2 0 (e) 20 O 88n 20 F X 9 8 23.6 Complete the following nuclear equations and identify X in each case: (a) 135 I 88n 135 Xe X 54 53 X (b) 40 K 88n 0 1 19 (c) 59 Co 1 n 88n 56 Mn X 27 0 25 Study Guide TOC Textbook Website MHHE Website 932 NUCLEAR CHEMISTRY (d) 235 92 U 1 0n 88n 99 Sr 40 135 52 Te Problems 2X 23.23 Fill in the blanks in the following radioactive decay series: NUCLEAR STABILITY Review Questions (a) 23.7 State the general rules for predicting nuclear stability. 23.8 What is the belt of stability? 23.9 Why is it impossible for the isotope 2 He to exist? 2 23.10 Define nuclear binding energy, mass defect, and nucleon. 23.11 How does Einstein’s equation, E mc2, allow us to calculate nuclear binding energy? 23.12 Why is it preferable to use nuclear binding energy per nucleon for a comparison of the stabilities of different nuclei? (b) H(g) H(g) 88n H2(g) H° 436.4 kJ calculate the change in mass (in kg) per mole of H2 formed. 23.18 Estimates show that the total energy output of the sun is 5 1026 J/s. What is the corresponding mass loss in kg/s of the sun? 23.19 Calculate the nuclear binding energy (in J) and the binding energy per nucleon of the following isotopes: (a) 7 Li (7.01600 amu) and (b) 35 Cl (34.95952 amu). 3 17 23.20 Calculate the nuclear binding energy (in J) and the binding energy per nucleon of the following isotopes: (a) 4 He (4.0026 amu) and (b) 184 W (183.9510 amu). 2 74 Th 88n _____ 88n _____ 88n 228Th 235 U 88n _____ 88n _____ 88n 227Ac (c) _____ 88n 233Pa 88n _____ 88n _____ 23.24 A radioactive substance undergoes decay as follows: TIME (DAYS) 23.25 23.26 23.27 23.28 23.29 23.30 MASS (g) 0 1 2 3 4 5 6 Problems 23.13 The radius of a uranium-235 nucleus is about 7.0 10 3 pm. Calculate the density of the nucleus in g/cm3. (Assume the atomic mass is 235 amu.) 23.14 For each pair of isotopes listed, predict which one is less stable: (a) 6 Li or 9 Li, (b) 23 Na or 25 Na, 3 3 11 11 (c) 48 Ca or 48 Sc. 20 21 23.15 For each pair of elements listed, predict which one has more stable isotopes: (a) Co or Ni, (b) F or Se, (c) Ag or Cd. 23.16 In each pair of isotopes shown, indicate which one you would expect to be radioactive: (a) 20 Ne and 10 17 40 45 95 92 10 Ne, (b) 20 Ca and 20 Ca, (c) 42 Mo and 43 Tc, 195 196 209 242 (d) 80 Hg and 80 Hg, (e) 83 Bi and 96 Cm. 23.17 Given that 232 500 389 303 236 184 143 112 Calculate the first-order decay constant and the halflife of the reaction. The radioactive decay of T1-206 to Pb-206 has a halflife of 4.20 min. Starting with 5.00 1022 atoms of T1-206, calculate the number of such atoms left after 42.0 min. A freshly isolated sample of 90Y was found to have an activity of 9.8 105 disintegrations per minute at 1:00 P.M. on December 3, 1992. At 2:15 P.M. on December 17, 1992, its activity was redetermined and found to be 2.6 104 disintegrations per minute. Calculate the half-life of 90Y. Why do radioactive decay series obey first-order kinetics? In the thorium decay series, thorium-232 loses a total of 6 particles and 4 particles in a 10-stage process. What is the final isotope produced? Strontium-90 is one of the products of the fission of uranium-235. This strontium isotope is radioactive, with a half-life of 28.1 yr. Calculate how long (in yr) it will take for 1.00 g of the isotope to be reduced to 0.200 g by decay. Consider the decay series A 88n B 88n C 88n D NATURAL RADIOACTIVITY Review Questions 23.21 Discuss factors that lead to nuclear decay. 23.22 Outline the principle for dating materials using radioactive isotopes. Back Forward Main Menu TOC where A, B, and C are radioactive isotopes with halflives of 4.50 s, 15.0 days, and 1.00 s, respectively, and D is nonradioactive. Starting with 1.00 mole of A, and none of B, C, or D, calculate the number of moles of A, B, C, and D left after 30 days. Study Guide TOC Textbook Website MHHE Website 933 QUESTIONS AND PROBLEMS NUCLEAR TRANSMUTATION USES OF ISOTOPES Review Questions Problems 23.31 What is the difference between radioactive decay and nuclear transmutation? 23.32 How is nuclear transmutation achieved in practice? 23.47 Describe how you would use a radioactive iodine isotope to demonstrate that the following process is in dynamic equilibrium: PbI2(s) 34 Pb2 (aq) Problems 23.33 Write balanced nuclear equations for the following reactions and identify X: (a) X(p, )12C, (b) 27Al(d, )X, (c) 55Mn(n, )X 13 25 6 23.34 Write balanced nuclear equations for the following reactions and identify X: (a) 80Se(d,p)X, (b) X(d,2p)9Li, (c) 10B(n, )X 34 3 5 23.35 Describe how you would prepare astatine-211, starting with bismuth-209. 23.36 A long-cherished dream of alchemists was to produce gold from cheaper and more abundant elements. This dream was finally realized when 198Hg was converted 80 into gold by neutron bombardment. Write a balanced equation for this reaction. NUCLEAR FISSION Review Questions 23.37 Define nuclear fission, nuclear chain reaction, and critical mass. 23.38 Which isotopes can undergo nuclear fission? 23.39 Explain how an atomic bomb works. 23.40 Explain the functions of a moderator and a control rod in a nuclear reactor. 23.41 Discuss the differences between a light water and a heavy water nuclear fission reactor. What are the advantages of a breeder reactor over a conventional nuclear fission reactor? 23.42 No form of energy production is without risk. Make a list of the risks to society involved in fueling and operating a conventional coal-fired electric power plant, and compare them with the risks of fueling and operating a nuclear fission-powered electric plant. NUCLEAR FUSION Review Questions 23.43 Define nuclear fusion, thermonuclear reaction, and plasma. 23.44 Why do heavy elements such as uranium undergo fission while light elements such as hydrogen and lithium undergo fusion? 23.45 How does a hydrogen bomb work? 23.46 What are the advantages of a fusion reactor over a fission reactor? What are the practical difficulties in operating a large-scale fusion reactor? Back Forward Main Menu TOC 2I (aq) 23.48 Consider the following redox reaction: IO4 (aq) 2I (aq) H2O(l) 88n I2(s) IO3 (aq) 2OH (aq) When KIO4 is added to a solution containing iodide ions labeled with radioactive iodine-128, all the radioactivity appears in I2 and none in the IO3 ion. What can you deduce about the mechanism for the redox process? 23.49 Explain how you might use a radioactive tracer to show that ions are not completely motionless in crystals. 23.50 Each molecule of hemoglobin, the oxygen carrier in blood, contains four Fe atoms. Explain how you would use the radioactive 59 Fe (t 1 46 days) to show 26 2 that the iron in a certain food is converted into hemoglobin. ADDITIONAL PROBLEMS 23.51 How does a Geiger counter work? 23.52 Nuclei with an even number of protons and an even number of neutrons are more stable than those with an odd number of protons and/or an odd number of neutrons. What is the significance of the even numbers of protons and neutrons in this case? 23.53 Tritium, 3H, is radioactive and decays by electron emission. Its half-life is 12.5 yr. In ordinary water the ratio of 1H to 3H atoms is 1.0 1017 to 1. (a) Write a balanced nuclear equation for tritium decay. (b) How many disintegrations will be observed per minute in a 1.00-kg sample of water? 23.54 (a) What is the activity, in millicuries, of a 0.500-g sample of 237 Np? (This isotope decays by -particle 93 emission and has a half-life of 2.20 106 yr.) (b) Write a balanced nuclear equation for the decay of 237 93 Np. 23.55 The following equations are for nuclear reactions that are known to occur in the explosion of an atomic bomb. Identify X. (a) 235 U 1 n 88n 140 Ba 31 n X 92 0 56 0 (b) 235 U 1 n 88n 144 Cs 90 Rb 2X 92 0 55 37 (c) 235 U 1 n 88n 87 Br 31 n X 92 0 35 0 (d) 235 U 1 n 88n 160 Sm 72 Zn 4X 92 0 62 30 23.56 Calculate the nuclear binding energies, in J/nucleon, for the following species: (a) 10B (10.0129 amu), Study Guide TOC Textbook Website MHHE Website 934 NUCLEAR CHEMISTRY 23.57 23.58 23.59 23.60 23.61 23.62 23.63 23.64 23.65 23.66 23.67 Back (b) 11B (11.00931 amu), (c) 14N (14.00307 amu), (d) 56 Fe (55.9349 amu). Write complete nuclear equations for the following decay; processes: (a) tritium, 3H, undergoes (b) 242Pu undergoes -particle emission; (c) 131I undergoes decay; (d) 251Cf emits an particle. The nucleus of nitrogen-18 lies above the stability belt. Write an equation for a nuclear reaction by which nitrogen-18 can achieve stability. Why is strontium-90 a particularly dangerous isotope for humans? How are scientists able to tell the age of a fossil? After the Chernobyl accident, people living close to the nuclear reactor site were urged to take large amounts of potassium iodide as a safety precaution. What is the chemical basis for this action? Astatine, the last member of Group 7A, can be prepared by bombarding bismuth-209 with particles. (a) Write an equation for the reaction. (b) Represent the equation in the abbreviated form as discussed in Section 23.4. To detect bombs that may be smuggled onto airplanes, the Federal Aviation Administration (FAA) will soon require all major airports in the United States to install thermal neutron analyzers. The thermal neutron analyzer will bombard baggage with low-energy neutrons, converting some of the nitrogen-14 nuclei to nitrogen-15, with simultaneous emission of rays. Because nitrogen content is usually high in explosives, detection of a high dosage of rays will suggest that a bomb may be present. (a) Write an equation for the nuclear process. (b) Compare this technique with the conventional Xray detection method. Explain why achievement of nuclear fusion in the laboratory requires a temperature of about 100 million degrees Celsius, which is much higher than that in the interior of the sun (15 million degrees Celsius). Tritium contains one proton and two neutrons. There is no proton-proton repulsion present in the nucleus. Why, then, is tritium radioactive? The carbon-14 decay rate of a sample obtained from a young tree is 0.260 disintegration per second per gram of the sample. Another wood sample prepared from an object recovered at an archaeological excavation gives a decay rate of 0.186 disintegration per second per gram of the sample. What is the age of the object? The usefulness of radiocarbon dating is limited to objects no older than 50,000 years. What percent of the carbon-14, originally present in the sample, remains after this period of time? Forward Main Menu TOC 23.68 The radioactive potassium-40 isotope decays to argon-40 with a half-life of 1.2 109 yr. (a) Write a balanced equation for the reaction. (b) A sample of moon rock is found to contain 18 percent potassium40 and 82 percent argon by mass. Calculate the age of the rock in years. 23.69 Both barium (Ba) and radium (Ra) are members of Group 4A and are expected to exhibit similar chemical properties. However, Ra is not found in barium ores. Instead, it is found in uranium ores. Explain. 23.70 Nuclear waste disposal is one of the major concerns of the nuclear industry. In choosing a safe and stable environment to store nuclear wastes, consideration must be given to the heat released during nuclear dedecay of 90Sr cay. As an example, consider the (89.907738 amu): 90 38Sr The 90 88n 90Y 39 23.72 23.73 23.74 23.75 t1 2 28.1 yr Y (89.907152 amu) further decays as follows: 90 39Y 23.71 0 1 88n 90Zr 40 0 1 t1 2 64 h Zirconium-90 (89.904703 amu) is a stable isotope. (a) Use the mass defect to calculate the energy released (in joules) in each of the above two decays. (The mass of the electron is 5.4857 10 4 amu.) (b) Starting with one mole of 90Sr, calculate the number of moles of 90Sr that will decay in a year. (c) Calculate the amount of heat released (in kilojoules) corresponding to the number of moles of 90Sr decayed to 90 Zr in (b). Which of the following poses a greater health hazard: A radioactive isotope with a short half-life or a radioactive isotope with a long half-life? Explain. [Assume same type of radiation ( or ) and comparable energetics per particle emitted.] As a result of being exposed to the radiation released during the Chernobyl nuclear accident, the dose of iodine-131 in a person’s body is 7.4 mC (1 mC N to cal1 10 3 Ci). Use the relationship rate culate the number of atoms of iodine-131 this radioactivity corresponds. (The half-life of I-131 is 8.1 d.) Referring to the Chemistry in Action essay on p. 930, why is it highly unlikely that irradiated food would become radioactive? From the definition of curie, calculate Avogadro’s number. Given that the molar mass of Ra-226 is 226.03 g/mol and that it decays with a half life of 1.6 103 yr. Since 1994, elements 110, 111, and 112 have been synthesized. Element 110 was created by bombarding 208Pb with 62Ni; element 111 was created by bom- Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 23.76 23.77 23.78 23.79 23.80 Back barding 209Bi with 64Ni; and element 112 was created by bombarding 208Pb with 66Zn. Write an equation for each synthesis. Predict the chemical properties of these elements. Use X for element 110, Y for element 111, and Z for element 112. Sources of energy on Earth include fossil fuels, geothermal, gravitational, hydroelectric, nuclear fission, nuclear fusion, solar, wind. Which of these have a “nuclear origin,” either directly or indirectly? A person received an anonymous gift of a decorative cube which he placed on his desk. A few months later he became ill and died shortly afterward. After investigation, the cause of his death was linked to the box. The box was air-tight and had no toxic chemicals on it. What might have killed the man? Identify two of the most abundant radioactive elements that exist on Earth. Explain why they are still present? (You may need to consult a handbook of chemistry.) (a) Calculate the energy released when an U-238 isotope decays to Th-234. The atomic masses are given by: U-238: 238.0508 amu; Th-234: 234.0436 amu; He-4: 4.0026 amu. (b) The energy released in (a) is transformed into the kinetic energy of the recoiling Th-234 nucleus and the particle. Which of the two will move away faster? Explain. Cobalt-60 is an isotope used in diagnostic medicine and cancer treatment. It decays with ray emission. Calculate the wavelength of the radiation in nanometers if the energy of the ray is 2.4 10 13 J/photon. Forward Main Menu TOC 935 23.81 Am-241 is used in smoke detectors because it has a long half-life (458 yr) and its emitted particles are energetic enough to ionize air molecules. Given the schematic diagram of a smoke detector below, explain how it works. Current 241Am Battery 23.82 The constituents of wine contain, among others, carbon, hydrogen, and oxygen atoms. A bottle of wine was sealed about 6 years ago. To confirm its age, which of the isotopes would you choose in a radioactive dating study? The half-lives of the isotopes are: 13C: 5730 yr; 15O: 124 s; 3H: 12.5 yr. Assume that the activities of the isotopes were known at the time the bottle was sealed. 23.83 Name two advantages of a nuclear-powered submarine over a conventional submarine. 23.1 78 Se. 23.2 2.63 34 J/nucleon. 23.3 106 Pd 4 88n 46 2 Answers to Practice Exercises: 10 10 J; 1.26 109 1 47 Ag 1 p. Study Guide TOC 10 12 Textbook Website MHHE Website C HEMICAL M YSTERY The Art Forgery of the Century H an van Meegeren must be one of the few forgers ever to welcome technical analysis of his work. In 1945 he was captured by the Dutch police and accused of selling a painting by the Dutch artist Jan Vermeer (1632 – 1675) to Nazi Germany. This was a crime punishable by death. Van Meegeren claimed that not only was the painting in question, entitled The Woman Taken in Adultery, a forgery, but he had also produced other “Vermeers.” To prove his innocence, van Meegeren created another Vermeer to demonstrate his skill at imitating the Dutch master. He was acquitted of charges of collaboration with the enemy, but was convicted of forgery. He died of a heart attack before he could serve the one-year sentence. For twenty years after van Meegeren’s death art scholars debated whether at least one of his alleged works, Christ and His Disciples at Emmaus, was a fake or a real Vermeer. The mystery was solved in 1968 using a radiochemical technique. White lead — lead hydroxy carbonate [Pb3(OH)2(CO3)2] — is a pigment used by artists for centuries. The metal in the compound is extracted from its ore, galena (PbS), which contains uranium and its daughter products in radioactive equilibrium with it. By radioactive equilibrium we mean that a particular isotope along the decay series is formed from its precursor as fast as it breaks down by decay, and so its concentration (and its radioactivity) remains constant with time. This radioactive equilibrium is disturbed in the chemical extraction of lead from its ore. Two isotopes in the uranium decay series are of particular importance in this process: 226Ra (t 1 1600 yr) and 210Pb 2 (t 1 21 yr). (See Table 23.3.) Most 226Ra is removed during the extraction of lead 2 from its ore, but 210Pb eventually ends up in the white lead, along with the stable isotope of lead (206Pb). No longer supported by its relatively long-lived ancestor, 226Ra, 210 Pb begins to decay without replenishment. This process continues until the 210Pb activity is once more in equilibrium with the much smaller quantity of 226Ra that survived the separation process. Assuming the concentration ratio of 210Pb to 226Ra is 100:1 in the sample after extraction, it would take 270 years to reestablish radioactive equilibrium for 210Pb. If Vermeer did paint Emmaus around the mid-seventeenth century, the radioactive equilibrium would have been restored in the white lead pigment by 1960. But this was not the case. Radiochemical analysis showed that the paint used was less than one hundred years old. Therefore, the painting could not have been the work of Vermeer. 936 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website “Christ and His Disciples at Emmaus,” a painting attributed to Han van Meegeren. CHEMICAL CLUES 1. 2. Write equations for the decay of 226Ra and 210Pb. Consider the following consecutive decay series: A 88n B 88n C where A and B are radioactive isotopes and C is a stable isotope. Given that the half-life of A is 100 times that of B, plot the concentrations of all three species versus time on the same graph. If only A was present initially, which species would reach radioactive equilibrium? 3. The radioactive decay rates for 210Pb and 226Ra in white lead paint taken from Emmaus in 1968 were 8.5 and 0.8 disintegrations per minute per gram of lead (dpm/g), respectively. (a) How many half lives of 210Pb had elapsed between 1660 and 1968? (b) If Vermeer had painted Emmaus, what would have been the decay rate of 210Pb in 1660? Comment on the reasonableness of this rate value. 4. To make his forgeries look authentic, van Meegeren re-used canvases of old paintings. He rolled one of his paintings to create cracks in the paint to resemble old works. X-ray examination of this painting showed not only the underlying painting, but also the cracks in it. How did this discovery reveal to the scientists that the painting on top was of a more recent origin? 937 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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