SGCh23 - Chapter Twenty-Three NUCLEAR CHEMISTRY • • •...

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Unformatted text preview: Chapter Twenty-Three NUCLEAR CHEMISTRY • • • • • • The Nature of Nuclear Reactions The Stability of Nuclei Natural Radioactivity Nuclear Transmutation Nuclear Fission and Fusion Biological Effects of Radiation THE NATURE OF NUCLEAR REACTIONS STUDY OBJECTIVES 1. 2. Write balanced nuclear equations for radioactive decay and nuclear transmutation processes. Compare chemical reactions and nuclear reactions. Nuclear Reactions and Nuclear Equations. This chapter emphasizes changes that occur within the nucleus of an atom. The textbook discusses two types of nuclear reactions: radioactive decay and nuclear transmutation. Radioactivity is described as the spontaneous emission of particles and/or radiation by unstable atomic nuclei. These processes often result in the formation of a new element. The kinds of particles emitted from various nuclei are shown in Table 23.1. Table 23.1 Particles From Radioactive Decay _____________________________________________________ Particle Mass (amu) Charge Symbol _____________________________________________________ alpha 4.0 +2 beta 0.0005 –1 positron 0.0005 +1 4 4 2 α or 2 He 0 0 β or –1 e –1 0 0 β or +1 e +1 0 0γ gamma 0 0 _____________________________________________________ A nucleus can also undergo change by nuclear transmutation. In this process a nucleus reacts with another nucleus, an elementary particle, or a photon (gamma particle) to produce one or more new nuclei. Radioactive decay and nuclear transmutation processes are described by nuclear equations. These equations use isotopic and elementary particle symbols to represent the reactants and products of nuclear reactions. For example, in the first nuclear transmutation ever observed (in 1919) alpha particles were used to bombard nitrogen-14 nuclei. The observed products were atoms of oxygen-17 and protons. The nuclear equation is 14 4 17 1 N + 2 He → 8 O + 1 H 7 445 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 446 / Nuclear Chemistry Note that the balancing rules given in the textbook are followed in this equation: 1. 2. The sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products (conservation of mass number). 14 + 4 = 18 = 17 + 1 The sum of the nuclear charges of the reactants must equal the sum of the nuclear charges of the products (conservation of atomic number). 7 + 2 = 9 = 8 + 1 These rules will be illustrated further in the example problems. Comparison of Chemical and Nuclear Reactions. Table 23.1 of the text lists a number of comparisons between chemical and nuclear processes. Keep in mind that in chemical reactions the number of atoms of each element is conserved. Only changes in chemical bonding occur. However, in nuclear reactions the composition of the atomic nucleus is altered, and so elements are converted from one to another. Since a nucleus is so extremely small, and in some cases contains large numbers of positively charged protons, the energy changes associated with nuclear changes are much greater than energy changes in chemical reactions. _______________________________________________________________________________ EXAMPLE 23.1 Nuclear Equations Complete the following nuclear equations. a. b. 14 1 14 N + 0 n → 6 C + ___ 7 226 Ra → 4 α + ___ 2 •Method of Solution a. According to rule 1, the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products. 14 + 1 = 15 = 14 + A. Therefore, the unknown product will have a mass number of 1. According to rule 2, the sums of the nuclear charges on both sides of the equation must be the same. 7 + 0 = 7 = 6 + Z. Therefore the nuclear charge of the unknown product must be 1, making it a proton. 14 1 14 1 N+0n → 6 C+1p 7 b. Note that the atomic number of radium is missing. The periodic table indicates that Ra is element number 88. Balancing the mass numbers first, we find that the unknown product must have a mass number of 222. Balancing the nuclear charges next, we find that the atomic number of the unknown must be 86. Element number 86 is radon. 226 4 222 88 Ra → 2 α + 86 Rn _______________________________________________________________________________ EXERCISES 1. 2. 3. Write symbols for alpha particles, beta particles, and gamma rays. How many protons (p), neutrons (n), and electrons (e) does an atom of the radioisotope phosphorus-32 contain? Complete the following nuclear reactions: 14 1 4 9 14 a. 7 N + 0 n → 6 C + _____ 210 4 b. Po → 2 α + ____ 4. Back 12 c. 2 He + 4 Be → 6 C + ____ 15 0 d. 7 N → ____ + –1 β When a nucleus decays by positron emission, the atomic number of the decay product will be (higher or lower) than the original radioisotope. Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 447 THE STABILITY OF NUCLEI STUDY OBJECTIVES 1. 2. 3. Compare relative stabilities of given nuclei by applying stability rules. Predict, using the n : p ratio, whether a given isotope will decay by beta decay or positron decay. Calculate the nuclear binding energies of given nuclei. Nuclear Stability. Little is known about the forces that hold the nucleus together. However, some interesting facts emerge if we examine the numbers of protons and neutrons found in those nuclei that are stable. Nuclei can be classified according to whether they contain even or odd numbers of protons and neutrons. The number of stable isotopes of each of the four types of nuclei classified in this way are shown in Table 23.2. Table 23.2 The Number of Stable Isotopes _______________________________________________ Number of protons even even odd odd Number of neutrons even odd even odd _______________________________________________ Number of stable isotopes 157 52 50 8 _______________________________________________ The following rules are useful in predicting nuclear stability: 1. 2. 3. Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of these particles. Nuclei that contain certain specific numbers of protons and neutrons within a nucleus ensure an extra degree of stability. These so-called magic numbers for protons and for neutrons are 2, 8, 20, 28, 50, 82, and 126. Nuclei with even numbers of protons or neutrons are generally more stable than those with odd numbers of these particles. All isotopes of elements after bismuth (Z = 83) are radioactive. The Belt of Stability. The principal factor for determining whether a nucleus is stable is the neutron to proton ratio. Figure 23.1 shows a plot of the number of neutrons versus the number of protons in various isotopes. Each dot represents a stable isotope. The stable nuclei are located in an area of the graph known as the belt of stability. In this figure we see that at low atomic numbers stable nuclei possess a neutron to proton ratio of about 1.0. Above Z = 20 the number of neutrons always exceeds the number of protons in stable isotopes. The n : p ratio increases to about 1.5 at the upper end of the belt of stability. If you were given the symbol of a radioisotope, without any experience, it would be impossible to tell its mode of decay. But with knowledge of the belt of stability, you can make accurate predictions of the expected mode of decay. Isotopes with too many neutrons lie above the belt of stability. The nuclei of these isotopes decay in such a way as to lower their n : p ratio. Thus one of its neutrons may decay into a proton and a beta particle. 0 1 0 1 n → 1 p + –1 β The proton remains in the nucleus, and the beta particle is emitted from the atom. The loss of a neutron and the gain of a proton produces a new isotope with two important properties. It has a lower n : p ratio, and thus is more likely to be stable. Also, the daughter product has an atomic number that is one greater than the decaying isotope due to the additional proton. Consider the decay of carbon-14 for example. Carbon-14 is continually produced in the upper atmosphere by the interaction of cosmic rays with nitrogen. Carbon-14 has a higher n : p ratio than either of carbon's stable isotopes (C-12 and C-13), and decays by beta decay. 14 14 0 6 C → 7 N + –1 β Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 448 / Nuclear Chemistry 14 Note that the product isotope 7 N is one atomic number greater than carbon. Also, it is stable. Its n : p ratio is 1.0. n/p 1.50 120 Number of neutrons 140 o 100 80 n/p 1.28 n/p=1 60 o n/p 1.14 40 o 20 0 20 40 60 80 Number of protons 100 Figure 23.1 A graph showing all the stable isotopes when plotted as neutron number versus proton number. The shaded area represents the belt of stability. In order for isotopes with Z = 20 to be stable, the ratio of n : p must be about 1.1. For isotopes with Z = 40 to be stable, the ratio of n : p must be about 1.3 : 1.0. For isotopes with Z = 80 to be stable, the ratio of n : p must be about 1.5 : 1.0. Isotopes with too many protons have a low n : p ratio and lie below the belt of stability. These isotopes tend to decay by positron emission because this process produces a new isotope with a higher n : p ratio. During 0 positron emission a proton emits a positron, +1 β, and becomes a neutron. 1 1 0 1 p → 0 n + +1 β The neutron remains in the nucleus, and the positron is ejected from the atom. Thus, a product nucleus will contain one less proton and one more neutron than the parent nucleus. The n : p ratio increases due to positron decay. Electron capture accomplishes the same end, that is, a higher n : p ratio. Some nuclei decay by capturing an orbital electron of the atom. 1 0 1 1 p + +1 e → 0 n Lanthanum-138, a naturally occurring isotope with an abundance of 0.089 percent, decays by electron capture. 138 138 0 57 La + –1 e → 58 Ba Electron capture is accompanied by X-ray emission. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 449 Nuclear Binding Energy. One of the important consequences of Einstein's theory of relativity was the discovery of the equivalence of mass and energy. The total energy content (E) of a system of mass m is given by Einstein's theory: E = mc2 where c is the velocity of light (3.0 × 10 8 m/s). Therefore, the mass of a nucleus is a direct measure of its energy content. It was discovered in the 1930s that the measured mass of a nucleus is always smaller than the sum of the separate masses of its constituent nucleons. This difference in mass is called the mass defect. When the mass defect is expressed as energy by applying Einstein's equation, it is called the binding energy of the nucleus. The binding energy is the energy required to break up a nucleus into its component protons and neutrons. The binding energy provides a quantitative measure of nuclear stability. The greater the binding energy the more stable the nucleus is toward decomposition. 17 In terms of the 8 O nucleus for instance, the binding energy (BE) is the energy required for the process 17 1 1 8 O + BE → 8 1 p + 9 0 n The mass defect ∆m is equal to the total mass of the products minus the total mass of the reactants. 17 ∆m = [8(proton mass) + 9(neutron mass)] – ( 8 O nuclear mass) Here we can substitute atomic masses, which include the electrons, for the nuclear masses. 1 17 ∆m = [8(1 H atomic mass) + 9(neutron mass)] – ( 8 O atomic mass) This works because the mass of 8 electrons in the 8 hydrogen atoms is canceled by the mass of the 8 electrons in the oxygen atom. 1 8(1 H atomic mass) = 8(proton mass) + 8(electron mass) 17 17 ( 8 O atom mass) = ( 8 O nuclear mass)+ 8(electron mass)] Continuing by using atomic masses given in Chapter 23 of the text and Table 23.3 below. 1 17 ∆m = [8(1 H atomic mass) + 9(neutron mass)] – ( 8 O atomic mass) ∆m = [8(1.007825 amu) + 9(1.008665 amu)] – (16.999131 amu) = 0.141454 amu The eight protons and nine neutrons have more mass than the oxygen-17 nucleus. The binding energy is: ∆E = ∆mc2 The calculation of binding energy will be illustrated in Example 23.4. In comparing the stability of any two different nuclei, we must account for the different numbers of nucleons per nucleus. A satisfactory comparison of nuclear stabilities can be made by using the binding energy per nucleon, that is, the binding energy of each nucleus divided by the total number of nucleons in the nucleus. This is one of the most important properties of a nucleus. When the BE per nucleon is plotted as a function of the atomic mass we get the curve of binding energy as shown in Figure 23.2 of the text. Note that at first it rises rapidly with increasing atomic mass, and then it reaches a maximum at mass 56. Above mass 56, the binding energy drops slowly as atomic mass increases. Table 23.1 compares the total binding energy and the binding energy per nucleon for several isotopes. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 450 / Nuclear Chemistry Table 23.3 Binding Energies of Selected Isotopes ______________________________________________________________ Mass (amu) Binding Energy (J) __ Total Per Nucleon ______________________________________________________________ 2 1H 3 2 He 4 2 He 16 O 8 17 O 8 56 26 Fe 238 U 92 2.01410 3.57 × 10–13 1.78 × 10–13 3.01603 1.24 × 10–12 4.13 × 10–13 4.00260 4.52 × 10–12 1.13 × 10–12 15.99491 2.04 × 10–11 1.28 × 10–12 16.999131 2.10 × 10–11 1.24 × 10–12 55.934939 7.90 × 10–11 1.41 × 10–12 238.0508 2.89 × 10–10 1.22 × 10–12 _______________________________________________________________ _______________________________________________________________________________ EXAMPLE 23.2 Nuclear Stability Using the stability rules, rank the following isotopes in order of increasing nuclear stability. 39 20 Ca 40 20 Ca 30 15 P •Method of Solution Phosphorus-30 should be the least stable because it has odd numbers of both protons and neutrons. Calcium-39 has an even number of protons (20) and an odd number of neutrons. With an even number of protons and a "magic number" at that, it should be more stable than P-30. Of the three isotopes Calcium-40 should be the most stable. It has an even number of protons and of neutrons. Both numbers are "magic numbers." 30 39 40 15 P < 20 Ca < 20 Ca _______________________________________________________________________________ EXAMPLE 23.3 Types of Radioactive Decay The only stable isotope of sodium is sodium-23. What type of radioactivity would you expect from sodium-25? •Method of Solution Sodium-23 has 11 protons and 12 neutrons and is in the belt of stability. Sodium-25 must have two more 0 neutrons, and so has a higher n : p ratio than the stable isotope. Sodium-25 will decay by –1 β emission. _______________________________________________________________________________ EXAMPLE 23.4 Nuclear Binding Energy 3 Calculate the nuclear binding energy of the light isotope of helium, helium-3. The atomic mass of 2 He is 3.01603 amu. •Method of Solution The binding energy is the energy required for the process 3 1 1 2 He → 2 1 p + 0 n Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 451 3 where ∆m = [2(proton mass) + (neutron mass)] – 2 He (nuclear mass) The mass difference is calculated using atomic masses: 1 3 ∆m = [2(1 H atomic mass) + (neutron mass)] – ( 2 He atomic mass) ∆m = [2(1.007825 amu) + 1.008665 amu] – 3.01603 amu ∆m = 8.29 × 10 –3 amu Using Einstein's equation: ∆E = (∆m)c2 = (8.29 × 10–3 amu) × (3.00 × 108 m s–1)2 = 7.46 × 10 14 amu m2 s–2 = 7.46 × 10 14 amu m2 s–2 × 1.00 kg 1J × 6.022 × 1026 amu 1 kg m2 s–2 The binding energy is: ∆E = 1.24 × 10–12 J/atom 3 Each 2 He atom contains three nucleons. The binding energy per nucleon is: BE per nucleon = 1.24 × 10–12 J/atom 3 nucleons/atom The binding energy per nucleon is: ∆E = 4.13 × 10–13 J/nucleon •Comment By combining the above conversion factors into one constant, the number of steps in future calculations can be lessened. ? J/amu = (3.00 × 10 8 m s–1)2 × 1.00 kg 1J × 6.022 × 1026 amu 1 kg m2 s–2 = 1.49 × 10 –10 J/amu This constant is a useful factor relating energy to mass in amu. Applying this to the mass defect (0.14145 amu) 17 calculated earlier in the discussion for 8 O, the binding energy is 2.10 × 10–11 J/atom. See Table 23.3. _______________________________________________________________________________ EXERCISES 5. For each pair of nuclei, predict which one is the more stable. 3 4 a. 2 He or 2 He. 26 27 b. 13 Al or 13 Al. 6. Fluorine has only one stable isotope, fluorine-19. a. The nucleus of fluorine-18 lies below the belt of stability. Write an equation for the decay of fluorine-18. b. The nucleus of fluorine-21 lies above the belt of stability. Write an equation for the decay of fluorine-21. 7. How many atomic mass units are in one kilogram? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 452 / Nuclear Chemistry 8. Which has more mass, the nucleus of aluminum–27 or 13 protons and 14 neutrons? 27 9. What is the binding energy of 13 Al? Calculate its binding energy per nucleon. The atomic mass of aluminum-27 is 26.981541 amu. 10. How much energy is released when one Po-214 atom decays by alpha emission? 214 210 4 84 Po → 2 α + 82 Pb Given the atomic masses: Pb-214 = 213.99519 amu, Po-210 = 209.98286 amu, and He-4 = 4.00260 amu. NATURAL RADIOACTIVITY STUDY OBJECTIVES 1. 2. 3. Describe general features of the uranium series. Use the first-order rate equation to determine the amount of a radioisotope remaining after a given time. Calculate the age of a rock sample given information about the amount of a particular radioisotope present in the rock. Natural Radioactivity. A number of isotopes exist in nature that have an n : p ratio that places them outside the belt of stability. Radioisotopes that occur naturally on Earth give rise to natural radioactivity. Uranium, thorium, radon, potassium-40, carbon-12, and tritium (H-3) are naturally occurring radioisotopes. Uranium, which is fairly abundant in Earth's crust, is the start of a radioactive decay series. The series is a sequence of decay reactions that change U-238 ultimately to a stable isotope of lead. The uranium series is shown in Table 23.3 of the textbook. The uranium decay series includes all the elements between lead and uranium. All of the radioisotopes in this series decay by alpha or beta emission. In the first step U-238 decays by alpha emission. 238 234 4 U → 90 Th + 2 α 92 The decay product thorium-234 is also radioactive and decays by beta emission. 234 234 0 Th → 91 Pa + –1 β 90 The product is also radioactive and the series continues by a number of steps to end at Pb-206. Rates of Decay. Radioactive decay rates obey first-order kinetics. decay rate = number of atoms disintegrating per unit time = λN where λ is the first-order rate constant, called the decay constant, and N is the number of atoms of the particular radioisotope present in the sample. Recall that the half-life is related to the rate constant by t1/2 = 0.693/λ The integrated first-order equation is N0 ln = λt N Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 453 Where N is the number of atoms of the radioisotope present in the sample after time t has elapsed, and N0 is the number of atoms of the radioisotope present initially. Normally N0 , N, and t are known, and this information is used to calculate the rate constant, λ. U 238 Loss of 4 α Th 230 Mass number 234 Loss of – β Th 2 0 Pa U 91 92 Ra 226 Rn 222 218 Po 214 Pb Bi Po 210 Pb Bi Po 206 Pb 83 84 81 82 85 86 87 88 89 90 Atomic number Figure 23.2. The uranium decay series involves 14 steps as uranium decays eventually into lead. However, the object of radioactive dating is to determine the age of geological and archaeological samples and specimens. The age (t in the calculation) of certain rocks, for instance, can be estimated from analysis of the number of atoms of a particular radioisotope present now (N), as compared to the number present originally when the rock was formed (N0 ). Rearranging the above equation, the age t is given by t= N0 1 ln λ N The value of the initial number of atoms N0 is the sum, N + D, where D is the number of daughter nuclei resulting from the decay of atoms of the radioisotope. The original number of atoms of a radioisotope present in a rock sample is equal to the number N remaining at time t, plus the number of daughter atoms (D). _______________________________________________________________________________ EXAMPLE 23.5 Radioactive Dating The rubidium-87/strontium-87 method of dating rocks was used to analyze lunar samples. The half-life of Rb87 is 4.9 × 10 10 yr. 87 87 0 37 Rb → 38 Sr + –1 β Estimate the age of moon rocks in which the mole ratio of Rb-87 to Sr-87 is 40. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 454 / Nuclear Chemistry •Method of Solution The age of the rock can be calculated from t= N0 1 ln λ N Since the decay constant k is not given, we must use the equation relating k and t1/2 : λ= 0.693 t1/2 This can be substituted for k, yielding: t= t1/2 N0 ln 0.693 N Since Rb decays to Sr, the initial number of Rb atoms is equal to the sum of the Rb atoms and Sr atoms present. NRb = NRb + NSr 0 Given that N Rb /NSr = 40, then: NSr = 1 NRb 40 Therefore, after substitution for NSr, we get: NRb = NRb + 0 1 NRb = 1.025 NRb 40 Substituting into the rate equation, we get: t= 4.9 × 1010 yr 1.025 NRb ln = (7.07 × 1010 yr) ln 1.025 0.693 NRb t = 1.7 × 10 9 yr _______________________________________________________________________________ EXERCISES 11. The uranium decay series starts with uranium-238 and ends with lead-206. Each step in the series involves the loss of either an alpha or a beta particle. In the entire series how many alpha particles and how many beta particles are emitted? 12. The rates of decay of all radioisotopes follow the same type of rate equation. What type of equation is it? 13. Cobalt-60 has a half-life of 5.26 years. a. Calculate the decay constant for this isotope. b. How much cobalt-60 will remain from a 20.0 mg sample after 8.75 years? 14. A 2.52 mg sample of pure uranium-238 has a decay rate (activity) of 31.2 dps due to 238 U. a. What is the decay constant for U-238? b. What is the half-life of 238 U? 15. The 14C activity of some ancient corn was found to be 10 disintegrations per minute per gram of C. If present day plant life gives 15.3 dpm/g C, how old is the corn? The half-life of 14C is 5730 y. 16. Estimate the age of rocks in which the mole ratio of U-238 to Pb-206 is 0.75. The half-life of U-238 is 4.5 × 109 yr. 17. Analysis of a sample of uranite ore yields 3.2 g of U-238 and 1.5 g Pb-206. Assuming there was no Pb-206 present initially, how old is the rock? The half-life of U-238 is 4.5 × 10 9 yr. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 455 NUCLEAR TRANSMUTATION STUDY OBJECTIVES 1. 2. Determine the products of nuclear transformations and use the abbreviated notation for transformation reactions. Predict the product resulting from neutron capture by a given isotope, which is then followed by beta decay. Artificial Radioactivity. Nuclear transmutation is the process of converting one element into another. Artificial radioactivity results when an unstable nucleus is produced by transmutation. Irene Curie and Frederic Joliot discovered this phenomenon in 1933 while bombarding light elements with α particles from radioactive sources. For example, when aluminum atoms are bombarded with alpha particles the product is phosphorus-30 which is radioactive. Phosphorus-30 decays by positron emission and has a half-life of 2.5 min. It does not occur naturally in phorphorus compounds. 27 13 Al + 30 15 P → 4 30 1 2 He → 15 P + 0 n 30 0 14 Si + +1 β Neutrons readily produce "artificial" radioactivity because they are easily captured by stable nuclei, with the result that a new nucleus is formed that has a higher n : p ratio. This leads to a product that decays by beta decay. Neutron capture by chlorine-37 yields chlorine-38. 1 37 38 0 n + 17 Cl → 17 Cl 38 38 0 17 Cl → 18 Ar + –1 β Chlorine-38 has a short half-life, and is not found naturally on Earth. Note that capture of a neutron, followed by beta decay, yields a new element (Ar) with an atomic number one greater than the original element. This procedure of neutron capture followed by beta decay of the product nucleus has been used to synthesize elements that were "missing" from the periodic table, such as Tc and Pm, elements number 43 and 61, respectively. All isotopes of these elements are radioactive. Using neutrons as bombarding particles is convenient because neutrons have no charge and therefore are not repelled by target nuclei. When projectiles are positively charged such as protons and alpha particles, they must have considerable kinetic energy in order to overcome the electrostatic repulsion as these particles approach and collide with the target nucleus. Nuclear transmutation processes can be abbreviated according to the following format. Target nucleus (bombarding particle, ejected particle) Product nucleus The alpha bombardment reaction above can be written: 27 30 13 Al ( α,n) 15 P The symbols n, p, d, α, e, and γ are used in this notation to represent neutron, proton, deuteron, alpha particle, electron, and gamma ray. _______________________________________________________________________________ EXAMPLE 23.6 Synthesizing a Transuranium Element The first transuranium element to be synthesized by scientists was neptunium, atomic number 93. Devise a means to produce neptunium starting with U-238 and neutrons. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 456 / Nuclear Chemistry •Method of Solution Neutron capture by a nucleus followed by beta decay produces a new nucleus of one atomic number higher than the original. Neptunium is one atomic number beyond uranium. So start with uranium and bombard it with neutrons. 238 1 239 U + 0 n → 92 U 92 239 239 0 U → 93 Np + –1 β 92 •Comment Uranium's other naturally occuring isotope, U-235, will not produce neptunium in the same way because it undergoes neutron-induced fission. _______________________________________________________________________________ EXERCISES 18. Predict the product of the following bombardment reactions. 27 1 1 a. 13 Al + 0 n → X + 1 H 11 4 1 b. 5 B + 2 He → X + 1 H 19. Predict the product of the following bombardment reactions. 11 a. 5 B(p, α )X 10 b. 5 B(n,p)X 20. Devise a scheme by which bombardment of molybdenum with neutrons and subsequent β decay of the 99 product could be used to prepare an isotope of the missing element technetium, 43 Tc. NUCLEAR FISSION AND FUSION STUDY OBJECTIVES 1. 2. Describe both fission and fusion, and relate these processes to the curve of binding energy per nucleon versus mass number. Compare the energy released by fission and fusion with that of an ordinary chemical reaction. Nuclear Fission. The curve of binding energy per nucleon versus mass number (Figure 23.2) shows that the most stable nuclei are those with masses close to iron-56, which is the most stable nucleus. During fission a heavy nucleus of mass greater than about 230 amu splits into two lighter nuclei whose masses are usually between 80 and 160 amu. Since the two smaller nuclei are more stable than the larger nucleus, energy is released in the process. Although many nuclei heavier than uranium can undergo fission, the most important ones are U-233, U-235, and Pu-239. These isotopes undergo fission upon capture of a neutron. It is important to realize that naturally occurring uranium consists of two isotopes, U-235 and U-238, but that only U-235 is fissionable with thermal neutrons. The term thermal neutrons refers to those existing at temperatures around 25°C. Since the Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 457 nucleus does not repel a neutron, as it does an alpha particle, for instance, neutron-induced fission will occur at ordinary temperatures. The reaction is quite complex because the same two products are not formed by all fissioning nuclei. The two reactions below show just two out of many possibilities. 141 92 1 137 96 1 → 56 Ba + 36 Kr + 3 0 n 235 1 92 U + 0 n — → 52 Te + 40 Zr + 2 0 n The actual distribution of product yields is shown in Figure 23.7 of the text. A significant feature of fission is that on the average 2 to 3 neutrons are released per fission event. Since neutrons are required to initiate fission, and because neutrons are also products of fission, a nuclear chain reaction is possible. The energy released during nuclear fission depends somewhat on just what products are formed. The energy released from the fission of one mole of U-235 atoms can be calculated from the equation ∆E = ∆mc2 . The calculation shows that about 2.0 × 10 10 kJ are released per mole of uranium. This amount of energy is 70 million times the amount released in the exothermic chemical reaction in which 1 mol of H 2 reacts with 1/2 mol of O2 to form 1 mol of water! Nuclear Fusion. Radioactivity and nuclear fission are processes in which matter "comes apart." The energy releasing processes that occur on the sun are ones in which matter is fused. Nuclear fusion is the combining of small nuclei, such as hydrogen, to form a larger, more stable nucleus. Such a nucleus will have a higher average binding energy per nucleon, and fusion reactions will be exothermic. Because all nuclei are positively charged they must collide with enormous force in order to combine (fusion). This means that the atoms that will undergo fusion must be heated to millions of degrees. Fusion reactions are called thermonuclear reactions because they occur only at very high temperatures, such as those in the sun. The reaction that accounts for the tremendous release of energy by the sun is believed to be the stepwise fusion of four hydrogen nuclei to produce one helium nucleus. The net process is 1 4 0 4 1 H → 2 He + 2 +1 e ∆E = – 4.3 × 10 –12 J One gram of hydrogen upon fusion releases the energy equivalent to the combustion of 20 tons of coal. The fusion of four moles of H atoms by the preceding equation releases 2.6 × 109 kJ of energy. Our sun is made up of mostly hydrogen (90 percent) and helium (9 percent). In its interior the temperatures are estimated to reach 15 million °C. At these temperatures hydrogen will fuse to form helium; but helium, with its greater nuclear charge, will not fuse to form the heavier elements. The heavier elements up to iron are formed by fusion reactions that occur in exploding stars, called nova and supernova, where the temperatures can reach 2 billion °C. _______________________________________________________________________________ EXAMPLE 23.7 Relationship of Mass and Energy Calculate the mass of hydrogen that must undergo nuclear fusion each day in order to provide just the fraction of the daily energy output of the sun that reaches the earth, which is 1.5 × 10 22 J. •Method of Solution Using the following equation: 1 4 0 4 1 H → 2 He + 2 +1 e ∆E = –4.3 × 10–12 J we see that fusion of 4 H atoms yields 4.3 × 10 –12 J. We can set up the calculation: g H atoms = Back Forward 1.5 × 1022 J 1 day Main Menu TOC Study Guide TOC Textbook Website MHHE Website 458 / Nuclear Chemistry g H atoms = 1.5 × 1022 J 4 H atoms 1.66 × 10–24 g × × –12 J 1 day H atom 4.3 × 10 g H atoms = 2.3 × 1010 g Expressing the answer in tons, we get: 2.3 × 1010 g = 25,500 tons of hydrogen _______________________________________________________________________________ EXAMPLE 23.8 Comparing Nuclear Fission and Fusion Compare fission and fusion with respect to the temperatures required and the nature of the by-products of these processes. •Method of Solution Neutron-induced fission occurs at ordinary temperatures, whereas nuclear fusion requires temperatures in the millions of degrees. Fission of heavy elements yields hundreds of isotopes of elements with intermediate atomic numbers. These isotopes, by and large, have an excess of neutrons and are beta emitters. On the other hand, fusion of "light" nuclei yields stable isotopes of low-to-medium atomic mass. _______________________________________________________________________________ EXERCISES 21. Which process, nuclear fusion or nuclear fission, requires temperatures of millions of degrees. 22. Complete the following fission reaction of plutonium. 239 144 1 1 94 Pu + 0 n → 58 Ce + X + 3 0 n 23. Complete the following nuclear fission reaction. 2 3 1 1H+1H → X+0 n BIOLOGICAL EFFECTS OF RADIATION STUDY OBJECTIVES 1. 2. Describe how radiation interacts with matter. Define radiation dose in units of rads and rems. Interaction of Radiation with Matter. In passing through matter alpha, beta, and gamma rays lose energy chiefly by interaction with electrons which make up most of the volume of matter. Alpha and beta particles, on colliding with electrons, forcefully eject these electrons from atoms and molecules, and thereby produce ions. These particles lose only a small fraction of their energy in a single collision with an electron. Because alpha and beta particles are extremely energetic, thousands of collisions are required to bring them to rest. These particles produce "tracks" of ionization. Alpha, beta, and gamma rays are known as ionizing radiation. Most devices for detecting radioactivity depend on the formation of ions. The best known instrument for detecting radiation is the Geiger counter in which ions produced by a particle trigger a pulse of electricity that is counted electronically. Darkening of photographic plates, discharging of electroscopes, and damage to biological tissue all involve ionization. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 459 Units of Radiation Dose. Two units used to measure radiation dose are the rad and the rem. The rad (radiation absorbed dose) is defined as 0.01 joule of energy absorbed per kilogram of any absorbing material. The millirad is one-thousandth of a rad. Because beams of different radiations cause very different biological damage even when the body absorbs the same amount of energy from each type, it is necessary to define a unit specifically for biological tissue. The unit of biologically effective dose is the rem (radiation equivalent in man), which is the absorbed dose in rads multiplied by the relative biological effectiveness factor, RBE. The millirem is one thousandth of a rem. For beta and gamma rays RBE = 1.0; for fast neutrons and alpha particles RBE = 10. Thus, a dose of one rad of alpha radiation is equivalent to 10 rem. dose (in rem) = RBE × dose (in rad) ______________________________________________________________________________ EXERCISES 24. How does a rem differ from a rad? ______________________________________________________________________________ PRACTICE TEST 1. 2. What similarities and differences exist between beta particles and positrons? Complete and balance the following nuclear equations. 239 4 a. 94 Pu → 2 α + _____ 4 6 b. _____ + 3 Li → 2 2 He 90 0 c. 38 Sr → _____ + –1 β 10 4 d. 5 B + 2 He → _____ + 1 n 0 56 e. 26 Fe + 1 n → _____ 0 3. Rank the following nuclides in order of increasing nuclear stability: 40 20 Ca 4. 13 6. Back 11 5B With reference to the belt of stability, state the modes of decay you would expect for the following: a. 7 N 5. 39 20 Ca 26 b. 13 Al 28 c. 13 Al 27 27 a. Calculate the binding energy and the binding energy per nucleon of 13 Al. The atomic mass of 13 Al is 26.98154 amu. 28 b. Compare this result to the binding energy of 14 Si which has an even number of protons and neutrons. 28 The atomic mass of 14 Si is 27.976928 amu. Cobalt-60 is used in radiation therapy. It has a half-life of 5.26 years. a. Calculate the rate constant for radioactive decay. b. What fraction of a certain sample will remain after 12 years? Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 460 / Nuclear Chemistry 7. Radioactive decay follows first-order kinetics. If 20 percent of a certain radioisotope decays in 4.0 years, what is the half-life of this isotope? 3 8. Estimate the age of a bottled wine that has a tritium (1 H) content that is 3/4 that of environmental water obtained from the area where the grapes were grown. t1/2 = 12.3 yr. 9. Analysis of a sample of uranite ore shows that the ratio of U-238 atoms to Pb-206 atoms is 3.8. Assuming there was no Pb-206 present initially, how old is the rock? t1/2 (U-238) = 4.5 × 10 9 yr. 10. Consider the following fusion reactions: 1 2 3 a. 1 H + 1 H → 2 He 3 3 4 1 b. 2 He + 2 He → 2 He + 21 H 2 3 4 c. 1 H + 1 H → 2 He + 1 n 0 3 4 3 Given the atomic masses: 2 He = 3.016029 amu; 2 He = 4.002603 amu; 1 H = 3.017005 amu. Which of the above has the largest change in energy as indicated by its change in mass ∆m? 11. One atom of element 109 was prepared by bombardment of a target of bismuth-209 with accelerated nuclei of iron-58. Write a balance nuclear equation to show the formation of the isotope of element 109 with a mass number of 266. 12. How does the fusion-type reaction discussed in problem 11 differ from fusion reactions discussed in the textbook? 13. Write nuclear equations that show how Pu-239 is formed in a "breeder" reactor. 137 235 14. 55 Cs is a fission product of 92 U. If it is formed along with three neutrons, what is the other isotope formed? 15. What is the mode of decay expected for "light" nuclei which are unstable because of low n : p ratio? 238 206 16. One natural radioactive series begins with 92 U and ends with 82 Pb. All steps in the series are either alpha or beta decay. How many α particles and β particles are emitted? 17. The β particles emitted by carbon-14 atoms have a maximum energy of 2.5 × 10 –14 J per particle. What is the dose in rads when 8.0 × 10 10 carbon-14 atoms decay, and all the energy from their decay is absorbed by 2.0 kg of matter? 18. A sample of biological tissue absorbs a dose of 1.0 rad of alpha radiation. How many rems is this? General Problem 19. Radium-226 is an α emitter with a half-life of 1600 years. If 2.0 g of radium were allowed to undergo decay for 10 years, and all of the α particles were collected over the that time as helium gas, what would be the mass and the volume of the He at STP? ANSWERS Exercises 1. 4 2α 0 –1 0 0 β γ 2. 3. Back 15 protons, 17 neutrons, and 15 electrons a. 1 H 1 Forward 206 b. 82 Pb c. 1 n 0 Main Menu 15 c. 8 O TOC Study Guide TOC Textbook Website MHHE Website Nuclear Chemistry / 461 4. lower 4 27 5. a. 2 He b. 13 Al 18 18 0 6. a. 9 F → 8 O + +1 β 21 21 0 b. 9 F → 10 Ne + -1 β 7. 6.022 × 1026 amu = 1.00 kg 8. 13 protons and 14 neutrons 9. 1.34 × 10–12 J/nucleon 10. 1.45 × 10–12 J/nucleon 11. 5 alpha particles and 6 beta particles 12. the integrated first-order rate equation 13. a. 0.132 y–1 b. 6.3 mg 14. a. 4.89 × 10 –18 s–1 b. 4.49 × 109 y 15. 3520 y 16. 3.6 × 109 y 17. 2.8 × 109 y 27 18. a. 12 Mg 8 19. a. 4 Be 14 b. 6 C 10 b. 4 Be 98 99 20. 42 Mo + 1 n → 42 Mo 0 99 99 0 42 Mo + → 43 Tc + –1 β 21. fusion 93 22. 36 Kr 4 23. 2 He 24. Both a rad and a rem refer to the absorption of the same amount of energy per kilogram of an object. However, for living matter the actual damage done varies for each type of particle absorbed. The rem is the dosage unit used for biological material. It is the absorbed dose in rads multiplied by the relative biological effectiveness factor, RBE. For beta and gamma rays RBE = 1.0; for fast neutrons and alpha particles RBE = 10. dose (rem) = RBE × dose (rad) Practice Test 1. Beta particles and positrons have the same mass, 0.00055 amu. Beta particles have one unit of negative charge while positrons have one unit of positive charge. 2. a. b. c. d. e. Back Forward 239 Pu → 4 α + 235 U 2 92 2 4 6 1 H + 3 Li → 2 2 He 90 90 0 38 Sr → 39 Y + –1 β 10 4 13 B + 2 He → 7 N + 1 n 0 5 56 57 1 26 Fe + 0 n → 26 Fe Main Menu TOC Study Guide TOC Textbook Website MHHE Website 462 / Nuclear Chemistry 3. 4. 5. 6. 7. 8. 9. 10. 11 39 40 B < 20 Ca < 20 Ca. Boron-11 has an odd number of protons and an odd number of neutrons (rule 2). 5 Calcium-39 has an even number of protons and an odd number of neutrons (rule 1). Calcium-40 has an even number of protons and an even number of neutrons. Is also has a "magic number" of protons and neutrons (rule 2). a. positron emission b. positron emission c. beta decay a. 3.6 × 10 –11 J/atoms; 1.3 × 10 –12 J/nucleon b. 3.8 × 10–11 J/atoms; 1.4 × 10 –12 J/nucleon a. λ = 0.132/yr b. N/N0 = 0.205 t1/2 = 12.4 yr t = 5.1 yr t = 1.5 × 10 9 yr c. ∆m = –1.98 × 10–2 amu 209 58 238 1 266 11 83 Bi + 26 Fe → 109 X + 1 n 0 12. The fusion reactions discussed in the textbook are those of the "light" elements, which are exothermic. In the synthesis of element-109 energy is not evolved; rather, energy is absorbed. 239 13. 92 U + 0 n → 92 U 239 239 0 U → 93 Np + –1 β 92 239 239 0 Np → 94 Pu + –1 β 93 239 137 96 14. 92 U + 1 n → 55 Cs + 37 Rb + 31 n 0 0 15. positron emission 16. 8 alpha particles account for the mass change; 8 alpha and 6 beta particles account for the change in positive charge. 17. 0.10 rad 18. 10 rem 19. 2.3 × 1019 Ra atoms decayed in 10 years; 2.3 × 1019 He atoms have a mass of 1.6 × 10 –4 g and occupy a volume of 0.87 cm 3 at STP. ___________________________________________________________________________________ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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