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Nuclear Chemistry Chapter 11 Problem-Set Solutions 11.1 To identify a nucleus or atom uniquely, both its atomic number and its mass number must be specified. In one notation, a superscript before the symbol for the element is the mass number, and a subscript is the atomic number. In the second notation, the mass number is placed immediately after the name of the element (which indicates the atomic number). a. 104 Be, Be - 10 38 20 Ca, b. 25 11Na, Na - 25 Ca-38 b. 80 37 Rb, Rb-80 c. 96 41Nb, Nb - 96 125 51 Sb, Sb-125 Es-256 a. 11.3 Use the alternate notation that also indicates the atomic number and the mass number. b. 197 79 Au 17 8O b. 212 82 Pb d. 256 99 Es, 11.2 a. 147 N c. d. 257 103 Lr, Lr - 257 c. tin–121 d. boron–10 c. rubidium–92 d. bismuth–201 11.4 a. 11.5 a. 24 An alpha particle consists of two protons and two neutrons. The superscript indicates a mass of 4 amu (two protons + two neutrons); the subscript indicates that the charge is a +2 (from two protons). b. -10 A beta particle has a mass (very close to zero amu) and charge (–1) identical to those of an electron. c. 00 A gamma ray is a form of high-energy radiation without mass or charge. 11.6 a. +2 and 4 amu b. –1 and 0 amu (0.00055 amu) 11.7 An alpha particle consists of two protons and two neutrons. 11.8 They are identical in charge and mass. 11.9 Alpha particle decay is the emission of an alpha particle from a nucleus. The product nucleus has a mass number that is four less than the original nucleus, and an atomic number that is two less than the original nucleus. In writing nuclear equations, make sure that the mass numbers and the atomic numbers balance on the two sides of the equation. a. 200 84 Po 24 + c. 244 96 Cm 24 + 11.10 a. 229 90 c. 152 64 418 196 82 b. 240 96 Cm 24 + Pu d. 238 92 U 24 + Pb 240 94 c. 0 and 0 amu 236 94 Pu 234 90 Th Th 24 + 225 88 Ra b. 210 83 Bi 24 + Gd 24 + 148 62 Sm d. 243 95 Am 24 + 206 81 Tl 239 93 Np Copyright © Houghton Mifflin Company. All rights reserved. 419 Problem-Set Solutions Chapter 11 11.11 In beta particle decay, the mass number of the new nuclide is the same as that of the original nuclide; however, the atomic number has increased by one unit. Balance the superscripts and subscripts on both sides of the equation. + 147 N a. 10 4 Be 0 -1 + 105 B b. 14 6 C c. 21 9 F 21 + 10 Ne d. 25 11 Na 0 -1 11.12 a. 77 32 Ge b. 235 92 U 0 -1 + c. 16 7 N d. 60 26 Fe 0 -1 + 60 27 Co 0 -1 0 -1 0 -1 + 77 33 As + 168 O 0 -1 + 25 12 Mg 235 93 Np 11.13 When an alpha particle decay occurs, the product nucleus has an atomic number that is two less than that of the original nucleus and a mass number that is four less than that of the original nucleus: A A – 4; Z Z – 2 11.14 A does not change; Z Z + 1 11.15 We know that the sum of the superscripts (mass numbers) and the sum of the subscripts (atomic numbers or particle charges) on the two sides of the equation must be equal. a. 0 -1 b. 28 12 Mg 11.16 a. 200 80 Hg 234 94 Superscripts: The particle has a zero mass number, so the reactant and product atoms have the same mass number (28). Subscripts: The sum of the atomic numbers on the product side is 12, so this is also the atomic number of the reactant (at. no. 12 = Mg). Superscripts: The mass number of the product is four less than the mass number of the reactant; the particle has a mass number of four. Subscripts: The atomic number of the product is two less than that of the reactant; the particle has an atomic number of two. c. 24 d. Superscripts: Reactant and product atoms have the same mass number, so the particle has a zero mass number. Subscripts: The product has an atomic number one greater than that of the reactant, so the particle has a charge of –1. Pu Superscripts: The mass number of the product must be four less than the mass number of the reactant (204 – 4 = 200). Subscripts: The atomic number of the product must be two less than that of the reactant (82 – 2 = 80). b. 188 76 Os c. 84 36 Kr d. 0 -1 11.17 To write each of these decay equations, put the parent nuclide on the reactant side and the daughter nuclide on the product side. Then put in the correct particle on the product side to balance the subscripts and the superscripts. a. 24 b. 0 -1 Superscripts (mass numbers): 190 (parent) – 186 (daughter) = 4 Subscripts (atomic number or charge): 78 (parent) – 76 (daughter) = 2 Superscripts (mass numbers): 19 (parent) – 19 (daughter) = 0 Subscripts (atomic number or charge): 8 (parent) – 9 (daughter) = –1 Copyright © Houghton Mifflin Company. All rights reserved. 420 Problem-Set Solutions Chapter 11 4 11.18 a. 2 b. 0 -1 11.19 The half-life (t) is the time required for one-half of a given quantity of a radioactive substance to undergo decay. To calculate the fraction of nuclide remaining after a given time, first determine the number of half-lives that have elapsed. Then use the equation below (n is the number of half-lives): Amount of radionuclide original amount 1 = x undecayed after n half-lives of radionuclide 2 n a. b. c. d. 11.20 a. 1 2 2 1 2 6 1 2 3 1 2 6 1 = = = = of original is undecayed. 4 1 of original is undecayed. 64 1 of original is undecayed. 8 1 of original is undecayed. 64 1 1 = 4 2 16 b. 1 1 = 6 2 64 c. 1 1 = 3 2 8 d. 1 1 = 8 2 256 11.21 First, determine the number of half-lives that have elapsed (using the equation in Problem 11.19). Then divide the time that has elapsed by the number of half-lives to find the length of one half-life. a. 1 1 = 4; 16 2 c. 1 1 = 8; 256 2 11.22 a. 1 1 = 3; 8 2 b. 1 1 = 7; 128 2 c. 1 1 = 5; 32 2 d. 1 1 = 9; 512 2 5.4 days = 1.4 days 4 5.4 days = 0.68 day 8 b. 1 1 = 6; 64 2 d. 1 1 = 10 ; 1024 2 5.4 days = 0.90 day 6 5.4 days = 0.54 day 10 3.2 days = 1.1 days 3 3.2 days = 0.46 day 7 3.2 days = 0.64 day 5 3.2 days = 0.36 day 9 Copyright © Houghton Mifflin Company. All rights reserved. 421 Problem-Set Solutions Chapter 11 11.23 First, determine the number of half-lives by dividing the total time elapsed (60.0 hr) by the length of one half-life (15.0 hr). Then use the number of half-lives to determine the fraction of undecayed nuclide remaining (equation in problem 11.19). 11.24 60.0 hr = 4; 15.0 hr 1 1 ; 4 = 2 16 112 yr = 4.0; 28 yr 1 1 = ; 4 2 16 1 x 4.00 g = 0.250 g 16 1 x 4.00 g = 0.250 g 16 11.25 Over 2000 bombardment-produced radionuclides that do not occur naturally are now known. 11.26 7 times greater 11.27 Uranium, element 92, has the highest atomic number of any naturally occurring element. 11.28 83, bismuth 11.29 To write each of these equations for bombardment reactions, put the parent nuclide and the bombardment particle on the reactant side and the daughter nuclides and decay particles on the product side. Balance the superscripts (mass numbers) and the subscripts (atomic numbers or charges). a. 24 b. 25 12 Mg Superscripts: 24 + x (reactant side) = 27 + 1 (product side); x = 4 Subscripts: 12 + y (reactant side) = 14 + 0 (product side); y = 2 Superscripts: 27 + 2 (reactant side) = x + 4 (product side); x = 25 Subscripts: 13 + 1 (reactant side) = y + 2 (product side); y = 12 c. 24 Superscripts: 9 + x (reactant side) = 12 + 1 (product side); x = 4 Subscripts: 4 + y (reactant side) = 6 + 0 (product side); y = 2 d. 11 p Superscripts: 6 + x (reactant side) = 4 + 3 (product side); x = 1 Subscripts: 3 + y (reactant side) = 2 + 2 (product side); y = 1 11.30 a. 249 98 Cr b. 17 8 O c. 1 0 n d. 24 11.31 We would expect lead-207 to be a stable nuclide, since it terminates the uranium-235 decay series; termination of a decay series requires a stable nuclide. 11.32 All isotopes of radon are radioactive. Copyright © Houghton Mifflin Company. All rights reserved. 422 Problem-Set Solutions Chapter 11 11.33 In each equation in this decay series, a parent nuclide produces a daughter nuclide and an alpha or beta particle. For each equation (after the first), the parent nuclide is the daughter nuclide from the previous equation. Balance the superscripts and the subscripts. 232 90 228 88 Superscripts: 232 (reactants) = 4 + x (products); x = 228 Subscripts: 90 (reactant) = 2 + y (product); y = 88 Th 24 + 228 88 Ra + 228 89 Ac Superscripts: 228 (reactants) = 0 + x (products); x = 228 Subscripts: 88 (reactant) = –1 + y (product); y = 89 Ra 0 -1 Th Superscripts: 228 (reactants) = 0 + x (products); x = 228 Subscripts: 89 (reactant) = –1 + y (product); y = 90 Superscripts: 228 (reactants) = 4 + x (products); x = 224 228 4 224 90Th 2 + 88 Ra Subscripts: 90 (reactant) = 2 + y (product); y = 88 11.34 + 228 89 Ac 235 92 U 24 + 231 90 Th 0 -1 0 -1 231 90 + 231 91 Pa 24 + 227 89 Ac 0 -1 228 90 + Th 231 91 227 89 Pa Ac 227 90 Th 11.35 An ion pair is the electron and the positive ion that are produced during an ionizing interaction between a molecule (or an atom) and radiation. 11.36 a chemical species that contains an unpaired electron 11.37 Draw the Lewis structure for each of the species; from the Lewis structure, determine whether there is an unpaired electron (a free radical). a. b. H O H O H H H Yes, there is an unpaired electron; H2O+ is a free radical. c. H O Yes, there is an unpaired electron; OH is a free radical. 11.38 a. H2O + radiation H2O+ + e– No, there is no unpaired electron; H3O+ is not a free radical. d. H O No, there is no unpaired electron; OH— is not a free radical. b. H2O+ + H2O H3O+ + OH 11.39 The fate of a radiation particle involved in an ion pair formation is to continue on, interacting with other atoms and forming more ion pairs. 11.40 (1) interaction with another free radical to form a non-free radical molecule; (2) interaction with another molecule to form another free radical Copyright © Houghton Mifflin Company. All rights reserved. Problem-Set Solutions Chapter 11 423 11.41 Alpha particles are the most massive and the slowest particles involved in natural radioactive decay processes; they are, therefore, less penetrating and are stopped by a thick sheet of paper. Beta particles and gamma rays, having much greater speed than alpha particles, go through a thick sheet of paper. 11.42 Alpha and beta are stopped; gamma goes through. 11.43 Alpha particle velocities are on the order of 0.1 the speed of light; beta particle velocities are up to 0.9 the speed of light; gamma rays have a velocity equal to the speed of light. 11.44 An alpha produces 40,000 ion pairs, and a beta produces 2000 ion pairs. 11.45 Table 11.3 gives the effects of short-term whole-body radiation exposure on humans. a. Whole-body radiation exposure of 10 rems has no detectable effects. b. Whole-body radiation exposure of 150 rems can cause nausea, fatigue, and lowered blood cell count. 11.46 a. short-term reduction in blood cell count b. appetite loss, general malaise, sore throat, pallor, diarrhea 11.47 Figure 11.11 shows the estimated annual radiation exposure of an average American. From this graph we can see that 19% of the radiation exposure that an average American receives is human-made, and 81% comes from natural sources. 11.48 radon seepage, cosmic radiation, rocks and soil, minerals in the body, medical x-rays 11.49 Film badges are used to monitor the extent of radiation exposure. 11.50 Radiation ionizes atoms and molecules; this produces a pulse of electricity which is detected. 11.51 Nearly all diagnostic radionuclides are gamma emitters because the penetrating power of alpha and beta particles is too low; it is necessary to be able to detect the radiation externally (outside the body). 11.52 to minimize exposure and increase radiation intensity to detectable levels 11.53 Table 11.4 lists selected radionuclides used in diagnostic procedures. a. Barium-131 is used to detect bone tumors. b. Sodium-24 is used to detect circulatory problems. c. Iron-59 is used in the evaluation of iron metabolism in blood. d. Potassium-42 is used to determine intercellular spaces in fluids. 11.54 a. b. c. d. fluid buildup in brain, location of cysts and blood clots, thyroid problems blood studies, breast carcinoma brain tumors, spleen and thyroid problems, location of blood clots blood studies, kidney activity 11.55 Radionuclides used for therapeutic purposes are usually or emitters instead of emitters. Therapeutic radionuclides are used to selectively destroy abnormal (usually cancerous) cells; an intense dose of radiation in a small localized area is needed. 11.56 cobalt-60, external radiation source; yttrium-90, implantation therapy Copyright © Houghton Mifflin Company. All rights reserved. 424 Problem-Set Solutions Chapter 11 11.57 In each of these nuclear fission equations, the difference in mass numbers between product and reactant sides gives the number of neutrons produced. a. Reactant side (235 + 1 = 236) – product side (135 + 97 = 232) = 4 neutrons b. Reactant side (235 + 1 = 236) – product side (72 + 160 = 232) = 4 neutrons c. Reactant side (235 + 1 = 236) – product side (90 + 144 = 234) = 2 neutrons d. Reactant side (235 + 1 = 236) – product side (142 + 91 = 233) = 3 neutrons 11.58 a. 3 b. 3 c. 2 d. 2 11.59 The mass number of the nuclide that undergoes nuclear fission plus the mass of the initiating neutron is equal to the sum of the mass numbers of the products: x + 1 = 143 + 94 + 3(1); x = 239 The atomic number (and thus the atomic identity) of the reacting nuclide is equal to the sum of the atomic numbers of the fission products: y = 56 + 37 = 93 Element number 93, with a mass number of 239, is neptunium-239. 11.60 234 92 U 11.61 a. Balance the mass numbers: products [2(3)] = reactants [2(1) + x] = 6; x = 4 Balance the atomic numbers: products (2 + x) = reactants [2(2)] = 4; x = 2 The particle having a mass number of 4 and an atomic number of 2 is: b. Balance the mass numbers: products [2(4) + 1] = reactants [7 + x] = 9; x = 2 Balance the atomic numbers: products [2(2) + 0] = reactants [3 + x] = 4; x = 1 The particle having a mass number of 2 and an atomic number of 1 is: 4 2 2 1 11.62 a. 2 1 H H b. 3 2 He 11.63 a. Nuclear fusion is a nuclear reaction in which two small nuclei are put together to make a larger one. For fusion to occur, an extremely high temperature is required. b. The process of nuclear fusion occurs on the sun. c. During transmutation a nuclide of one element is changed into another nuclide of another element. Transmutation occurs in both fission and fusion reactions. d. Bombardment of a nuclide with neutrons may cause it to split into two fragments; this process is called nuclear fission. 11.64 a. both b. fission c. fission d. fusion 11.65 During a nuclear fission reaction, a large nucleus splits into two medium-sized nuclei. During a nuclear fusion reaction, two small nuclei are put together to make a larger one. a. This is a fusion reaction because two small nuclei come together to form a larger one. b. This is a fission reaction because one large nucleus splits into two medium sized nuclei. c. This reaction is neither fission nor fusion because the large nucleus does not split into two medium-sized nuclei. d. This reaction is neither fission nor fusion because the large nucleus does not split into two medium-sized nuclei. Copyright © Houghton Mifflin Company. All rights reserved. 425 Problem-Set Solutions Chapter 11 11.66 a. neither b. neither c. fusion d. fission 11.67 Write these products and/or reactants of some radioactive decay processes in the form of equations. Balance the mass numbers and the atomic numbers to identify the missing parts of the equations. 11.68 Tl Superscripts: x (reactant side) = 0 + 206 (product side); x = 206 Subscripts: y (reactant side) = –1 + 81 (product side); y = 80 + 109 47 Ag Superscripts: 109 (reactant side) = 0 + x (product side); x = 109 Subscripts: 46 (reactant side) = –1 + y (product side); y = 47 + a. 206 80 Hg 0 -1 b. 109 46 Pd 0 -1 c. 245 96 Cm 24 + 241 94 Pu Superscripts: x (reactant side) = 4 + 241 (product side); x = 245 Subscripts: y (reactant side) = 2 + 94 (product side); y = 96 d. 249 100 Fm 24 + 245 98 Cf Superscripts: 249 (reactant side) = 4 + x (product side); x = 245 Subscripts: 100 (reactant side) = 2 + y (product side); y = 98 206 81 1 1 a. left = 3 8 2 3 half-lives x 18 hrs = 54 hrs 1 half-life b. 1 1 left = 5 32 2 5 half-lives x 18 hrs = 90 hrs 1 half-life c. 1 1 left = 6 64 2 6 half-lives x 18 hrs = 108 hrs 1 half-life d. 1 1 left = 7 128 2 7 half-lives x 18 hrs = 126 hrs 1 half-life 11.69 In a bombardment reaction, the bombarding particle is written on the reactant side of the nuclear equation. Write and balance the mass numbers and the atomic numbers to identify the missing parts of the equations. 11.70 Pu + 24 a. 243 94 b. 246 96 Cm c. 27 13 Al + 24 30 15 d. 23 11 Na + 21 H 21 10 142 60 246 96 + 126 C Nd + 01 n 143 61 Cm + 01 n 254 102No Superscripts: x + 4 (react.) = 246 +1 (prod.); x = 243 Subscripts: y + 2 (react.) = 96 + 0 (prod.); y = 94 + 410 n Superscripts: 246 + x (react.) = 254 +4(1) (prod.); x = 12 Subscripts: 96 + y (react.) = 102 + 4(0) (prod.); y = 6 P + 01 n Superscripts: 27 + 4 (react.) = x + 1 (prod.); x = 30 Subscripts: 13 + 2 (react.) = y + 0 (prod.); y = 15 Ne + 24 Superscripts: 23 + 2 (react.) = 21 + x (prod.); x = 4 Subscripts: 11 + 1(react.) = 10 + y (prod.); y = 2 Pm + -10 11.71 Table 11.2 lists the most stable nuclides of transuranium elements. Fourteen of these elements have nuclides with half-lives less than 1.0 day. 11.72 four neutrons Copyright © Houghton Mifflin Company. All rights reserved. 426 Problem-Set Solutions Chapter 11 11.73 When we consider the decay series E F G H, we can see that the half-lives of E and F are both very short as compared with the total time (50 days). Therefore, we can say that there are no (or a negligible amount of) E or F atoms left at the end of 50 days. The half-life of G is not short (12.5 days), so we can use the half-life decay equation to calculate the fraction of G atoms remaining at the end of 50 days (50 days/12.5 days = 4 half-lives): 1 2n = 1 2 4 = 1 16 Multiply the fraction of G atoms remaining by the total atoms (1/16 x 1000 = 63 G atoms). Subtract the number of G atoms from the number of total atoms to find the number of H atoms (1000 – 63 = 937 H atoms). At the end of 50 days there are the following numbers of atoms: E = 0 (negligible amount), F = 0 (negligible amount), G = 63 atoms, H = 937 atoms 11.74 228 Ac, 228Th, 224Ac 11.75 Statement c. is incorrect. A gamma ray is a form of high-energy radiation without mass or charge. 11.76 b 11.77 The daughter nuclide for the alpha decay of polonium-212 is lead-208. 212 84 Po 24 + 208 82 Pb 11.78 c 11.79 After 3 half-lives have elapsed, the amount of radioactive sample which is not decayed is 1/8 the original amount. 1 2 n = 1 2 3 = 1 8 11.80 c 11.81 The general production process for synthetic elements involves both a transmutation reaction and a bombardment reaction. 11.82 a 11.83 Statement d. is incorrect. A piece of aluminum foil will stop alpha particles but not gamma radiation. 11.84 a Copyright © Houghton Mifflin Company. All rights reserved.