2. No, coal-fired power plants also pose risks. A partial list of risks is:

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1 CHAPTER 9 THE UCLEUS: A CHEMIST'S VIEW Questions. Characteristic frequencies of energies emitted in a nuclear reaction suggest that discrete energy levels exist in the nucleus. The extra stability of certain numbers of nucleons and the predominance of nuclei with even numbers of nucleons suggest that the nuclear structure might be described by using quantum numbers.. o, coal-fired power plants also pose risks. A partial list of risks is: Coal Air pollution Coal mine accidents Health risks to miners (black lung disease) uclear Radiation exposure to workers Disposal of wastes Meltdown Terrorists Public fear 3. Beta-particle production has the net effect of turning a neutron into a proton. Radioactive nuclei having too many neutrons typically undergo β-particle decay. Positron production has the net effect of turning a proton into a neutron. uclei having too many protons typically undergo positron decay.. Annihilation is collision of matter and antimatter resulting in the change of particulate matter into electromagnetic radiation. 5. The transuranium elements are the elements having more protons than uranium. They are synthesized by bombarding heavier nuclei with neutrons and positive ions in a particle accelerator. 6. All radioactive decay follows first-order kinetics. A sample is analyzed for the 76 Lu and 76 Hf content, from which the first-order rate law can be applied to determine the age of the sample. The reason 76 Lu decay is valuable for dating very old objects is the extremely long half-life. Substances formed a long time ago that have short half-lives have virtually no remaining nuclei. On the other hand, 76 Lu decay hasn t even approached one half-life when dating 5-billion-year-old objects. 7. E mc ; the key difference is the mass change when going from reactants to products. In chemical reactions, the mass change is indiscernible. In nuclear processes, the mass change is discernible. It is the conversion of this discernible mass change into energy that results in the huge energies associated with nuclear processes. 737

2 738 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 8. Effusion is the passage of a gas through a tiny orifice into an evacuated container. Graham s law of effusion says that the effusion of a gas in inversely proportional to the square root of the mass of its particle. The key to effusion, and to the gaseous diffusion process, is that they are both directly related to the velocity of the gas molecules, which is inversely related to the molar mass. The lighter 35 UF 6 gas molecules have a faster average velocity than the heavier 38 UF 6 gas molecules. The difference in average velocity is used in the gaseous diffusion process to enrich the 35 U content in natural uranium. 9. The temperatures of fusion reactions are so high that all physical containers would be destroyed. At these high temperatures, most of the electrons are stripped from the atoms. A plasma of gaseous ions is formed that can be controlled by magnetic fields.. The linear model postulates that damage from radiation is proportional to the dose, even at low levels of exposure. Thus any exposure is dangerous. The threshold model, on the other hand, assumes that no significant damage occurs below a certain exposure, called the threshold exposure. A recent study supported the linear model. Exercises Radioactive Decay and uclear Transformations. All nuclear reactions must be charge balanced and mass balanced. To charge balance, balance the sum of the atomic numbers on each side of the reaction, and to mass balance, balance the sum of the mass numbers on each side of the reaction a. H He + e c. Be e Li b. Li Be + e d. B Be + e 8 Be He e. 5 P 6S + e 8 Li He + e a. Co i + e b. Tc e Mo c. Tc Ru + e d. Pu U He 3. a Ga + e 68 Zn b. 6 Cu e i c. Fr 87 He + 8 At d Sb e + 9 Te 5. a Ga 73 Ge + 3 e b Pt 88 Os + 76 He c Bi 5 Pb e d. Cm + 96 e Am 95

3 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW U 7 8 Pb +? He +? e From the two possible decay processes, only alpha-particle decay changes the mass number. So the mass number change of 8 from 35 to 7 must be done in the decay series by seven alpha particles. The atomic number change of from 9 to 8 is due to both alpha-particle production and beta-particle production. However, because we know that seven alpha-particles are in the complete decay process, we must have four beta-particle decays in order to balance the atomic number. The complete decay series is summarized as: 35 9 U 7 8 Pb + 7 He + e Bk 7 8 Pb +? He + e; the change in mass number (7! 7 ) is due exclusively to the alpha-particles. A change in mass number of requires He particles to be produced. The atomic number only changes by 97! 8 5. The alpha-particles change the atomic number by, so 5 e (5 beta-particles) are produced in the decay series of 7 Bk to 7 Pb. 7. a. Am 95 He + 37 p 93 b. c. 95 Am 8 He + e Bi; the final product is 83 Bi Am p + α 9Pa + α 9 U + β Th + α 88 Ra + α 3 7 Po + β 83 Bi + α Pb + α 83 Bi + β 5 At + α 87 Fr + α 89 Ac + β The intermediate radionuclides are: p, Pa, 9 U, 9 Th, 5 88 Ra, Ac, 87 Fr, 85 At, Bi, 8 Po, and 8 Pb 8. The complete decay series is: 3 9 Th 8 88 Ra + 8 He 89 Ac + e 8 9 Th + e 88 Ra + He e + 8 Po e + 83 Bi He + 8 Pb 6 He + 8 Po 86 Rn + He 8 8 Pb + He Fe has too many protons. It will undergo either positron production, electron capture, and/or alpha-particle production Fe has too many neutrons and will undergo beta-particle production. (See Table 9. of the text.)

4 7 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW. Reference Table 9. of the text for potential radioactive decay processes. 7 F and 8 F contain too many protons or too few neutrons. Electron capture and positron production are both possible decay mechanisms that increase the neutron to proton ratio. Alpha-particle production also increases the neutron-to-proton ratio, but it is not likely for these light nuclei. F contains too many neutrons or too few protons. Beta-particle production lowers the neutronto-proton ratio, so we expect F to be a beta-emitter.. a Cf O 63 6 Sg + n b Rf ; 6 Sg 59 He + Rf. a. 95 Am + 3 He 97 Bk + n b U + 6 C 98 Cf + 6 n c Cf Db + n d Cf + 5 B 57 3 Lr + n Kinetics of Radioactive Decay 3. All radioactive decay follows first-order kinetics where t / (ln )/k.. k ln.693 t / 3 k. h 69 h ln t/.6935 yr d h yr 365 d h 36 s s Rate k 5.8 s 5. g mol g 3 6. mol nuclei 6.35 decays/s 6.35 alpha particles are emitted each second from a 5.-g Am sample. 5. Kr-8 is most stable because it has the longest half-life, while Kr-73 is hottest (least stable) because it has the shortest half-life. 6. a. k.5% of each isotope will remain after 3 half-lives: % 5% 5%.5% t / t/ t/ For Kr-73: t 3(7 s) 8 s; for Kr-7: t 3(.5 min) 3.5 min For Kr-76: t 3(.8 h). h; for Kr-8: t 3(. 5 yr) yr ln t/.693 d h 6.7.8d h 36 s 7 s

5 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 7 b. Rate k s 3 mol g 6. g mol 3 nuclei Rate.65 decays/s c. 5% of the 6 Cu will remain after half-lives (% decays to 5% after one half-life, which decays to 5% after a second half-life). Hence (.8 days) 5.6 days is the time frame for the experiment. 7. Units for and are usually number of nuclei but can also be grams if the units are the same for both and. In this problem, m the initial mass of 7 Ca + to be ordered. k ln ; t / ln kt (.693)t, t / 5. µg Ca ln m +.693(. d).5 d.3 5. m e.3.73, m 6.8 µg of 7 Ca + needed initially 6.8 µg 7 Ca + 7. µg 7. µg 7 7 CaCO Ca µg 7 CaCO 3 should be ordered at the minimum. 8. a.. Ci 3.7 decays/ s Ci decays/s; k ln t / Rate k, decays s h h 36 s, 5.5 atoms of 38 S 5.5 atoms 38 S mol S atoms mol a mol S SO 9. mol a 38 SO 9. mol a 38 SO g a SO mol a SO.3 9 g.3 ng a 38 SO. b % decays,.% left; ln!kt (.693)t.87 h, t 38. hours hours ln 9. t 6. yr; k ; t ln!kt /.7% of the 9 Sr remains as of July 6, 9. (.693)6. yr 8.9 yr!.53,. 53 e.7

6 7 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 3. Assuming significant figures in /: ln(/ )!kt; (.) ; t / (ln )/k 3. ln(.) ln!kt (ln )t t / (ln )t, t / (.693)t, t 53 days 8. d. g ln m h d d 3. min 6 min h. g ln!3., m 3.. m e, m. g 8 Br needed 8 8. g 8 mol Br mol a Br 5. g a Br Br 6 g a 8 Br g mol Br mol a Br 3. Assuming the current year is 9, t 63 yr. 8 ln!kt (.693)t t /.693(63 yr), ln, yr.6 decay events min. g water ln 33. k ; t ln!kt / (.693)t, t / ln (5, yr)! yr. e 8.7, counts per minute per g of C 3.6 If we had. mg C, we would see:. mg g mg.3 counts min g.3 counts min It would take roughly min to see a single disintegration. This is too long to wait, and the background radiation would probably be much greater than the C activity. Thus C dating is not practical for very small samples. 3. ln!kt (.693)t t /., ln 3.6 (.693)t, t. yr 573 yr 35. Assuming. g 38 U present in a sample, then.688 g 6 Pb is present. Because mol 6 Pb is produced per mol 38 U decayed: 38 U decayed.688 g Pb mol Pb mol U 38 g U.795 g 38 U 6 g Pb mol Pb mol U

7 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 73 Original mass 38 U present. g g.795 g 38 U ln!kt (ln )t. g.693(t), ln t.795 g, t yr 9 /.5 yr 36. a. The decay of K is not the sole source of Ca. b. Decay of K is the sole source of Ar and no Ar is lost over the years. c..95 g. g Ar current mass ratio K.95 g of K decayed to Ar..95 g of K is only.7% of the total K that decayed, or:.7(m).95 g, m 8.9 g total mass of K that decayed Mass of K when the rock was formed was. g g 9.9 g.. g 9.9 g ln K K!kt (ln )t t (.693)t, t. 9 years old 9.7 yr / d. If some Ar escaped, then the measured ratio of Ar/ K is less than it should be. We would calculate the age of the rock to be less than it actually is. Energy Changes in uclear Reactions 37. E mc, m 3 E 3.9 kg m /s kg c (3. m/s) The sun loses.3 6 kg of mass each second. ote: J kg m /s s kj J kj 36 s h h day.6 J/day E mc E.6 J, m 8 c (3. m/s).8 5 kg of solar material provides day of solar energy to the earth.6 kj g kg J 5. kg of coal is needed to provide the J 3 kj g same amount of energy 39. We need to determine the mass defect m between the mass of the nucleus and the mass of the individual parts that make up the nucleus. Once m is known, we can then calculate E (the binding energy) using E mc. ote: J kg m /s.

8 7 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW For 3 9 Pu (9 e, 9 p, 38 n): mass of 3 Pu nucleus g! mass of 9 electrons mass of 3 Pu nucleus g! 9(9.939 ) g g m g! (mass of 9 protons + mass of 38 neutrons) m g! [9(.676 ) + 38(.6793 )] g!3.68 g For mol of nuclei: m E mc (!.98 For 3 9 Pa (9 e, 9 p, n): mass of 3 Pa nucleus g/nuclei 6. 3 nuclei/mol!.98 g/mol 3 kg/mol)( m/s)!.75 J/mol g - 9( ) g g m g - [9(.676 ) + (.6793 )] g!3.66 g E mc 3.66 nuclei 7 kg 6. mol 3 nuclei.9979 s 8 m.7 J/mol. From the text, the mass of a proton.78 amu, the mass of a neutron.866 amu, and the mass of an electron 5.86 amu. Mass of 56 6 Fe nucleus mass of atom mass of electrons (.586) amu 56 6 H + 3 n 6 Fe; m amu [6(.78) + 3(.866)] amu.585 amu E mc.585 amu.665 amu 7 kg ( m/s) J Binding energy ucleon J 56 nucleons.8 J/nucleon

9 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 75. Let m e mass of electron; for C (6e, 6p, and 6n): Mass defect m [mass of C nucleus] [mass of 6 protons + mass of 6 neutrons]. ote: Atomic masses given include the mass of the electrons. m. amu 6m e [6(.78 m e ) + 6(.866)]; mass of electrons cancel. m. [6(.78) + 6(.866)].9888 amu E mc.9888 amu.665 amu 7 kg ( m/s).76 J Binding energy ucleon.76 J nucleons.3 J/nucleon For 35 U (9e, 9p, and 3n): m m e [9(.78 m e ) + 3(.866)].939 amu E mc amu 7 kg ( m/s).8563 J Binding energy ucleon.8563 J 35 nucleons.5 J/nucleon Because 56 Fe is the most stable known nucleus, the binding energy per nucleon for 56 Fe (.8 J/nucleon) will be larger than that of C or 35 U (see Figure 9.9 of the text).. For H : Mass defect m mass of H nucleus mass of proton mass of neutron. The mass of the H nucleus will equal the atomic mass of H minus the mass of the electron in an H atom. From the text, the pertinent masses are m e 5.9 amu, m p.78 amu, and m n.866 amu. m. amu.59 amu (.78 amu amu).39 3 amu E mc amu amu Binding energy ucleon J nucleons J/nucleon kg ( m/s) J For 3 H : m [.78 + (.866)] 9. 3 amu E 9. 3 amu.665 amu 7 kg ( m/s).36 J Binding energy ucleon.36 J 3 nucleons.53 3 J/nucleon

10 76 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 3. Let m Li mass of 6 Li nucleus; an 6 Li nucleus has 3p and 3n.!.33 amu m Li! (3m p + 3m n ) m Li![3(.78 amu) + 3(.866 amu)] m Li 6.38 amu Mass of 6 Li atom 6.38 amu + 3m e (5.9 amu) 6.53 amu (includes mass of 3 e ). Binding energy.36 nucleon J 7 nucleons 3.58 J for each 7 Mg nucleus E mc E 3.58 J, m 8 c (.9979 m/s)! kg m! kg amu kg!.399 amu mass defect Let m Mg mass of 7 Mg nucleus; an 7 Mg nucleus has p and 5 n.!.399 amu m Mg! (m p + 5m n ) m Mg! [(.78 amu) + 5(.866 amu)] m Mg amu Mass of 7 Mg atom amu + m e, (5.9 amu) amu (includes mass of e ) 5. H + H H + + e; m (. amu! m e + m e )! (.78 amu! m e ) m.! (.78) + (.59)!. When mol of protons undergoes fusion, m!. E mc!. amu for two protons reacting g. 7 kg (3. 8 m/s)!. J. J mol protons mol.g!. J/g of hydrogen nuclei 6. H + 3 H He + n; using atomic masses, the masses of the electrons cancel when determining m for this nuclear reaction. m [ ! ( )] amu!.889 amu

11 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 77 For the production of mol of He : m!.889 g!.889 E mc kg ( m/s).698 J/mol 5 kg For nucleus of He.698 mol J mol 6. 3 nuclei!.8 J/nucleus Detection, Uses, and Health Effects of Radiation 7. The Geiger-Müller tube has a certain response time. After the gas in the tube ionizes to produce a "count," some time must elapse for the gas to return to an electrically neutral state. The response of the tube levels off because at high activities, radioactive particles are entering the tube faster than the tube can respond to them. 8. ot all of the emitted radiation enters the Geiger-Müller tube. The fraction of radiation entering the tube must be constant. 9. Water is produced in this reaction by removing an OH group from one substance and H from the other substance. There are two ways to do this: O O i. CH 3 C OH 8 + H OCH 3 8 CH 3 C OCH 3 + HO H O O ii. CH 3 CO H H O CH 3 CH 3 CO CH 3 + H OH Because the water produced is not radioactive, methyl acetate forms by the first reaction in which all the oxygen-8 ends up in methyl acetate. 5. The only product in the fast-equilibrium step is assumed to be 6 O 8 O, where is the central atom. However, this is a reversible reaction where 6 O 8 O will decompose to O and O. Because any two oxygen atoms can leave 6 O 8 O to form O, we would expect (at equilibrium) one-third of the O present in this fast equilibrium step to be 6 O and twothirds to be 8 O. In the second step (the slow step), the intermediate 6 O 8 O reacts with the scrambled O to form the O product, where is the central atom in O. Any one of the three oxygen atoms can be transferred from 6 O 8 O to O when the O product is formed. The distribution of 8 O in the product can best be determined by forming a probability table. 6 O (/3) 8 O (/3) 6 O (/3) from 6 O 8 O 8 O (/3) from 6 O 8 O 6 O (/9) 8 O 6 O (/9) 6 O 8 O (/9) 8 O (/9)

12 78 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW From the probability table, /9 of the O is 6 O, /9 of the O is 8 O, and /9 of the O is 6 O 8 O (/9 + /9 /9). ote: 6 O 8 O is the same as 8 O 6 O. In addition, 6 O 8 O is not the only O 3 intermediate formed; 6 O 8 O and 8 O 3 can also form in the fast-equilibrium first step. However, the distribution of 8 O in the O product is the same as calculated above, even when these other O 3 intermediates are considered U + n 9 58 Ce + 38 Sr +? n +? e; to balance the atomic number, we need beta-particles, and to balance the mass number, we need neutrons. 5. So 38 9 U + n 39 9 U e p e + 39 Plutonium-39 is the fissionable material in breeder reactors. 53. Release of Sr is probably more harmful. Xe is chemically unreactive. Strontium is in the same family as calcium and could be absorbed and concentrated in the body in a fashion similar to Ca. This puts the radioactive Sr in the bones; red blood cells are produced in bone marrow. Xe would not be readily incorporated into the body. 9 Pu The chemical properties determine where a radioactive material may be concentrated in the body or how easily it may be excreted. The length of time of exposure and what is exposed to radiation significantly affects the health hazard. (See Exercise 5 for a specific example.) 5. (i) and (ii) mean that Pu is not a significant threat outside the body. Our skin is sufficient to keep out the alpha-particles. If Pu gets inside the body, it is easily oxidized to Pu + (iv), which is chemically similar to Fe 3+ (iii). Thus Pu + will concentrate in tissues where Fe 3+ is found. One of these is the bone marrow, where red blood cells are produced. Once inside the body, alpha-particles cause considerable damage. Connecting to Biochemistry 55. All nuclear reactions must be charge balanced and mass balanced. To charge balance, balance the sum of the atomic numbers on each side of the reaction, and to mass balance, balance the sum of the mass numbers on each side of the reaction a. Cr + e V b. I e Xe c. P e S 56. a. Cobalt is a component of vitamin B. By monitoring the cobalt-57 decay, one can study the pathway of vitamin B in the body. b. Calcium is present in the bones in part as Ca 3 (PO ). Bone metabolism can be studied by monitoring the calcium-7 decay as it is taken up in bones. c. Iron is a component of hemoglobin found in red blood cells. By monitoring the iron-59 decay, one can study red blood cell processes.

13 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 79 ln (ln )t 57. k ; ln kt ; t / t/ ln (.693)(8. h) h e 5. 5.; the fraction of 99 Tc that remains is., or.% lb 53.6 g lb 8 g C g body.6 g g C C mol g C C nuclei C. mol C nuclei C Rate k; k ln t / yr yr d h d h 36 s s Rate k; k 3.8 s (. 5 C nuclei) 38 decays/s A typical 8 lb person produces 38 beta particles each second mg a 3 3 PO 3. mg 65. mg a 3 P 3 3 PO 33.9 mg 3 P; k ln t / ln (.693)t m.693(35. d) kt, ln t / 33.9 mg ; carrying extra sig. figs.:.3 d ln(m) , m e mg 3 P remains 6. Total activity injected cps Activity withdrawn. cps/. ml. cps/ml Assuming no significant decay occurs, then the volume of the animal s blood multiplied by. cps/ml blood withdrawn must equal the total activity injected.. cps V cps, V 37 ml ml 6. All evolved oxygen in O comes from water and not from carbon dioxide. 6. Sr-9 is an alkaline earth metal having chemical properties similar to calcium. Sr-9 can collect in bones, replacing some of the calcium. Once embedded inside the human body, betaparticles can do significant damage. Rn- is a noble gas, so one would expect Rn to be unreactive and pass through the body quickly; it does. The problem with Rn- is the rate at which it produces alpha-particles. With a short half-life, the few moments that Rn- is in

14 75 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW the lungs, a significant number of decay events can occur; each decay event produces an alpha-particle that is very effective at causing ionization and can produce a dense trail of damage. Additional Exercises 63. The most abundant isotope is generally the most stable isotope. The periodic table predicts that the most stable isotopes for exercises a-d are 39 K, 56 Fe, 3 a, and Tl. (Reference Table 9. of the text for potential decay processes.) a. Unstable; 5 K has too many neutrons and will undergo beta-particle production. b. Stable c. Unstable; a has too few neutrons and will most likely undergo electron capture or positron production. Alpha-particle production makes too severe of a change to be a likely decay process for the relatively light a nuclei. Alpha-particle production usually occurs for heavy nuclei. d. Unstable; 9 Tl has too few neutrons and will undergo electron capture, positron production, and/or alpha-particle production. 5. (ln )t 6. t / 573 yr; k (ln )/t / ; ln(/ )!kt; ln, t 9 yr yr o; from C dating, the painting was produced during the late 8s or early 9s. 65. The third-life will be the time required for the number of nuclides to reach one-third of the original value ( /3). (.693)t (.693)t ln!kt, ln, t / 3 3. yr The third-life of this nuclide is 9.8 years. t 9.8 yr 66. ln(/ )!kt; k (ln )/t / ;.. (ln )t ln,, yr ln(.)!(.88 5 )t, t 5 yr, yr 67. (ln )t ln!kt,.3 yr.7 ln!(5.6 )t, t 3. yr It takes 3. years for the tritium to decay to 7% of the original amount. Hence the watch stopped fluorescing enough to be read in 975 (9 + 3.).

15 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW m!( amu)!.97 amu E mc 3! amu amu E photon /( kg 3 J) 8.85 J hc/λ ( m/s)! J λ E hc J s J 8 m/s.7 3 m.7 nm 69., ton TT 9 J ton TT 35 mol 3 U J 35 g mol U U 9 g 35 U 9 g 35 U This assumes that all of the 35 U undergoes fission. 7. In order to sustain a nuclear chain reaction, the neutrons produced by the fission must be contained within the fissionable material so that they can go on to cause other fissions. The fissionable material must be closely packed together to ensure that neutrons are not lost to the outside. The critical mass is the mass of material in which exactly one neutron from each fission event causes another fission event so that the process sustains itself. A supercritical situation occurs when more than one neutron from each fission event causes another fission event. In this case, the process rapidly escalates and the heat build up causes a violent explosion. 7. Mass of nucleus atomic mass mass of electron. amu.59 amu.355 amu 3 RT u rms M / 7 3(8.35 J/K mol)( K).355 g( kg/ g) / 7 5 m/s KE avg mu amu amu 7 kg (7 5 m/s) 8 6 J/nuclei We could have used KE ave (3/)RT to determine the same average kinetic energy. 7. H + n H + n + H; mass H mass H.78 amu mass of proton m p m 3m p + m n (m p + m n ) m p (.78).56 amu E mc.56 amu.6656 amu 7 kg ( m/s) E 3.66 J of energy is absorbed per nuclei, or.86 J/mol nuclei.

16 75 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW The source of energy is the kinetic energy of the proton and the neutron in the particle accelerator. Challenge Problems 73. k ln ; t ln / kt (.693)t t / For 38 (.693)(.5 yr).693 U: ln.693, e yr For 35 (.693)(.5 yr).39 U: ln.39, e yr 9 9 If we have a current sample of, uranium nuclei, 998 nuclei of 38 U and 7 nuclei of 35 U are present. ow let s calculate the initial number of nuclei that must have been present.5 9 years ago to produce these, uranium nuclei. For 38 U: 998 nuclei.5, U nuclei For 35 U: 7 nuclei U nuclei So.5 billion years ago, the,-nuclei sample of uranium was composed of. 38 U nuclei and U nuclei. The percent composition.5 billion years ago would have been: ( U nuclei +. ) total nuclei 7. Total activity injected Ci 77% 38 U and 3% 35 U Activity withdrawn 3.6 Ci. ml H O Ci ml H O Assuming no significant decay occurs, then the total volume of water in the body multiplied by.8 6 Ci/mL must equal the total activity injected. V 6.8 Ci ml H O 8.65 Ci, V.8 ml H O Assuming a density of. g/ml for water, the mass percent of water in this 5-lb person is:

17 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW 753. g HO.8 ml HO ml 5 lb lb 53.6 g 7% 75. Assuming that the radionuclide is long-lived enough that no significant decay occurs during the time of the experiment, the total counts of radioactivity injected are: 3 5. cpm. ml 5. cpm ml Assuming that the total activity is uniformly distributed only in the rat s blood, the blood volume is: V 8 cpm 5. cpm, V. ml. ml ml 76. a. From Table 8.: H O + e H + OH E!.83 V E o cell E E!.83 V +.36 V.53 V o HO o Zr Yes, the reduction of H O to H by Zr is spontaneous at standard conditions because o E cell >. b. ( H O + e H + OH ) Zr + OH ZrO CH O + H O + e 3 H O(l) + Zr(s) H (g) + ZrO CH O(s) c. G!nFE! ( mol e )(96,85 C/mol e )(.53 J/C)!5.9 5 J!59. kj E E! E.59 log Q; at equilibrium, E and Q K. n.59 log K, log K n (.53).59, K d.. 3 kg Zr g mol Zr mol H.9 mol H kg 9. g Zr mol Zr.9 mol H.6 g H mol H. g H V nrt (.9 mol)(.86 L atm/mol K)(73 K).3 6 L H P. atm

18 75 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW e. Probably yes; less radioactivity overall was released by venting the H than what would have been released if the H had exploded inside the reactor (as happened at Chernobyl). either alternative is pleasant, but venting the radioactive hydrogen is the less unpleasant of the two alternatives. 77. a. b. C; it takes part in the first step of the reaction but is regenerated in the last step. C is not consumed, so it is not a reactant. 3, 3 C,, 5 O, and 5 are the intermediates. c. H He + + e ; m.6 amu! m e + m e! [(.78 amu m e )] m.6! (.78) + (.59)!.68 amu for four protons reacting For mol of protons, m!.68 g, and E for the reaction is: E mc!.68 5 kg ( m/s)!.38 J For mol of protons reacting:.38 mol H J!5.95 J/mol H 78. a U 86 Rn +? He +? e ; to account for the mass number change, four alphaparticles are needed. To balance the number of protons, two beta-particles are needed. 86 Rn 8 He + 8 Po; polonium-8 is produced when Rn decays. b. Alpha-particles cause significant ionization damage when inside a living organism. Because the half-life of Rn is relatively short, a significant number of alpha-particles will be produced when Rn is present (even for a short period of time) in the lungs. c. 86 Rn He Po; 8 8 Po He + 8 Pb; polonium-8 is produced when radon- decays. 8 Po is a more potent alpha-particle producer since it has a much shorter half-life than Rn. In addition, 8 Po is a solid, so it can get trapped in the lung tissue once it is produced. Once trapped, the alpha-particles produced from polonium- 8 (with its very short half-life) can cause significant ionization damage. d. Rate k; rate. pci L pci Ci 3.7 decays/ sec Ci.5 decays/scl k ln t/.639 d h. 3.8 d h 36 s 6 s rate.5 decays/s L K 6 7. Rn atoms/l. s

19 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW L Rn atoms mol 6. 3 Rn atoms. 9 mol Rn/L 79. Mol I 33 counts min mol I min 6.6 mol I 5. counts [I ] 6.6 mol I.5 L. mol/l Hg I (s) Hg + (aq) + I (aq) K sp [Hg + ][I ] Initial s solubility (mol/l) Equil. s s From the problem, s. K sp (s)(s) (. mol/l, s. mol/l. )(. ) H + H He; Q for H.6 9 C; mass of deuterium amu. E 9 9. J m/c (QQ ) r KE / mv ; 9 9. J m/c (.6 5 m 9 C) 3 J per alpha particle 3 7 J / ( amu.66 kg/amu)v, v 8 6 m/s From the kinetic molecular theory discussed in Chapter 5: / 3RT u rms, where M molar mass in kilograms M 8 6 m/s 3(8.35 J/K mol)(t) 3 kg /, T 5 9 K 3 kg/mol for deuterium Integrative Problems Bk + e 67 7 Bh +?; this equation is charge balanced, but it is not mass balanced. The products are off by mass units. The only possibility to account for the mass units is to have neutrons produced. The balanced equation is: Bk + e 7 Bh + n

20 756 CHAPTER 9 THE UCLEUS: A CHEMIST S VIEW ln!kt (.693)t t /, ln 99 (.693)t, t 6.7 s (Assuming is exact.) 5. s Bh: [Rn]7s 5f 6d 5 is the expected electron configuration Fe + n 6 7 Co +?; in order to balance the equation, the missing particle has no mass and a charge of!; this is an electron. An atom of 6 7 Co has 7 e, 7 p, and 33 n. The mass defect of the 6 Co nucleus is: m ( m e ) [7(.78 m e ) + 33(.866)]!.563 amu E mc!.563 amu.665 amu 7 kg ( m/s)!8.3 J Binding energy ucleon 8.3 J 6 nucleons. J/nucleon 3 The emitted particle was an electron, which has a mass of 9.9 kg. The debroglie wavelength is: h λ mv J s kg ( m/s).7 m