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Chapter 21 Solutions

Chapter 21 Solutions - CHAPTER TWENTY-ONE THE N UCLE US A...

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Unformatted text preview: CHAPTER TWENTY-ONE THE N UCLE US: A CHEMIST'S VIEW Radioactive Decay and Nuclear Transformations 11 A11 nucicar reactions must be charge balanced and mass baianccd. Tc charge balance, balance the sum ofthc atornic numbers on each side ofthc reaction and to mass balance, baiance the sum ofthe mass numbers on each. side of the reaction. 51 0 1.1, 51 131I O 131 a. MC: + _1e 23V b. 531M413 54 Xc c. $21122: + is d. 25251} w» 3119, 9&1'1'11 2. a 3263 ~> gee + fie . b. 2982?: 286805 +41% 1:. 280333131 2;}ng + £6 :1. ZgéCm + “23 1., 2945121211 3. a. (363 + 0e 232:; b. 33cm .1, £3 + 3:111 c. 2813A 1, gm + 23581 d. 13535 A 33 133A 4. a zgéAm ~+ :He + 23381;: b. ZgéAm ”1 8 31%: + 4 fife + 2:381; The final product is 23381. c. zgéAm —23§Np + a — zgiPa + a 23311 -+ 13 2331114» +1: -+ Zggaa + a / Afl‘ 54‘ 644 CENTER 2}. THE NUCLEUS: A CHEMISTS VIEW 249 15 . .. 260 2 249 . .10 c. 98cm 7h +105Db+40n 61. 98cm 53 :0 v+ 252,}; +2511 Kinetics of Radioactive Decay 11, kzgggze.693lsx lyr ><1d X 1h: am 432.2yr 365d 24h 36005 = 5.086 x 10“ s41 lmol x 6.022x 1029‘ nuclei Rate “—‘kN 2 5.886 X 10415" X 5.80 g x 241g moi = 6.35 X £0” decaysls 6.35 X 10” alpha particies axe emitted each second {tom 3 5.00 g MAJ]: sample. 12. Kr—Si is most stable since it has the longest haif—Iife while KP73 is hottest {least stabie) since ithas the shortest half-«life. 12.5% of each isotope Win remain afier 3 half—fives: 100% “f" 50% Te- 25% “we 12,5 :2 HZ ‘vz For Kr-73: t = 3(27 5) m 818 For Kit—74: t W 3(115 min) =2 34.5 min For Kr-'?6: t = 304.8 h) x 44.4 h For Kx~31: tm 30.1 X 105‘ yr) m 6.3 X105 yr m2:__o.6931>< 1d X 1h {”2 12.8d 24?: 36005 13. a. k a = 6.27 >< 10'? s“ 1 moi x 6.022 x 1023nuciei b. Rate ”RN W 6.27 X 10" s" >< 28.0 x 10'3 g x 64.0 g mo}. Rate = 1.65 x 10” decays J’s e. 25% of the r"‘Cu will remain after 2 half»1ives (100% decays to 50% after one haif—Iife which decays to 25% afier a second haif-iife). Renee, 2(12.8 days) = 25.6 days is the time flame for the experiment. ID 14. a. OBEDOCiXW m 3.7 x IDs-decaysfs; k m ESE. C1 t1}? RatemkN, x 2.87 h 3600 s B e 1 3.?x18 decays :( 0.6931 1h ] KN, 1%ij 10” atoms was 5 CHAPTER 21 THE NUCLEUS: A CHEMISTS VIEW 645 1 m; 333 1m: NaSSSO X 5.5 x 10” atoms 3"3 x 4 m 9.1 x 10'” moi NaZESSOd 6.02 x 1023 atoms moi 338 148.0 \2‘ 38so 9.: x 10*12 mo} Na238804 >< #flffiwi 2: L3 x 109 g m 1.3 ng Nafisqt moi Najgso4 b. 99.99% decays, 0.01% lefi; In 953 v-kt = 393%, 1:33.1 hours ,_. 49 hours 100 2.37 h 15. Man yr; 1cm 3‘3; In 33 ~16 m mm; 3:1. ne'l'“=0.242 the N0 28.8 yr No - 24.2% of the ”Sr remains as of Juiy 16, 2004. 16. 11%»:me 49:; N m 0.810 No; rm : (In 2y}: 1n(0.010) = ‘0“ 2” 2 $49335, tn 54 days - Em 8% 32 ' 17. 175mg 1~Ia13323304 x M = 33.9 mg 32P; km 3313 165.0 mg N33 ”FDA to: _ N _ —0.6931 t m ~v0.6931(35.0d) . . in ——— ”mist“ ,ha — w;Carrym extrasn.fis.: [NJ tU2 (33.9 mg) 14.36 g g g 111011) m -1.696 + 3.523 3 1.827, :11 == em? m 6-22 mg 37‘1? remains 18. The assumptions are that the :4C ievei in the atmosphere is constant or that the i‘*(3 iévei at the time the piant died can be caieuiated. A constant ”C Ievel is a poor assumption and accounting for variation is complicated. Another problem is that some of the material must be destroyed to determine the 1"C love}. 15.; ~(1n 2}t 19. t 2«”5780 ;km 1112!! ;h1 KN 2:- ‘Inm: m YT ( )m (N a) 1‘1", 15.3 5730y1‘ No; From i4C dating, the painting was produce/(1m; the earliest) during the late 18005. , tr-‘109yr 20. am 513; 11: t1)? 8 N t 15.3 5730 yr 0 3g“) =~—1<;t= 416931;, 1% N ] w ‘0.693(15.{}00yr) ___1‘ if? 13% m 6"” m (11?, N 315.3 X 0.17 =~“- 26 counts per minute per g ofC Ifwe had 10. mg C, we wouid see: 10. mg x 3 g x 2.6 Counts :4 0.026 eounts 1000 mg min g mm CHAPTER 21 THE NUCLEUS: A CHEMISTS VIEW 647 d. If some “A: escaped, then the measured ratio of ”AIWK would be less than it should be. We would calculate the age of the rocks to be iess than it aetuaily is. Energy Changes in Nuclear Reactions 23 . 2 . 2 25_ Agmmcl’mmfizmaggxmékg c 2 (3.00 x 103 ma’s)2 The sun loses 4.3 >< 10‘ kg of mass each second. Note: 1 I m 1 kg mzfs2 1.8 ><._10M Id 1000} 3600 s 24 h x x ~ x 8 k3 h day 26. "' 1.6 X 1022 deay 6 (3-00 X 108 INS)2 soia: energy to the earth. 1141 x 1g X 1kg 1000: 32k} 1000g 1.6 new >< e 5.0 X10” kg of coal is needed to provide the same amount of energy. 27. From the table at the back ofthe text, the mass of a. proton 3 1.0038 emu, the mass of a neutron = 1.00366 arm, and the mass of an electron : 5.436 x 10“ mo. Mass of nuciens = mass of atom - mass of electrons = 55.9349 - 26(03005486) : 55.9206 aim: 56 Fe; Am m 55.9206 emu — [26(1.00728) + 30(1.09866)} mm: a «0.5285 emu 26 -27 AF. a Am2 a '0.5235 mm x ~w kg 1 1 I... 26:3 + 300m >< (2.9979 X 108 noisy: 51.887 X 10'“ J amu . . ~12 binding energy m 7.88? x 10 3 m 1.403 x 10.” Jinncieon noeieon - 56 nucleons ' ' 28. For g H: mass defect : Am 2 mass of ? H necieus - mass of proton 4 mass of neutron. The mass of the ”H nucleus Wili equal the atomic mass of 2H minus the mass of the electron in a 2H atom. From the back of the text, the pertinent masses are: m.; m 5.49 x 10“ emu, mJP = 1.06328 me, 113,, e 1.00866 emu. Am a 2.0mm emu — 0.000549 amu .. [1.00723 mm + 1.00366 mm] = «2.39 x 10-3 emu X 3.6605 x 10““ kg amu AB m :3ch x -2.39 x 10-3 mm x (2.993 x 10* my e .357 x 10'13 I BE 3 357x13”! nucleon 2 nucleons : 1.?9 X 10'” anucleon .4..\.:.".7«..'-'."—1.."_"._." 21“,“: 648 29. 30. 3E. CHAP’EER 21 THE NUCLEUS: A CHEMISTS VIEW For 3 H: Am = 3.01605 — 0.000549 - [1.00023 + 20 00305)} 9 910 x 10'3 am 1 '37 AB 9 -910 x 10301111: x ”3605 x 19mg; x (2.993 x 103 mm)2 x -136 >< 10-” I amu BE 2 1.30x10'"123 nucleon 3 nucieons : 453 x 10‘” Jinucleon Let in: m mass of electron; For 12C (68, 6p, 621): mass defect 2 Am m mass of ”G nucieus {mass of 6 protons + mass of 6 neuzrons]. Note: Atomic masses given include the mass of the electrons. Am m 12.00000 amu - 6 me - [6(100782 - me) + 6(1.00365)]; Mass of Eiccfions canoe}. And = 12.00000 ~ [6(1.00782}+ 6(1.00866)} 3 —0.09838 amu ' ‘2'? .033; = Am2 2 009383 mm x Mil? x (2.9909 x 10* my 9 “1.475 x 10“ I 311111 BE z 1.4?6x 10m”! nucieon 12 nucleons m 1.230 X 10“” Nancie-on For ”5U (920, 9213, 1433): 0m 9 235.0439 — 92 me ~ {92900782 — m9 91430003663 «9 «1.9139 3.11111 1.06054 x 10"?” kg amu AB 9 Am! = 3.9139 mm: x x (2.99792 x :03 mfs)’ = 9.3553 x 104": BE 2 “2.8563 >< 10*“: w 3 1.2154 x 10”” Ifnucieon nucieon 235 nucleons Since 2‘SFe is the most stabie know nucleus, than the binding energy per nucieon for SfiF0. (1.408 X 10'” anucleon) W01 be Larger than that-of 12C or 22'51] (see Figure 21.9 of the text). Am = «26.486 >< 10“4 mm) = 91.097 X 10’3 amu £0005 x .2043“; amt: AE “ Ame: m 4.097 X 1032:3111 x X (2.9979 x 103 1:015):E : 4.637 X 10‘” J Emma 1.0.0.63? x 3.0“” 3)... 8.185 x 10*” J 90090 ha _ 6.6261x10“34} 5 x 2.9909 x103 m/s E 8.185 x10”“‘1 9L: '" 2.427 >< 20'” m == 2.427 X 10”3 1111: iii + iii *9 3H0 9. $0; Mass of eicctrons cancel when determming Am for this nuclear reaction. 11m 3 [4.00260 +1.00366 - (2.02410 + 3.01605)} emu“; «1.889 X 10‘2 amu ."\p(.'|\.uwM.uI-— mew-M aw—......-..m. . . .......... CHAPTER 21 THE NUCLEUS: A CfflEMEST'S VIEW 653 c. z 4 o 41H 21-19, +2 +1 Am 3 4.00260 — 4(130782) + 4(0000549) : -0.02648 21:11:: for 4 protons reacting e; Am m 4.00260 emu — 2 noB + 2 noa — {4(1.DO782 amu n ma] For 4 moi of protons, Am 3 4102648 g and AB for the reaction is: AB = 1.311162 == ~2.648 X 10‘5 kg K (2.9979 X 108 mfs 3 fl 0.380 X 10” J __ 5-2 - For 1 mol of protons reacting: Mi m "5.950 x 10” Elmo! ‘H 4 mol ’3 Chalienge Problems 55. 56. moII= [1’3 33 counts x 1 mom-min min 5.0 x 10‘1 counts m 6.6 >< 10‘“ mol 1 -g} _ m 6.6x10 moi! 54.4X10‘mmob’L {HSOL Holes) o Hsflaq) + 21136:) mefiisflfi? Initial s * solubility (moi/I.) 0 0 Equil. s 25' From the probiem, 23 m 4.4 >< 10"" molfL, s = 2.2 x 10“" room; Kw == (SXZSY m (2.2 X 10“”)(4.4 X 10‘1")2 x 4.3 X 10'” 8.. b. 29?fo -+ zgfizkn + ? 3H6 + ? fie; To account for the mass number change, 4 alpha pafiioles an: needed. To baianoe the number of protons, 2 beta particies are needed. 282: Rn —> :He + 2;: 1’0; Polonium—Sé is produced when mRn decays. Aipha panieies cause significant ionization damage when inside a iiving organism. Since the haltllife of mkn is relatively short, a significant number of alpha particles wilt be produced when mRn is present (even for a short period of time) in the lungs. 2;: Po; 2;: Po -»+ 31% + 2;: Pb; Polonium~218 is produced when radon-222 decays. MP0 is a more potent alpha partioie producer since it has a much shorter half—life than 222 4 861m W 2He + ”2111:. En addition, 2131% is a solid, so it can get trapped in the hing tissue once it is produced. Once trapped, the alpha. particles produced from polonium—Zi 8 (with its very short halfwiife) can cause significant ionization damage. ...
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