Chapter 23. Nuclear Chemistry

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Chapter 23. Nuclear ChemistryWhat we will learn: Nature of nuclear reactions Nuclear stability Nuclear radioactivity Nuclear transmutation Nuclear fission Nuclear fusion Uses of isotopes Biological effects of radiationGCh23-1

Nuclear ReactionsReactions involving changes in lProton11H (11p)1.67252x10-271.007276 1.6022x10-191Neutron1n1.67496x10-271.008665 00Electron0e (0-1b)9.1095x10-310.000549 -1.6022x10-19-1Positron0e (01b)9.1095x10-310.000549 1.6022x10-191a particle4He (42a) 6.64465 10 274.001506 3.2044x10-1920-112kgamuAtomic mass unit (amu)It is defined as one twelfth of the rest mass of an unbound neutral atom ofcarbon-12 in its nuclear and electronic ground state, and has a value of1.660538921 10 27 kgGCh23-2

Analogy with elementsAtomic number (Z)Number of protons in nucleus number of electrons in a neutral atomMass number (A)Sum of number of protons plus number of neutrons in nucleusAZXExample11H126Chydrogen atom having 1 proton in the nucleuscarbon atom having 6 protons and 6 neutrons in the nucleusNucleonNucleon is one of the particles that makes up the atomic nucleusGCh23-3

ElementA form of matter in which all of the atoms have the same atomic numberIsotopesTwo atoms of the same element with different mass numbersThey have the same number of protons but a different number of neutronsExamples of isotopes146Ccarbon with 6 protons and 8 neutrons126Ccarbon with 6 protons and 6 neutrons23892Uuranium with 92 protons and 146 neutrons23592Uuranium with 92 protons and 143 neutronsGCh23-4

Examples of H isotopesName Protons Neutrons Half-life timeMass (2) years3.0160414H131.39(10) 10 22 s4.0278115H149.1 10 22 s5.0353116H152.90(70) 10 22 s6.0449417 23H162.3(6) 10 s7.052751Balancing nuclear equationsTotal of Z and A must be same for products as for reactantsU(uranium) Th(thorium)23492Ug23090Th 42HeGCh23-5

ProblemBalance the following nuclear equations Po(polonium) Tb(terbium)a)21284Pog20882Tb XThe mass difference is 212-84 4, it means X is a particlea21284Pog208Tb 4b)g137Ba X13755Cs82562The mass difference is 0, it means X is b particle with an atomic number -13755Csg13756Ba 0-1bGCh23-6

Comparison of chemical and nuclear reactionsChemical reactionNuclear reactionAtoms are rearanged by the breakingand forming chemical bondsElements are converted from one toanotherOnly electrons in atomic or molecularorbitals are involvedProtons, neutrons, electrons andother particles may be involvedRelative small energy is formed orconsumedA very big amount of energy isformedRates of reactions are influencedby temperature, pressure, orconcentrationsRates of reactions are generally notaffectedGCh23-7

Nuclear densityThe calculation of a nucleous densityRadius of a nucleusr 5 x 10-13 cmVolume of a nucleusV 4/3 p r3 4/3 p (5 x 10-13 cm)3Mass of a nucleus 1 x 10-22 g (30 protons and 30 neutrons)d m/V 2 x 1014 g/cm3(this is a very big density)The highest density of an element is 22.6 g/cm3GCh23-8

Nuclear stabilityGeneral facts and rules Neutrons stabilize the nucleus More protons require more neutrons to stabilize the nucleus Up to Z 20, most stable nuclei have (neutron/proton) ratio 1 larger Z g larger n/p ratio (2.5 for Bi-209) All isotopes with Z 83 are radioactive - undergo spontaneous decay togive a stable isotope Magic numbers - 2, 8, 20, 50, 82, 126 protons or neutrons Even number of p and n - generally more stable Belt of stability nuclei outside are radioactiveGCh23-9

Belt of stabilityThe stable nuclei are located in anarea of the beltThe most radioactive nuclei areoutside this beltThe ratio of (n/p) 1 stablilityWhen a nucleus is above the belt and it has the (n/p) ration bigger than 1 itundergous the following process (called -b-particle emission)10n g11p 0-1bwhich increases the number of protons in the nucleus making the ratio (n/p)closer to 1GCh23-10

Examples146Cg144019Kg40Ca 0b9740Zrg97Nb 0b7N 20410-1b-1-1When a nucleus is below the belt and it has the (n/p) ration smaller than 1 itundergous the following process (called b-particle emission )11p g1on 0 1bor undergos electron capture, which decreases the number of protons, andhence moves up toward the belt of stabilityExamples3819Kg3818Ar 0 1bGCh23-11

Nuclear binding energy Energy required to break up a nucleus into protons and neutrons Mass defect - difference in mass of an atom and the sum of the masses ofits protons, neutrons, and electrons E (Dm)c2m mass difference (kg)c speed of light (3 x 108 m/s)E energy liberated (Joules)GCh23-12

Mass defectDifference in mass of an atom and the sum of the masses of its protons,neutrons, and electronsExampleF isotope has an atomic mass 18.9984 amu. Mass of 11H atom (1.007825amu), mass of the neutron (1.008665 amu).199The mass of 9 11H atoms (9 protons and 9 electrons) is9 x 1.007825 amu 9.070425 amu 10.08665 amuand the mass of 10 neutrons is10 x 1.008665 amuTherefore, the atomic mass of a 199F calculated from known numbers ofelectrons, protons and neutrons isGCh23-13

9.070425 amu 10.08665 amu 19.15705 amuwhich is larger than 18.9984 amu (the measured mass of 199F) by 0.1587 amu.The difference between the mass of an atom and the sum of masses of individualparticles is called mass defect.Based on the mass-energy equivalence relationship we can calculate the amountof energyDE (Dm)c2DE energy of the product - energy of the reactantDm mass of the product - mass of the reactantDm -0.1587 amuDE (-0.1587 amu) (3.0 x 108 m/s)2 -1.43 x 1016 amu m2 / s2GCh23-14

1 kg 6.022 x 1026 amu1 J 1 kg m2 / s2DE -2.37 x 10-11 JThis is the amount of energy released when one fluorine nucleus is formed from9 protons and 10 neutrons. Therefore in the formation of 1 mole of fluorine nucleithis energy need to be multiplied by the Avogadro numberDE (-2.37 x 10-11) x (6.022 x 1023 /mol)DE -1.43 x 1010 kJ / mol(which is a very big number)The average chemical reaction releases about 200 kJ / mol energy.GCh23-15

Nuclear binding energy per nucleonnuclear binding energy number of nucleonsGCh23-16

ProblemThe mass of a 73Li nucleus is 7.016005 amu. Given that the mass of a proton is1.007276 amu and that of a neutron is 1.008665 amu, calculate the mass defect,the binding energy per nucleon, and the binding energy per molemass of protons 3 (1.007276 amu)mass of neutrons 4 (1.008665 amu) 3.021828 amu 4.034660 amuTotal mass 7.056488 amuActual mass 73Li nucleus 7.016005Mass defect7.056488 - 7.016005 0.040483 amu(0.040483 amu) / (6.023 x1026 amu/kg) 6.721 x 10-29 kgGCh23-17

Binding energyE (Dm)c2- (6.721 x 10-29 kg) ( 2.998 x 108 m/s)2 6.04x10-12 J/nucleus-(6.04 x 10-12 J) /(nucleus) x (nucleus) /(7 nucleons) 8.628 X 10-13 J/nucleonBinding energy per mole- (6.04 x 10-12 J/nucleon) x 6.02 x 1023 nuclei/mole - 3.64 x 1012 J/molGCh23-18

ProblemThe mass of a 12753I nucleus is 126.9004 amu. Given that the mass of a proton is1.007276 amu and that of a neutron is 1.008665 amu, calculate the mass defect,and the binding energy per nucleonThere are 53 protons and 74 neutrons the mass of 53 11H atoms is53 x 1.007276 amu 53.41473 amu 74.64121 amuThe mass of 74 neutrons is74 x 1.008665 amuTherefore the predicted mass for 12753I is 53.41473 74.64121 128.05594 amuThe mass defectDm 126.9004 - 128.05594 -1.1555 amuGCh23-19

DE (Dm)c2 (-1.1555 amu) (3.0 x 108 m/s)2 -1.04 x 1017 amu m2 / s2 -1.73 x 10-10 JThus the nuclear binding energy is -1.73 x 10-10 J. The nuclear binding energy pernucleon is (-1.73 x 10-10 J) / (127 nucleons) 1.36 x 10-12 J/nucleonGCh23-20

Natural radioactivityRadioactive decaySequence of nuclear reactions that ultimately result in the formation of a stableisotopeTypes of decay Alpha decay23892 U gemission of a particle (42He or42a 23490Beta decay146C g0-11472a)Themission of b particle (0-1e orb 40-1b)NGCh23-21

The decay series of naturally occuring uranuim 238 which involves 14 stepsGCh23-22

Naturally occurring isotopes U(Uranium), Th (Thorium), Pa(Protactinium), Pb(Lead) Uranium SeriesHe t1/2 4.5 x 109 yr.23892U g234Th 23490Th g234Pa 0e t1/2 24.4 da.23491Pa g234U 0e t1/2 1.14 min.23492U g90919223090Th 42-1-142He t1/2 2.7 x 105 yr.Net decay23892U g20682Pb 8 42He 6 0-1et1/2 4.5 x 109 yr.GCh23-23

Half-life and rates of decayAll radioactive decays obey first-order kineticsrate of decaylN l - the first order rate constantN - the number of radioactive nuclei present at time tln( Nt / N0 ) - l tThe corresponding half-time of the reaction ist1/2 0.693 / lThe value of t1/2 depends on a particular nucleus23892U g21484Po g23490210Th 824Pb 2He t1/2 42Het1/2 4.51 x 109 year1.60 x 10-4 sGCh23-24

Radioactive datingThe half-live of radioactive isotopes have ben used as “atomic clock” todetermine the ages of certain objectsCarbon dating 146C146C is produced in atmosphere and this isotope decays according to the equation146C g147N 0-1bt1/2 5740 yearsCO2 in living plants has fixed amount of 146C taken from the atmosphere. Whenplant dies, 146C decay begins, and the concentration of this isotope decreaseswith time. If we know a current concentration of 146C in a dead plant with theconcentration of this isotope in a living plant, we can use the formulaln( Nt / N0 ) - l tto estimate the time of plant deadGCh23-25

Dating using 23892UU is a naturally occuring mineral which is used to estimate a time of rocks inthe earth. The decay of this isotope is very long2389223892U g20682Pb 8 42a 6 0-1bt1/2 4.5 x 109 yearsThe time is estimated from the ratio of concentrations of 23892U and 20682Pb in asample.If only half of a mole of uranium undergone decay, the mass ratio of20682Pb / 23892U 206 g / 238 g 0.866Ratios lower than 0.866 mean that the rocks are less than 4.51 x 109 year old.GCh23-26

Dating using 4019KDating using 4019K is based on the reaction of electron capture4019K 0-1e g4018Ar t1/2 1.2 x 109 yearsThe concentration of gaseous argon is used to measure the age of an object.This technique is used in geochemistry. A mineral is melted and the concentrationof argon is measured using a mass spectrometer.GCh23-27

Nuclear transmutationsBombardment of nuclei in particle accelerator - may form new isotopesHistorically, the first reaction of nuclear transmutation was performed byRutherford in 1919147N 42ag178O 11pwhere oxygen-17 was produced with the emission of a proton. This reactionopened a possibility for making new elementsProblemWrite the balanced equation for the nuclear reaction 5626Fe(d,a)5425Mn, where thed symbol represents the deuterium nucleus 21H5626Fe 21Hg42a 5425MnGCh23-28

Nucleosynthesis10527B 13Al 42He g4213He g7N 30151nP 100nTransuranium elementsElements with atomic numbers greater than 92, made in particle acceleratorsZSymbolSynthetic Method93Np94Pu23995Am23996Cm23999Md25323892U 10n g23993Half-lifeNp 0-1e86.4 yr93Np g 23994Pu 94Pu g 10n g 24095Am 0-1e458 yr94Pu 42He g 24296Cm 10n4.5 hr99Es 42He g 256101 Md 10n1.5 hr0-1e2.35 daGCh23-29

Nuclear FissionProcess in which heavy nuclei are split into smaller nuclei and neutrons23592U 10n g [ 23692U] g9038Sr 14354Xe 3 10n ENERGYThe heavy nucleus is less stable than its products, and this process releasses alarge amount of energyNuclear bindingenergyU2.82 x 10-10 JSr1.23 x 10-10 J238929038143541.92 x 10-10 JXeThe difference in binding energy of the reactants and products per nucleus (1.23 x 10-10 1.92 x 10-10)J - (2.82 x 10-10)J 3.3 x 10-11 JGCh23-30

The difference in binding energy of the reactants and products for 1 mole (3.3 x 10-11)J x (6.02 x 1023) / mol 2.0 x 1013 J/molFor comparison, 1 ton of coal is only 5 x 107 JDuring this reaction neutrons are produced and next are originally captured in theprocess making the reaction self-sustainingChain reactionSelf-sustaining sequence of nuclear fission reactionsCritical massMinimum amount of fissionable material needed for chain reactionGCh23-31

Atomic BombThe first application of nuclear fission was in development of the atomic bomb.A small atomic bomb is equilvalent to 20,000 ton of TNT (trinitrotoluene)1 ton TNT releases20,000 ton TNT releases4 x 109 J of energy8 x 1013 J of energy1 mole (235 g) of uranium-235 releases 2 x 1013 J of energy, therefore the massof the isotope present in a small atomic bomb is(235 g) (8 x 1013 J) / (2 x 1013 J) 1 kgThe critical mass if formed in a bomb using a conventional explisive like TNTUranium-235 was used in HiroshimaPlutonium-239 was used in NagasakiGCh23-32

Nuclear ReactorLight water reactors (H2O)Most of the nuclear reactors in US are light water reactors (H2O)Water is used to transport heat of the nuclear reaction outside the reaction and togenerate electricityUranium-235 is used as a reactor fuleThe nuclear reaction is controled bycadmium or boron rodes which captureneutrons11348Cd 105B 1010n gn g1144873Cd gLi 42aGCh23-33

Heavy water reactors (D2O)Heavy water absorbs neutrons less efficiently than light water, and thereforeheavy water reactors do not require enriched uranium, which is needed in lightwater reactorsHeavy water is produced from light water by fractional distillation or electrolysis ofordinary water which includes of very small amount of heavy water.Canada is currently the only nation producing heavy water for nuclear reactorsBreeder reactorsThe reactor using uranium fuel, from which it produces more fissionable materialsthan it uses23892U 23992U g23993Np g10n g2399323923992UNp 94Pu 0-10b-1bTherefore nonfissionable uranium-238 is converted into fissonable plutonium-239GCh23-34

Nuclear FusionCombining of small nuclei into large onesProduces even more energy that fissionNuclear fusion occurs constantly in the sun. The sun is made up mostly ofhydrogen and hellium. At the temperature 15 million C the following reactions arebelieved to take place21H 1132He 311H H g21132HeHe gH g4221He 2 11HH 01bGCh23-35

Fusion reactorsThe fusion reaction requires a very high temperature to proceed21H 2H g3H 21H 3H g4He 63Li 21111211H E 6.3 x 10-13 J10n E 2.8 x 10-12 JH g 2 42H E 3.6 x 10-12 JThose reactions require a temperature of about 100 million C. At this temperaturemolecules can not exist losing electrons, and forming a new state of matter whichis called plasmaPlasmaThe state of matter formed at very high temperature (million of degrees Celsius)from a gaseous mixture of positive ions and electronsTechnically plasma can be formed inside a very strong magnetic field keeping theparticles togetherGCh23-36

Hydrogen bombThermonuclear bomb containing solid lithium deuteride (LiD) which can bepacked very tightlyThe detonation of a hydrogen bomb occurs in two stages- fission reaction- fusion reactionThe required high temperature for fusion reaction is achieved from an atomicbomb63Li 2H g 2 42a21H 2H g1131H 11HThere is no critical mass of the fusion bombGCh23-37

Uses of isotopesStructure determinations (helps locate atoms via radioactive emission)Mechanism studies (trace steps via radioactivity)ExampleThe formula of thiosulfate ion is S2O32-. The ion is prepared from the reactionSO32- (aq) S (s) g S2O32- (aq)When the ion is treated with an acid, the reaction is reversedS2O32- (aq) g SO32- (aq) S (s)and the elemental sulfur is precipitated. Chemists were uncertain whether thetwo sulfur atoms occpied equivalence position in the ionGCh23-38

Two possible structures were proposedThe first reaction started with elemental surfur enriched with the radioactivesulfur-35. After precipitation the label sulfur is found in the solid elemental sulfor,and not in the solution, which indicates that nonlinear structure is valid, wheresulfur atoms occupy non equivalent positionsGCh23-39

Medicine Diagnosis (measure radioactivity)-Sodium-24 - traces blood flow-Iodine-125 goes to thyroid and gives way to view activity-Technetium-99 - traces organs such as heartTherapy (destroying tumor sites)-Cobalt-60 - g radiation directed at tumor sites-Iodine-131 destroys overactive thyroidGCh23-40

Biological effects of radiationThe fundamental unit of radioactivity is curie (Ci)1 Ci 3.7 x 1010 nuclear disintegrations (results of radioactive decay) persecondThis decay rate is equivalence to that of 1 g of radium. A milicurie (mCi) is onethousand of a curie.Example10 mCi (carbon-14) (10 x 10-3) (3.7 x 1010) 3.7 x 108disintegrations per secondThe unit of the absorbed dose of radiation is the rad (radiation absorbed dose)1 rad amount of radiation that results in the absorption of 1 x 10-2 Jenergy per kilogram of irradiated materialGCh23-41

The biological effect of radiation also depends on the part of the body and thetype of radiationRBE-relative biological effectivness 1 for beta and gamma 10 for alphaGenerallya particlesusually have the least penetration powerb particlesusually have stronger penetration powerg raysusually have the strongest penetration powerRadiation can remove electrons from atom and molecules forming ions andradicalsGCh23-42

RadialsA molecular fragment having one or more unpaired electronsThese are short-lived species and they are very reactiveH2OH 2OgH2O H2O ge-H3O .OHThe electron can also react with molecular oxygene - O2 g.O-2forming the superoxide ion .O2- which reacts with organic compounds of enzymeand DNA molecules, leading to cancerGCh23-43

ProblemComplete the following nuclear equations and indetify X(a)13553Ig13554Xe XThe sum of masses must be conserved, A 0The atomic number must be conserved, Z -1Ig135Kg013553(b)401954-1Xe 0-1bb XThe sum of masses must be conserved, A 40The atomic number must be conserved, Z 204019Kg0-1b 4020CaGCh23-44

(c)5927Co 10n g5625Mn XThe sum of masses must be conserved, A 4The atomic number must be conserved, Z 25927(d)Co 1n g56U 1n g99235920025Mn40Zr 42135a52Te 2XThe sum of masses must be conserved, A 1The atomic number must be conserved, Z 023592U 10n g9940Zr 13552Te 210nGCh23-45

ProblemEstimations show that the total energy output of the sun is 5 x 1026 J/s. What isthe corresponding mass loss in kg of the sun.We use the DE (Dm)c2 equation1 J 1 kg m2 / s2Dm DE / c2 Dm 6 x 109 kg(5 x 1026 kg m2 / s2) / (3 x 108 m/s)2ProblemCalculate the nuclear binding energy (in J) and the binding energy per nucleon ofthe following isotopesa)42He (4.0026 amu)The binding energy is the energy required for the processGCh23-46

42He g 2 11p 2 10nThere are two protons and 2 neutrons in the helium nucleus. The mass of 2protons is2 (1.007825 amu) 2.015650 amuand the mass of 2 neutrons2 (1.008665 amu) 2.017330 amuTherefore the predicted mass of 42He ism(42He) 2.015650 2.017330 amu 4.032980 amuThe mass defect isDm 4.032980 - 4.0026 0.0304 amuGCh23-47

The energy change isDE (Dm)c2 (0.0304 amu) (3 x 108 m/s)2 2.74 x 1015 amu m2 / s2Conversion the energy into J1 J 1 kg m2 / s2 (2.74 x 1015 amu m2 / s2) / (6.022 x 1023 amu / g) (1 kg /1000 g) 4.55 x 10-12 JThis is the nuclear binding energy required to break up one helium-4 nucleusinto 2 protons and 2 neutronsGCh23-48

Nuclear binding energy per nucleonnuclear binding energy number of nucleonsb)18474 (4.55 x 10-12 J ) / (4 nucleon) 1.14 x 10-12 J/nucleonW (183.9510 amu)The binding energy is the energy required for the process18474W g 74 11p 110 10nThere are 74 protons and 110 neutrons in the wolfram nucleus. The mass of74 protons is74 (1.007825 amu) 74.57905 amuGCh23-49

and the mass of 110 neutrons is110 (1.008665 amu) 110.9532 amuTherefore the predicted mass of 18474W ism(18474W) 74.57905 110.9532 amu 185.5323 amuThe mass defect isDm 185.5323 - 185.9510 1.5813 amuThe energy change isDE (Dm)c2 (1.5813 amu) (3 x 108 m/s)2 1.42 x 1017 amu m2 / s2GCh23-50

Conversion the energy into J1 J 1 kg m2 / s2 (1.42 x 1017 amu m2 / s2) / (6.022 x 1023 amu / g) (1 kg /1000 g) 2.36 x 10-10 JThis is the nuclear binding energy required to break up one wolfram-184nucleus into 74 protons and 110 neutronsNuclear binding energy per nucleonnuclear binding energy number of nucleons (2.36 x 10-10 J) / (184 nucleons) 1.28 x 10-12 J/nucleonGCh23-51

ProblemA freshly isolated sample of 9039Y was found to have an activity of 9.8 x 105disintegrations per minute at 1:00 p.m. on December 3, 2003. At 2:15 p.m. onDecember 17 2003, its activity was redetermined and found to be 2.6 x 104disintegrations per minute. Calculate the half-life time of 9039Y.The radioactive decay process has fist-order rate law, thereforet1/2 0.693 / lwhere l is the rate constantln( Nt / No ) - l tThe time interval is:(2:15 p.m. 12/17/2003) - (1:00 p.m. 12/3/2003) 14d 1hr 15min 20,235 minGCh23-52

ln [(2.6 x 104 dis/min) / (9.8 x 105 dis/min) ] - l (20,235 min)l 1.8 x 10-4 /mint1/2 0.693 /( 1.8 x 10-4 min) 3.9 x 103 minProblemWhat is the activity (rate), in millicuries, of a 0.5 g sample of 23793Np. The isotopedecays by a-particle emision and has a half-life of 2.2 x 106 yr. Write a balancednuclear eqation for the decay of 23793NpOne millicurie represents 3.7 x 107 disintegrations/s. The rate of decay of thisisotope is given by the rate lawrate l Nwhere N is the number of atoms in the sample.GCh23-53

t1/2 0.693 / ll 0.693 / t1/2 0.693 / (2.2 x 106 yr) (1 yr / 365 d) (1 d / 24 h) (1 h /3600 s) 9.99 x 10-15 /sThe number of atoms (N) in 0.5 g sample of neptunium-237 is:0.5 g / (237 g) (6.022 x 1023 ) 1.27 x 1021 atomsThe rate of decay 1.27 x 107 dis/sThe activity of the sample (in millicuries) is the rate of decay / (1 millicurie)The activity is (1.27 x 107 dis/s) (1 millicurie / 3.7 x 107 dis/s)GCh23-54

0.343 millicuries(b) The decay equation is:23793Np g23391Pa 42aGCh23-55

ProblemStrontium-90 is one of the product of fission of uranium-235. This strontiumisotope is radiactive, with a half-life 28.1 yr. Calculate how long (in yr) it will takefor 1.0 g of the isotope to be reduced to 0.2 g by dacayThe radioactive decay process has a fist-order rate law, thereforeln (Nt / No ) - l twhere l is the rate constant, and a half-life time ist1/2 0.693 / lThereforel 0.693 / t1/2The half-life time in secondst1/2 (28.1 yr ) ( 365 dy / yr) (24 h / dy ) (3600 s /h) 8.86 x 108 sGCh23-56

l 0.693 / t1/2 7.82 x 10-10 /sln (Nt / No ) - l tt - ln (Nt / No ) / lt - ln(0.2/1.0) / 7.82 x 10-10 /st 2.06 x 109 s (2.06 x 109 s) / [ ( 365 dy / yr) (24 h / dy ) (3600 s /h) ] 65.2 yrGCh23-57

ProblemCobalt-60 is used in radiation therapy. It has a half-life time of 5.26 years.(a) Calculate the rate constant for radioactive decay(b) What fraction of a certain sample will remain after 12 yearsThe radioactive decay process has fist-order rate law, thereforeln (Nt / No ) - l twhere l is the rate constant, and a half-life time ist1/2 0.693 / lThereforel 0.693 / t1/2 0.693 / (5.26 yr) 0.132 yrGCh23-58

The fraction of a certain sample that will remain after 12 years isNt / Nowhere t 12 yrRearrange the equationln (Nt / No ) -ltNt / No e- l t e- (0.132 yr)( 12 /yr) 0.205GCh23-59

Nuclear Chemistry What we will learn: Nature of nuclear reactions Nuclear stability Nuclear radioactivity Nuclear transmutation Nuclear fission Nuclear fusion Uses of isotopes Biological effects of radiation. GCh23-2 Nuclear Reactions Reactions involving changes in nucleus Particle Symbol Mass Charge

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