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Introduction to Nuclear Engineering


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  • 1. Introduction to Nuclear Engineering Third Edition John R. Lamarsh Late Professor with the New York Polytechnic Institute Anthony J. Baratta Pennsylvania State University Prentice Hall -------- Prentice Hall Upper Saddle River, New Jersey 07458
  • 2. Library of Congress Cataloging-in-Publication Data is on file. Vice President and Editorial Director, ECS: Marcia J. Horton Acquisitions Editor: Laura Curless Editorial Assistant: Erin Katchmar VIce President and Director of Production and Manufacturing, ESM: David W. Riccardi ".ecutive Managing Editor: Vince O'Brien Managing Editor: David A George Production Editor: Leslie Galen Director of Creative Services: Paul Belfanti Creative Director: Carole Anson Art Director: Jayne Conte Art Editor: Adam Velthaus Cover Designer: Bruce Kenselaar Manufacturing manager: Trudy Pisciotti Marketing Manager: Holly Stark Marketing Assistant: Karen Moon Cover image: Courtesy ofFramatome Technologies © 2001 by Prentice-Hall, Inc. Upper Saddle River, New Jersey 07458 The author and publisher of this book have used their best efforts ir preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. Printed in the United States of America 10 9 8 7 6 5 4 3 ISBN 0-201-82498-1 Prentice-Hall International (UK) Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall of Canada Inc., Toronto Prentice-Hall Hispanoamericana, S.A., Mexico Prentice-Hall of India Private Limited, New Delhi Prentice-Hall of Japan, Inc., Tokyo Pearson Education Asia Pte. Ltd., Singapore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeiro
  • 3. Preface to Third Edition This revision is derived from personal experiences in teaching introductory and advanced level nuclear engineering courses at the undergraduate level. In keeping with the original intent of John Lamarsh, every attempt is made to retain his style and approach to nuclear engineering education. Since the last edition, however, considerable changes have occurred in the industry. The changes include the devel­ opment of advanced plant designs, the significant scale-back in plant construction, the extensive use of high speed computers, and the opening of the former Eastern Block countries and of the Soviet Union. From a pedagogical view, the World Wide Web allows access to many resources formerly only available in libraries. Attempts are made to include some of these resources in this edition. In an attempt to update the text to include these technologies and to make the text useful for the study of non-western design reactors, extensive changes are made to Chapter 4, Nuclear Reactors and Nuclear Power. The chapter is revised to include a discussion of Soviet-design reactors and technology. The use, projection, and cost of nuclearpower worldwide is updated to the latest available information. In Chapter 11, Reactor Licensing and Safety, the Chemobyl accident is dis­ cussed along with the latest reactor safety study, NUREG 1150. A section is also included that describes non-power nuclear accidents such as Tokai-Mura. iii
  • 4. iv Preface to Third Edition The basic material in Chapters 2-7is updated to include newer references and to reflect the author's experience in teaching nuclear engineering. Throughout the text, the references are updated were possible to include more recent publications. In many topic areas, references to books that are dated and often out of print had to be retained, since there are no newer ones available. Since these books are usually available in college libraries, they should be available to most readers. Chapter 9 is retained in much its same form but is updated to include a more complete discussion of the SI system of units and of changes in philosophy that have occurred in radiation protection. Since many of these changes have yet to reach general usage, however, the older discussions are still included. As in the second edition, several errors were corrected and undoubtedly new ones introduced. Gremlins never sleep!
  • 5. Preface to Second Edition At his untimely death in July 1 98 1 , John R. Lamarsh had almost completed a revi­ sion of the first edition of Introduction to Nuclear Engineering. The major part of his effort went into considerable expansion of Chapters 4, 9, and 11 and into the addition of numerous examples and problems in many of the chapters. However, the original structure of that edition has been unchanged. Chapter4, Nuclear Reactors and Nuclear Power, has been completely restruc­ tured and much new material has been added. Detailed descriptions of additional types of reactors are presented. Extensive new sections include discussion of the nuclear fuel cycle, resource utilization, isotope separation, fuel reprocessing, and radioactive waste disposal. In Chapter 9, Radiation Protection, considerable new material has been added on the biological effects of radiation, and there is a new section on the calculation of radiation effect. The section on the sources of radiation, both artificial and nat­ ural, has been expanded, and the sections on standards of radiation protection and computation of exposure have been brought up to date. A section on standards for intake of radionuclides has also been added. In Chapter 1 1 , Reactor Licensing, Safety, and the Environment, the sections on dispersion of effluents and radiation doses from nuclear facilities have been con­ siderable expanded to cover new concepts and situations. Included in this chapter is v
  • 6. vi Preface to Second Edition a discussion of the 1979accident at Three Mile Island. The structure of this chapter has been kept as it was in the first edition in spite of the earlier suggestion that it be broken up into two chapters treating environmental effects separately from safety and licensing. Several errors that were still present in thelast printing ofthe first edition have been corrected, including those in Example 6.7and in the table of Bessel functions in Appendix V. We are indebted to many of John Lamarsh's friends and colleagues who helped in many ways to see this revision completed. Particularly, we wish to thank Nonnan C. Rasmussen, Raphael Aronson, Marvin M. Miller, and Edward Melko­ nian for their assistance in the final stages of this revision. Finally, we are grateful for comments and suggestions received from users of the earlier edition of this book. Although all their suggestions could not be incor­ porated, the book is greatly improved as a result of their comments. November 1982 Addison-Wesley Publishing Company Reading, Massachusetts
  • 7. Preface to First Edition Thisbookisderivedfromclassroomnoteswhichwerepreparedforthreecoursesofferedby theDepartmentofNuclearEngineering atNew YorkUniversity andthePolytechnic Institute ofNew York. Theseareaone-yearintroductorycourse in nuclearengineering (Chapters 1-8), aone-tenn course in radiationprotection(Chapters 9 and 10), and aone-tenncourse inreactorlicensing, safety, andtheenvironment(Chapter 11).ThesecoursesareofferedtojuniorsandseniorsintheDepartment'sundergraduateprogramandtobeginninggraduatestudentswhohavenothadprevioustraininginnuclearengineering.Nuclearengineering isanextremely broadfield,anditis notpossiblein abook of finite size and reasonable depth to cover all aspects ofthe profession.Needlesstosay,thepresentbookislargelyconcernedwithnuclearpowerplants,sincemostnuclearengineersarecurrentlyinvolvedintheapplicationofnuclearenergy.Nevertheless,IhaveattemptedinChapter1toconveysomefeelingfortheenonnousbreadthofthenuclearengineeringprofession.Inmyexperience,thecoursesinatomicandnuclearphysicsgivenbyphysicsprofessorsarebecomingincreasinglytheoretical.Ihavefounditnecessary,there- vii
  • 8. viii Preface to First Edition fore, to review these subjects at some length for nuclear engineering students.Chapters2and3arethesubstanceofthisreview.Chapter4 beginstheconsidera­tionofsomeofthepracticalaspectsofnuclearpower, andincludes adescriptionofmostofthereactorscurrentlyinproductionorunderdevelopment.NeutrondiffusionandmoderationarehandledtogetherinChapter5.Moder­ationistreatedinasimplewaybythegroupdiffusionmethod, which avoidstheusually tedious andrelatively difficultcalculations ofslowing-down density andFermi age theory. While such computations areessential, in myjudgment, forathoroughunderstandingofthefundamentalsofneutronmoderation,theyareprob­ablynotnecessaryinabookatthislevel.Chapters6, 7,and8 areintendedtogivesufficientbackgroundinreactordesignmethodstosatisfytheneedsofnuclearen­gineersnotspecificallyinvolvedindesignproblemsandalsotoprovideabaseformoreadvancedcoursesinnuclearreactortheoryanddesign.Chapters9 and10dealwiththepracticalaspectsofradiationprotection.Bothchaptersrelyheavilyontheearlierpartsofthebook.Chapter11 wasoriginallyintendedtobe twochapters-one on safety andlicensing,andasecondonalloftheenvironmentaleffectsofnuclearpower.How­ever, inordertomeetapublicationdeadline,thediscussionofenvironmentalef­fectshadtobeconfinedtothoseassociatedwithradioactiveeffluents.Whenthebookwasfirstconceived,Ihadplannedtoutilizeonlythemodernmetric system(theSIsystem).However,theU.S. Congress hasbeenmorereluc­tanttoabandontheEnglishsystemthanhadbeenexpected. Especially,therefore,inconnection with heattransfercalculations, I have feltcompelled to introduceEnglish units. Adiscussion ofunits andtables ofconversion factors is given inAppendixI.Mostofthe datarequiredtosolvetheproblemsattheendofeach chapteraregiveninthebodyoftherelevantchapterorintheappendixesattherearofthebook. Datawhich are too voluminous tobe included inthe appendixes, such asatomicmasses,isotopicabundances,etc., will generallybefoundontheChartofNuclides,availablefromtheU.S.GovernmentPrintingOffice.Itis alsohelpfulifthereaderhasaccesstothesecondandthirdeditionsof"NeutronCrossSections,"Brookhaven National LaboratoryReportBNL-325. Title10 oftheCodeofFed­eralRegulations shouldbe obtainedfromthe Printing Officeinconnection withChapter11. Iwouldliketoacknowledgetheassistanceofseveralpersonswhoreadandcommented upon various parts ofthemanuscript. I especially wish to thankR.Aronson,C.F. Bonilla, H.Chelemer,W.R.Clancey,R.J. Deland, H. Goldstein,H.C.Hopkins,F. R.HubbardIII,R.W.Kupp,G.Lear,BrunoPaelo,R.S.Thorsen,andM.E.Wrenn.IalsowishtothankthepersonneloftheNationalNeutronCross SectionCenteratBrookhavenNationalLaboratory,especially D. I.Garber,B.A.
  • 9. Preface to First Edition ix Magurno,andS.Pearlstein,forfurnishingvariousnucleardataandplotsofneutroncrosssections.Finally, I wish to express my gratitude to my wife Barbara and daughterMichelefortheirpricelessassistanceinpreparationofthemanuscript. Larchmont,NewYorkJanuary 1975 J.R.L.
  • 10. Acknowledgement Theauthorwishestothankthefollowingfortheircontribution:NuclearEnergyIn­stitute,WestinghouseElectricCorporation,Framatome,DivisionofNavalReactors,thePennStatestudentsinNuclearEngineering3011302, R.Guida, G. Robinson,F. Remmick, S.Johnson,L.Curless,L.Galen,K.Almenas,R.Knief,E.Klevans,B.Lamarsh,V.Whisker,T.Beam,A.Barnes,L.Hochreiter,K.Ivanov,C.Caldwell,C.Wilson,F. Ducceschi,F. Duhamel,E. Supko,R.Scalise,F. Castrillo,R.McWaid,L.Pasquini,J.Kie1I,L.Prowett,andE.Sartori. Dedication Toallthosewhohadfaithinthisprojectparticularlymyfamily. Tony
  • 11. Contents 1 NUCLEAR ENGINEERING 1 2 ATOMIC AND NUCLEAR PHYSICS 5 2.1 FundamentalParticles 5 2.2 AtomicandNuclearStructure 7 2.3 AtomicandMolecularWeight 8 2.4 AtomicandNuclearRadii 11 2.5 MassandEnergy 11 2.6 ParticleWavelengths 14 2.7 ExcitedStatesandRadiation 15 2.8 NuclearStabilityandRadioactiveDecay 18 2.9 RadioactivityCalculations 22 2.10 NuclearReactions 26 2.11 BindingEnergy 29 2.12 NuclearModels 33 2.13 Gases,Liquids,andSolids 37 2.14 AtomDensity 40 xi
  • 12. xii References 44 Problems 45 3 INTERACTION OF RADIATION WITH MATTER 3.1 NeutronInteractions 52 3.2 Cross-Sections 54 3.3 NeutronAttenuation 57 3.4 NeutronFlux 60 3.5 NeutronCross-SectionData 62 3.6 EnergyLossinScatteringCollisions 68 3.7 Fission 74 3.8 y-RayInteractionswithMatter 90 3.9 ChargedParticles 100 References 109 Problems 110 4 NUCLEAR REACTORS AND NUCLEAR POWER 4.1 TheFissionChainReaction 117 4.2 NuclearReactorFuels 119 4.3 Non-NuclearComponentsofNuclearPowerPlants 129 4.4 ComponentsofNuclearReactors 133 4.5 PowerReactorsandNuclearSteamSupplySystems 136 4.6 NuclearCycles 185 4.7 IsotopeSeparation 201 4.8 FuelReprocessing 217 4.9 RadioactiveWasteDisposal 219 References 223 Problems 224 5 NEUTRON DIFFUSION AND MODERATION 5.1 NeutronFlux 230 5.2 Fick'sLaw 231 5.3 TheEquationofContinuity 235 5.4 TheDiffusionEquation 237 5.5 BoundaryConditions 238 5.6 SolutionsoftheDiffusionEquation 240 5.7 TheDiffusionLength 246 5.8 TheGroup-DiffusionMethod 248 5.9 ThennalNeutronDiffusion 252 5.10 Two-GroupCalculationofNeutronModeration 257 Contents 52 117 230
  • 13. Contents xiii References 260 Problems 260 6 NUCLEAR REACTOR THEORY 266 6.1 One-GroupReactorEquation 266 6.2 TheSlabReactor 271 6.3 OtherReactorShapes 274 6.4 TheOne-GroupCriticalEquation 282 6.5 ThennalReactors 286 6.6 ReflectedReactors 297 6.7 MultigroupCalculations 308 6.8 HeterogeneousReactors 309 References 320 Problems 321 7 THE TIME-DEPENDENT REACTOR 327 7.1 ClassificationofTimeProblems 328 7.2 ReactorKinetics 330 7.3 ControlRodsandChemicalShim 348 7.4 TemperatureEffectsonReactivity 365 7.5 FissionProductPoisoning 376 7.6 CorePropertiesduringLifetime 389 References 397 Problems 398 8 HEAT REMOVAL FROM NUCLEAR REACTORS 403 8.1 GeneralThennodynamicConsiderations 404 8.2 HeatGenerationinReactors 408 8.3 HeatFlowbyConduction 417 8.4 HeatTransfertoCoolants 428 8.5 BoilingHeatTransfer 441 8.6 ThennalDesignofaReactor 450 References 457 Problems 459 9 RADIATION PROTECTION 466 9.1 HistoryofRadiationEffects 467 9.2 RadiationUnits 468 9.3 SomeElementaryBiology 476 9.4 TheBiologicalEffectsofRadiation 479
  • 14. xiv Contents 9.5 QuantitativeEffectsofRadiationontheHumanSpecies 485 9.6 CalculationsofRadiationEffects 495 9.7 NaturalandMan-MadeRadiationSources 499 9.8 StandardsofRadiationProtection 506 9.9 ComputationsofExposureandDose 511 9.10 StandardsforIntakeofRadionuclides 526 9.11 Exposurefromy-RaySources 535 Glossary 539 References 542 Problems 544 10 RADIATION SHIELDING 548 10.1 Gamma-RayShielding:BuildupFactors 549 10.2 InfinitePlanarandDiscSources 559 10.3 TheLineSource 566 10.4 InternalSources 571 10.5 MultilayeredShields 573 10.6 NuclearReactorShielding:PrinciplesofReactorShielding 576 10.7 RemovalCross-Sections 578 10.8 TheReactorShieldDesign:Removal-AttenuationCalculatons 584 10.9 TheRemoval-DiffusionMethod 588 10.10 ExactMethods 590 10.11 Shieldingy-Rays 595 10.12 CoolantActivation 599 10.13 DuctsinShields 604 References 605 Problems 606 11 REACTOR LICENSING, SAFETY, AND THE ENVIRONMENT 612 11.1 GovernmentalAuthorityandResponsibility 613 11.2 ReactorLicensing 614 11.3 PrinciplesofNuclearPowerPlantSafety 623 11.4 DispersionofEffluentsfromNuclearFacilities 631 11.5 RadiationDosesfromNuclearPlants 650 11.6 ReactorSiting 669 11.7 ReactorAccidents 681 11.8 AccidentRiskAnalysis 701 11.9 EnvironmentalRadiationDoses 710 References 721 Problems 723
  • 15. Contents APPENDIXES I UnitsandConversionFactors 731 II FundamentalConstantsandData 737 III VectorOperationsinOrthogonalCurvilinearCoordinates 745 IV ThermodynamicandPhysicalProperties 751 V BesselFunctions 757 INDEX xv 761
  • 16. 1 Nuclear Engineering Nuclear engineering is an endeavor that makes use ofradiation and radioactivematerialforthe benefitofmankind. Nuclearengineers, liketheircounterparts inchemicalengineering,endeavortoimprovethequalityoflifebymanipulatingbasicbuildingblocksofmatter.Unlikechemicalengineers,however,nuclearengineersworkwithreactionsthatproducemillionsoftimesmoreenergyperreactionthananyotherknownmaterial.Originatingfromthenucleusofanatom,nuclearenergyhasprovedtobeatremendoussourceofenergy.Despite its association with the atomic bombs dropped during World WarII andthearmsrace ofthecoldwar,nuclearenergytodayprovides asignificantamountofenergyonaglobalscale.Manyarenowheraldingitasasourcefreefromthe problems offossil fuels-greenhouse gas emissions. Despite these benefits,thereisstilltheassociationofnuclearpowerwiththetremendousdestructiveforceexhibitedbythebombingsinJapan.Withtheendofthecoldwar,nuclearengineeringislargelyfocusedontheuseofnuclearreactionstoeithergeneratepoweroronitsapplicationinthemedicalfield.Nuclearpowerapplicationsgenerallyinvolvetheuseofthefissionreactionsinlargecentralpowerstationsandsmallermobilepowerplantsusedprimarilyforshippropulsion. The worlddemandforelectricity is againincreasing and with ittheneedfornewgenerationfacilities.Forthoseareasoftheworldthathavelittle 1
  • 17. 2 Nuclear Engineering Chap. 1 inthe wayoffossilfuels orhavechosento use theseforfeedstockinthepetro­chemicalindustry,nuclearpowerisconsideredthesourceofchoiceforelectricitygeneration.IntheUnitedStatesalone,nuclearpowergeneratesnearly22.8%oftheelectricity.Inothercountries,notablyFrance,theproportionapproaches100%.Recentconcernsovertheemissionofnitrousoxidesandcarbondioxidehaveincreased the concern about continued use offossil fuels as a source ofenergy.The Kyoto accords, developed in 1997, require a reduction in emissions belowcurrentvalues.ThesetargetscanbereachedintheUnitedStatesonlybyloweringtheliving standards orby continuing use ofnuclearpowerforthegenerationofelectricity. A typical 1,000-megawatt coal-burning plant may emit in 1 year asmuchas100,000tonsofsulphurdioxide,75,000tonsofnitrogenoxides,and5,000tonsofflyash.Nuclearpowerplantsproducenoneoftheseairpollutantsandemitonlytraceamountsofradioactive gasses. Asaresult,in 1999,theuseofnuclearpowertogenerate20%oftheelectricityintheUnitedStatesavoidedtheemissionof150milliontonnesofCO2.Todate,thewidestapplicationofnuclearpowerinmobilesystemshasbeenforthepropulsionofnavalvessels,especiallysubmarinesandaircraftcarriers.Herethetremendous advantagesofnuclearpowerareutilizedtoallow extendedopera­tionswithoutsupportships.Inthecaseofthesubmarine,theabilitytocruisewith­outlargeamountsofoxygenforcombustionenables the submarinetoremainatseaunderwaterforalmostlimitlesstime.Inthecaseofanaircraftcarrier,thelargequantityofspacethatwastakenupbyfueloilinaconventionally-poweredaircraftcarriercan be devoted to aviation fuel and other supplies on anuclear-poweredaircraftcarrier.In addition tonaval vessels, nuclear-powered merchantshipswere also de­veloped. The U.S. ship Savannah, whichoperatedbriefly in the late 1960s andearly 1970s, showedthatnuclearpowerforamerchantship while practical wasnoteconomical.OthercountriesincludingJapan,Gennany,andthefonnerSovietUnionhavealsousednuclearpowerforciviliansurfaceshippropulsion.Ofthese,theGennanorecarrier,Otto Hahn, operatedsuccessfullyfor 10yearsbutwasre­tiredsinceittooproveduneconomical.TheicebreakerLenin ofthefonnerSovietUniondemonstratedanotherusefulapplicationofnuclearpower.ThetrialwassosuccessfulthattheSovietsbuiltadditionalshipsofthistype.Nuclearpowerhas alsobeendevelopedforaircraftandspace applications.From 1949to 1961,whentheprojectwastenninated,theUnitedStatesspentap­proximately$1 billiontodevelopanuclear-poweredairplane.Theproject,theAir­craft Nuclear Propulsion Project (ANP), was begun at a time when the UnitedStatesdidnothaveaircraftthatcould flyroundtripfromtheUnitedStatesmain­landtoadistantadversary.Becauseoftheenonnousrangethatcouldbeexpectedfromanuclear-powered airplane, therangeproblem wouldhavebeen easily re­solved.Withtheadventoflong-rangeballisticmissiles,whichcouldbefiredfrom
  • 18. Chap. 1 Nuclear Engineering 3 themainlandorfromsubmarines,theneedforsuchanaircraftdisappearedandtheprogramwastenninated.Nuclear-poweredspacecrafthavebeendevelopedandareinusetoday.Typ­ically,anuclearreactionisusedtoprovideelectricityforprobesthatareintendedforuseindeepspace.There,photovoltaicsystemscannotprovidesufficientenergybecauseoftheweaksolarradiationfoundindeep space. Typically, aradioactivesourceisusedandtheenergyemittedisconvertedintoheatandthenelectricityus­ingthennocouples.Nuclear-poweredrocketsareunderconsiderationaswell.ThelongdurationofamannedflighttoMars,forexample,suggeststhatnuclearpowerwouldbeusefulifnotessential.Thedesirabilityofanuclearrocketforsuchlong­distancemissions stemsfromthefactthatthetotalvehicularmassrequiredforalong-distancemission is considerably less ifthevehicle ispoweredby anuclearrocket,ratherthanbyaconventionalchemicalrocket.Forinstance, theestimatedmassofachemicalrocketrequiredforamannedmissionfromastationarypark­ing orbitto anorbitaroundMars is approximately4,100,000kg. The mass ofanuclearrocketforthe same mission is estimatedtobe only 430,000kg.NuclearrocketshavebeenunderactivedevelopmentintheUnitedStatesformanyyears.Theapplicationofradiationandnuclearreactionsisnotlimitedtonuclearex­plosivesandnuclearpower.Radiationandradioactiveisotopesareusefulinawiderange ofimportant applications. The production ofradioisotopes, whether from reactorsoraccelerators,is amajorindustryinits ownright. Theapplications ofradiationandradioisotopesrangefromlife-savingmedicalprocedurestomaterialcharacterizationtofoodpreservation.Radioactivetracingisonesuchmethod.Inthistechnique,oneoftheatomsinamoleculeisreplacedbyaradioactiveatomofthesameelement.Forexample,aradioactivecarbonatommaybesubstitutedforanonnalcarbonatomataparticularlocationinamoleculewhenthemoleculeissynthesized.Later,afterthemoleculehasreactedchemically,eitherinalaboratoryexperimentorabiologicalsystem,itispossibletodetenninethedispositionoftheatominquestionbyobservingtheradiationemanatingfromtheradioactiveatom. Thistechniquehasprovedtobeofenonnous valuein studies ofchemicalreactionprocesses andinresearch in thelifesciences.Asimilarprocedureisusedinindustrytomeasure,andsometimestocontrol,theflowandmixingoffluids. Asmallquantityofradioactivematerialisplacedinthemovingfluidandtheradiationismonitoreddownstream.Bypropercalibration, itispossibletorelatethe downstream radiation level with thefluid'srateofflowortheextentofits dilution.Inasimilarway,radioactiveatomsmaybeincorporatedatthetimeoffabricationintovariousmovingpartsofmachinery,suchaspistons,toolbits, andsoon. Theradioactivityobservedinthelubricatingfluidthenbecomesanaccuratemeasureoftherateofwearofthepartunderstudy. Arelatedtechnique, knownasactivation analysis, isbasedonthefactthateveryspeciesofradioactiveatomemitsitsowncharacteristicradiations.Thechem-
  • 19. 4 Nuclear Engineering Chap. 1 icalcompositionofasubstancecanthereforebedetenninedbyobservingtheradia­tionemittedwhenasmallsampleofthesubstanceiscausedtobecomeradioactive.Thismaybedonebyexposingthesampletobeamsofeitherneutronsorchargedparticles. Becauseitispossibletodetermineextremelyminuteconcentrations inthisway(in somecases, onepartin1012), activationanalysishasprovedtobea valuabletool inmedicine,lawenforcement, pollutioncontrol,andotherfields inwhichtraceconcentrationsofcertainelementsplayanimportantrole.
  • 20. 2 Atomic and Nuclear Physics A knowledge ofatomic and nuclear physics is essential to the nuclear engineerbecausethese subjects form the scientific foundation onwhich thenuclearengi­neeringprofessionis based. Therelevantpartsofatomicandnuclearphysics arereviewedinthischapterandthenext. 2.1 FUNDAMENTAL PARTICLES Thephysical worldiscomposedofcombinationsofvarious subatomicorfunda­mental particles. A number offundamental particles have been discovered. Thisledtothediscoverythatthesefundamentalparticlesareintummadeupofquarksboundtogetherbygluons.Incurrenttheory,particlesofinteresttothenuclearengineermaybedividedintoleptonsandhadrons.Theelectron,positron,andneutrinoareleptons.Hadronsofinterestaretheprotonandneutron,whichbelongtoasubclassofhadronscalledbaryons. Theleptonsaresubjecttotheweaknuclearforces,whereashadronsandbaryonsinparticularexperienceboththeweakandstrongnuclearforces.Itisthehadronsthatarecomposedofquarks, andit is the exchange ofgluons betweencollectionsofquarksthatisresponsibleforthestrongnuclearforce. 5
  • 21. 6 Atomic and Nuclear Physics Photon e Photon Figure 2.1 The annihilation ofa negatron and positron with the release of two photons. Chap. 2 Fortheunderstandingofnuclearreactorsandthereactionsofinteresttotheiroperation,itisonlyimportanttoconsideraclassoftheseparticlesandnotexploretheirstructure.Ofthese,onlythefollowingareimportantinnuclearengineering.1 Electron The electron has arest-mass2 me = 9.10954 X 10-31 Kg3 andcarriesachargee = 1.60219 x 10-19 coulombs.Therearetwotypesofelectrons:onecarryinganegativecharge-e, theothercarryingapositivecharge+e. Exceptforthedifferenceinthesignoftheircharge,thesetwoparticlesareidentical.Thenegativeelectrons,ornegatrons astheyaresometimescalled,arethenonnalelec­tronsencounteredinthisworld.Positiveelectrons,orpositrons, arerelativelyrare.When,underthepropercircumstances,apositroncollideswithanegatron,thetwoelectrons disappear andtwo (and occasionally more) photons (particles ofelec­tromagneticradiation)areemittedasshowninFig.2.1. Thisprocessisknownaselectron annihilation, andthephotonsthatappeararecalledannihilation radiation. Proton Thisparticlehasarestmassmp = 1.67265 x 10-27Kgandcarriesapositivechargeequal inmagnitude tothe chargeonthe electron. Protons withnegativechargehavealsobeendiscovered,buttheseparticlesareofnoimportanceinnuclearengineering. I A discussion ofquarktheory may be found in several of the particle physics references at the end of this chapter. 2According to the theory of relativity, the mass of a particle is a function of its speed relative to the observer. In giving the masses of the fundamental particles, it is necessary to specify that the particle is at rest with respect to the observer-hence, the term rest mass. 3A discussion of units, their symbols, and abbreviations, together with tables of conversion factors, are found in Appendix I at the end of this book. Tabulations of fundamental constants and nuclear data are given in Appendix I.
  • 22. Sec. 2.2 Atomic and Nuclear Structure 7 Neutron The mass oftheneutron is slightly largerthanthe mass oftheproton-namely, mn = 1.67495 x 10-27 Kg, and it is electrically neutral. Theneutronisnotastableparticleexceptwhenitisboundintoanatomicnucleus.Afreeneutrondecaystoaprotonwiththeemissionofanegativeelectron(t3-decay;seeSection2.8)andanantineutrino,aprocessthattakes,ontheaverage,about12minutes. Photon Itis acuriousfactthatallparticlesinnaturebehavesometimeslikeparticlesandsometimeslikewaves.Thus,certainphenomenathatarenormallythoughtofasbeingstrictlywavelikeincharacteralsoappeartohaveanassociatedcorpuscularorparticlelikebehavior.Electromagneticwavesfallintothiscategory.Theparticleassociatedwithelectromagneticwavesiscalledthephoton. Thisisaparticlewithzerorestmassandzerocharge, whichtravelsinavacuumatonlyonespeed-namely,thespeedoflight,c = 2.9979 X 108m1sec. Neutrino This is another particle with zero restmass andno electricalchargethatappearsinthedecayofcertainnuclei. Thereareatleastsix types ofneutrinos,onlytwoofwhich(calledelectron neutrinos andelectron antineutrinos) areimportantintheatomicprocessandareofinterestinnuclearengineering.Formostpurposes,itisnotnecessarytomakeadistinctionbetweenthetwo,andtheyarelumpedtogetherasneutrinos. 2.2 ATOMIC AND NUCLEAR STRUCTURE Asthe reader is doubtless aware, atoms are the building blocks ofgross matteras itis seen andfelt. The atom, in tum, consists ofa small butmassive nucleussurroundedbyacloudofrapidlymoving(negative)electrons.Thenucleusiscom­posedofprotonsandneutrons.Thetotalnumberofprotonsinthenucleusiscalledtheatomic number ofthe atom and is given the symbol Z. Thetotal electricalchargeofthenucleusistherefore+Ze.Inaneutralatom,thereareasmanyelec­trons as protons-namely, Z-moving about the nucleus. The electrons arere­sponsibleforthechemicalbehaviorofatomsandidentifythevariouschemicalel­ements.Forexample,hydrogen(H)hasoneelectron,helium(He)hastwo,lithium(Li)hasthree,andsoon.Thenumberofneutronsinanucleusisknownastheneutron number andisdenotedby N Thetotalnumberofnucleons-that is, protonsandneutronsinanucleus-isequalto Z+ N = A, where A is called theatomic mass number ornucleon number. Thevariousspeciesofatomswhosenucleicontainparticularnumbersofpro­tonsandneutronsarecallednuclides. Eachnuclideisdenotedbythechemicalsym­boloftheelement(this specifies Z) withtheatomicmassnumberas superscript
  • 23. 8 Atomic and Nuclear Physics Chap. 2 (thisdeterminesN, sinceN= A - Z). Thus,thesymbol 1Hreferstothenuclideofhydrogen(Z = 1)withasingleprotonas nucleus; 2Histhehydrogennuclidewithaneutronaswellasaprotoninthe nucleus eHis alsocalleddeuterium orheavy hydrogen); 4Heisthehelium(Z = 2)nuclidewhosenucleusconsistsoftwoprotonsandtwoneutrons;andsoon.Forgreaterclarity,Z issometimeswrittenasasubscript,asin�H,iH, iHe,andsoon.Atomssuchas1Hand2H,whosenucleicontainthesamenumberofprotonsbutdifferentnumbersofneutrons(sameZ butdifferentN-thereforedifferentA), areknownasisotopes. Oxygen, forinstance,hasthree stableisotopes, 160, 170,180 (Z = 8;N = 8, 9, 10), andfive known unstable (i.e., radioactive) isotopes,l30, 140, 150,190,and2°0(Z = 8; N= 5,6,7, 11, 12).The stableisotopes (andafew oftheunstableones) aretheatomsthatarefound inthenaturallyoccurringelementsinnature. However, they arenotfoundinequalamounts; some isotopesofagivenelementaremore abundantthanoth­ers.Forexample, 99.8%ofnaturallyoccurringoxygenatomsaretheisotope160,.037% aretheisotope 170,and.204%are 180. Atableofsomeofthemoreim­portantisotopes andtheirabundanceis given inAppendix II. It shouldbe notedthatisotopicabundancesaregiveninatompercent-thatis,thepercentagesoftheatomsofanelementthatareparticularisotopes.Atompercentisoftenabbreviatedasa/o. Example 2.1 Aglass ofwater is known to contain 6.6 x 1024 atoms ofhydrogen. How many atoms of deuterium eH) are present? Solution. According to Table 11.2 in Appendix II, the isotopic abundance of 2H is 0.015 a/o. The fraction of the hydrogen, which is 2H, is therefore 1.5 x 10-4 The total number of 2H atoms in the glass is then 1.5 x 10-4 x 6.6 X 1024 = 9.9 X 1020 [Ans.] 2.3 ATOMIC AND MOLECULAR WEIGHT Theatomic weight ofanatomis defined asthe mass oftheneutralatomrelativetothemassofaneutral 12Catomonascalein whichtheatomic weightof12Cisarbitrarilytakentobeprecisely12.Insymbols,letmeZ) bethemassoftheneutralatomdenotedbyAZ andm(12C)bethemassofneutral12C.ThentheatomicweightofAZ, M(AZ), isgivenby A meZ) M( Z) = 12 X 12 . m( C) (2.1)
  • 24. Sec. 2.3 Atomic and Molecular Weight 9 Supposethatsomeatomwasprecisely twice asheavyas 12C. Then accordingtoEq.(2.1),thisatomwouldhavetheatomicweightof12 x 2= 24.AsnotedinSection2.2,theelementsfoundinnatureoftenconsistofanum­berofisotopes.Theatomicweightoftheelementisthendefinedastheaverage atomicweightofthemixture.Thus,ifYi istheisotopicabundanceinatompercentoftheithisotopeofatomicweightMi, thentheatomicweightoftheelementis (2.2) Thetotal mass ofamolecule relative tothe mass ofaneutral 12C atomiscalledthemolecular weight. Thisismerelythesumoftheatomicweightsoftheconstituent atoms. Forinstance, oxygen gasconsistsofthe molecule O2, anditsmolecularweightistherefore2 x 15.99938= 31.99876. Example 2.2 Using the data in the following table, compute the atomic weight of naturally occur­ ring oxygen. Isotope Abundance (a/o) 99.759 0.037 0.204 Solution. From Eq. (2.2), it follows that Atomic weight 15.99492 16.99913 17.99916 M(O) = 0.01[ye60)M(170) + y(I70)M(170) + yesO)MesO)] = 15.99938. [Ans.] Itmustbeemphasizedthatatomicandmolecularweightsareunitlessnum­bers, being ratios ofthe masses of atoms or molecules. By contrast, the gram atomic weight and gram molecular weight are defined as the amount ofa sub­stancehaving amass, ingrams, equal to theatomicormolecularweightofthesubstance.Thisamountofmaterialisalsocalledamole. Thus,onegramatomicweightoronemoleof12C is exactly 12gofthisisotope, onemoleof02 gas is31.99876g,andsoon.Since atomic weightis a ratio ofatomic masses and amole is an atomicweightingrams,itfollowsthatthenumberofatomsormoleculesinamoleofanysubstance is aconstant,independentofthenatureofthesubstance. Forinstance,supposethatahypotheticalnuclide has anatomicweightof24.0000. Itfollowsthattheindividualatomsofthis substanceareexactlytwice as massiveas 12C.
  • 25. 10 Atomic and Nuclear Physics Chap. 2 Therefore,there mustbethe samenumberofatomsin24.0000gofthis nuclideasin 12.0000gof12C.ThisstateofaffairsisknownasAvogadro's law, andthenumberofatomsormoleculesinamoleiscalledAvogadro's number. ThisnumberisdenotedbyNA andisequaltoNA = 0.6022045 X 10244 UsingAvogadro'snumber,itispossibletocomputethemassofasingleatomormolecule. Forexample, since one grammole of12C has a mass of 12 g andcontainsNA atoms,itfollowsthatthemassofoneatomin12Cis m(12C) = 12 = 1.99268 x 1023g.0.6022045 X 1024 Thereis,however, amorenaturalunitintennsofwhichthemassesofindi­vidualatomsareusuallyexpressed.Thisistheatomic mass unit, abbreviatedamu,whichisdefinedasonetwelfththemassoftheneutral12Catom,thatis Invertingthisequationgives Introducingm12Cfromtheprecedingparagraphgives 1 amu= -12 x 1.99268 X 10-23g= I/NA g = 1.66057 X 10-24g. AlsofromEq. (2.1), sothat Thus,the mass ofanyatominamu is numericallyequaltotheatomicweightoftheatominquestion. 40rdinarily a number of this type would be written as 6.0222045 x 1023 However, in nuclear engineering problems, forreasons given in Chap. 3 (Example 3.1), Avogadro's number should always be written as the numerical factor times 1024•
  • 26. Sec. 2.5 Mass and Energy 1 1 2.4 ATOMIC AND NUCLEAR RADII The sizeofanatomis somewhatdifficult to define because the atomic electronclouddoesnothaveawell-definedouteredge. Electronsmayoccasionallymovefarfromthenucleus,whileatothertimestheypassclosetothenucleus.Areason­ablemeasureofatomicsizeisgivenbytheaveragedistancefromthenucleusthattheoutermostelectronistobefound.Exceptforafewofthelightestatoms,theseaverageradiiareapproximatelythesameforallatoms-namely,about2 x la-10m.Sincethenumberofatomicelectronsincreaseswithincreasingatomicnumber,itisevidentthattheaverageelectrondensityintheelectroncloudalsoincreaseswithatomicnumber.The nucleus, like the atom, does nothave a sharp outerboundary. Its sur­face,too,isdiffuse,althoughsomewhatless thanthatofanatom. Measurementsinwhichneutronsarescatteredfrom nuclei (see Section 3.5) showthattoafirstapproximationthenucleusmaybeconsideredtobeaspherewitharadiusgivenbythefollowingformula: R = 1.25frn x A1/3 (2.3) whereRisinfemtometers(fm)andA istheatomicmassnumber.Onefemtometeris 10-l3centimeters.Since the volume ofa sphere is proportional to the cube ofthe radius, itfollowsfromEq. (2.3)thatthevolume V ofanucleus is proportional toA. ThisalsomeansthattheratioA/ V-thatis,thenumberofnucleonsperunitvolume­is aconstantforall nuclei. Thisuniform density ofnuclearmatter suggests thatnucleiare similartoliquiddrops, which alsohavethesamedensity whethertheyarelargeorsmall.Thisliquid-dropmodelofthenucleusaccountsformanyofthephysicalpropertiesofnuclei. 2.5 MASS AND ENERGY OneofthestrikingresultsofEinstein'stheoryofrelativityisthatmassandenergyareequivalentandconvertible,onetotheother. Inparticular,thecompleteanni­hilationofaparticleorotherbodyofrestmassrno releasesanamountofenergy,Eresb whichisgivenbyEinstein'sfamousformula (2.4) wherec isthespeedoflight.Forexample,theannihilationof1 gofmatterwouldleadtoareleaseofE = 1x (2.9979X 1010)2 = 8.9874x 1020ergs= 8.9874x 1013 JOUles.Thisisasubstantialamountofenergy,whichinmorecommonunitsisequaltoabout25 millionkilowatt-hours.
  • 27. 12 Atomic and Nuclear Physics Chap. 2 Anotherunitofenergythatisoftenusedinnuclearengineeringistheelectron volt, denoted by eV. This is defined as the increase in the kinetic energy of anelectronwhen itfallsthroughanelectricalpotentialofone volt. This, intum, isequaltothechargeoftheelectronmultipliedbythepotentialdrop-thatis, 1 eV= 1.60219 x 10-19 coulomb x 1 volt = 1.60219 x 10-19 joule. Other energy units frequently encountered are the MeV (106 eV) andthekeV(103 eV). Example 2.3 Calculate the rest-mass energy of the electron in MeV. Solution. From Eq. (2.4), therest-mass energy ofthe electron is mec2 = 9.1095 x 10-28 x (2.9979 X 1010)2 = 8.1871 X 10-7 ergs = 8.1871 x 10-14 JOUle. Expressed in MeV this is 8.1871 X 10-14 joule -;- 1 .6022 x 10-13 joule/MeV = 0.51 10 MeV. [Ans.] Example 2.4 Compute the energy equivalent of the atomic mass unit. Solution. This can most easily becomputed using the result ofthe previous example. Thus, since according to Section 2.3, 1 amu = 1.6606 x 10-24g, it follows that 1 amu is equivalent to 1 .6606 x 10-24 g/amu --------:-:28=----- X 0.5 1 10 MeV/electron = 931.5 MeV. [Ans.] 9.1095 x 10- g/electron Whenabody is inmotion,its mass increasesrelativetoan observeratrestaccordingtotheformula (2.5) whererna isitsrestmassandv isitsspeed.FromEq.(2.5), itisseenthatrn reducestorna asv goestozero.However,asv approachesc, rn increaseswithoutlimit.Thetotalenergy ofaparticle,thatis,itsrest-massenergyplusitskineticenergy,isgivenby (2.6)
  • 28. Sec. 2.5 Mass and Energy 13 where m is as given in Eq. (2.5). Finally, the kinetic energy E is the differencebetweenthetotalenergyandtherest-massenergy.Thatis, 2 [ 1 ]=moc - 1 J1 - v2jc2 (2.7) (2.8) TheradicalinthefirstterminEq.(2.8) canbeexpandedinpowersof(vjc)2 usingthebinomialtheorem.Whenv « c, theseriesmaybetruncatedafterthefirstterm.TheresultingexpressionforE is (2.9) whichisthefamiliarformulaforkineticenergyinclassicalmechanics.ItshouldbenotedthatEq.(2.9) maybeusedinsteadofEq.(2.8) onlywhenthekineticenergycomputedfromEq.(2.9) issmallcomparedwiththerest-massenergy.Thatis,Eq.(2.9) isvalidprovided (2.10) Asapracticalmatter,Eq. (2.9) isusuallyaccurateenoughformostpurposespro­videdv ::: 0.2c or E ::: 0.02Erest• (2.11) AccordingtoExample2.3, therest-massenergyofanelectronis0.511 MeV.FromEq.(2.11), itfollowsthattherelativisticformulaEq. (2.8) mustbeusedforelec­tronswithkineticenergiesgreaterthanabout0.02x 0.51 1 MeV=0.010MeV= 10 keV. Sincemanyoftheelectronsencounteredinnuclearengineeringhavekineticenergiesgreaterthanthis,itisoftennecessarytouseEq.(2.8) forelectrons.Bycontrast,therestmassoftheneutronisalmost1,000MeVand0.02Erest= 20MeV.Inpractice,neutronsrarelyhavekineticenergiesinexcessof20MeV.Itispermissible,therefore,inallnuclearengineeringproblemstocalculatethekineticenergyofneutronsfromEq.(2.9). WhentheneutronmassisinsertedintoEq.(2.9), thefollowinghandyformulaisobtained: v = 1.383 x 106JE, (2.12) wherev isincm/secandE isthekineticenergyoftheneutronineV.ItisimportanttorecognizethatEqs.(2.8) and(2.9) arevalidonlyforparticleswithnonzerorestmass;forexample, theydonotapplytophotons. (Itshouldbeunderstoodthatphotonshavenorest-massenergy,anditisnotpropertousethe
  • 29. 14 Atomic and Nuclear Physics Chap. 2 termkinetic energy inreferringtosuchparticles.)Photons onlytravelatthespeedoflight,andtheirtotalenergyisgivenbyquiteadifferentformula-namely, E = hv, (2.13) whereh isPlanck'sconstantandv isthefrequencyoftheelectromagneticwaveassociatedwiththephoton.Planck'sconstanthasunitsofenergy x time;ifE istobeexpressedineV,h isequalto4.136 x 10-15eV-sec. 2.6 PARTICLE WAVELENGTHS ItwaspointedoutinSection2.1thatalloftheparticlesinnaturehaveanassociatedwavelength.ThewavelengthA associatedwithaparticlehavingmomentump is h A = - , (2.14)P whereh isagainPlanck'sconstant.Forparticlesofnonzerorestmass,p isgivenby p = mv, (2.15) wherem isthemassoftheparticleandv isitsspeed.Atnonrelativisticenergies,p canbewrittenas p = J2mo E, whereE isthekineticenergy.WhenthisexpressionisintroducedintoEq.(2.14),theparticlewavelengthbecomes h A = . -J2mo E (2.16) This formulais validforthe neutrons encounteredin nuclearengineering. Intro­ducingthevalueoftheneutronmassgivesthefollowingexpressionfortheneutronwavelength: 2.860X 10-9A = ----- -JE (2.17) whereA isincentimetersandE isthekineticenergyoftheneutronineV.Fortherelativisticcase,itis convenientto compute p directly by solvingtherelativisticequationsintheprecedingsection.Thisgives 1 / 2 2 P = -V Etotal - Erest'c (2.18)
  • 30. Sec. 2.7 andso Excited States and Radiation hc A = ---;::::::======= .JEtotal - Erest 15 (2.19) ThemomentumofaparticleofzerorestmassisnotgivenbyEq. (2.15), butratherbytheexpression E p = -, c (2.20) inwhichE istheenergyoftheparticle.WhenEq.(2.20) isinsertedintoEq.(2.14), theresultis hc A - ­ - E · (2.21) Introducingnumericalvaluesforhandc intheappropriateunitsgivesfinally 1.240 x 10-6 A = ----­ E (2.22) whereA is inmeters and E is ineV.Equation(2.22) is validforphotons andallotherparticlesofzerorestmass. 2.7 EXCITED STATES AND RADIATION TheZ atomicelectronsthatclusteraboutthenucleusmoveinmoreorlesswell­definedorbits. However, some ofthese electrons aremoretightlybound in theatomthanothers. Forexample, only7.38eVis requiredtoremovetheoutermostelectronfromaleadatom(Z = 82), whereas 88 keV (88,000eV) isrequiredtoremovetheinnermostorK-electron. Theprocessofremovinganelectronfromanatomiscalledionization, andtheenergies7.38eVand 88 keVareknownastheionization energies fortheelectronsinquestion.Inaneutralatom,itispossiblefortheelectronstobeinavarietyofdifferentorbitsorstates.Thestateoflowestenergyistheoneinwhichanatomisnormallyfound,andthisiscalledtheground state. Whentheatompossessesmoreenergythanitsgroundstateenergy, itissaidtobeinanexcited state oranenergy level. Thegroundstateandthe various excited statescan convenientlybe depictedbyanenergy-level diagram, liketheoneshowninFig.2.2 forhydrogen.Thehighestenergystatecorrespondstothesituationinwhichtheelectronhasbeencompletelyremovedfromtheatomandtheatomisionized.Anatomcannotremaininanexcitedstateindefinitely;iteventuallydecaystooneoranotherofthestatesatlowerenergy,andinthiswaytheatomeventuallyreturnstothegroundstate.Whensuchatransitionoccurs,aphotonisemittedby
  • 31. 16 Atomic and Nuclear Physics 13.58-���� 12.07 10.19 o Figure 2.2 The energy levels ofthe hydrogen atom (not to scale). Chap. 2 theatomwith anenergyequaltothedifferenceinthe energies ofthetwostates.Forexample,whenahydrogenatominthefirstexcitedstateat10.19 eV(seeFig.2.2) decaystothegroundstate, aphotonwithanenergyof10.19 eVisemitted.From Eq. (2.22), this photonhas awavelength of).. = 1.240 X 10-6/10.19 = 1.217 x 10-7 m. Radiationofthis wavelength lies intheultravioletregionoftheelectromagneticspectrum. Example 2.5 A high-energy electron strikes a lead atom and ejects one of the K-electrons from the atom. What wavelength radiation is emitted when an outer electron drops into the vacancy? Solution. The ionization energy of the K-electron is 88 keV, and so the atom minus this electron is actually in an excited state 88 keV above the ground state. When the outer electron drops into the K position, the resulting atom still lacks an electron, but now this is an outer, weakly bound electron. In its final state, therefore, the atom is excited by only 7.38 eV, much less than its initial 88 keY. Thus, the photon in this transition is emitted with an energy of slightly less than 88 keY. The corresponding wavelength is A = 1.240 X 10-6/8.8 X 104 = 1 .409 x 10-"m. [Ans.] Such a photon is in the x-ray region of the electromagnetic spectrum. This process, the ejection of an inner, tightly bound electron, followed by the transition of another electron, is one way in which x-rays are produced. Thenucleonsinnuclei,liketheelectronsinatoms,canalsobethoughtofasmoving aboutinvariousorbits, although thesearenotaswelldefinedandunder-
  • 32. Sec. 2.7 Excited States and Radiation 17 stoodasthoseinatoms.Inanycase,thereisastateoflowestenergy,thegroundstate;exceptfortheverylightestnuclei,allnucleihaveexcitedstatesaswell.ThesestatesareshowninFig.2.3 forl2C.AcomparisonofFigs. 2.2 and2.3 showsthattheenergiesoftheexcitedstatesandtheenergiesbetweenstatesareconsiderablygreaterfornuclei thanforatoms. Although this conclusion is based only onthe 24 tetc 22 20 18 16 14 > Il.) :E :>. 12 rfc:: U.l 10 8 6 4 2 Figure 2.3 The energy levels of 0 carbon 12.
  • 33. 18 Atomic and Nuclear Physics Chap. 2 states ofhydrogen and 12C, it is found to be true in general. This is due to thefactthatthe nuclearforces acting between nucleons aremuch strongerthan theelectrostaticforcesactingbetweenelectronsandthenucleus.Nuclei inexcited states may decay toalowerlying state, asdo atoms, byemittingaphotonwithanenergyequaltothedifferencebetweentheenergiesoftheinitialandfinalstates.Theenergiesofphotonsemittedinthiswayfromanucleusare usually much greater than the energies ofphotons originating in electronictransitions,andsuchphotonsarecalledy-rays. Anucleus inanexcitedstatecanalsoloseitsexcitationenergybyinternalconversion.Inthisprocess,theexcitationenergyofthenucleusistransferredintokineticenergyofoneoftheinnennostatomicelectrons.Theelectronisthenejectedfromtheatomwithanenergyequaltothatofthenucleartransitionlesstheioniza­tionenergyoftheelectron.Internalconversionthuscompeteswithy-rayemissioninthedecayofnuclear-excitedstates.Theholeremainingintheelectroncloudafterthedepartureoftheelectronininternalconversionislaterfilledbyoneoftheouteratomicelectrons.Thistransi­tionisaccompaniedeitherbytheemissionofanx-rayortheejectionofanotherelectron ina process similarto internal conversion. Electrons originating inthiswayarecalledAuger electrons. 2.8 NUCLEAR STABILITY AND RADIOACTIVE DECAY Figure2.4 shows aplot oftheknownnuclides as afunctionoftheiratomic andneutronnumbers.Onalargerscale,withsufficientspaceprovidedtotabulatedataforeachnuclide,Fig. 2.4 isknownasaSegre chart orthechart a/the nuclides. ThefiguredepictsthattherearemoreneutronsthanprotonsinnuclideswithZ greaterthanabout20, thatis,foratomsbeyondcalciumintheperiodic table. Theseextra neutronsarenecessaryforthestabilityoftheheaviernuclei.Theexcessneutronsactsomewhatlikenuclearglue, holdingthenucleustogetherbycompensatingfortherepulsiveelectricalforcesbetweenthepositivelychargedprotons.ItisclearfromFig. 2.4 thatonlycertaincombinationsofneutronsandpro­tonsleadtostable nuclei. Althoughgenerallythereareseveralnuclides withthesameatomicnumberbutdifferentneutronnumbers (thesearetheisotopesoftheelement),ifthereareeithertoomanyortoofewneutronsforagivennumberofpro­tons,theresultingnucleusisnotstableanditundergoes radioactive decay. Thus,asnotedinSection2.2, theisotopesofoxygen(A = 8) withN= 8, 9, and 10arestable,buttheisotopes with N = 5, 6, 7, 11, and 12 areradioactive. InthecaseoftheisotopeswithN = 5, 6,and7, therearenotenoughneutronsforstability, whereastheisotopeswithN= 11 and12 havetoomanyneutrons.Nucleisuchas150,whicharelackinginneutrons,undergof3+-decay. Inthisprocess, one oftheprotons in the nucleus is transfonned into aneutron, and a
  • 34. Sec. 2.8 Nuclear Stability and Radioactive Decay 19 1 10 r--.---.--.---.---r--.--�--.---.---.--.---.--�--.---.--. 100 I----+----+-----+----+--f----+---+---+---+--+-----+--+----+--_+_- )_ �, I ::d����,� � :f---+---+----+--+--+---+---+----+--.-,+,u.-•.•-::t.-.�+-��- : �-� � --,�, � W ��l:i,.r-tAiF�� j 60 ."i'����I'lPi��'- J 50 �_'lilil::::W���:���g!� ,-'---I--+-----t----+--+---If---+----t40 f---+---+----+--+.-.:Fp:: I .::� • S�abl� nuclei 30 f-----+---f----:�'t�ttil8go 0 13 emltters . :;i�t��r06g 13+emitters (or electron capture) 20 .�'Jgg.o ® 13+, 13 - emitters (or electron capture) ::::Hgooj � } a-emitters (with � -, �+, or electron capture) 10 -�, 0 a-emitters (pure or �-stable) o � , � ol I I I I I I J I o 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 Neutron number (N) Figure 2.4 The chart of nuclides showing stable and unstable nuclei. (Based on S. E. Liverhant, Elementary Introduction to Nuclear Reactor Physics. New York: Wiley, 1960.) positronandaneutrinoareemitted.Thenumberofprotonsisthusreducedfrom8to7 sothattheresultingnucleusisanisotopeofnitrogen, 15N,whichisstable.Thistransformationiswrittenas 150 � 15N+v, wherefJ+signifiestheemittedpositron,whichinthiscontextiscalledafJ-rayandv denotestheneutrino.Bycontrast,nucleilike190,whichareexcessivelyneutron­rich,decaybyfJ-decay,emittinganegativeelectronandanantineutrino: 190 � 19F+v, wherev standsfortheantineutrino.Inthis case, aneutronchangesintoaprotonandtheatomicnumberincreasesbyoneunit.It shouldbenotedthatinbothfJ+­decayandfJ--decaytheatomicmassnumberremainsthesame. InbothformsoffJ-decay,theemittedelectronsappearwithacontinuousen­ergyspectrumlikethatshowninFig.2.5. Theordinateinthefigure,N(E), isequaltothenumberofelectronsemittedperunitenergy,whichhaveakineticenergyE.
  • 35. 20 Atomic and Nuclear Physics E Figure 2.5 A typical energy spectrum of electrons emited in beta de­ cay. Chap. 2 Thus,theactualnumberofelectronsemittedwithkineticenergiesbetween E andE +dE isN(E) dE. Itshouldbenotedinthefigurethatthereisadefinitemaxi­mumenergy,Emax, abovewhichnoelectronsareobserved.IthasbeenshownthattheaverageenergyoftheelectronsE isapproximatelyequaltoO.3Emax inthecaseoff3--decay.Inf3+-decay,E ::: O.4Emax. Frequently,thedaughter nucleus, thenucleusformedinf3-decay,is alsoun­stableandundergoesf3-decay.Thisleadstodecay chains likethefollowing: Anucleuswhichislackinginneutronscanalsoincreaseitsneutronnumberby electron capture. Inthisprocess, an atomic electron interacts with oneoftheprotonsinthenucleus,andaneutronisfonnedoftheunion.Thisleavesavacancyintheelectroncloudwhichislaterfilledbyanotherelectron,whichintumleadstotheemissionofy-rays,whicharenecessarilycharacteristicofthedaughterelementortheemissionofanAugerelectron. Usuallytheelectronthatiscapturedbythenucleus isthe innennostorK-electron, andsothismodeofdecay is alsocalledK-capture. Sincethedaughternucleusproducedinelectroncaptureisthesameasthenucleusfonnedinf3--decay,thesetwodecayprocessesoftencompetewithoneanother.Anotherwayby which unstable nuclei undergo radioactive decay is bytheemissionofana-particle. This particle isthehighly stablenucleusoftheisotope4He, consisting oftwoprotons andtwo neutrons. The emissionofana-particlereducestheatomicnumberbytwoandthemassnumberbyfour.Thus,forinstance,thea-decayof��8U(uranium-238)leadsto�64Th(thorium-234)accordingtotheequation 238U �234Th+4He92 90 2 ·
  • 36. Sec. 2.8 Nuclear Stability and Radioactive Decay TABLE 2.1 ALPHA-PARTICLE SPECTRUM OF 226Ra a-particle energy 4.782 4.599 4.340 4. 194 Relative number of particles (%) 94.6 5.4 0.005 1 7 x 10-4 21 Decaybya-particleemissioniscomparatively rareinnuclideslighterthanlead,butitiscommonfortheheaviernuclei.Inmarkedcontrastto,8-decay,a-particlesareemittedinadiscrete(line)energyspectrumsimilartophotonlinespectrafromexcitedatoms.ThisisshowninTable2.1,wheredataisgivenforthefourgroupsofa-particlesobservedinthedecayof226Ra(Radium-226).Thenucleusfonnedastheresultof,8-decay (+ or -), electroncapture,ora-decayisoftenleftinanexcitedstatefollowingthetransfonnation.Theexcited(daughter) nucleus usually decays5 bytheemissionofoneormorey-raysinthemannerexplainedinSection2.7.AnexampleofasituationofthiskindisshowninFig. 2.6fordecay of60Co-anuclide widely usedin nuclearengineering. Adiagram like that shown inthe figure is known as adecay scheme. It should beespeciallynotedthatthemajory-raysareemittedbythedaughternucleus,inthiscase6oNi,althoughtheyarefrequentlyattributedto(andariseastheresultof)thedecayoftheparentnucleus,60Co.Mostnucleiinexcitedstatesdecaybytheemissionofy-raysinanimmea­surablyshorttime afterthese statesarefonned. However,owing topeculiaritiesintheirinternalstructure,thedecayofcertainexcitedstatesisdelayedtoapointwherethenucleiinthesestatesappeartobesemistable. Suchlong-livedstatesarecalled isomeric states ofthenuclei in question. The decay by y-rayemissionofoneofthesestatesiscalledanisomeric transition andisindicatedasITinnucleardatatabulations.Insomecases,isomericstatesmayalsoundergo,8-decay.Figure 2.6 shows theisomeric statefoundat58 keVabovethegroundstateof60Co. Asarule,isometricstatesoccuratenergiesveryclosetothegroundstate.This stateis fonnedfromthe,8--decayof60Pe, notshowninthe figure. Itis observedthatthisisomeric state decays intwo ways: eitherto the ground stateof60Co orby,8--decaytothefirsttwoexcitedstatesof6oNi.Ofthesetwodecaymodes,thefirstisbyfarthemoreprobable (> 99%) andoccurslargelybyinternalconversion. SStrictly speaking, the tenn decay should not be used to describe the emission of y-rays from nuclei in excited states since only the energy and not the character of the nucleus changes in the process. More properly, y-ray emission should be referred to as nuclear dependent-excitation, not decay. However, the use of the tenn decay is well established in the literature.
  • 37. 22 0.0586 IT (10.5 m) 99 + % 0.0 99+ % 0.013 % 0.12 % 0.009 % 0.24 % Atomic and Nuclear Physics 2.506 2.158 1.332 --_........._....I....-- O.O 60 Ni Figure 2.6 Decay scheme of cobalt 60, showing the known radiation emitted. The numbers on the side ofthe excited states are the energies of these states in MeV above the ground state. The relative occurrence of competing decays is indicated by the various percentages. Chap. 2 Insummary,anucleuswithoutthenecessarynumbersofprotonsandneutronsforstability will decayby the emissionofa-rays orf3-rays orundergo electroncapture,allofwhichmaybeaccompaniedbythesubsequentemissionofy-rays.Itmustbeemphasizedthatradioactivenucleiordinarilydonotdecaybyemittingneutronsorprotons. 2.9 RADIOACTIVITY CALCULATIONS Calculationsofthedecayofradioactivenucleiarerelativelystraightforward,owingtothefactthatthereisonlyonefundamentallawgoverningalldecayprocesses. Thislaw states thattheprobabilityperunittime that anucleus will decay is aconstantindependent oftime. This constant is called thedecay constant and isdenotedbyA.6 Consider the decay ofa sample ofradioactive material. Ifat time t therearenet) atomsthatasyethavenotdecayed, An(t) dt ofthese willdecay, ontheaverage, inthe time interval dt between t and t + dt. Therate atwhich atomsdecay in the sample is thereforeAn(t) disintegrations per unit time. This decay 6By tradition, the same symbol, A, is used forboth decay constant and the wavelength defined earlier, as well as formean free path, defined later. Because of their very different uses, no confusion should arise.
  • 38. Sec. 2.9 Radioactivity Calculations 23 rateiscalledtheactivity ofthesampleandisdenotedbythesymbola. Thus,theactivityattimet isgivenby aCt) = An(t). (2.23) Activityhastraditionallybeenmeasuredinunitsofcuries, whereonecurie,denotedasCi, isdefined asexactly 3.7 x 1010 disintegrationspersecond. Unitstodescribesmallactivitiesarethemillicurie, 10-3 curie,abbreviatedasmCi,themicrocurie, 10-6 curie,denotedbytLCi,andthepicocurie, 10-12 curie, whichiswrittenaspCi.TheSIunitofactivityisthebecquerel, Bq,whichisequaltoexactlyonedisintegrationpersecond.OneBq= 2.703 x 10-11 Ci:::: 27 pCi.SinceAn(t) dt nucleidecayinthetimeintervaldt,itfollowsthatthedecreaseinthenumberofundecayednucleiinthesampleinthetimedtis -dn(t) = An(t) dt. Thisequationcanbeintegratedtogive n(t) = noe->..t (2.24) whereno isthenumberofatomsatt = O. MultiplyingbothsidesofEq.(2.24) by A givestheactivityofthesampleattimet-namely, (2.25) whereao istheactivityatt = O. Theactivitythusdecreasesexponentiallywithtime.Thetimeduringwhichtheactivityfallsby afactoroftwo is knownasthehalf-life andisgiventhesymbol T1/2• Introducingthisdefinition intoEq.(2.25) gives ThentakingthelogarithmofbothsidesofthisequationandsolvingforTI/2 gives In 2 0.693 TI/2 = T = - A - ' SolvingEq.(2.26) forA andsubstitutingintoEq. (2.25) gives aCt) = aoe-O.693tITI/2 (2.26) (2.27) ThisequationisofteneasiertouseincomputationsofradioactivedecaythanEq.(2.25), especiallywiththeadventofpocket-sizeelectroniccalculatorsandspread
  • 39. 24 Atomic and Nuclear Physics Chap. 2 sheetprograms,becausehalf-livesaremorewidelytabulatedthandecayconstants.Equation(2.27) eliminatestheneedtocomputeA. Itisnotdifficulttoshowthattheaveragelifeexpectancyormean-life, t, ofaradioactivenucleusisrelatedtothedecayconstantbytheformula t = I/A. FromEq.(2.25), itcanbeseenthatinonemean-lifetheactivityfallsto l/e ofitsinitialvalue.InviewofEq.(2.26), themean-lifeandhalf-lifearerelatedby - TI/2t = 0.693 = 1.44Tl/2. (2.28) Theexponential decay ofaradioactive sample is shown in Fig. 2.7, where thehalf-lifeandmean-lifearealsoindicated.Itisfrequentlynecessarytoconsiderproblemsinwhichradioactivenuclidesareproducedinanuclearreactororinthetargetchamberofanaccelerator.LetitbeassumedforsimplicitythatthenuclideisproducedattheconstantrateofRatoms/sec. As soonasitisformed,ofcourse,aradioactive atommaydecay. Thechangeinthenumberofatomsofthenuclideinthetimedt isgivenbyasimplerateequation thetimerateofchangeofthenuclide= orsymbolically c :E � a.J2 a.Je therateofproduction-therateofloss T1/2 t Time (afbitfary units) Figure 2.7 The decay of a radioactive sample.
  • 40. Sec. 2.9 Radioactivity Calculations 25 dnldt = -An +R. Thisequationcanbeintegratedwiththeresult Rn =noe-M +-(1 - e -A.t ), A (2.29) whereno isagainthenumberofradioactiveatomspresentatt = O. MultiplyingthisequationthroughbyAgivestheactivityofthenuclide (2.30) Ifao = 0, then,accordingtothisresult,a increasessteadilyfromzeroandast ---+ 00, a approachesthemaximumvalueamax = R. Similarly,n approachesaconstantvaluenmax givenbyRIA.Ifao =F 0, thentheactivityduetodecayoftheatomsoriginallypresentisaddedtotheactivityofthenewlyproducednuclide.Inbothcases,theactivityapproachesthevalueamax = Rast ---+ 00. Example 2.6 Gold-198 (T1/2 = 64.8 hr) can be produced by bombarding stable 197Au with neu­ trons in a nuclear reactor. Suppose that a 197Au foil weighing 0.1 g is placed in a certain reactor for 12 hrs and that its activity is 0.90 Ci when removed. (a) What is the theoretical maximum activity due to 198Au in the foil? (b) How long does it take for the activity to reach 80 percent ofthe maximum? Solution 1. The value of R in Eq. (2.30) can be found from the data at 12 hrs. From Eq. (2.26), A = 0.693/64.8 = 1 .07 x 10-2 hr-1 Then substituting into Eq. (2.30) gives or from Eq. (2.26) 0.90 = R[1 _ e-O.693xI2/64.8]. Solving either of these equations yields R = 7.5 Ci. (To get R in atoms/sec, which is not required in this problem, it is merely necessary to mUltiply the prior value of R by 3.7 x 1010.) According to the previous discussion, the theoretical maximum activity is also 7.5 Ci. [Ans.] 2. Thetime to reach 80% of O!max can also be found from Eq. (2.30): 0.8R = R(1 - eA.t) . Solving for t gives t = 150 hrs. [Ans.]
  • 41. 26 Atomic and Nuclear Physics Chap. 2 AnotherproblemthatisoftenencounteredisthecalculationoftheactivityoftheradioactivenuclideBinthedecaychain, A ---+ B ---+ C ---+ Itis clearthatanatomofB is formedwith eachdecayofanatomofA.Wecanwriteasimplerateequationforthisbehavior. thetimerateofchangeofB = therateofproduction,fromA-therateofdecayofBtoC. SinceAAnA istherateatomsofAdecayintoatomsofB,therateatomsofBareproducedisAAnA. TherateofdecayofBatomsisABnB, sotherateofchangeofB,dnB/dt is, dnB -- = -ABnB + AAnA. dt SubstitutingEq.(2.24) fornA givesthefollowingdifferentialequationfornB: dnB -AAt -- = -ABnB + AAnAOe dt wherenAO isthenumberofatomsofAatt = O. IntegratingEq.(2.31) gives -ABt + nAoAA ( - nB = nBOe e AB - AA Intennsofactivity,thisequationmaybewrittenas -ABt + aAoAB ( -AM -ABt )aB = aBOe e - e , AB - AA (2.31) (2.32) (2.33) whereaAO andaBO aretheinitialactivitiesofAandB,respectively.GeneralizationsofEq. (2.33) forcomputingtheactivityofthenthnuclideinalongdecaychainhavebeenderivedandarefoundinthereferencesattheendofthechapter. 2.1 0 NUCLEAR REACTIONS Anuclear reaction is saidtohavetakenplace when two nuclearparticles-twonucleioranucleusandanucleon-interacttoproducetwoormorenuclearparticlesory-rays.Iftheinitialnucleiaredenotedbya andb, andtheproductnucleiand/ory-rays(forsimplicityitisassumedthatthereareonlytwo)aredenotedbyc andd, thereactioncanberepresentedbytheequation a + b ---+ c + d. (2.34)
  • 42. Sec. 2. 10 Nuclear Reactions 27 Thedetailedtheoreticaltreatmentofnuclearreactionsisbeyondthescopeofthisbook.Forpresentpurposes,itissufficienttonotefourofthefundamentallawsgoverningthesereactions: 1. Conservation ofnucleons. Thetotal numberofnucleonsbeforeandafterareactionarethesame. 2. Conservation of charge. The sumofthecharges onalltheparticles beforeandafterareactionarethesame. 3. Conservation ofmomentum. Thetotalmomentumoftheinteractingparticlesbeforeandafterareactionarethesame. 4. Conservation ofenergy. Energy,includingrest-massenergy,isconservedinnuclearreactions. Itisimportanttonotethat,aswesee,conservationofnucleonsandconser­vationofchargedonotimplyconservationofprotonsandneutronsseparately.Theprincipleoftheconservationofenergycanbeusedtopredictwhetheracertainreactionisenergeticallypossible. Consider,forexample, areactionofthetype given inEq. (2.34). The total energybeforethe reaction is the sum ofthekineticenergiesoftheparticlesa andb plustherest-massenergyofeachparticle.Similarly,theenergyafterthereactionisthesumofthekineticenergiesofparticlesc andd plustheirrest-massenergies.Byconservationofenergyitfollowsthat (2.35) where Ea, Eb, andsoon,arethekineticenergiesofparticlesa, b, etc. Equation(2.35) canberearrangedintheform (2.36) Sincethequantitiesontheleft-handsiderepresentsthekineticenergyofthepar­ticles,itisevidentthatthechangeinthekineticenergiesoftheparticlesbeforeandafterthereactionisequaltothedifferenceintherest-massenergiesoftheparticlesbeforeandafterthereaction.Theright-handsideofEq.(2.36) isknownastheQ-value ofthereaction;thatis, (2.37) Inallcomputationsandtabulations, Q isalwaysexpressedinMeV.Recallingthat1amuisapproximately931 Mev,wecanwritethe Q valueas (2.38)
  • 43. 28 Atomic and Nuclear Physics Chap. 2 FromEq. (2.36), itis clearthatwhen Q ispositivethereis anetincrease inthekineticenergiesoftheparticles. Suchreactionsarecalledexothermic. When Qisnegativethereis anetdecreasein the energies ofthe particles, andthereactionissaidtobeendothermic. Sinceinexothennicreactionsthereisanetdecreaseinmass,nuclearmassisconvertedintokineticenergy,whileinendothennicreactionsthereisanetincrease,thuskineticenergyisconvertedintomass.Equation (2.37) gives Q in tenns ofthe masses ofthe nuclei a, b, and soon.However,the Qvaluecanalsobeintennsofthemassesoftheneutralatomscontainingthesenuclei. Thus,inviewoftheconservationofcharge, (2.39) whereZa, Zb, andsoon,aretheatomicnumbersofa, b, andsoon,andEq.(2.37) canbeputinthefonn Q= {[(Ma + Zame) + (Mb + Zbme)] - [(Me + Zeme) + (Md + Zdme)]}931 Mev, (2.40) whereme istheelectronrestmassinamu. ButMa + Zame isequaltothemassoftheneutralatomofa, Mb + Zbme isthemassoftheatomofb, andsoon. ItfollowsthatEq. (2.37) is avalidfonnulafor Q, where Ma, Mb, and soon, areinterpretedasthemasses (in amu) ofthe neutral atoms inquestion, althoughtheactual nuclearreactioninvolves only the atomic nuclei. Itis fortunatethat Qcanbecomputedfromneutralatomicmassdatasincethemassesofmostbarenucleiarenotaccuratelyknown.Incidentally,intheusualexperimentalarrangement,oneoftheparticles,saya, isatrestinsomesortoftarget,andtheparticleb isprojectedagainstthetarget.Inthiscase,Eq.(2.34) isoftenwrittenintheabbreviatedfonn a(b, c)d or a(b, d)c, whichever is the more appropriate. Forexample, when oxygen is bombardedbyenergeticneutrons,oneofthereactionsthatoccursis 160 + n ---+ 16 N+ IH. Inabbreviatedfonn,thisis
  • 44. Sec. 2. 1 1 Binding Energy 29 wherethesymbolsnandprefertotheincidentneutronandemergentproton,re­spectively. Example 2.7 Complete the following reaction: 14N + n ---+? + IH. Solution. The atomic number of 14N is 7, that of the neutron is O. The sum of the atomic numbers on the left-hand side of the reaction is therefore 7, and the sum on the right mustalso be 7. Since Z = 1 for hydrogen, it follows that Z of the unknown nuclide is 7 - 1 = 6 (carbon). The total number of nucleons on the left is the sum of the atomic mass numbers-namely, 14 + 1 = 15. Since the mass number of IH is 1, the carbon isotope fonned in this reaction must be 14C. Thus, the reaction is Example 2.8 One of the reactions that occurs when 3H (tritium) is bombarded by deuterons eH nuclei) is where d refers to the bombarding deuteron. Compute the Q value of this reaction. Solution. The Q value is obtained from the following neutral atomic masses (in amu): MeH) = 3.016049 MeH) = 2.014102 M(4He) = 4.002604 M(n) = 1.008665 M(4He) + M(n) = 5.01 1269 Thus, from Eq. (2.37), the Q value in amu is Q = 5.030151 -5.01 1269 = 0.018882 amu. Since 1 amu = 931.502 MeV (see Ex. 2.4), Q = 0.018882 x931.502 = 17.588 MeV, which is positive and so this reaction is exothennic. This means, for instance, that when stationary 3H atoms are bombarded by I-MeV deuterons, the sum of the kinetic energies ofthe emergent a-particle (4He) and neutron is 17.588+ 1 = 18.588 MeV. 2.1 1 BINDING ENERGY Whenalow-energyneutronandaprotoncombinetoformadeuteron,thenucleusof2H,a2.23-MeVy-ray,isemittedandthedeuteronrecoilsslightlywithanenergyofabout 1.3keV.Thereactioninquestionis
  • 45. 30 Atomic and Nuclear Physics Chap. 2 p + n � d + y, or,intermsoftheneutralatoms, IH+n�2H + y. Sincethey-rayescapesfromthesiteofthereaction,leavingthedeuteronbehind,itfollowsfromtheconservationofenergythatthemassofthedeuteroninenergyunitsisapproximately2.23MeVlessthanthesumofthemassesoftheneutronandproton.Thisdifferenceinmassbetweenthedeuteronanditsconstituentnucleonsiscalledthemass defect ofthedeuteron.Inasimilarway,themassesofallnucleiaresomewhatsmallerthanthesumofthemassesoftheneutronsandprotonscontainedinthem.Thismassdefectforanarbitrarynucleusisthedifference (2.41) whereMA isthemassofthenucleus.Equation(2.41)canalsobewrittenas (2.42) whereme isthemassofanelectron.ThequantityMp + me isequaltothemassofneutral1 H,whileMA + Zme isequaltothemassM oftheneutralatom.Themassdefectofthenucleusistherefore (2.43) whichshowsthat�canbecomputedfromthetabulatedmassesofneutralatoms.Equations(2.41)and(2.43)arenotpreciselyequivalentowingtoslightdifferencesinelectronicenergies,butthisisnotimportantformostpurposes.When � isexpressedinenergyunits,itisequaltotheenergythatisneces­sarytobreakthenuleusintoitsconstituentnucleons.Thisenergyisknownasthebinding energy ofthesystemsinceitrepresentstheenergywithwhichthenucleusisheldtogether.However,whenanucleusisproducedfromA nucleons,�isequaltotheenergyreleasedintheprocess.Thus,inthecaseofthedeuteron,thebindingenergyis2.23MeV.Thisistheenergyreleasedwhenthedeuteronisformed,anditis alsotheenergyrequiredtosplitthedeuteronintoaneutronandproton.ThetotalbindingenergyofnucleiisanincreasingfunctionoftheatomicmassnumberA. However,itdoesnotincreaseataconstantrate.Thiscanbeseenmostconvenientlybyplottingtheaveragebindingenergypernucleon, �/A, versusA, asshowninFig.2.8.ItisnotedthatthereareanumberofdeviationsfromthecurveatlowA, whileaboveA = 50 thecurveisasmoothbutdecreasingfunctionofA. This behaviorofthe binding energy curve is importantin determining possiblesourcesofnuclearenergy.
  • 46. Sec. 2.1 1 Binding Energy 9 8 2 o �Y'� .... r-_.... r--• .. .. .. � '-- . f' ----- --------. f I o 20 40 60 80 1 00 1 20 140 1 60 1 80 200 220 240 Atomic mass number Figure 2.8 Binding energy per nucleon as a function of atomic mass number. 31 Thosenucleiinwhichthebindingenergypernucleonishighareespeciallystableortightlybound,andarelativelylargeamountofenergymustbesuppliedtothesesystemstobreakthemapart.However,whensuchnucleiarefonnedfromtheirconstituentnucleons,arelativelylargeamountofenergyisreleased.Bycon­trast,nucleiwithlowbindingenergypernucleoncanbemoreeasilydisruptedandtheyreleaselessenergywhenfonned.The Qvalueofanuclearreactioncanbeexpressedintennsofthebindingenergies ofthe reactingparticles or nuclei. Consider the reaction given by Eq.(2.34).FromEq.(2.43),thebindingenergyofa,inunitsofmass,is BE(a) = ZaMeH) + NaMn - Ma. Themassofacanthenbewrittenas Similarly,
  • 47. 32 Atomic and Nuclear Physics Chap. 2 andso on. Substituting theseexpressionsforMa, Mb, andsoon, into Eq. (2.37)andnotingthat gives Za + Zb = Zc + Zd Na + Nb = Nc + Nd, Q = [BE(c) + BE(d)] - [BE(a) - BE(b)]. Here,thec2 hasbeendroppedbecauseQandthebindingenergiesallhaveunitsofenergy.ThisequationshowsthatQispositive-thatis,thereactionisexothennic­when the total binding energy ofthe product nuclei is greaterthan the bindingenergyoftheinitialnuclei.Putanotherway,wheneveritis possibletoproduceamorestableconfigurationbycombiningtwoless stablenuclei,energyisreleasedintheprocess.Suchreactionsarepossiblewithagreatmanypairsofnuclides.Forinstance, whentwodeuterons, each with abinding energyof2.23MeV,reacttofonn3H,havingatotalbindingenergyof8.48MeV,accordingtotheequation 22H---+ 3H+ 1H, (2.44) thereisanetgaininthebindingenergyofthesystemof8.48-2 x 2.23 = 4.02MeV. Inthis case,theenergyappearsaskineticenergyofthe product nuclei 3HandIH.ReactionssuchasEq. (2.44),inwhichatleastoneheavier,morestablenu­cleus is producedfromtwolighter, lessstablenuclei,areknownasfusion reac­ tions. Reactionsofthistypeareresponsiblefortheenonnousreleaseofenergyinhydrogenbombsandmaysomedayprovideasignificantsourceofthennonuclearpower.IntheregionsoflargeA inFig. 2.8,itisseenthatamorestableconfigura­tionisfonnedwhenaheavynucleussplitsintotwoparts.Thebindingenergypernucleonin238U,forinstance,isabout7.5MeV,whereasitisabout8.4MeVintheneighborhoodofA = 238/2 = 119. Thus, ifauraniumnucleusdividesintotwolighternuclei,eachwithabouthalftheuraniummass,thereisagaininthebindingenergyofthesystemofapproximately0.9 MeVpernucleon,whichamountstoatotalenergyreleaseofabout238 x 0.9= 214MeV.Thisprocessiscallednuclear fission, andisthesourceofenergyinnuclearreactors.ItmustbeemphasizedthatthebindingenergypernucleonshowninFig.2.8isanaverageoverallofthenucleonsinthenucleusanddoesnotrefertoanyonenucleon. Occasionally itis necessary toknow the binding energy ofaparticularnucleoninthenucleus-thatis,theamountofenergyrequiredtoextractthenu­cleonfromthe nucleus. Thisbindingenergyis alsocalledtheseparation energy andisentirelyanalogoustotheionizationenergyofanelectroninanatom.Con-
  • 48. Sec. 2. 12 N uclear Models 33 sidertheseparationenergy Es oftheleastboundneutron-sometimescalledthelast neutron-inthenucleusAZ. Sincetheneutronis boundinthenucleus,itfol­lowsthatthemassofthenucleus(andtheneutralatom)AZ islessthanthesumofthemassesoftheneutronandtheresidual nucleus A-I Z byanamount,inenergyMev,equaltoEs. Insymbols,thisis Es = [Mn + M(A-l Z) - M(AZ)]931 Mev/amu. (2.45) TheenergyEs isjustsufficienttoremoveaneutronfromthenucleuswithoutprovidingitwithanykineticenergy.However, ifthisprocedureisreversedandaneutronwithnokineticenergyis absorbedbythenucleusA-I Z, theenergy Es isreleasedintheprocess. Example 2.9 Calculate the binding energy of the last neutron in l3C. Solution. If the neutron is removed from l3C, the residual nucleus is l2C. The bind­ ing energy or separation energy is then computedfrom Eq. (2.45) as follows: Met2C) = 12.00000 Mn = 1.00866 Mn + Met2C) = 13.00866 - M(l3C) = 13.00335 Es = 0.00531 amu x 931 Mev = 4.95 MeV [Ans.] Beforeleavingthediscussionofnuclearbindingenergy, itshouldbenotedthatnuclei containing 2, 6, 8, 14, 20, 28, 50, 82, or 126 neutronsorprotons areespeciallystable.Thesenucleiaresaidtobemagic, andtheirassociatednumbersofnucleons are known as magic numbers. These correspond to the numbers ofneutronsorprotonsthatarerequiredtofill shells (orsubshells)ofnucleonsinthenucleusinmuchthesamewaythatelectronshellsarefilledinatoms.Theexistenceofmagicnucleihasanumberofpracticalconsequencesinnu­clearengineering.Forinstance, nucleiwithamagicneutronnumberabsorbneu­trons to only avery small extent,and materials ofthis type can be used whereneutronabsorption mustbe avoided. Forexample,Zirconium,whosemostabun­dantisotopecontains50neutrons,hasbeenwidelyusedasastructuralmaterialinreactorsforthisreason. 2.12 NUCLEAR MODELS Two models ofthe nucleus are useful in explaining the various phenomena ob­servedin nuclearphysics-the shell model andtheliquiddropmodel. Although
  • 49. 34 Atomic and Nuclear Physics Chap. 2 neitherofthese models cancompletely explain the observed behavior ofnuclei,theydoprovidevaluableinsightintothenuclearstructureandcausemanyofthenuclearreactionsofinteresttothenuclearengineer. Shell Model Theshellmodelmaybethoughtofasthenuclearanaloguetothemanyelectronatom.Inthismodel,thecollectiveinteractionofthenucleonsinthenucleusgener­ateapotentialwell.Onecanthenthinkofasinglenucleonasifitismovinginthewellcreatedbytheaverageeffectoftheothernucleons.Asinthecaseofothersuchpotentialwells,suchawellcanhaveoneormorequantizedstates.Thesestatesarethenpopulatedinthesamewaythatthe atomic orbitalsofanatomarepopulatedbyelectrons.Justasintheatom,thereisamaximumnumberofnucleonsthatmayoccupyashell.Whenthisnumberisreached,aclosedshellresults.Althoughadetaileddiscussionofthismodelisbeyondthescopeofthistext,afewremarksarenecessarytounderstandthe stability ofcertain nuclei andtheoriginofthemagicnumbers.Theneutronsandprotonsfilleachlevelinapotentialwell according tothe Pauli exclusion principle. Accounting forthe angularmo­mentumforeachstate,thereare2j+ 1possiblesubstatesforeachlevelwithtotalangularmomentumj.Since we are dealing with two sets ofidentical particles-neutrons on theonehandandprotonsontheother-therearereallytwosuchwells,oneforeach.Theydifferby the coulomb interaction oftheprotons. The levelsarethenfilledaccording to the exclusion principle. The differing mj values will splitapartinenergybecauseofthespinorbitinteraction. Asinthecaseofthemanyelectronatoms,thismayresultinreorderingofthelevels anddevelopmentofwidergapsinbetweentheenergylevelsthanotherwise wouldbeexpected.Sincetheneutronandprotonwellscaneachhaveclosedshells,the nuclei canbeextremely stablewhenbothwellshaveclosedshellsandlesssowhenneitherdo.Thisphenomenongivesrisetothemagicnumbersdiscussedearlier. Liquid Drop Model FromSection2.11,thebindingenergyisthemassdefectexpressedinenergyunits.Theliquid-dropmodelofthenucleusseekstoexplainthemassdefectintennsofabalancebetweentheforcesbindingthenucleonsinthenucleusandthecoulombicrepulsionbetweentheprotons.Thenucleusmaybethoughtofas adropofnuclearliquid.Justasawaterdropletexperiences anumberofforcesactingtoholdittogether,sodoesthenu­cleardroplet.Toafirstapproximation,themassofanucleardropletisjustthemassofthecomponents-theneutronsandprotons.Theseareinteractinginthenucleusandareboundbythenuclearforces.Thebindingofeachnucleontoitsneighbors
  • 50. Sec. 2. 1 2 Nuclear Models 35 meansthatenergyandmassmustbeaddedtotearthenucleusapart.Themassmaythenbeapproximatedby: M = NMn +ZMp - aA. (2.46) Equation 2.46 overestimates the effect ofthe bondsbetweenthe nucleons sincethosenearthesurfacecannothavethesamenumberofbondsasthosedeepinsidethenucleus.Tocorrectforthis,asurfacecorrectiontermmustbeadded: (2.47) whereT denotesthesurfacetension.SincetheradiusRofthenucleusisproportionaltoA1/3, wecanrewritethisterm: M = NMn + ZMp - aA +fJA2/3 (2.48) The coulombic repulsion tends to increase the energy andhence mass ofthe nucleus. Usingthepotential energy associated withtherepulsive force, theexpressionbecomes: (2.49) Thereareadditional,strictlynucleareffectsthatmustbeaccountedforinthemassequation.TheseaccountforapreferenceforthenucleonstopairtogetherandfortheeffectofthePauliexclusionprinciple.7 Intheshellmodel,thenucleonswerethoughtofasfillingtwopotentialwells.Thelowestenergysystemwouldthenbeoneinwhichthenumberofprotonswouldequalthenumberofneutronssince, inthis case,thewells wouldbefilledtothesameheight,witheachlevelfilledaccordingtothePauliexclusionprinciple.ThenucleushavingN = Z shouldthenbemorestablethanthenucleuswithN # Z. Toaccountforthiseffect,acorrectiontermmustbeaddedtothemassequation: (2.50) Finally,ifoneexaminesthestablenuclides,onefindsapreferencefornucleiwithevennumbersofneutronsandprotons.Thepreferencereflectsthat,experimen­tally,thebondbetweentwoneutronsortwoprotonsisstrongerthanthatbetweenaneutronandproton. Nuclei withoddnumbersofneutrons andoddnumbersofprotonswouldthusbelessstronglyboundtogether.WheneitherZ orN isoddandtheothereven, onewouldexpectthebindingtobe somewhereinbetweenthesetwocases. Toaccountforthiseffect, apairingtermdenotedby8 is addedtotheexpression: 7For a discussion of the Pauli exclusion principle, see references on modem physics.
  • 51. 36 Atomic and Nuclear Physics Chap. 2 Theterm8 is0ifeitherN orZ isoddandtheothereven,positiveifbothareodd,andnegativeifbothareeven.Equation2.51 isthemassequation.Thecoefficientsforthemassequationareobtainedbyfittingtheexpressiontotheknownnuclei.Whenthisisdone,thesemi-empiricalmassequationisobtained.Thevaluesforeachofthecoefficientsaretypicallytakenas: MassofneutronMassofprotona {3 y � 8 939.573MeV938.280MeV15.56MeV17.23MeV0.697MeV23.285MeV12.0MeV TheformulacanaccuratelypredictthenuclearmassesofmanynuclideswithZ > 20.Theabilitytopredictthesemassessuggeststhatthereissometruthtothewaytheliquid-dropmodelstheinteractionsofthenucleonsinthenucleus.Foratomicmasses,themassofahydrogenatom(938.791MeV)maybesub­stitutedforthemassoftheprotontoaccountforthemassoftheatomicelectrons. Example 2.10 Calculate the mass and binding energy of!�7Ag using the mass equation. Solution. The mass equation may be used to calculate the binding energy by noting that the negative of the sum of the last five terms represents the binding energy of the constituent nucleons. The atomic mass of the !�7Ag is first obtained by using the mass formula and noting that N is even and Z is odd. The term involving 8 is thus taken as zero. N x mn Z x mH -ex x A +fJ X A2/3 Y X Z2/A1/3 � x (A - 2 x Z)2/A Mass (MeV) Mass (u) 60 x 939.573 MeV 47 x 938.791 MeV - 15.56 x 107 MeV 17.23 x 1072/3 MeV 0.697 x 472/1071/3 MeV 23.285 x (107 - 2 x 47)2/107 MeV 99548.1 173 MeV 106.8684 u The measured mass of !�7Ag is 106.905092 u or within 0.034% of the calcu­ lated value. Summing the last fourterms gives a total binding energy of949.44 MeV or g.9 MeV! nucleon=a value slightly higher than the me(l�ured value of approxi­ mately 8.6 MeV.
  • 52. Sec. 2.13 Gases, Liquids, and Solids 37 2.13 GASES, LIQUIDS, AND SOLIDS Beforeconcludingthisreviewofatomicandnuclearphysics,itis appropriatetoconsiderthenatureofgrossphysicalmattersince this isthematerialencounteredinallpracticalproblems.Classically,therearethreeso-calledstatesofmatter:gas,liquid,andsolid.Theprincipalcharacteristicsoftheseareasfollows. Gases Thenoblegases-helium,neon,argon,krypton,xenon,andradon-andmostmetallicvaporsaremonatomic-thatis,theyarecomposedofmoreorlessfreelymoving,independentatoms.Virtuallyallothergasesconsistofequallyfreelymovingdiatomicorpolyatomicmolecules. Therandom,disorderedmotionoftheseparticlesisoneofthecharacteristicfeaturesofallgases. Solids Mostofthesolidsusedin nuclear systems-namely, metals andceramics-are crystalline solids. Such solids are composed oflarge numbers ofmicrocrystals, eachofwhichconsistsofanorderedthree-dimensionalarrayorlat­ticeofatoms.Eachmicrocrystalcontainsanenormousnumberofindividualatoms.SincetheregUlarityinthearrangementoftheatomsinthelatticeextendsoversomanyatoms (oftenovertheentiremicrocrystal), suchcrystals aresaidtoexhibitlong-rangeorder.Thereareanumberofothermaterialsthatarecalledsolids be­causetheyarerigidbodies,thatdonotexhibitlong-rangeorder.Examplesofsuchmaterialsareplastics,organicmaterials,glasses,andvariousamorphoussolids. Liquids Themicroscopic structure ofliquids is considerably morecom­plicatedthanisusuallyassumed.Theatomsand/ormoleculesinaliquidinteractstronglywithoneanother;asaresult,theytendtobeorderedastheyareinacrys­tal,butnotoversuchlongdistances.Theorderedarrangementbreaksdown,sotospeak,overlongdistances.Forthisreason,liquidsaresaidtoexhibitshort-rangeorder. The Maxwellian Distribution Inagas,theenergies ofthe atoms ormolecules aredistributedaccordingtotheMaxwelliandistribution function. IfN(E) is thedensityofparticlesperunit en­ergy,then N(E) dE is the number ofparticles per unit volume having energiesbetweenE andE + dE. AccordingtotheMaxwelliandistribution,N(E) isgivenbytheformula (2.52)
  • 53. 38 Atomic and Nuclear Physics Chap. 2 InEq.(2.52),N isthetotalnumberofparticlesperunitvolume;thatis,theparticledensity;k isBoltzmann'sconstant,whichhasunitsofenergyperdegreeKelvin: k = 1.3806X 10-23joule/oK = 8.6170x 10-5eV/oK; andT istheabsolutetemperatureofthegasindegreesKelvin.ThefunctionN(E) isplottedinFig.2.9.Forsolidsandliquids,theenergydistributionfunctionsaremorecomplicatedthantheonegiveninEq.(2.52).However,ithasbeenshownthat,toafirstapprox­imation, N(E) forsolidsandliquidscanalsoberepresentedbyEq.(2.52),buttheparameterT differs somewhatfromtheactualtemperatureofthe substance.Thedifferenceissmallfortemperaturesaboveabout300° K.Atthesetemperatures,itcanoftenbeassumedthatEq.(2.52)appliestosolids,liquids,andgases.Themostprobable energy inadistributionsuchastheonegiveninEq.(2.52)is definedastheenergycorrespondingtothemaximumofthecurve. Thiscanbecalculated by placing the derivative of N(E) equal to zero. The most probableenergy,EP' inaMaxwellianenergydistributionistheneasilyfoundtobe Ep = �kT However,theaverage energy, E, isdefinedbytheintegral 1 100E = - N(E)E dE. N 0 2 ElkT 3 4 Figure 2.9 The Maxwellian distribution function. (2.53) (2.54)
  • 54. Sec. 2. 1 3 Gases, Liquids, and Solids SubstitutingEq.(2.52) intoEq.(2.54) andcarryingouttheintegrationgives- 3 E = 2kT 39 (2.55) ThecombinationofparameterskT inEqs.(2.53) and(2.55) oftenappearsintheequationsofnuclearengineering.Calculationsinvolvingtheseparametersarethenexpeditedbyrememberingthat,forTo = 293.61°K, kT hasthevalue kTo = 0.0253 eV:::: to eV (2.56) Example 2.11 What are the most probable and average energies of air molecules in a New York City subway in summertime at, say, 38°C (about 100°F)? Solution. It is first necessary to compute the temperature in degrees Kelvin. From the formula OK = °C + 273. 15, it follows that the temperature of the air is 31 1.15°K. Then using Eqs. (2.53) and (2.55) gives 1 3 1 1 . 15 Ep = "2 x 0.0253 x 293.61 = 0.0134 eY. [Ans.] Finally, E = 3Ep, so that E = 0.0402 eY. [Ans.] The Gas Law Toafirstapproximation,gasesobeythefamiliarideal gas law (2.57) where P isthegaspressure, V isthevolume,nM isthenumberofmolesofgascontainedin V, R isthegasconstant,andT istheabsolutetemperature.Equation(2.57) canalsobewrittenas whereNA is Avogadro'snumber. Sincethereare NA atomspermole, itfollowsthattherearenM NA atomsinV ThefirstfactoristhereforeequaltoN-thetotalnumberofatomsormoleculesperunitvolume.Atthesametime,thefactorRjNA isthedefinitionofk, Boltzmann'sconstant.Theideal gaslawcanthusbeputintheconvenientform P = NkT. (2.58)
  • 55. 40 Atomic and Nuclear Physics Chap. 2 FromEq.(2.58) itisseenthatgaspressurecanbe(and,incertainapplications,is)expressedinunitsofenergyperunitvolume. 2.14 ATOM DENSITY In nuclearengineeringproblems,itis oftennecessarytocalculatethenumberofatoms ormolecules contained in 1 cm3 ofasubstance. Consider first amaterialsuch as sodium, which is composed ofonly one type ofatom. Then ifp is itsphysicaldensitying/cm3andM isitsgramatomicweight,itfollowsthatthereare p/M grammolesofthe substance in 1 cm3 Since each grammolecontains NA atoms,whereNA isAvogadro'snumber,theatomdensityN, inatomspercm3,issimply Example 2.12 pNA N = ­ M The density of sodium is 0.97 g/cm3 Calculate its atom density. Solution. The atomic weight of sodium is 22.990. Then from Eq. (2.59), 0.97 x 0.6022 x 1024 24N = 22.990 = 0.0254 x 10 [Ans.] (It is usual to express the atom densities as a factor times 1024.) (2.59) Equation(2.59) alsoappliestosubstancescomposedofindividualmolecules,exceptthat N is the molecule density (molecules per cm3) and M is the grammolecularweight.Tofindthenumberofatomsofaparticulartypepercm3,itismerely necessarytomultiplythemolecular density by the number, ni, ofthoseatomspresentinthemolecule, (2.60) ThecomputationofatomdensityforcrystallinesolidssuchasNaCIandforliquids isjustaseasyasforsimpleatomic andmolecularsubstances, but theex­planation is more complicated. The problemhere is thatthereareno recogniz­ablemolecules-anentiremicrocrystalofNaCIis, sotospeak,amolecule.Whatshouldbe done in this case is to assumethatthematerial consists ofhypothet­ ical molecules containing appropriate numbers ofthe constituent atoms. (Thesemoleculesareinfactunitcellsofthecrystallinesolidandcontaintheappropriate
  • 56. Sec. 2.14 Atom Density 41 numberofatoms.)ThenbyusingthemolecularweightforthispseudomoleculeinEq.(2.59), thecomputedvalueofN givesthemoleculardensityofthismolecule.Theatomdensities canthenbecomputedfromthisnumberintheusualway asillustratedinthefollowingexample. Example 2.13 The density of a NaCI crystal is 2.17 g/cm3 Compute the atom densities of Na and Cl. Solution. The atomic weight of Na and CI are 22.990 and 35.453, respectively. The molecular weight for a pseudomolecule ofNaCI is therefore 58.443. Using Eq. (2.59) gives 2. 17 x 0.6022 x 1024 N = = 0.0224 X 1024 molecules/cm3 58.443 Since there is one atom each of Na and CI per molecule, it follows that this is also equal to the atom density ofeach atom. [Ans.] Frequentlyitisrequiredtocomputethenumberofatomsofaparticulariso­topepercm3 Since, aspointedoutinSection 2.2, the abundanceofisotopesis alwaysstatedinatompercent,theatomdensityofanisotopeisjustthetotalatomdensityoftheelementasderivedearlier,multipliedbytheisotopicabundanceex­pressedasafraction.Thus,theatomdensityNi fortheithisotopeis N. _ YipNA I - 100M ' (2.61) whereYi istheisotopicabundanceinatampercent, abbreviateda/a. Thechemicalcompositionsofmixturesofelements suchasmetallic alloysareusuallygivenintermsofthepercentbyweightofthevariousconstituents.Ifp isthephysicaldensityofthemixture,thentheaveragedensityoftheithcomponentis (2.62) where Wi istheweight percent, abbreviated W/a, ofthecomponent. FromEq.(2.59), itfollowsthattheatomdensityofthiscomponentis (2.63) whereMi isitsgramatomicweight.
  • 57. 42 Atomic and Nuclear Physics Chap. 2 Withasubstancewhosecompositionisspecifiedbyachemicalfonnula,thepercentbyweightofaparticularelementisequaltotheratioofitsatomicweightinthecompoundtothetotalmolecularweightofthe compound. Thus, withthecompoundXm Yn, themolecularweightismMx + nMy, whereMx andMy aretheatomicweightsofX andY, respectively,andthepercentbyweightoftheelement X is mMx wjo(X) = x 100. mMx + nMy (2.64) Insomenuclearapplications,theisotopiccompositionofanelementmustbechangedbyartificialmeans.Forinstance,theuraniumusedasfuelinmanynuclearreactorsmustbeenrichedintheisotope235U(theenrichmentprocessisdiscussedinChap.4). Inthiscase,itisalsothepracticetospecifyenrichmentinweightpercent.Theatomicweightoftheenricheduraniumcanbecomputedasfollows.Thetotalnumberofuraniumatomspercm3isgivenby N = LNi , whereNi istheatomdensityoftheithisotope.IntroducingN fromEq.(2.59) andNi fromEq. (2.63) gives 1 1 Wi - = - L ­ M 100 Mi (2.65) The application ofthis fonnula in computations involving enriched uranium isillustratedinExample2.16. Example 2.14 For water of normal (unit) density compute: (a) the number of H20 molecules per cm3, (b) the atom densities of hydrogen and oxygen, (e) the atom density of 2H. Solution 1. The molecular weight of H20 is 2 x 1.00797 + 15.9994 moleculardensity is therefore, 18.0153. The 1 x 0.6022 X 1024 N(H20) = 18.0153 = 0.03343 x 1024 molecules/cm3. [Ans.]
  • 58. Sec. 2.14 Atom Density 43 2. There are two atoms of hydrogen and one atom of oxygen per H20 molecule. Thus, the atom density ofhydrogen N(R) = 2 x 0.03343 X 1024 = 0.06686 X 1024 atoms/cm3, and N(O) = 0.03343 X 1024 atoms/cm3 [Ans.] 3. The relative abundance of 2H is 0.015 alo so that NeH) = 1.5 x 10-4 x N(H) = 1.0029 x 10-5 x 1024 atoms/cm3 [Ans.] Example 2.15 A certain nuclearreactor is fueled with 1,500 kg ofuraniumrods enriched to 20 wi0 235V. The remainder is 238V. The density of the uranium is 19.1 g/cm3 (a) How much 235V is in the reactor? (b) What are the atom densities of 235V and 238V in the rods? Solution 1. Enrichment to 20 wi0 means that 20% of the total uranium mass is 235V. The amount of 235V is therefore 0.20 x 1500 = 300kg. [Ans.] 2. The atomic weights of235V and 238V are 235.0439 and 238.0508, respectively. From Eq. (2.63), 235 20 x 19.1 x 0.6022 x 1024 N( V) - 100 x 235.0439 = 9.79 x 10-3 x 1024atoms/cm3 [Ans.] The 238V is present to the extent of80 wi0, and so Example 2.16 238 80 x 19. 1 x 0.6022 x 1024 N( V) - 100 x 238.0508 = 3.86 x 10-2 x 1024atoms/cm3 [Ans.] The fuel for a reactor consists of pellets of uranium dioxide (V02), which have a density of 10.5 g/cm3 Ifthe uranium is enriched to 30 wlo in 235V, what is the atom density of the 235V in the fuel? Solution. It is first necessary to compute the atomic weight of the uranium. From Eq. (2.65), 1 1 ( 30 70 ) M = 100 235.0439 + 238.0508 '
  • 59. 44 Atomic and Nuclear Physics Chap. 2 which gives M = 237. 141. The molecular weight ofthe V02 is then 237. 141 + 2 x 15.999 = 269.139. In view of Eq. (2.64), the percent by weight of uranium in the V02 is (237. 141/269.139) x 100 = 88.1 w/o. The average density ofthe uranium is therefore 0.881 x 10.5 = 9.25 g/cm3, and the density of 235V is 0.30 x 9.25 = 2.78 g/cm3 The atom density of235V is finally REFERENCES General 235 2.78 x 0.6022 x 1024 N( U) - 235.0439 = 7. 1 1 x 10-3 x 1024 atoms/cm3 [Ans.] Arya, A. P., Elementary Modern Physics, Reading, Mass.: Addison-Wesley, 1974. Beiser, A., Concepts ofModern Physics, 5th ed. New York: McGraw-Hill, 1994, Chapters 3, 4, 5, 1 1, 12, 13, and 14. Burcham, W. E., Nuclear Physics: An Introduction, 2nd ed. Reprint Ann Arbor. Foster, A. R., and R. L. Wright, Jr., Basic Nuclear Engineering, 4th ed. Paramus: Prentice­ Hall, 1982, Chapter 2 and 3. Goble, A. T., and K. K. Baker, Elements ofModern Physics, 2nd ed. New York: Ronald Press, 1971, Chapters 8-10. Kaplan, I., Nuclear Physics, 2nd ed. Reading, Mass.: Addison-Wesley Longman, 1962. Krane, K. S., Nuclear Physics, 3rd ed. New York: John Wiley, 1987. Krane, K. S., Modern Physics, 2nd ed. New York: John Wiley, 1995. Lapp, R. E., and H. L. Andrews, Nuclear Radiation Physics, 4th ed. Englewood Cliffs, N.J.: Prentice-Hall, 1972, Chapters 1-7. Liverhant, S. E., Elementary Introduction to Nuclear Reactor Physics. New York: Wiley, 1960. Meyerhof, W. E., Elements ofNuclear Physics, New York: McGraw-Hill, 1967, Chapters 2 and 4. Oldenberg, 0., and N. C. Rasmussen, Modern Physicsfor Engineers, reprint, Marietta, Technical Books, 1992, Chapters 12, 14, and 15. Semat, H" and J. R. Albright, Introduction to Atomic and Nuclear Physics, 5th ed. New York: Holt, Rinehart & Winston, 1972. Serway, R. A., Moses, C. J., and Moyer, C. A., Modern Physics, 3rd ed. Philadelphia: Saunders, 1990. Tipler, P. A., Modern Physics, 2nd ed. New York: Worth, 1977. Wehr, M. R., Richards, J. A., andAdair, T.W., Physics oftheAtom, 4th ed. Reading, Mass.: Addison-Wesley, 1984.
  • 60. Problems 45 Weidner, R. T., and R. L. Sells, Elementary Modern Physics, 3rd ed. Boston, Mass.: Allyn & Bacon, 1980. Williams, W. S. C., NuclearandParticle Physics, Oxford, Eng.: Clarendon Press, 199 1 . Nuclear Data The Chart ofthe Nuclides. Thisvaluablechart,amustforeverynuclearengineer,isavailablefromtheLockheedMartinDistribution Services 10525 ChesterRoadCincinnati,OH45215, Lederer,C.M.,Hollander,J.M.,andPerlman,I.,Tables ofIsotopes, 7thed.NewYork:Wiley, 1978.Thisisanextraordinarycollectionofnucleardata,whichincludesamongotherthings masses ofthenuclides (given intennsofthemassexcess,denotedinthetablesas�;seeProblem2.47),nuclearenergylevels,decayschemes,andsoon.TheNational NuclearDataCenteratBrookhaven NationalLaboratory, Up­ton,NewYork,collects,evaluates,anddistributesawiderangeofnucleardata.Acomprehensivecollectionofdatais available throughthe National PROBLEMS 1. How many neutrons and protons are there in the nuclei ofthe following atoms: (a) 7Li, (b) 24Mg, (c) 135Xe, (d) 209Bi, (e) 222Rn? 2. The atomic weight of 59CO is 58.933 1 9. How many times heavier is 59Co than 12C? 3. How many atoms are there in 10 9 of 12C? 4. Using the data given next and in Example 2.2, compute the molecular weights of (a) H2 gas, (b) H20, (c) H202. Isotope Abundance, a/0 99.985 0.015 Atomic weight 1 .007825 2.01410 5. When H2 gas is formed from naturally occurring hydrogen, what percentages of the molecules have molecular weights of approximately 2, 3, and 4?
  • 61. 46 Atomic and Nuclear Physics Chap. 2 6. Natural uranium is composed of three isotopes: 234U, 235U, and 238U. Their abun­ dances and atomic weights are given in the following table. Compute the atomic weight of natural uranium. Isotope Abundance, a/0 0.0057 0.72 99.27 Atomic weight 234.0409 235.0439 238.0508 7. A beaker contains 50 g of ordinary (i.e., naturally occurring) water. (a) How many moles of water are present? (b) How many hydrogen atoms? (c) How many deuterium atoms? 8. The glass in Example 2. 1 has an inside diameter of 7.5 cm. How high does the water stand in the glass? 9. Compute the mass of a proton in amu. 10. Calculate the mass of a neutral atom of 235U (a) in amu; (b) in grams. 11. Show that 1 amu is numerically equal to the reciprocal of Avogadro's number. 12. Using Eq. (2.3), estimate the radius of the nucleus of 238U. Roughly what fraction of the 238U atom is taken up by the nucleus? 13. Using Eq. (2.3), estimate the density of nuclear matter in g/cm3; in Kg/m3 Take the mass of each nucleon to be approximately 1 .5 x 10-24 g. 14. The planet earth has a mass of approximately 6 x 1024 kg. If the density of the earth were equal to that of nuclei, how big would the earth be? 15. The complete combustion of 1 kg of bituminous coal releases about 3 x 107 J in heat energy. The conversion of 1 g of mass into energy is equivalent to the burning ofhow much coal? 16. The fission of the nucleus of 235U releases approximately 200 MeV. How much en­ ergy (in kilowatt-hours and megawatt-days) is released when 1 g of 235U undergoes fission? 17. Compute the neutron-proton mass difference in MeV. 18. An electron starting from rest is accelerated across a potential difference of 5 million volts. (a) What is its final kinetic energy? (b) What is its total energy? (c) What is its final mass? 19. Derive Eq. (2. 1 8). [Hint: Square both sides ofEq. (2.5) and solve for mv.J 20. Show that the speed of any particle, relativistic or nonrelativistic, is given by the fol­ lowing fonnula:
  • 62. Problems v = c 1 _ E;est 2 ' Etotal 47 where Erest and Etotal are its rest-mass energy and total energy, respectively, and c is the speed oflight. 21. Using theresultderived inProblem2.20, calculate the speed ofa I-MeV electron, one with a kinetic energy of 1 MeV. 22. Compute the wavelengths of a I-MeV (a) photon, (b) neutron. 23. Show that the wavelength of a relativistic particle is given by where AC = h/mec = 2.426 x 10-10 cm is called the Compton wavelength. 24. Using the formula obtained in Problem 2.23, compute the wavelength of a I-MeV electron. 25. An electron moves with a kinetic energy equal to its rest-mass energy. Calculate the electron's (a) total energy in units of mec2; (b) mass in units ofme; (c) speed in units of c; (d) wavelength in units ofthe Compton wavelength. 26. According to Eq. (2.20), a photon carries momentum, thus a free atom or nucleus recoils when it emits a photon. The energy of the photon is therefore actually less than the available transition energy (energy between states) by an amount equal to the recoil energy of the radiating system. (a) Given that E is the energy between two states and Ey is the energy ofthe emitted photon, show that Ey � E (I - �) ' 2Mc where M is the mass of the atom or nucleus. (b) Compute E - Ey forthe transitions from the first excited state ofatomic hydrogen at 10. 1 9 eV to ground and the first excited state of I2C at 4.43 MeV to ground (see Figs. 2.2 and 2.3). 27. The first three excited states ofthe nucleus of 199Hg are at0. 158 MeV, 0.208 MeV, and 0.403 MeV above the ground state. If all transitions between these states and ground occurred, what energy y-rays would be observed? 28. Using the chart ofthe nuclides, complete the following reactions. Ifa daughternucleus is also radioactive, indicate the complete decay chain.
  • 63. 48 (a) 18N� (b) 83y� (c) 135Sb� (d) 219Rn� Atomic and Nuclear Physics Chap. 2 29. Tritium eH) decays by negative beta decay with a half-life of 12.26 years. The atomic weight of 3H is 3.016. (a) To what nucleus does 3H decay? (b) What is the mass in grams of 1 mCi of tritium? 30. Approximately what mass of 90Sr (TI/2 = 28.8 yr.) has the same activity as 1 g of 60Co (TI/2 = 5.26 yr.)? 31. Carbon tetrachloride labeled with 14C is sold commercially with an activity of 10 millicuries per millimole (10 mCilmM). What fraction of the carbon atoms is 14C? 32. Tritiated water (ordinary water containing some 1H3HO) for biological applications can be purchased in 1-cm3 ampoules having an activity of 5 mCi per cm3 What frac­ tion ofthe water molecules contains an 3H atom? 33. After the initial cleanup effort at Three Mile Island, approximately 400,000 gallons of radioactive water remained in the basement of the containment building of the Three Mile Island Unit 2 nuclear plant. The principal sources of this radioactivity were 137Cs at 156 J,tCilcm3 and 134Cs at 26 J,tCilcm3 How many atoms per cm3 of these radionuclides were in the water at that time? 34. One gram of 226Ra is placed in a sealed, evacuated capsule 1.2 cm3 in volume. (a) At what rate does the helium pressure increase in the capsule, assuming all ofthe a-particles are neutralized and retained in the free volume of the capsule? (b) What is the pressure 10 years after the capsule is sealed? 35. Polonium-210 decays to the ground state of 206Pb by the emission of a 5.305-MeV a-particle with a half-life of 138 days. What mass of 21Opo is required to produce 1 MW ofthennal energy from its radioactive decay? 36. The radioisotope generator SNAP-9 was fueled with 475 g of 238puC (plutonium-238 carbide), which has a density of 12.5 g/cm3 The 238Pu has a half-life of 89 years and emits 5.6 MeV per disintegration, all of which may be assumed to be absorbed in the generator. The thennal to electrical efficiency of the system is 5.4%. Calculate (a) the fuel efficiency in curies per watt (thennal); (b) the specific power in watts (thennal) per gram offuel; (c) the power density in watts (thennal) per cm3; (d) the total electrical power of the generator. 37. Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has been steadily decreasing since the earth was fonned about 4.5 billion years ago. How long ago was the isotopic abundance of 235U equal to 3.0 a/o, the enrichment of the uranium used in many nuclear power plants? 38. The radioactive isotope Y is produced at the rate of R atoms/sec by neutron bombard­ ment of X according to the reaction
  • 64. Problems 49 X(n, y)Y. If the neutron bombardment is carried out for a time equal to the half-life of Y, what fraction of the saturation activity of Y will be obtained assuming that there is no Y present at the start of the bombardment? 39. Consider the chain decay A � B � C � , with no atoms of B present at t = O. (a) Show that the activity of B rises to a maximum value at the time tm given by at which time the activities ofA and B are equal. (b) Show that, for t < tm, the activity of B is less than that ofA, whereas the reverse is the case for t > tm. 40. Show that ifthe half-life ofB is much shorter than the half-life ofA, then the activities ofA and B in Problem 2.39 eventually approach the same value. In this case, A and B are said to be in secular equilibrium. 41. Show that the abundance of 234U can be explained by assuming that this isotope orig­ inates solely from the decay of 238U. 42. Radon-222, a highly radioactive gas with a half-life of 3.8 days that originates in the decay of234U (see the chart ofnuclides), may bepresent in uranium mines in dangerous concentrations ifthe mines are not properly ventilated. Calculate the activity of222Rn in Bq per metric ton of natural uranium. 43. Accordingto U.S. NuclearRegulatory Commission regulations, the maximum pennis­ sible concentration of radon-222 in air in equilibrium with its short-lived daughters is 3 pCiIliter for nonoccupational exposure. This corresponds to how many atoms of radon-222 per cm3? 44. Consider again the decay chain in Problem 2.39 in which the nuclide A is produced at the constant rate ofR atoms/sec. Derive an expressionforthe activity ofB as a function of time. 45. Complete the following reactions and calculate their Q values. [Note: The atomic weight of 14C is 14.003242.] (a) 4He(p, d) (b) 9Be(a, n) (c) 14N(n, p) (d) 1 15In(d, p) (e) 207Pb(y, n) 46. (a) Compute therecoil energyoftheresidual, daughternucleus following theemission of a 4.782-MeV a-particle by 226Ra. (b) What is the total disintegration energy for this decay process?
  • 65. 50 Atomic and Nuclear Physics Chap. 2 47. In some tabulations, atomic masses are given in terms of the mass excess rather than as atomic masses. The mass excess, �, is the difference � = M - A, where M is the atomic mass and A is the atomic mass number. For convenience, �, which may be positive or negative, is usually given in units ofMeV. Show that the Q value for the reaction shown in Eq. (2.38) can be written as 48. According to the tables of Lederer et al. (see References), the mass excesses for the (neutral) atoms in the reaction in Example 2.8 are as follows: �eH) = 14.95 MeV, �eH) = 13. 14 MeV, �(n) = 8.07 MeV, and �(4He) = 2.42 MeV. Calculate the Q value of this reaction using the results of Problem 2.47. 49. The atomic weight of 206Pb is 205.9745. Using the data in Problem 2.35, calculate the atomic weight of 21Opo. [Caution: See Problem 2.46] 50. Tritium eH) can be produced through the absorption of low-energy neutrons by deu­ terium. The reaction is 2H + n -+3 H + y, where the y-ray has an energy of 6.256 MeV. (a) Show that the recoil energy of the 3H nucleus is approximately 7 keV. (b) What is the Q value of the reaction? (c) Calculate the separation energy of the last neutron in 3H. (d) Using the binding energy for2H of2.23 MeV and the result from part (c), compute the total binding energy of 3H. 51. Consider the reaction Using atomic mass data, compute: (a) the total binding energy of 6Li, 9Be, and 4He; (b) the Q value ofthe reaction using the results of part (a). 52. Using atomic mass data, compute the average binding energy per nucleon of the fol­ lowing nuclei: (a) 2H (b) 4He (c) 12C (d) 5lV (e) 138Ba (0 235U 53. Using the mass formula, compute the binding energy per nucleon for the nuclei in Problem 2.52. Compare the results with those obtained in that problem.
  • 66. Problems 54. Compute the separation energies of the last neutron in the following nuclei: (a) 4He (b) 7Li (c) 170 (d) 5ly (e) 208Pb (f) 235U 51 55. DeriveEq. (2.53). [Hint: Try taking the logarithm ofEq. (2.52) before differentiating.] 56. What is 1 atmosphere pressure in units of eY/cm3? [Hint: At standard temperature and pressure (ODC and 1 atm), 1 mole of gas occupies 22.4 liters.] 57. Calculate the atom density ofgraphite having density of 1.60 glcm3 58. Calculate the activity of 1 gram of natural uranium. 59. What is the atom density of 235U in uranium enriched to 2.5 alo in this isotope if the physical density of the uranium is 19.0 g/cm3? 60. Plutonium-239 undergoes a-decay with a half-life of 24,000 years. Compute the ac­ tivity of 1 gram ofplutonium dioxide, 239PU02. Express the activity in terms ofCi and Bq. 61. It has been proposed to use uranium carbide (UC) for the initial fuel in certain types ofbreeder reactors, with the uranium enriched to 25 wlo. The density of UC is 13.6 g/cm3 (a) What is the atomic weight of the uranium? (b) What is the atom density ofthe 235U? 62. Compute the atom densities of 235U and 238U in U02 of physical density 10.8 g/cm3 if the uranium is enriched to 3.5 wi0 in 235U. 63. The fuel for a certain breeder reactor consists of pellets composed of mixed oxides, U02 and PU02, with the PU02 comprising approximately 30 wi0 ofthe mixture. The uranium is essentially all 238U, whereas the plutonium contains the following isotopes: 239Pu (70.5 wlo), 240pU (21 .3 wlo), 24lpU (5.5 wlo), and 242Pu (2.7 wlo). Calculate the number of atoms of each isotope per gram of the fuel.
  • 67. 3 Interaction of Radiation with Matter Thedesignofallnuclearsystems-reactors,radiationshields,isotopicgenerators,andsoon-dependsfundamentallyonthewayinwhichnuclearradiationinteractswithmatter.Inthischapter, theseinteractionsarediscussedforneutrons, y-rays,andvariouschargedparticleswithenergiesuptoabout20MeV.Mostoftheradi­ationencounteredinpracticalnucleardevicesliesinthisenergyregion. 3.1 NEUTRON INTERACTIONS Itisimportanttorecognizeattheoutsetthat,sinceneutronsareelectricallyneutral,theyarenotaffectedbytheelectrons inanatomorbythepositivechargeofthenucleus. As aconsequence, neutrons pass throughtheatomicelectroncloud andinteractdirectlywiththenucleus. Inshort,neutrons collide withnuclei,notwithatoms.Neutronsmayinteractwithnucleiinoneormoreofthefollowingways. Elastic Scattering Inthisprocess,theneutronstrikesthenucleus,whichisalmostalwaysinitsgroundstate(seeSection2.7),theneutronreappears,andthe 52
  • 68. Sec. 3.1 Neutron Interactions 53 nucleusisleftinitsgroundstate.Theneutroninthiscaseissaidtohavebeenelas­ tically scattered bythenucleus. Inthenotationofnuclearreactions (see Section2.10),thisinteractionisabbreviatedbythesymbol(n,n). Inelastic Scattering Thisprocessisidenticaltoelasticscatteringexceptthatthenucleusisleftinanexcitedstate.Becauseenergyisretainedbythenucleus,this is clearly an endothermic interaction. Inelastic scattering is denoted by thesymbol (n, n'). Theexcitednucleus decays, asexplainedin Section 2.7, bytheemissionofy-rays.Inthiscase,sincethesey-raysoriginateininelasticscattering,theyarecalledinelastic y-rays. Radiative Capture Heretheneutroniscapturedbythenucleus,andoneormorey-rays--calledcapture y-rays-are emitted. Thisisanexothermicinter­actionandisdenotedby(n,y).Sincetheoriginalneutronisabsorbed,thisprocessisanexampleofaclassofinteractionsknownasabsorption reactions. Charged-Particle Reactions Neutronsmayalsodisappearastheresultofabsorptionreactionsofthetype(n,ex) and(n,p). Suchreactionsmaybeeitherexothermicorendothermic. Neutron-Producing Reactions Reactionsofthetype(n,2n)and(n,3n)occurwithenergeticneutrons.Thesereactionsareclearlyendothermic sinceinthe(n, 2n) reaction one neutronandinthe (n, 3n) reaction 2 neutrons areextractedfromthe strucknucleus. The (n, 2n) reaction is especially important inreactorscontainingheavywaterorberylliumsince2Hand9Behavelooselyboundneutronsthatcaneasilybeejected. Fission Neutrons colliding withcertainnuclei may causethe nucleus tosplitapart-toundergofission. Thisreaction,asnotedinChap. 2, istheprincipalsourceofnuclearenergyforpracticalapplications.Manyofthesereactionsmaybeviewedasatwo-stepprocessinvolvingtheformation of a compound nucleus. For example, the scattering reactions, bothelastic andinelastic, maybethoughtofasaprocess in which theneutronis firstabsorbedbythetargetnucleustoformanewnucleuswhoseatomicnumberisun­changedbutwhosemassnumberisincreasedby 1.Thendependingonthespecificprocess,thenucleusdecaysvianeutronemission toproducetheoriginalnucleusplusaneutron.Theproductnucleusisleftinthegroundstateoranexcitedstateaccording to the type ofscattering reaction involved. This model is particularlyusefulinunderstandingthefissionprocess.Eachoftheseinteractionsisdiscussedinthischapter. However,todescribequantitativelythevariousinteractions,itisnecessarytointroducecertainparame­ters.
  • 69. 54 Interaction of Radiation with Matter Chap. 3 3.2 CROSS-SECTIONS Theextenttowhichneutronsinteractwithnucleiisdescribedintermsofquantitiesknownascross-sections. Thesearedefinedbythefollowingtypeofexperiment.Supposethatabeamofmonoenergetic(single-energy)neutronsimpingesonathintargetofthickness X andareaA asshowninFig.3.1 . Iftherearen neutronspercm3inthebeamandv isthespeedoftheneutrons,thenthequantity 1 = nv (3.1) iscalledtheintensity ofthebeam. Sincetheneutronstravelthedistance v cmin1 sec,alloftheneutronsinthevolumevA infrontofthetargetwillhitthetargetin 1 sec. Thus, nvA = IA neutrons strike the entire target per second, and itfollowsthatIA/A = I isequaltothenumberofneutrons strikingthetargetpercm2/sec. Sincenuclei are small andthetargetis assumedtobethin, most oftheneutrons strikingthetargetinanexperimentlikethatshowninFig.3.1 ordinarilypassthroughthetargetwithoutinteractingwithanyofthenuclei.Thenumberthatdocollidearefoundtobeproportionaltothebeamintensity, totheatomdensityN ofthetarget,andtotheareaandthicknessofthetarget.Theseobservationscanbesummarizedbytheequation Numberofcollisionspersecond(inentiretarget)= a INAX, (3.2) wherea, theproportionalityconstant,iscalledthecross-section. ThefactorNAX inEq.(3.2) isthetotalnumberofnucleiinthetarget.Thenumberofcollisionspersecondwithasinglenucleus isthereforejusta I. Itfollowsthata isequaltothenumberofcollisionspersecondwithonenucleusperunitintensityofthebeam.Thereisanotherwaytoviewtheconceptofcross-section.Asalreadynoted,IA neutrons strikethe target per second. Ofthese, a I interactwith any givennucleus.Therefore,itmaybeconcludedthat a I a = AI A Neutrons in beam ,- - , - I -' I " ' I , / / �I.---------- v----------� Figure 3.1 Neutron beam striking a target. (3.3) -- Area A - - Thickness X
  • 70. Sec. 3.2 Cross-Sections 55 isequaltotheprobabilitythataneutroninthebeamwillcollidewiththisnucleus.ItisclearfromEq.(3.3) thata hasunitsofarea.Infact,itisnotdifficulttoseethata isnothingmorethantheeffectivecross-sectionalareaofthenucleus,hencethetenncross-section. Neutroncross-sectionsareexpressedinunitsofbarns, where 1 barn,abbre­viated b, is equal to 10-24 cm2 One thousandth ofabam is called amillibarn, denotedasmb.Uptothispoint,ithasbeenassumedthattheneutronbeamstrikestheentiretarget. However, inmany experiments, the beam is actually smaller in diameterthanthetarget. Inthiscase,thepriorfonnulasstill hold, butnow.A referstotheareaofthebeaminsteadoftheareaofthetarget.Thedefinitionofcross-sectionremainsthesame,ofcourse.Each ofthe processesdescribedin Section 3.1 by whichneutrons interactwith nuclei is denoted by a characteristic cross-section. Thus, elastic scatteringis describedbythe elastic scatteringcross-section, ae ; inelastic scatteringbytheinelastic scattering cross-section, ai ; the (n, y) reaction (radiative capture) bythecapturecross-section, ay ; fissionby the fissioncross-section, af ; andsoon.The sumofthecross-sections forallpossible interactions is known asthetotal cross-section andisdenotedbythesymbolat ; thatis (3.4) Thetotalcross-sectionmeasurestheprobabilitythataninteractionofanytypewilloccurwhenneutronsstrikeatarget.Thesumofthecross-sectionsofallabsorptionreactionsisknownastheabsorption cross-section andisdenotedbyaa . Thus, (3.5) where ap and aa are the cross-sections for the (n, p) and (n, a) reactions. AsindicatedinEq. (3.5), fission,byconvention, istreatedasanabsorptionprocess.Similarly,thetotal scatteringcross-sectionisthesumoftheelasticandinelasticscatteringcross-section.Thus, and Example 3.1 A beam of I-Me V neutrons of intensity 5 x 108 neutrons/cm2-sec strikes a thin 12C target. The area of the target is 0.5 cm2 and is 0.05 cm thick. The beam has a cross­ sectional area of 0.1 cm2 At 1 MeV, the total cross-section of 12C is 2.6 b. (a) At what rate do interactions take place in the target? (b) What is the probability that a neutron in the beam will have a collision in the target?
  • 71. 56 Interaction of Radiation with Matter Chap. 3 Solution 1. According to Table I.3 in Appendix II, N = 0.080 X 1024 for carbon. Then from Eq. (3.2), the total interaction rate is atINAX = 2.6 X 10-24 x 5 X 108 x 0.080 X 1024 x 0.1 x 0.05 = 5.2 x 105 interactions/sec. [Ans.] It should be noted that the 10-24 in the cross-section cancels with the 1024 in atom density. This is the reason for writing atom densities in the form of a number x 1024 2. In 1 sec, IA = 5 X 108 x 0.1 = 5 x 107 neutrons strike the target. Of these, 5.2 x 105 interact. The probability that a neutron interacts in the target is there­ fore 5.2 x 105/5 X 107 = 1 .04 X 10-2 Thus, only about 1 neutron in 100 has a collision while traversing the target. Example 3.2 Thereareonly two absorption reactions-namely, radiative capture and fission-that can occur when 0.0253-e V neutronsl interact with 235U. The cross-sections for these reactions are 99 b and 582 b, respectively. When a 0.0253-e V neutron is absorbed by 235U, what is the relative probability that fission will occur? Solution. Since ay and af are proportional to the probabilities of radiative capture and fission, it follows that the probability of fission is af/(ay + af) = af/aa = 582/681 = 85.5%. [Ans.] ToreturntoEq.(3.2), thiscanbewrittenas Numberofcollisionspersecond(inentiretarget)= INat x AX, (3.6) whereat hasbeen introducedbecausethis cross-sectionmeasurestheprobabilitythatacollisionofanytypemayoccur. Since AX isthetotalvolumeofthetarget,itfollowsfromEq. (3.6) thatthe numberofcollisionspercm3/secinthetarget,whichiscalledthecollision density F, isgivenby (3.7) TheproductoftheatomdensityN andacross-section,asinEq.(3.7), occursfrequentlyintheequationsofnuclearengineering;itisgiventhespecialsymbolb andis calledthemacroscopic; cross-set;tiQn. III particular,theproduct Nat = bt iscalledthemacroscopic total cross-section, Nas = bs iscalledthemacroscopic scattering cross-section, andsoon. Since N anda haveunitsofcm-3 andcm2, I For reasons explained in Section 3.6, neutron cross-sections are tabulated at an energy of 0.0253 e V.
  • 72. Sec. 3.3 Neutron Attenuation 57 respectively, L hasunitsofcm- Intermsofthemacroscopiccross-section, thecollisioninEq. (3.7)reducesto (3.8) Example 3.3 Referring to Example 3.1, calculate the (a) macroscopic total cross-section of 12C at Me V; (b) collision density in the target. Solution 1. From the definition given previously, bt = 0.08 X 1024 x 2.6 X 10-24 = 0.21 cm-l [Ans.] 2. Using Eq. (3.8) gives F = 5 X 108 x 0.21 = 1 .05 x 108 collisions/cm3-sec. [Ans.] These collisions occur, ofcourse, only in the region ofthe target that is struck by the beam. 3.3 NEUTRON ATTENUATION Theprecedingsectionreferstoexperimentsinvolvingthintargets.Supposeathicktargetofthickness X is placed in amonodirectionalbeam ofintensity 10 and aneutron detector is located at some distance behind the target as shown in Fig.3.2. Itis assumedthatboth thetargetand detectorare sosmallthatthe detectorsubtends a small solid angle atthe target. Foraneutron to reach the detector, itmustbetravelingdirectlytowardboththetargetanddetector.Consequently,everyneutron that has a collision in the target is lostfromthe beam, and only thoseneutronsthatdonotinteractenterthedetector. Incident neutrons Scattered Uncollided neutrons Detector Figure 3.2 Measurement of neutrons that have not collided in a target.
  • 73. 58 Interaction of Radiation with Matter Chap. 3 LetI(x) betheintensityoftheneutronsthathavenot collidedafterpenetrat­ingthedistancex intothetarget.Thenintraversinganadditionaldistance dx, theintensityoftheuncollidedbeamis decreasedbythenumberofneutronsthathavecollidedinthethin sheetoftargethaving anareaof1 cm2 andthethickness dx. FromEq. (3.2), thisdecreaseinintensityisgivenby -dI(x) = NatI(x) dx = L-tI(x) dx. (3.9) Thisequationcanbeintegratedwiththeresult I(x) = loe-Etx (3.10) Theintensityoftheuncollidedneutronsthusdecreasesexponentiallywiththedis­tanceinsidethetarget.Theintensityofthebeamofuncollidedneutronsemergingfromthetargetisthen (3.1 1) andthisistheintensitymeasuredbythedetector.Ifthetarget is very thick, as with a radiation shield, then almostall oftheincidentneutrons willhaveatleastonecollisioninthetarget,sothatmostoftheemerging neutrons will have undergone scattering within the target. Since theseneutronsarespecificallyexcludedinthederivationofEq.(3.1 1), thisequationmustnotbeusedtocalculatetheeffectivenessofashield.Todosowouldignoreamostimportantcomponentoftheemergentradiation-namely,thescatteredneutrons.WhenEq. (3.9) isdividedbyI(x), theresultis dl(x) --- = L-t dx. I(x) (3.12) SincethequantitydI (x) is equaltothenumberofneutronsoutofatotalofI(x) thatcollidein dx, itfollowsthatdI(x)/I(x) istheprobabilitythataneutronwhichsurvivesuptox withoutacollisioninteractsinthenext dx. Therefore,fromEq.(3.12), L-t dx is equaltotheprobabilitythataneutronwillinteractin dx, anditmaybeconcludedthatL-t istheprobabilityperunitpathlengththataneutronwillundergosomesortofacollisionasitmovesaboutinamedium.Itshouldalsobenotedthat,sinceI(x) referstothoseneutronsthathavenotcollidedinpenetratingthe distance x, the ratio I(x)/Io = e-Etx is equal totheprobabilitythataneutroncanmovethroughthisdistancewithouthavingacolli­sion.Nowletthequantity p(x) dx betheprobabilitythataneutronwillhaveitsfirstcollisionin dx intheneighborhoodofx. Thisisevidentlyequaltotheproba­bilitythattheneutronsurvivesuptox withoutacollisiontimestheprobabilitythatitdoesinfaeteollideintheadditionaldistance dx. Since L-t istheprobabilityofinteractionperpathlength,p(x) dx isgivenby
  • 74. Sec. 3.3 Neutron Attenuation p(x) dx = e-I:tx x �t dx = �te-I:tX dx. 59 Theaveragedistancethataneutronmovesbetweencollisions is calledthemeanfree path. Thisquantity,whichisdesignatedbythesymbolA, isequaltotheaveragevalueofx, thedistancetraversedbyaneutronwithoutacollision.ThevalueofA isobtainedfromtheprobabilityfunctionp(x) bydeterminingtheaveragedistancetraveled,thatis, A = 100xp(x) dx = �t 100xe-E,x dx (3.13) Example 3.4 Calculate the mean free path of 100-keV neutrons in liquid sodium. At this energy, the total cross-section of sodium is 3.4 b. Solution. From Table 11.3, the atom density of sodium is 0.0254 x 1024 The macro­ scopic cross-section is then bt = 0.0254 X 1024 x 3.4 X 10-24 = 0.0864 cm-l The mean free path is therefore A = 1/0.0864 = 1 1 .6 cm [Ans.J Considernextahomogeneousmixtureof2nuclearspecies, X and Y, con­tainingNxandNy atoms/cm3ofeachtype,andletax andaybethecross-sectionsofthe2nucleiforsomeparticularinteraction.Theprobabilityperunitpaththataneutroncollideswithanucleusofthefirsttypeis �x = Nxax andwithanucleusofthesecondtypeis�y = Nyay. Thetotalprobabilityperunitpaththataneutroninteractswitheithernucleusistherefore (3.14) Ifthe nuclei arein atoms thatarebound together in amolecule, Eq. (3.14) canbe used to define anequivalent cross-sectionforthe molecule. This is donesimplybydividingthemacroscopiccross-sectionofthemixturebythenumberofmoleculesperunitvolume.Forexample,ifthereareN moleculesXm Yn percm3,thenNx = mN, Ny = nN, andfromEq.(3.14) thecross-sectionforthemoleculeis � a = - = max + nay. N (3.15) Equations (3.14) and (3.15) are based on the assumptionthatthe nuclei X and Y actindependently ofone another when they interact with neutrons. This
  • 75. 60 Interaction of Radiation with Matter Chap. 3 is correctforall neutron interactions except elastic scattering by molecules andsolids.Low-energyscatteringcross-sectionsforsuchsubstancesmustbeobtainedfromexperiment. Example 3.5 The absorption cross-section of 235U and 238U at 0.0253 eV are 680.8 b and 2.70 b, respectively. Calculate I:a for natural uranium at this energy. Solution. By use of the methods of Section 2.14, the atom densities of 235U and 238U in natural uranium are found to be 3.48 x 10-4 x 1024 and 0.0483 x 1024, respectively. Then from Eq. (3.14) I:a = 3.48 X 10-4 x 680.8 + 0.0483 x 2.70 = 0.367 cm-1 [Ans.] Example 3.6 The scattering cross-sections (in barns) of hydrogen and oxygen at 1 Me V and 0.0253 eV are given in the following table. What are the values of as for the wa­ ter molecule at these energies? Solution. Equation (3.15) applies at 1 Me V, so that as(H20) = 2asCH) +as(O) = 2 x 3 + 8 = 14 b. [Ans.] Equation (3.15) does not apply at 0.0253, and as(H20) =I=- 2 x 21 + 4 = 46 b. The experimental value of as(H20) at 0.0253 eV is 103 b. [Ans.] 3.4 NEUTRON FLUX H o I Me V 3 8 0.0253 e V 21 4 Itwas showninSection3.2 thatwhenabeamofneutronsofintensity I strikes athintarget,thenumberofcollisionspercm3/secisgivenby F = L.tI, (3.16) whereL.t isthemacroscopictotalcross-section.ConsideranexperimentofthetypeshowninFig.3.3, inwhichasmalltargetisexposedsimultaneouslytoseveral neutronbeams.Theintensitiesofthebeamsaredifferent,butitisassumedthattheneutronsinallofthebeamshavethesameenergy.Inviewofthefactthattheinteractionofneutronswithnucleiisindepen­dentoftheangleatwhichtheneutronscollidewiththenuclei,thetotalinteractionrateisclearly F = L.t(IA + IE + Ie + . . ) . (3.17)
  • 76. Sec. 3.4 Neutron Flux :�--'[-ar-ge--: Figure 3.3 Neutron beams striking a target. 61 Theneutronshavebeenassumedtobemonoenergetic.SofromEg.(3.1), thismayalsobewrittenas )v, (3.18) wherenA , nB , andsoonarethedensitiesoftheneutronsinthevarious beamsand v istheneutronspeed.SincenA +nB + nc + isequalton, thetotaldensityofneutronsstrikingthetarget,Eq.(3.18) becomes (3.19) Thesituationatanypointinareactorisageneralizationofthisexperiment,butwiththeneutronsmovinginall directions.ItfollowsthatEq.(3.19) isvalidforareactor,wheren istheneutrondensityatthepointwhereF iscomputed.Thequantitynv inEq. (3.19) iscalledtheneutronflux, inthiscaseformo­noenergeticneutrons,andisgiventhesymbol¢;thus ¢= nv. (3.20) It is evident that the units ofneutron flux are the same as the units of beamintensity-namely,neutrons/cm2-sec.Intenusoftheflux,thecollisiondensityis (3.21) Tounderstandtheimportanceofthefluxandthisrelationship, considerthefollowingexample. Example 3.7 A certain research reactor has a flux of 1 x 1013 neutrons/cm2-sec and a volume of 64,000 cm3 If the fission cross-section, :E[, in the reactor is 0.1 cm-1 , what is the power of the reactor? Solution. The power may be obtained from the fission rate using the relationship between the energy released per fission (200 Me V) and the rate at which fissions are occuring:
  • 77. 62 Power Interaction of Radiation with Matter Chap. 3 1 MW 1 W-sec 1 .60 x 10-13 joule 200 Me V fi ' --- --- x x x SSlon rate 106 watt joules Me V fission I MW 6 x 3.2 X 10-11 watts/fission/sec x fission rate 10 watt = 3.2 x 10-17 MW/fissionlsec x fission rate From Eq. (3.21), the fission rate is Fission rate = br<l> = 0. 1 cm-I x 1 x 1013 neutrons/cm2-sec. The reactor power/cm3 is then Power/cm3 = 3.2 x 10-17 MW/fission/sec x 1 x 1012 fission/sec-cm3 The total power is the power/cm3 times the volume of the active core. Power = 3.2 x 10-5 MW/cm3 x 64, 000 cm3 = 2 MW [Ans.] 3.5 NEUTRON CROSS·SECTION DATA Allneutroncross-sectionsarefunctionsoftheenergyoftheincidentneutronsandthenatureofthetargetnucleus.Thesefactorsmustbetakenintoconsiderationinthechoiceofmaterialsforuse innucleardevices. Mostofthecross-sectiondataneededforsuchpurposesarefoundintheBrookhavenNationalLaboratoryreportBNL-325 and other source, which are discussed in the references atthe end ofthe chapter. Before turning to the data, however, it is ofinterest to considerthemechanismsbywhichneutronsinteractwithnuclei. Compound Nucleus Formation Mostneutroninteractionsproceedin2 steps.Theincidentneutron,onstrikingthetargetnucleus,firstcoalesceswithittoformacompoundnucleus.IfthetargetnucleusisA Z, thecompoundnucleusisA+l Z. Thecompoundnucleusmaythendecayinanumberofways.Porexample,when IO-MeVneutronsstrikean56Petarget,thecompoundnucleusis57Pe,andthisnucleusmaydecaybyemittinganelasticorinelasticneutron,ay-ray,ortwoneutrons.Insymbols,th€S€proc€ssesare 56Pe+ n(elasticscattering) 56 57 *�56Fe+ n' (inelasticscattering)Fe+ n---+ ( Fe) . . 57Pe+ y (radIatIvecapture) 55Pe+ 2n(n,2nreaction).
  • 78. Sec. 3.5 Neutron Cross-Section Data 63 Oneofthestrikingfeaturesofinteractionsthatproceedbywayofcompoundnucleus formation is thattheircross-sections exhibit maxima atcertainincidentneutronenergies. Suchmaximaarecalled resonances andarise inthe followingway.ItisrecalledfromSection 2.7 thatnucleihavevariousexcitedstatescorre­spondingtodifferentconfigurationsofnucleonsinthenucleus.Itturnsoutthattheincidentneutronandtargetnucleusaremorelikelytocombine andformacom­poundnucleusiftheenergy oftheneutronis suchthatthecompoundnucleus isproducedinoneofitsexcitedstates.Theresonancesshowupinthecross-sectionbecauseitisnecessarytoformthecompoundnucleusbeforetheinteractioncanproceed.It is recalledfrom Section 2.1 1 that it takes energy-namely, the neutronseparationenergy-toremove aneutronfromanucleus. This separation energyreappears, however, whentheneutronreentersthenucleus.Therefore,itfollowsthatwhenaneutroncollides withanucleus, thecompoundnucleus isformedinanexcitedstatehavinganenergyequaltothekineticenergyoftheincidentneu­tronplustheseparationenergyorbindingenergyoftheneutroninthecompoundnucleus.2 Elastic Scattering The elastic scattering cross-section as afunction oftheenergyoftheincidentneutroncanbedividedintothreedistinctregions.Inthefirst,low-energyregion,rre isapproximatelyconstant.Thescatteringinthisregiondoesnotoccurbycompoundnucleusformation,butmerelybecauseoftheforcesexertedby thetarget nucleus on the passing neutron. The cross-sectionforthispotentialscatteringisgivenby rre (potentialscattering)= 4nR2, (3.22) whereRisthenuclearradius.Beyondthepotentialscatteringregion,thereisaregionofresonancesthatisduetocompoundnucleusformation.Atstillhigherenergies,theresonancescrowdtogethertosuchanextentthattheindividualresonancescannolongerberesolved;inthisregion,rre isasmoothandslowlyvaryingfunctionofenergy.Figure3.4 showsthesethreeregionsforthetargetnucleuscarbon.Carbonisarelativelylightnucleus.Withheaviernuclei,theregionofresonancesisfoundatlowerenergies.Forexample,theresonanceregionof238Ubeginsatonly6 eVandendsatroughly 1 keV. Example 3.8 Using experimental elastic scattering data, estimate the radius ofthe C nucleus. 2This discussion is somewhat simplified, center-of-mass effects having been ignored. For a more complete discussion, see Introduction to Nuclear Reactor Theory, noted in the references.
  • 79. 64 �e 0 ,S; U 11) rn '"'" 2 u Interaction of Radiation with Matter Chap. 3 Elastic and Total Cross Sections C MT = 1 2 Constant to 0.015 eV Potential scattering 10° Resonance region Smooth region 10-1 ����������������������������� 10-I 100 10I 102 103 104 105 106 107 108 Neutron Energy (eV) Figure 3.4 The elastic scattering and total cross-section of carbon. (Plotted from data received over the Internet from the Korean Atomic Energy Research Institute using ENDFPLOT and ENDFIB 6.1.) Solution. From Fig. 3.4, it is observed that ae has the constant value of 4.8 b from about 0.02 eV to 0.01 Me V. This is due to potential scattering. Then from Eq. (3.22), 41l' R2 = 4.8 X 10-24 and R = 6.2 X 10-13 cm. [Ans.] Inelastic Scattering Thisprocessdoesnotoccurunlesstheneutronhassufficientenergytoplacethetargetnucleusinitsfirstexcitedstate.3 Asaresult,ai iszeroupto somethresholdenergy. Generallyspeaking,theenergyatwhichthefirstexcitedstateisfounddecreaseswithincreasingmassnumber.Asaconse­quence,ai isnonzerooveralargerenergyregionfortheheaviernucleithanforthelighternuclei.Thethresholdforinelasticscatteringis4.80 MeVforC,whereasitisonly44 keYfor238U.Atenergieswellabovethreshold,ai isroughlyequaltoas . Radiative Capture As inthe case ofelastic scattering, it is convenienttodividetheradiativecapturecross-sectionintothreeregions. Inthelow-energy 3Because of center-of-mass effects, the threshold energy for inelastic scattering is actually somewhat higher than the energy of the first excited state. Except for the very light nuclei, however, this can be ignored.
  • 80. Sec. 3.5 104 I 103 e c 0 102.;: (.) � CI) rJ'J rJ'J 2 101U 10° Neutron Cross-Section Data Radiative Capture Cross Section Au- 197 MT = 102 65 10-1 ������--����������--�������� 10-3 10-2 10-1 100 101 Neutron Energy (eV) Figure 3.5 The radiative capture cross-section ofAu-197 atlow energy. (From ENDFIB 6 plotted over the Internet using ENDFPLOT from the Korean Atomic Energy Research Institute.) 10 regionofmostnuclei,ay variesat1/y'E, whereE istheneutronenergy.Sincethe neutronspeedv isproportionaltoy'E, thismeansthatay variesas 1/v. Thelow­energyregionofay is thereforeknownasthe l/v region. Neutroncross-sectionsareoftenplottedonalog-logscale,andacross-sectionthatis 1/ v thenappearsasastraightlinewithaslopeof-1/2. ThiscanbeseeninFig.3.5, inwhichthe 1/v regionandthefirstresonanceareshownfor197 Au.Forafewimportantnuclei,cry doesnotshowexact1/v behavioratlowenergy,andsuchnucleiarecallednon-1/v absorbers. Abovethe 1/v region,thereisaregionofresonancesthatoccursatthesameenergiesastheresonancesinas . NearanisolatedresonanceattheenergyEr, ay isgivenbytheBreit-Wignerone-levelfonnula: Yr2g rn rg a - -- ----�--- y - 4Jr (E - Er)� + r2/4 . (3.23) Inthisexpression,Yr isthewavelengthofneutronswithenergyEr, g isaconstantknownas the statisticalfactor, rn and rg areconstants called,respectively,theneutron width andtheradiation width, andr = rn + ry iscalledthetotal width. Itis easy to show thatay falls toone halfofits maximumvalue attheenergies
  • 81. 66 Interaction of Radiation with Matter Chap. 3 Er ± r/2. Inshort, r isthewidthoftheresonanceatonehalfitsheight,andthisistheoriginofthetermwidth. Abovetheresonancesregion, whichendsatabout1 keVintheheavynucleiandatincreasinglyhigherenergiesinlighternuclei,O"y dropsrapidlyandsmoothlytoverysmallvalues. Charged-Particle Reactions As a rule, the (n, p), (n, a), and othercharged-particlereactionsareendothermicanddonotoccurbelowsomethresholdenergy.Theircross-sectionsalsotendtobesmall,evenabovethreshold,especiallyfortheheaviernuclei.However,therearesomeimportantexothermicreactionsinlightnuclei.Oneoftheseisthereaction lOB(n,a)7Li,thecross-sectionofwhichisshowninFig.3.6. ItisobservedthatO"a isverylargeatlowenergy;forthisreason, lOB isoftenusedtoabsorblow-energyneutrons. Itshould alsobenotedinFig. 3.6 thatO"a is1/v overseveralordersofmagnitudeinenergy.A similar exothermic reaction that also shows a strong 1/v behavior is6Li(n,a)3H.Thisreactionisusedfortheproductionoftritium,3H.Someendothermiccharged-particlereactionsareimportantinreactorseventhoughtheirthresholds arehigh.Inwaterreactors,forexample, the 160(n,p)16N �e- o 00;:l u � tI:l '" '" 0 .... U 103 102 101 n- alpha Cross Section B-lO MT 107 100 �����������L-�������������� 10-2 10-1 100 101 102 103 104 105 Neutron Energy (eV) Figure 3.6 The cross-section for B-I0, n-alpha reaction from O.Ol eV to 10,000 eY. (From ENDFIB 6 using ENDFPLOT from the Korean Atomic Energy Research Institute over the Internet.)
  • 82. Sec. 3.5 Neutron Cross-Section Data 67 reaction istheprincipal sourceoftheradioactivity ofthewater(the 16N under­goes�-decay, withahalf-lifeofapproximately7 secs, which is accompaniedbytheemissionof6- to7-MeVy-rays),despitethatordinarilyonlyoneneutroninseveralthousandhasanenergygreaterthanthe9-MeVthresholdforthisreaction. Total Cross-Section Sinceat isthesumofalltheothercross-sections,thevariationofat withenergyreflectsthebehavioroftheindividualcomponentcross-sections.Inparticular,atlowenergy,at behavesas 2 C at = 4rrR + ,..fE' (3.24) where C is a constant. The first tenn in this expression is the cross-section forelasticscattering; the secondtenn gives thecross-sectionforradiativecaptureorwhateverotherexothennicreaction is possible atthisenergy. IfthefirsttenninEq.(3.24)ismuchlargerthanthesecond,thenat isaconstantatlowenergy;ifthesecondtenndominates,at is lIvinthisenergyregion.Intheresonanceregion,at exhibitstheresonancesfoundinas andaj , allofwhichoccuratthesameenergiesineachofthesecross-sections.Athigherenergiesabovetheresonanceregion, at becomes asmoothandrollingfunctionofenergy,asshowninFig.3.4. Hydrogen and Deuterium The nuclei 1H and 2H, which arepresentinlargeamounts inmanynuclearreactors, interactwithneutrons inasomewhatdifferent manner from other nuclei. For one thing, interactions with 1H and 2Hdonotinvolvethefonnationofacompoundnucleus. They also donothaveanyresonances. Thecross-section, as , is constantup to 10 KeV, anday is l/v at allenergies.Furthennore, these nucleihavenoexcited states e His,afterall, only asingleproton),andsoinelasticscatteringdoesnotoccur. Example 3.9 The value of O" y for 1H at 0.0253eV is 0.332b. What is O" y at 1 eV? Solution. Since O" y is 1/v, it can be written as O" y (E) = O" y (Eo) !Eo 'I E' where Eo is any energy. In this problem, O" y is known as 0.0253 eV and so it is reasonable to take this to be the value of Eo. Then )0.0253O" y (Ie V) = 0.332 x -1- = 0.0528b. [Ans.]
  • 83. 68 E, p O�---i� Incident neutron Interaction of Radiation with Matter " E ):X/_ _ _ _ - Scattered neutron Tillge�;���----------------------' nucleus ' " " Recoiling nucleus E A' P ' � Figure 3.7 Elastic scattering of a neutron by a nucleus. 3.6 ENERGY LOSS IN SCATTERING COLLISIONS Chap. 3 Whenaneutroniselasticallyscatteredfromanucleusatrest, thenucleusrecoilsfromthesiteofthecollision.Thekineticenergyofthescatteredneutronisthere­ fore smallerthan the energy ofthe incident neutron by an amount equal to theenergyacquiredbytherecoilingnucleus.Inthisway,neutronsloseenergyinelas­ticcollisionseventhoughtheinternalenergyofthenucleusdoesnotchange.Theenergy lossinelastic scatteringcanbefoundfromthelawsofconser­vationofenergyandmomentum. Let E, p and E', p ' bethekineticenergy and(vector)momentumoftheneutronbeforeandafterthecollision,respectively,andletEA andP betheenergyandmomentumoftherecoilingnucleus.TheneutronisscatteredattheangleiJ; thenucleusrecoilsattheangle({J (seeFig.3.7). Inviewofthefactthatthecollisioniselastic,itfollowsthat E = E' + EA. (3.25) Theconservationofmomentum,namely, p = p ' + P, can be depicted by the vector diagram shown in Fig. 3.8. Then by the law ofcosines, p2 = p2 + (p')2 _ 2pp' cosiJ. p ... Figure 3.8 Vector diagram for conservation of momentum. (3.26)
  • 84. Sec. 3.6 Energy Loss in Scattering Collisions 69 Fromclassicalmechanics, p2 = 2MEA, p2 = 2mE, and p,2 = 2mE', where M andm arethemassesofthenucleusandneutron,respectively.Equation(3.26) canthenbewrittenas MEA = mE + mE' - 2m-JEE' cosiJ. (3.27) SinceM/m isapproximatelyequaltoA, theatomicmassnumberofthenucleus,Eq.(3.27) isequivalentto AEA = E + E' - 2-JEE' cosiJ. Next,introducingEA fromEq.(3.25) gives,whenarranged, (A + l)E' - 2-JEE' cosiJ -(A - 1)E = O. Thisequationisquadraticiny'Ei andhasthesolution E' = E 2[cosiJ + jA2 _ sin2iJ] 2 (A + 1) (3.28) ItisofsomeinteresttoconsidertheconsequencesofEq.(3.28). Inagrazingcollision,iJ isapproximatelyequaltozero,andEq.(3.28) givesE' = E. Aswouldbe expected, there is no energy loss in such a collision. Except forthe special caseofhydrogen,whichmustbeconsideredseparately,itfollowsfromEq.(3.28) thattheminimumvalueofE', (E')min, occurswhen iJ = JT. Theneutronisthenscattereddirectlybackwardandsuffersthelargestpossiblelossinenergy.ForiJ = JT , Eq.(3.28) gives where (A - 1 )2 (E')min = -- E = ex E, A + 1 a = e��r iscalledthecollision parameter. Valuesofex aregiveninTable3.1. (3.29) (3.30) Thescatteringofneutrons byhydrogenisuniquebecausethemassesoftheneutron andhydrogen nucleus (proton) areessentially equal. Itis notdifficulttoshow from classical mechanics, and, indeed, it is acommon observation, that aparticlestrikinganotherparticleofthesamemass,whichisinitiallyatrest,cannotbescatteredthroughananglegreaterthan90° TheminimumenergyofneutronsscatteredfromhydrogenmustthereforebefoundbyplacingiJ = JT /2 inEq.(3.28), andthisgives (E')min = O.
  • 85. 70 Interaction of Radiation with Matter TABLE 3 . 1 COLLISION PARAMETERS Nucleus Mass No. ex ; Hydrogen 0 1.000 H2O * 0.920t Deuterium 2 0.1 1 1 0.725 020 * 0.509t Beryllium 9 0.640 0.209 Carbon 12 0.716 0. 158 Oxygen 16 0.779 0. 120 Sodium 23 0.840 0.0825 Iron 56 0.931 0.0357 Uranium 238 0.983 0.00838 *Not defined. tAn appropriate average value. Since this result could have also been obtained by placing A (3.29), itmaybeconcludedthat (E')min = aE isvalidforall nuclei,includinghydrogen. Chap. 3 1 in Eq. (3.31) Itis alsoofinteresttoknow the averageenergy ofanelastically scatteredneutron. This computationis moredifficultthanthatofmaximumandminimumenergiesandis notgiven here. It canbe shown, however, thatforscattering bylightnuclei,includinghydrogen,andatmostoftheneutronenergiesofinterestinnuclearreactors,theaverageenergyofthescatteredneutronisgivenapproximatelyby E' = i(1 +a)E. Theaverageenergyloss,�E then �E = E - E' = i(1 - a)E, andtheaveragefractionalenergylossis �E 1 - = 2(1 - a).E (3.32) Thisequationisalsovalidfortheheaviernuclei,butnotforhigh-energyneutrons.With238U, forexample,Eq. (3.32) isnotaccuratemuchabove anenergyof100 keV.Athigherenergies,theenergylossincollisionswiththeheaviernucleiislessthanthatpredictedbyEq.(3.32).
  • 86. Sec. 3.6 Energy Loss in Scattering Collisions 71 FromEq. (3.30) andTable3.1,itisobservedthatex iszeroforA = 1 (hy­drogen) andincreasesmonotonically tounity withincreasing A. In view ofEq.(3.30), itfollowsthattheaverage fractional energy loss decreasesfrom � inthecaseofhydrogentoalmostzerofortheheavynuclei.Thus,ontheaverage,aneu­tronlosesone-halfofitsenergyinacollisionwithhydrogen. Whenscatteredbycarbon,sinceex =0.716,itlosesabout14%ofitsenergy,whileinacollisionwithuranium,ex =0.983,aneutronloses lessthan 1%ofitsenergy.Inshort,neutronslose less and less energy the heavier the target nucleus is. It is often necessarytoslowdownfastneutrons-aprocessknownasmoderation. Fromtheforegoingdiscussion,itisclearthatmaterialsoflowmassnumberaremosteffectiveforthispurposesincetheneutronsslowdownmostrapidlyinsuchmedia.Neutronsalsoloseenergythroughinelasticcollisions,asaresultofbothre­coilandinternalexcitationofthetargetnucleus.Sincethethresholdenergyforin­elasticscatteringissohighinlightnuclei(usuallyontheorderofseveralMeV-itdoesnotoccuratallinhydrogen), moderationby inelastic scattering islessim­portantthanbyelasticscatteringinthesenuclei.Withtheheaviernuclei,however,theinelasticthresholdismuchlower,andinelasticscatteringisoftentheprincipalmechanismforneutronmoderation.Inmanyreactorcalculations, especiallyinconnectionwithneutronmodera­tion,itisconvenienttodescribeneutroncollisionsintermsofanewvariablecalledlethargy. Thisquantityisdenotedbythesymbolu andisdefinedas u =In(EMIE), (3.33) where EM isanarbitrary energy-usually thatofthe highestenergy neutron inthesystem. FromEq. (3.33), it shouldbe notedthat, athighenergy, aneutron'slethargyislow;asitslowsdownandEdecreases,itslethargyincreases.Theaveragechangeinlethargyinanelasticcollision, /)"u, liketheaverage fractionalenergyloss[seeEq.(3.32)], isindependentoftheenergyoftheincidentneutron.Thequantity/)"u appearsinmanynuclearengineeringcalculationsandisdenotedbythesymbol�.Byaderivationthatistoolengthytobegivenhere,itcanbeshownthat�isgivenby (A_l)2 (A+l)�= l- In -- 2A A-I ' (3.34) whereAisthemassnumberofthetargetnucleus.ExceptforsmallvaluesofA,�iswellapproximatedbythesimpleformula 2�::::'--2'A+3 (3.35) EvenforA=2,Eq.(3.35)isoffbyonlyabout3%.Exactvaluesof�aregiveninTable3.1.
  • 87. 72 Interaction of Radiation with Matter Chap. 3 Example 3.10 A I-MeV neutron is scattered through an angle of 45° in a collision with a 2H nu­ cleus. (a) What is the energy of the scattered neutron? (b) What is the energy of the recoiling nucleus? (c) How much of a change in lethargy does the neutron undergo in this collision? Solution 1. Substituting E = 1 Me V, A = 2, and 11 = 45° into Eq. (3.28) gives immedi­ ately E' = 0.738 Me V. [Ans.] 2. Since the collision is elastic (inelastic scattering does not occur with 2H), the recoil energy is EA = 1.000 - 0.738 = 0.262 Me V. [Ans.] 3. The lethargy going into the collision is u = In(EM/E) and coming out is u' = In(EM/E'). The change in lethargy is therefore D.u = u' - u = In(E/E') - In(1/0.738) = 0.304 (a unitless number). [Ans.] Polyenergetic Neutrons InSections3.2 and3.3, therateatwhichneu­tronsundergocollisioninatargetwascalculatedontheassumptionthattheinci­dentneutronsweremonoenergetic.Thiscaneasilybegeneralizedtoneutronsthatarenotmonoenergetic,buthaveadistributioninenergy.Forthis purpose, letnee) dE bethe numberofneutrons percm3 with en­ergies between E and E + dE in aneutronbeam incidenton athintarget. Theintensityoftheseneutronsis dI(E) = n(E)v(E) dE, (3.36) where vee) isthe speedcorrespondingtotheenergy E. AccordingtoEq. (3.8), thisbeaminteractsinthetargetatarateof dF(E) = n(E)v(E)"Et(E) dE collisions percm3/sec, where "Et(E) isthe macroscopic totalcross-section. Thecollisiondensityisthen F = fn(E)v(E)"E,(E) dE, (3.37) inwhichtheintegrationiscarriedoverallenergiesinthebeam.Tocomputetheinteractionrateforaparticulartypeofinteraction,itismerelynecessarytoreplace "Et(E) inEq. (3.37) bytheappropriatecross-section. Anes­peciallyimportantcase is the absorption ofthermal neutrons-that is, neutrons whoseenergiesaredistributedaccordingtotheMaxwellianfunctiondescribedinSection2.13. Theseneutronsarefoundincertaintypesofnuclearreactorscalledthermal reactors, whicharediscussedinChapters4 and6. Theabsorptionratein
  • 88. Sec. 3.6 Energy Loss in Scattering Collisions 73 abeamofthermalneutronsis Fa = fn(E)v(E)'Ea(E) dE, (3.38) where La(E) isthemacroscopicabsorptioncross-sectionandtheintegraliseval­uatedatthermalenergiesuptoabout0.1 eV.In Section 3.5, itwas pointed outthat at low energies mostnuclei exhibit1/v absorption,eitherastheresultofradiativecaptureorsomeotherabsorptionreaction.Attheseenergies, La(E) canbewrittenas (3.39) whereEo isanarbitraryenergyandVo isthecorrespondingspeed.WhenEq.(3.39) isintroducedintoEq.(3.38), theveE) cancelssothat Fa = 'Ea(Eo)vo fn(E) dE. (3.40) Theremainingintegralisequaltothetotaldensityofthermalneutron,n, andEq.(3.40) reducesto (3.41) Equation (3.41) shows thatfora l/v absorber, the absorption rate is inde­pendentoftheenergydistributionoftheneutronsandisdeterminedbythecross­sectionatanarbitraryenergy. Furthermore, itmaybeconcludedfromEq. (3.41)that,althoughtheneutronshaveadistributionofenergies,theabsorptionrateisthesameasthatforamonoenergeticbeamofneutronswitharbitraryenergy Eo andintensitynVo. Inviewoftheseresults,ithasbecomestandardpracticetospecifyallabsorp­tioncross-sections lIv ornot,atthesingleenergyofEo = 0.0253 eV.Thecorre­spondingspeedisVo = 2,200 meters/sec.Valuesofcross-sectionsat0.0253 eVarelooselyreferredtoasthermal cross-sections. Thesearetabulatedin anumberofplaces,includingthechartofnuclides;anabridgedtableisfoundinAppendixII.Thisquantitynvo inEq.(3.41) iscalled2,200 meters-per-secondfluxandisdenotedby4>0;thatis 4>0= nvo· (3.42) Theabsorptionrateisthensimply (3.43)
  • 89. 74 Interaction of Radiation with Matter Chap. 3 Althoughonlyacomparativelyfewnucleiarenon-l/vabsorbers,thesenucleiareusuallyimportantinnuclearsystemssincetheircross-sectionstendtoberatherhigh. The absorption rate forsuch nuclei is again given by Eq. (3.38), but nowthe integral cannotbe simplified as itwas in the 1/v case. Inparticular, Fa nowdependsonthefunctionneE) aswellas �a(E). However,byassumingthatneE) is the Maxwellian function, C. H. Westcottcomputed Fa numerically for all ofthe importantnon-1/ v absorbers. Theresulting value ofFa is afunction ofthetemperatureoftheneutrondistributionandisgivenintheform (3.44) wherega(T), whichiscalledthenon-l/vfactor, isatabulatedfunctionand�a(Eo) isagaintheabsorptioncross-sectionat0.0253 eV.Ashorttableofnon-1/vfactorsisgiveninTable3.2. Althoughthepriorresultswerederivedforabeamofneutronsincidentonathintarget,theyapplyequallywelltothemorecomplicatedsituationfoundinmanynuclearsystems in which theneutrons are moving inalldirections. Inparticular,the2,200 meters-per-secondfluxisdefinedatanypointwheretherearen thermalneutronspercm3,andEq. (3.43) or(3.44) canbeusedtocomputetheabsorptionrateatsuchapoint. Example 3.11 A small indium foil is placed at a point in a reactor where the 2,200 meters-per­ second flux is 5 X 1012 neutrons/cm2-sec. The neutron density can be represented by a Maxwellian function with a temperature of 600°C. At what rate are the neutrons absorbed per cm3 in the foil? Solution. From Table 11.3 in Appendix II, N = 0.0383 X 1024 and aa(Eo) = 194 b so that ba(Eo) = 0.0383 x 194 = 7.43 cm-I However, indium is non-l/v and from Table 3.2, ga(600°C) = 1 . 15. From Eq. (3.44), it follows that Fa = 1 . 15 x 7.43 x 5 X 1012 = 4.27 X 1013neutrons/cm3-sec. [Ans.] 3.7 FISSION ItwasshowninSection2.11 thatthebindingenergies ofnucleipernucleonde­crease with increasing atomic mass number, for A greaterthan about 50. Thismeansthatamorestableconfigurationofnucleons isobtainedwheneveraheavynucleus splitsintotwoparts-thatis, undergoesfission.Theheavier,moreunsta­blenuclei mightthereforebeexpectedtofissionspontaneously withoutexternalintervention. Suchfissionsdooccur.ItisinterestingtoexaminetheoriginofthedecreaseinthebindingenergypernucleonwithincreasingA. Figure3.9 showsthetermsinthebindingenergyper
  • 90. ..... U'I TABLE3.2NON-1/VFACTORS* CdInI35Xe149Sm233U235U238U239Pu T,oCgagagJgagagfgagfgagagf 201.32031.01921.15811.61700.99831.00030.97800.97591.00171.07231.0487 1001.59901.03501.21031.88740.99721.00110.96100.95811.00311.16111.1150 2001.96311.05581.23602.09030.99731.00250.94570.94111.00491.33881.2528 4002.55891.10111.18642.18541.00101.00680.92940.92081.00851.89051.6904 6002.90311.15221.09142.08521.00721.01280.92290.91081.01222.53212.2037 8003.04551.21230.98871.92461.01461.02010.91820.90361.01593.10062.6595 10003.05991.29150.88581.75681.02261.02840.91180.89561.01983.53533.0079 *BasedonC.H.Westcott,"EffectiveCross-SectionValuesforWell-ModeratedThennalReactorSpectra,"AtomicEnergyCommission reportAECL-1101,January1962. tBasedonE.C.Smithetai.,Phys.Rev.115,1693(1959).
  • 91. 76 > Q.) :5 = 0 Q.) U ;:l Z ...Q.) 0... ;>, e!lQ.) = � OJ) = � = iE 18 1 6 14 12 10 Interaction of Radiation with Matter --Surface Tenn . . . Coulomb Tenn ,.. . _ - Assymetry Tenn Chap. 3 8 ,............-.----.----.----._ - - - Binding Energy/Nucleon ( -�-�"'--.=""' 6 4 2 o +-----+-----+-----+-----+-----+-- 4 44 84 125 175 225 Atomic Mass Number Figure 3.9 Components of the binding energy per nucleon based on the semi-empirical mass formula. nucleoncurveasdetenninedfromtheliquiddropmodeloftheatom.Thedecreaseinbindingenergypernucleonislargelyduetothecoulombtennovercomingthevolume tenn. The volume tennrepresents the binding duetothe strong nuclearforces.Itisthecoulombrepulsion,then,thatislargelyresponsibleforfission.Althoughheavynucleidospontaneouslyfission,theydosoonlyrarely.Forfissiontooccurrapidlyenoughtobeusefulinnuclearreactors,itisnecessarytosupplyenergytothenucleus.This,intum,isduetothefactthatthereareattractiveforcesactingbetweenthenucleonsinanucleus,andenergyisrequiredtodefonnthenucleustoapointwherethesystemcan beginto split in two. This energyiscalledthecritical energy a/fission andisdenotedbyEcrit. ValuesofEcrit aregiveninTable3.3 forseveralnuclei.AnymethodbywhichenergyEcrit isintroducedintoanucleus,therebycaus­ingittofission,issaidtohaveinduced thefission.Themostimportantoftheseisneutronabsorption.ItisrecalledfromSections2.1 1 and3.5 thatwhenaneutronisabsorbedtheresulting compoundnucleusisfonnedinanexcitedstateatanen­ergyequaltothekineticenergyoftheincidentneutronplustheseparationenergyorbindingenergyoftheneutroninthecompoundnucleus. Ifthisbindingenergyaloneisgreaterthanthecriticalenergyforfissionofthecompoundnucleus,thenfissioncanoccurwithneutronshavingessentiallynokineticenergy.Forexample,accordingtoTable3.3, thebindingenergyofthelastneutronin236U is6.4 MeV,
  • 92. Sec. 3.7 Fission TABLE 3.3 CRITICAL ENERGIES FOR FISSION, IN Me V Fissioning Binding Energy of Nucleus A Z Critical Energy Last Neutron in AZ 232Th 5.9 * 233Th 6.5 5.1 233U 5.5 * 234U 4.6 6.6 235U 5.75 * 236U 5.3 6.4 238U 5.85 * 239U 5.5 4.9 239Pu 5.5 * 240pU 4.0 6.4 *Neutron binding energies are not relevant for these nuclei since they cannot be formed by the absorption ofneutrons by the nuclei A-1 Z. 77 where�s Ecrit is only 5.3 MeV. Thus, when aneutron ofzero kinetic energy isabsorbedby235U,thecompoundnucleus,236U,isproducedwith 1.1 MeVmoreenergythanitscriticalenergy, andfissioncanimmediatelyoccur.Nuclei suchas235U, thatlead to fission following the absorptionofazero-energyneutron, arecalledfissile. Note,however,thatalthoughitisthe235Uthatissaidtobefissile,thenucleusthatactuallyfissionsinthis caseisthe236U.FromTable3.3 itisevidentthat233Uand239pu(plutonium-239)arealsofissile.Inaddition,241Puandseveralothernucleinotindicatedinthetablearealsofissile.Withmostheavynucleiotherthan233U,235U,239pu,and241Pu,thebindingenergy ofthe incident neutron is not sufficientto supply the compoundnucleuswiththecriticalenergy,andtheneutronmusthavesomekineticenergytoinducefission. Inparticular,thisis alwaysthecasewhenthestrucknucleuscontainsanevennumberofnucleons since thebindingenergy ofthe incidentneutrontoaneven-A nucleusisalwayslessthantoanodd-A nucleus. (Thisodd-evenvariationinbindingenergyisevidentinTable3.3.)Forinstance, thebindingenergyofthelastneutronin239Uis only4.9MeV,andthis istheexcitationofthecompoundnucleusformedwhenaneutronofzerokineticenergyisabsorbedby238U. SinceEcrit for239Uis 5.5MeV, itisclearthatfissioncannotoccurunlesstheneutronincidentonthe238Uhas anenergy greaterthan about0.6 MeV. Nuclei suchas238U, which do not fission unless struckby an energetic neutron, are said to befissionable but nonfissile. Forreasonsthatareclearlaterinthisbook,thenonfissileisotopes suchas238Ucannotalonebeusedtofuelnuclearreactors, anditisthefissile isotopes, especially 235U and239pu, that are the practical fuels ofnuclearpower. Thesemi-empiricalmassformuladiscussedinChapter2 maybeusedtoex­aminethedependenceofthecriticalenergygiveninTable3.3.Forthecaseof235U,
  • 93. 78 Interaction of Radiation with Matter Chap. 3 thecriticalenergyisobtainedbycalculatingtheQ-valueforthereactionusingthesemi-empiricalmassfonnula.Theoriginofthevariationincriticalenergyisevi­dentbycomparingthevalueoftheodd-eventennasA is increasedfrom233 to238. Fission Cross-Sections Thecross-sectionsoffissilenucleiforneutron­inducedfission resemble radiative capture cross-sections in theirdependence ontheenergyoftheincidentneutron.Thus,asseeninFig.3.10, whereOJ isgivenfor235U, therearethreedistinctregionstothecross-section.Atlowenergy,OJ is l/v ornearlyso;thisisfollowedbyaregionofresonances;finally,abovetheresonanceregion,af issmoothandrolling. Itshouldbenotedthataf is especiallylargeinthe l/v region.Thefissioncross-sectionsoffissionablebutnonfissilenuclei,bycontrast,arezero up toathreshold energy, which always occurs above theresonanceregion.Asaresult,af isrelativelysmoothatallenergies.ThisisillustratedinFig.3.1 1, whereaf isshownfor238U. �e c 0 .� u C!) CI) en en 8U Uranium 235 Fission Cross Section MT = 18 103 102 101 �__�����U-__������__�__����____����� 10-4 10-3 10-2 Neutron Energy (eV) 10-1 Figure 3.10a The fission cross-section of U-235; see continuation on next three pages. (Plotted from ENDFIB 6 using ENDFPLOT over the Internetfrom the Korean Atomic Energy Research Institute.)
  • 94. Sec. 3.7 Fission 79 Uranium 235 Fission Cross Section MT = 1 8 '" 102 c ta e c ,:2 U CI) IZl '" '" 2 U 101 10° �������������������������������� 1 2 3 4 5 6 7 8 9 10 Neutron Energy (eV) Figure 3.10b Uranium 235 Fission Cross Section MT = 1 8 102 t- �, � - , � VE .�� � «j ,V �e c �0 101'';::: t- t) CI) IZl '" '" 2 u lOo t- - 10- 1 ���_�I-L��I���_I�����I���I���_�I-L��I���_I���� 20 30 40 50 60 70 80 90 Neutron Energy (eV) Figure 3.10c
  • 95. 80 CJJ C � e c .S: U (1) '" C/O C/O 2 u Interaction of Radiation with Matter Chap. 3 Uranium 235 Fission Cross Section MT = 1 8 102 101 10° 10-1 �______�____�__�__������________L-____�__���__��� 102 103 104 Neutron Energy (eV) Figure 3.10d Uranium 235 Fission Cross Section MT = 1 8 100 �__�__���-wWU____��__�-L�LL____�-L�-L���__�__�-L� 1� 1� 1� 1� Neutron Energy (eV) Figure 3.10e
  • 96. Sec. 3.7 1.40 1.20 1.00 �� = 0.80 .S; u � 0.60 8u 0.40 0.20 0.00 Fission 0 2 4 6 8 10 12 14 Energy MeV Figure 3.11 The fission cross-section of U-238. (Plotted using ENDFIB 6 and ENDFPLOT over the Internet from Korean Atomic En­ ergy Research Institute.) 81 16 Itshouldnotbeimpliedfromtheforegoingdiscussionthat,whenaneutroncollides withafissilenucleus, orwith afissionablebutnonfissile nucleus abovethefissionthreshold, theresultisalwaysfission.Thisisnotthecase;neutronsin­teractingwiththesenucleimaybescattered,elasticallyorinelastically,theymaybe absorbed in radiative capture, and so on. Thecross-sections for all oftheseprocesseshavebeenmeasuredandarefoundinENDFIB6. However,withfissilenucleiatlowenergies,onlythreeinteractionsarepossible:elasticscattering,radia­tivecapture,and,ofcourse,fission.Thevalueofas ismuchsmallerthaneitheray oraf' sothatradiativecaptureandfissionarebyfarthemoreprobableevents.Theratio ofthe cross-sections ofthese twoprocesses is called the capture-to-fission ratio andisdenotedbythesymbola, thatis, ay a = ­ af (3.45) Thisparameter,whichisafunctionofenergy,hasanimportantbearingonthedesignofmanyreactors.Valuesofa forthefissilenuclei at0.0253eVaregiveninTable3.4,alongwiththecross-sectionsforthesenuclei.
  • 97. 82 Interaction of Radiation with Matter Chap. 3 TABLE 3.4 THER MAL (0.0253 e V) DATA FOR THE FISSILE NUCLlDES* a ta af ex 17 v 233U 578.8 531 . 1 0.0899 2.287 2.492 235U 680.8 582.2 0. 169 2.068 2.418 239pU 101 1.3 742.5 0.362 2. 108 2.871 241Pu 1377 1009 0.365 2. 145 2.917 *From Neutron Cross-Sections, Brookhaven National Laboratory report BNL-325, 3rd ed., 1973. taa = ay + af· Fission Products Itshouldbeexpectedonintuitivegrounds, anditcanbeshownfromelementarycalculationsoftheenergies involved, thatafissioningnucleusshouldsplitmoreorlessinhalf. Infact, suchsymmetricfissionisarareevent.Fissionisalmostalwaysasymmetric,sothatthemassesofthetwofragmentsaresubstantiallydifferent.ThisisindicatedinFig.3.12, wherethefission-productyield,thatis,thepercentofthefissionfragmentsproducedwithagivenmassnum­ber,isshownasafunctionofAforfissioninducedbythermalneutronsin235U. Itshouldbenotedthatthefigureisplottedonthelogarithmicscalesothatthefission­productdistributionismorestronglyasymmetricthanitwouldatfirstappear.Withthe increasing energy ofthe incidentneutron, fission becomes more symmetric.ThisisillustratedinFig.3.12 bytheyieldoffissionproductsarisingfromfissioninducedby 14 MeVneutrons. 1 0 �--�--�----�--�� � -d' 0. 1 "0 ' ;;' c:: .� 0.01 tE 0.001 0.0001 LL-.--'------'------'----'__ -'-----'-----I-----I.-----' 70 90 l lO 1 30 150 Mass number Figure 3.12 Fission-product yields for thermal and 14-MeV fission neutrons in U-235.
  • 98. Sec. 3.7 Fission 83 Whenthefissionproductsareinitiallyfonned,theyareexcessivelyneutronrich;theycontainmoreneutronsthanarenecessaryfortheirstability. Asaresult,theydecaybyemittingasequenceofnegativefJ-rays,whichareaccompaniedbyvarious y-rays. Forexample, theisotope 115Pd (palladium-I15) is produced di­rectlyinfissionanddecaysbythechain 115Pd� 115Ag� 115Cd� 115In(stable). Manyfission-productdecay chains ofthis kindhavebeenidentifiedandcanbededucedfromthedataonthechartofthenuclides.Theradioactivityofthefissionproductsisthecauseofanumberofproblemsintheutilizationofnuclearenergy.Foronething,fissionproductsaccumulateinanoperatingreactorasthefuelundergoesfission,andelaborateprecautionsmustbetakentoensurethattheydonotescapetothesurroundingenvironment.Further­more,theheatreleasedbydecayingfissionproductsmaybesogreatthatareactormustbecooledaftershutdowntopreventdamagetothefuel.Thecontinuingemis­sion ofradiationfromthefission products also tends tomakeparts ofareactorhighlyradioactive.Whenremovedfromareactor,thesepartsmustbecooledwhilebeingstoredpriortodisposalorprocessing.Thequantitativeaspectsoffission-productdecayarecomplicatedbythefactthathundredsofdifferentradioactivenuclidesareproducedinfission,eachwithitsowncharacteristichalf-lifeanddecayradiation.Formanypurposes,however,thefollowingexpressionsmaybeusedtorepresent approximatelytheoveralldecayofthefissionproducts. Thus, theratesatwhichfJ-raysandy-raysareemittedinthetimeintervalfromabout10secondstoseveralweeksafterasinglefissionaregivenby RateofemissionoffJ-rays::: 3.8 x 10-6t-1.2fJ-rays/sec, Rateofemissionofy-rays::: 1.9 x 10-6t-1.2y-rays/sec, wheretisthetimeafterfissionindays. (3.46) (3.47) Toexpressthepriordisintegrationratesinunitsofcuries,itismerelyneces­sarytonotethateachfJ-rayoriginatesinthedecayofanuclide. Thensince I Ci= 3.7 x 101Odisintegrations/sec,thefission-productactivitytdaysafterI fissionis Fissionproductactivity:::3.8 x 10-6t-1.2/3.7 x lOw = 1.03 x 10-16t-1.2Ci. (3.48) Itisoftennecessarytocalculatethetotalfission-productactivitythataccu­mulatesinthefissilefuelofanoperatingreactor. Suppose, forexample, thatthereactorhasbeenoperatingataconstantpowerofP megawatts(MW)forT daysandisthenshutdown.Todetenninetheactivityofthefissionproductstdaysafter
  • 99. 84 I nteraction of Radiation with Matter dS-! I--I· -- s -----1 t 1-1· ----T ,I, - t ---J Startup Shutdown Observation time ....1·-- Reactor in operation ---, 1...·-- Reactor off -----l Figure 3.13 Diagram for computing fission-product activity. Chap. 3 shutdown, it is necessary to integrate Eq. (3.48) overthe appropriate time span.Lets betheintervalindaysfromthetimewhenafissionoccurstothetimethatthe activityofits fissionfragments is measured (see Fig. 3.13). Itis shownlaterinthissectionthatatapowerlevelofP MW thetotalfissionrateis2.7 x 1021 P fissionsperday.Thenumberoffissionsoccurringintimeds isthen2.7 x 1021 Pds fissions.FromEq.(3.48), theactivitys dayslateris 2.7 x 1021 Pds x 1.03 x 10-16S-1.2 = 0.28 X 106Ps-1.2 ds Ci. Thetotalactivityatt isthen jt+T Fissionproductactivity = 0.28 x 106P t s-1.2 ds = 1 .4 X 106p[t-O.2 - (t +T)-O.2] Ci. (3.49) Thepriorequationsmaybeusedtocomputetheactivityofasinglefuelrodthathasbeenleftinareactorfort daysandthenremoved. Inthiscase, P isthepowerproducedby the fuel rodin megawatts andt is the time in days since itsremovalfromthereactor.It is somewhat more difficult to analytically calculate the total energy re­leasedbyfissionproducts since,foronething,theenergyspectrumoftheemittedradiationchangesintimeasthesenuclidesdecay.Forroughestimates,theaverageenergiesofthef3-andy-raysaresometimestakentobe0.4 MeVand0.7 MeV,re­spectively.Therateofenergyreleasefromthedecayingfissionproductsfollowingonefissionisthen Decayenergyrate� 2.8 x 10-6t-1.2Me V/sec, (3.50) wheret isagainindays. However, afarbetterprocedureistouseactualexperi­mentaldataonenergyrelease.ThismethodisdiscussedindetailinChapter8 (seeSection8.1). Example 3.12 The total initial fuel loading of a particular reactor consists of 120 fuel rods. After the reactor has been operated at a steady power of 100 MW for 1 year, the fuel is
  • 100. Sec. 3.7 Fission 85 removed. Assuming that all rods contribute equally to the total power, estimate the activity of a fuel rod 1 day after removal. Solution. In Eq. (3.49), P = 100, t = 1, and t + T = 1 + 365 = 366. The activity of all fuel rods 1 day after removal is then 1.4 X 106 x 100 X [(1 )-0.2 - (366)-0.2] = 1 .4 x 108 (1 - 0.307) = 9.7 x 107 Ci. One rod would have an activity of 9.7 X 107/120 = 8.1 x 105 Ci. [Ans.] Fission Neutrons Mostoftheneutronsreleasedinfission(usuallymorethan99%) areemittedessentiallyattheinstantoffission.Thesearecalledprompt neutrons, in contrasttodelayed neutrons, whicharereleasedcomparatively longafterthefissionevent.Theaveragenumberofneutrons,bothpromptanddelayed,releasedperfis­sionisgiventhesymbolv. Valuesofv forfissioninducedbyO.0253-eVneutronsaregiveninTable3.4.Astheenergyoftheincidentneutronisraised, v increasesslowly.Oneadditionalneutronisemittedforevery6- to7-MeVincreaseinneutronenergy.Forlateruseinreactorcalculations,itisconvenienttodefinetheparameter1}, whichisequaltothenumberofneutronsreleasedinfissionperneutronabsorbedby a fissile nucleus. Since radiative capture competes with fission, 1} is alwayssmallerthan v . Inparticular,1} isequalto v multipliedbytherelativeprobability(afjaa) thatanabsorptionleadstofission(seeExample3.2)or af af 1} = v- = v-.........:.....- aa ay + af (3.51) Intermsofa, thecapture-to-fissionratio(seeEq.3.45),thiscanbewrittenas v 1} = l + a · (3.52) Foramixtureoffissileorfissileandnonfissilenuclides, 1} is defined astheaveragenumberofneutronsemittedperneutronabsorbedinthe mixture. In thiscase,1} isgivenby 1 1} = - L v(i)�f(i), �a i (3.53) where v(i) and�f(i) arethevalueofv andthemacroscopicfissioncross-sectionfortheithnuclide,respectively, and �a isthemacroscopiccross-sectionforthemixture.Itshouldbenotedthatv(i), �f(i) and�a inEq.(3.53)mustbecomputedattheenergy ofthe neutrons inducing the fission. Forexample, ifthe fuel is a
  • 101. 86 Interaction of Radiation with Matter Chap. 3 mixtureof235U and238U andthefissionsareinducedbylow-energyneutrons,then v(235)L1 (235) TJ = La(235) + La (238) (3.54) Therearenotennsinvolving238U inthenumeratorbecausethisnuclidedoesnotfission withlow-energy neutrons. However, ifthis samefuelwere used in afastreactor (see Section 4.2), in which the fissions are induced by highly energeticneutrons,TJ wouldbe v(235)L1(235) + v(238)L1(238) TJ = La (235) + La (238) . Inthisexpression,allquantitiesarecomputedattheelevatedenergies. Example 3.13 Calculate the value of TJ for natural uranium at 0.0253 e V. Solution. Written out in detail, Eq. (3.54) is v(235)!V(235)aj(235) TJ = !V(235)aa(235) + !V(238)aa(238) (3.55) According to Eq. (2.61), the atom density ofan isotope is proportional toits isotopic abundance y, so that v(235)y(235)aj(235) TJ = -------=------ y(235)aa(235) + y(238)aa(238) Introducing data from Table 3.4 and Table 11.2 ofAppendix II gives 2.418 x 0.72 x 582.2 TJ = = 1 .34. [Ans.J 0.72 x 680.8 + 99.26 x 2.70 [!Vote: For reasons discussed in Chap. 6, this is notprecisely the value used in reactor problems involving natural uranium; see Example 6.1 1.] ThepromptfissionneutronsareemittedwiththecontinuousenergyspectrumshowninFig. 3.14. Thisspectrumiswelldescribedbythefunction X (E) = 0.453e-1.036E sinhJ2.29E, (3.56) whereX (E) isdefinedsothatX (E) dEisthefractionofthepromptneutronswithenergiesbetweenE andE+dE andE isinMeV.ThefunctionX (E) isnonnalizedsothat [0X (E) dE = 1.
  • 102. Sec. 3.7 Fission I ;;-(t) 6 Gf x 0.3 0.2 0.1 2 3 4 5 6 E, MeV Figure 3.14 The prompt neutron spectrum. 87 7 TheaverageenergyE ofthepromptneutronscanbefoundfromtheintegral E = 100E X (E)dE = 1.98 MeV The most probable energy, correspondingtothepeakofthe X (E) curve, is 0.73 MeV.Although delayedneutrons ordinarilycompriselessthan 1% oftheneutronsreleasedinfission,theyplayanimportantroleinthecontrolofnuclearreactors,asisseeninChapter7. Theseneutronsoriginateinthedecaybyneutronemissionofnucleiproducedinthe,B-decayofcertainfissionproducts.Forexample,whenthefission-product87Brdecaysto87Kr,thelattermaybeformedinanexcitedstate.Inthiscase,theleastboundneutroninthe87Krisnotboundatallandisejectedfromthenucleuswithanenergyofabout0.3 MeV.Thisneutron isemittedassoonastheexcited state is formed. Therefore, itappearstobeemittedwith the 54.5-sechalf-lifeofthe87Br.Nucleisuchas87Brarecalleddelayed-neutronprecursors. Therearebelievedtobeabout20 suchprecursors,mostofwhichhavenowbeenpositivelyidentified.Theprecursorscanbedividedinto6 groups,eachwithitsowncharacteristichalf­life.Thegrouphalf-livesanddecayconstantsaregiveninTable3.5 forlow-energy(thermal)fissionin235U. Alsoincludedinthetablearetheobservedyields (neu­tronsperfission)ofthedelayedneutronsineachgroup,togetherwiththedelayed­neutronfractionsf3i.Thequantity,Bi isdefinedasthefractionofallofthefissionneutronsreleasedin fissionthatappearasdelayedneutrons inthe ith group. Inotherwords, ,Bi is the absoluteneutronyieldofthe ithgroupdividedby v. Thetotaldelayedfraction,B isthesumofallthe,Bi.
  • 103. 88 Interaction of Radiation with Matter Chap. 3 TABLE 3.5 DELAYED NEUTRON DATA FOR THERMAL FISSION IN 235 U * Group 2 3 4 5 6 Half-Life (sec) 55.72 22.72 6.22 2.30 0.610 0.230 Decay Constant (Ii. sec-I ) 0.0124 0.0305 0. 1 1 1 0.301 1.14 3.01 Energy Yield, Neutrons Fraction (ke V) per Fission (f3i) 250 0.00052 0.000215 560 0.00346 0.001424 405 0.003 10 0.001274 450 0.00624 0.002568 0.00182 0.000748 0.00066 0.000273 Total yield: 0.0158 Total delayed fraction (f3): 0.0065 *Based in part on G. R. Keepin, Physics ofNuclear Kinetics, Reading, Mass.: Addison-Wesley, 1965. Prompt ,-Rays Attheinstantoffission,anumberofy-raysareemittedfromthefissioningnucleus.Thesearereferredtoasprompt y-rays todistinguishthemfromthefission-producty-rays.Theenergyspectrumoftheprompty-raysisapproximatelythesameasthespectrumofthefission-producty-rays. The Energy Released in Fission Indiscussingtheenergyoffission,itisimportanttodistinguishbetweenthetotalenergyreleasedintheprocessandtheenergythatcanberecoveredinareactorandis thereforeavailablefortheproductionofheat. Therecoverableenergy andthetotalenergy,ingeneral,aredifferent.ThisisillustratedinTable3.6, whichgivesabreakdownofthecomponentenergiesastheyoccurintheneutron-inducedfissionof235U. TABLE 3.6 EMITTED AND RECOVERABLE E NERGIES FOR FISSION OF 235 U Emitted Energy, Recoverable Energy, Form Me V Me V Fission fragments 168 168 Fission-product decay f3-rays 8 8 y-rays 7 7 neutrinos 12 Prompt y-rays 7 7 Fission neutrons (kinetic energy) 5 5 Capture y -rays 3-12 Total 207 198-207
  • 104. Sec. 3.7 Fission 89 Asindicatedinthetable,most(almost85%)oftheenergyreleasedinfissionappears asthekineticenergyofthe fissionfragments. These fragmentscometorestwithinabout10-3cmofthefissionsitesothatalloftheirenergyisconvertedinto heat. The energies ofthefission-productf3-raysand y-rays,theprompt anddelayedneutrons,andtheprompty-raysarealsorecoverablesincealmostnoneoftheseradiationseverescapefromanuclearpowersystem.However,theneutrinosthataccompanyf3-decayinteractonlyslightlywithmatterandescapecompletelyfromeverynucleardevice.Theirenergyofalmost12MeVperfissionisthereforeirrevocablylostforpracticalpurposes.Becausemostfissionneutronsremainwithintheconfinesofareactor,theseneutronsareeventuallycapturedbythenucleiinthesystem.Itisshowninthenextchapter,however,thatoneofthev-neutronsemittedperfissionmustbeabsorbedbyafissionablenucleus andproduceanotherfissioninorderforanuclearreactortoremaininoperation.Therefore,itfollowsthattheremaining(v-1)neutronsperfissionmustbeabsorbedparasiticallyinthereactor,thatis,absorbedinanonfissionreaction.Eachabsorptionusuallyleadstotheproductionofoneormorecapturey-rays, whose energiesdependonthebinding energyoftheneutrontothecom­poundnucleus. Since visapproximately2.42 for235U(itsprecisevaluedependsontheenergyoftheneutronscausingthefission),thismeansthat,fromabout3 to12 MeVofcapture,y-radiationisproducedperfissiondependingonthematerialsusedinthereactor.Allthisy-rayenergyis,ofcourse,recoverable.ItisobservedinTable3.6 thattheenergyofthecapturey-rayscompensatestosomeextentfortheenergy lost by neutrinoemission. Inanycase,therecoverable energyperfissionisapproximately200MeV.Intheabsenceofmoreaccuratedata,thisisthevaluethatisnormallyusedatleastinpreliminarycalculations.Considernowareactorinwhichtheenergyfromthefission235UisreleasedattherateofP megawatts. In otherwords, thereactorisoperatingatathermalpowerofPmegawatts.Witharecoverableenergyperfissionof200MeV,therateatwhichfissionsoccurpersecondintheentirereactoris . 106joules fissionFIssionrate= PMWx x ---- MW-sec 200MeV MeVx x 1.60X 10-13joule 86,400sec day = 2.70 x 1021Pfissions/day. Toconvertthis togramsperdayfissioned, whichis also calledthe burnup rate, itismerelynecessarytodividebyAvogadro'snumberandmultiplyby235.0, thegramatomicweightof235U.Thisgivessimply Bumuprate= 1.05Pg/day. (3.57)
  • 105. 90 Interaction of Radiation with Matter Chap. 3 Thus, ifthereactoris operating atapowerof1 MW,the235U undergoes fissionat the rate ofapproximately 1 g/day. To put this anotherway, therelease of 1megawatt/dayofenergyrequiresthefissionof1 gof235U. Itmustbe remembered, however, that fissile nuclei are consumedboth infissionandradiativecapture. Sincethetotalabsorptionrateis (fa/(f! = (1 + a) timesthefissionrate,itfollowsfromEq. (3.57)that235U isconsumedattherateof Consumptionrate= 1.05(1 + a)P g/day. (3.58) For235U, thethennalvalueofa is0.169.Equation(3.58)showsthatthisisotopeisconsumedattherateofabout 1.23g/daypermegawattofpowerifthefissionsareinducedprimarilybythennalneutrons. Example 3.14 The energy released by the fissioning of 1 g of 235U is equivalent to the combustion of how much (a) coal with a heat content of 3 x 107 J/kg (13,000 Btullb) and (b) oil at 4.3 x 107 J/kg (6.5 x 106 Btulbarrel)? Solution. According to the prior discussion, the fissioning of 1 g of 235U releases approximately 1 megawatt/day = 24,000 kWh = 8.64 x 1010 1. 1. This energy is also released by 8 64 X 1010. 7 = 2.88 X 103 kg = 2.88 metric tons 3 x 10 = 3.17 ST (short tons) ofcoal. [Ans.] 2. In terms ofoil this is also 8.64 x 1010 3 4.3 X 107 = 2.00 x 10 kg = 2.00 t = 12.6 barrels. [Ans.] 3.8 ,·RAY INTERACTIONS WITH MATTER Althoughthetenny-rayisnonnallyreservedforradiationemittedbynucleiandx-rayreferstoradiationoriginatingintransitionsofatomicelectrons,bothfonnsofradiationarecalledy-raysinthepresentsection.Thereis,ofcourse,nofundamen­taldifferencebetweenthetworadiations,perse,astheyarebothelectromagneticradiation.Gammaraysinteractwithmatterinseveralways.Ordinarily,however,onlythreeprocessesmustbetakenintoaccountinnuclearengineeringproblems.Thesearethephotoelectriceffect,pairproduction,andComptoneffect.
  • 106. Sec. 3.8 y-Ray Interactions with Matter 91 The Photoelectric Effect Inthephotoelectriceffect,theincidenty-ray interactswithanentireatom,they-raydisappears,and1 oftheatomicelectronsisejectedfromtheatom.Theatomrecoilsinthisprocess,butcarrieswithitverylittlekineticenergy.Thekineticenergyoftheejectedphotoelectronisthereforeequaltotheenergyofthephotonlessthebindingenergyoftheelectrontotheatom-thatis,theionizationenergyfortheelectroninquestion.Ifay-raysucceedsinejectinganinneratomicelectron,theholeintheelec­tronic structure is laterfilled by atransition of 1 ofthe outerelectronsintothevacantposition.Thistransitionisaccompaniedbytheemissionofx-rayscharac­teristicoftheatomorbytheejectionofanAugerelectron(seeSection2.7). Thecross-sectionperatomforthephotoelectriceffectisdenotedbythesym­bol ape. This cross-section can be used in the same way as the neutron cross­sections discussedinthepreceding sections. Thus,ifI istheintensityofy-raysincidentonathintargetcontainingN atomspercm3,thenINape isthenumberofphotoelectricinteractions/cm3-sec. 1 0 5 .-----l----.,.sl-r,---+-----+----+---------lr,II,� 1 0 4 �--+__+-�--_+---�� 1 0 3 �-�--�_4�K-E-d-ge�--�-� � b'6.. 1 0 2 1----+-------+-----------+---+------1 1 0 �-- -+---�--�----�� '�0. 1 "--__...i.....-__-'-__--'-___.l..---.I 0.00 1 0.0 1 0. 1 10 Energy MeV Figure 3.15 The photoelectric cross-section of lead as a function of gamma ray energy.
  • 107. 92 Interaction of Radiation with Matter Chap. 3 Thecross-sectionape depends bothontheenergy E oftheincidentphotonandtheatomicnumberZ oftheatom.Figure3.15showsape forleadasafunctionofE. Itshouldbenotedthatape risestoverylargevaluesatlowenergy-lessthan1 MeV.Photonsinthisenergyregion obviously donotpenetratefarintoaleadtarget(orshield).AsalsoindicatedinFig.3.15,thereareanumberofdiscontinuitiesinape atlowenergy.Thesearecalledabsorption edges andcorrespondtoenergiesbelowwhichitisnotpossibletoejectcertainelectronsfromtheleadatom.Forinstance,belowtheK-edge,theincidentphotondoesnothavesufficientenergytoejectaK-electron-themosttightly boundelectron.Thenextmosttightlyboundelectronsafterthe K-electronsarethe L-electrons. Forreasons unimportanttothepresentdiscussion, these electrons havethree slightly different ionization energies. ThethreeedgesdenotedinthefigureasLI , LII , andLIII correspondtotheminimumphotonenergiesrequiredtoejectthe 3 differentlyboundL-electrons.Abovetheedges,thatis,abovetheK-edge,ape dropsoffroughlyasE-3 Thephotoelectriccross-sectiondepends stronglyonZ, varyingas (3.59) wheren isthefunctionofE showninFig.3.16.Becauseofthestrongdependenceofape onZ, thephotoelectriceffectisofgreatestimportancefortheheavieratoms,suchaslead,especiallyatlowerenergies. Pair Production Inthis process,thephotondisappears andanelectronpair-apositronandanegatron-iscreated.Sincethetotalrest-massenergyofthe2 electrons is 2mec2 = 1.02MeV, this effectdoesnotoccurunless the photonhas atleast this much energy. Above this threshold, the cross-sectionfora pairproduction, app, increases steadily with increasing energy, asshowninFig. 3.17,wherethepairproductioncross-sectionisshownforlead.Sincepairproductionisanelectromagneticinteraction,itcantakeplaceonlyinthevicinity ofaCoulombfield. Atmosty-rayenergiesofinterest,this isthe 4.6 4.4n 4.2 0.2 0.3 0.5 E, MeV 2 3 Figure 3.16 The constant n in Eq. (3.53) as a function of gamma ray energy. (From R. D. Evans, The Atomic Nucleus, New York: McGraw­ Hill, 1955.)
  • 108. Sec. 3.8 y-Ray Interactions with Matter VJ c: � 40 �------------�-------------. 30 r-------------�----------�� .0. 20 r-------------�--_____;�------____j "" b"" 10 I------------,.,c- 10 Energy, MeV 100 Figure 3.17 The pair production cross-section of lead as a function of y-ray energy. 93 fieldofthenucleus,notthesurroundingelectrons.Asaresult,app isafunctionofZ, and,inparticular,variesasZ2, thatis, a "-' Z2pp (3.60) Thetotalkineticenergyofthenegatron-positronpairisequaltotheenergyofthephotonless 1.02 MeV.Onceformed,these electrons moveaboutandloseenergy asaresultofcollisionswithatomsinthesurroundingmedium.Afterthepositronhas slowed downtoverylowenergies, itcombineswithanegatron, thetwo particles disappear, and two photons are produced (annihilation radiation),eachhavinganenergyof0.5 1 1 MeV. The Compton Effect TheComptoneffect, or Compton scattering as itissometimescalled,issimplytheelasticscatteringofaphotonbyanelectron, inwhichbothenergyandmomentumareconserved.AsshowninFig.3.18, theinci­dentphotonwithenergyE andwavelength}..isscatteredthroughtheanglef} and E, I A ' photon ;;{scattered � _ E, _ A ___......... _ _ _ _ � _ _ _ _ _ Recoiling Incident photon electron Figure 3.18 The Compton effect.
  • 109. 94 Interaction of Radiation with Matter Chap. 3 thestruckelectronrecoils. Sincetherecoilingelectron acquires some kinetic en­ergy,theenergyE' ofthescatteredphotonislessthanE, andsincethewavelengthofaphotonisinverselyproportionaltoitsenergy(seeEq.2.22), thewavelengthA' ofthescatteredphotonislargerthanA. Bysettinguptheequationsfortheconser­vationofenergyandmomentum,itisnotdifficulttoderivethefollowingrelation: , EEeE = , E(l -cos1J) + Ee (3.61) whereEe = mec2 = 0.51 1 MeVistherest-massenergyoftheelectron.AformulaequivalenttoEq.(3.61) is where A'-A= Ac(l -cos1J), h 10AC = - = 2.426 x 10- cmmec iscalledtheCompton wavelength. (3.62) (3.63) InComptonscattering,thephotoninteractswithindividualelectrons,anditisthereforepossibletodefineaComptoncross-sectionperelectrone(JC. Thiscross­sectiondecreases monotonically withincreasingenergy from amaximumvalue0.665 b(essentially � ofabarn)atE = 0, whichisknownastheThompson cross­ section, (JT . Figure3.19 showse(Jc asafunctionofphotonenergy.Incidentally,forE » Ee, e(JC behavesroughlyasE-1 TheComptoncross-sectionperatom,(Jc,isequaltothenumberofelectronsintheatom-namely, Z-multipliedbye(JC. Thus, (3.64) Fromapracticalstandpoint,theComptoneffectisthecauseofmanydiffi­cultproblems encounteredinthe shieldingofy-rays. This isbecausethephotondoesnotdisappearintheinteractionasitdoesinthephotoelectriceffectandinpairproduction.TheCompton-scatteredphotonisfreetointeractagaininanotherpartofthesystem.Althoughitistruethatx-raysandAugerelectronsareemittedfollowingthephotoelectriceffectandthatannihilationradiationaccompaniespairproduction, theseradiationsarealwaysmuchlessenergeticthantheinitialphotonanddonottendtopropagateinmattertothe sameextent asCompton-scatteredphotons.Thismultiplescatteringofy-raysisconsideredagaininChapter 10. Attenuation Coefficients Thetotalcross-sectionperatomfory-rayin­teractionisthesumofthecross-sectionsforthephotoelectriceffect,pairproduc-
  • 110. Sec. 3.8 �� 1 .0 0.8 0.6 0.4 0.2 G 0. 1 � 0.08 0.06 0.04 0.02 0.01 0.01 y-Ray Interactions with Matter e U c = U T = 0.665 b at E = 0 -r---- ---- ----......... �j'..... �p...,. 0.02 0.04 0.06 0.08 0. 1 0.2 0.4 0.6 0.8 1 .0 Energy, MeV � � 2 i'... ", f' 4 6 8 10 95 Figure 3.19 The Compton cross-section per electron as a function of gamma ray energy. tion,andComptonscattering: (3.65) A macroscopic cross-sectioncanalsobe defined, like the macroscopic neutroncross-section,bymultiplyinga inEq.(3.65) bytheatomdensity N. Bytradition,suchmacroscopic y-raycross-sectionsarecalledattenuation coefficients andaredenotedbythesymbolJ-L. Thus, J-L = Na = J-Lpe + J-Lpp + J-Lc, (3.66) whereJ-L isthetotalattenuationcoefficientandJ-Lpe, J-Lpp, andJ-Lc aretheattenuationcoefficients forthe three interaction processes. Like macroscopic cross-sectionsforneutrons, the various J-L'S have units ofem-I. It is alsoconvenienttodefinethequantityJ-L/p, whichiscalledthemass attenuation coefficient,4 wherep isthephysicaldensity.FromEq.(3.66), thisisgivenby � = J-Lpe + J-Lpp + J-Lc . P P P P (3.67) 4A new standard is now in use in which I-L is usedto designate the mass attenuation coefficient and I-L* the attenuation coefficient. In this text, the older convention is used.
  • 111. 96 Interaction of Radiation with Matter 0.07 j!l 0.06 Eu C 0.0511) u S 11) 0.040 u c: .� 0.03� ::s c: 2 0.02� rJ:! rJ:! � ::is 0.01 0 0 2 3 4 5 6 7 8 9 10 Energy, MeV Figure 3.20 The mass attenuation coefficients of lead as a function of y-ray energy. Chap. 3 SinceJ-landphaveunitsofcm-1 andg/cm3,respectively,itfollowsthatJ-l/phastheunitscm2/g.Figure3.20 showsthemassattenuationcoefficients,onalinearscale,forlead.Thereisminimumin J-l/p atabout3.5 MeVbecauseapeandac decreasewithincreasing y-rayenergy, whereasappincreasesfromits threshold at 1.02 MeV.Also,asshowninthefigure,Comptonscatteringisthedominantmodeofinterac­tionfromabout0.5 MeVto5 MeV.BecauseapeandappdependmorestronglyonZthandoesac,theenergyrangeoverwhichComptonscatteringisdominantin­creaseswithdecreasingZ.5 Thus,inthecaseofaluminum,forinstance,Comptonscatteringpredominatesallthewayfrom0.06 MeVto20 MeV.AtenergieswhereComptonscatteringistheprincipalmodeofinteraction, J-l P J-lc P Nac p Introducingtheusualfonnulaforatomdensity(seeEq.2.59), N= pNAM ' whereNAisAvogadro'snumberandM isthegramatomicweight,gives � � N:c = NA(�)eac. 5This is also illustrated in Fig. 3.21 .
  • 112. Sec. 3.8 y-Ray Interactions with Matter 97 whereusehasbeenmadeofEq.(3.64).Acheckofthechartofthenuclidesshowsthat,exceptforhydrogenandtheveryheavyelements,theratioZ/M isapproxi­matelyequalto!. Thismeansthat,atthoseenergieswhereComptonscatteringisthedominantprocess,valuesofJ.L/p tendtoberoughlythesameforallelements.Thisis illustratedinFig. 3.21,whereJ.L/p is shownforanumberofelements asafunction ofy-rayenergy. Numerical values ofJ.L/p aregiven in Table 11.4 inAppendixII.Sinceattenuationcoefficientsareessentiallymacroscopiccross-sections,thevalueofJ.Lforamixtureofelementsisgivenbythe sameformulaasEq. (3.14). � Eu C Il.l .(3 !E Il.l 0 U c: 0.':: C'O ;:s c: � � � C'O � 5 1 .0 0.5 0. 1 0.05 0.01 L--_----'_---'-----'--'----'--L.....l--'--' __----'_---'-----'-----'----'---'---'--.l....I 0. 1 0.5 5 10 Gamma-ray energy, MeV Figure 3.21 The mass attenuation coefficients of several elements. (From S. Glasstone and A. Sesonske,NuclearReactorEngineering. New York: Van Nostrand, 1967; by permission, US DOE.)
  • 113. 98 Interaction of Radiation with Matter Chap. 3 Thus, /1- =/1-1 + /1-2 + (3.68) whereIL l , /1-2 andsoonarethevaluesof/1- forthevariousconstituents.Also,itisnotdifficulttoshowthatthemass absorption coefficientforthemixtureisrelatedtothemassabsorptioncoefficientsoftheconstituentsbythefonnula � = _1 [WI (�) +W2(�) + p 100 P I P 2 J. (3.69) whereWI,W2,andsoonarethepercentsby weightofthevariouselements, and(/1-/p)1 , (/1-/Ph, andsoonarethemassabsorptioncoefficientsoftheelements,asgiveninTable11.4. Equations(3.68)and(3.69)arevalidatallenergies.BythesameargumentthatledtoEq.(3.13),itiseasytoshowthat/1- isequaltotheprobabilityperunitpaththatay-raywillhaveacollisioninamediumandthat 1A= -/1- (3.70) isthemeanfreepathofthey-ray.Furthennore,if10 istheintensity(y-rays/cm2-sec) ofthe monoenergetic y-ray beam striking atargetofthickness X, then theintensityofthephotonsthatpenetratethetargetwithouthavingacollisionis (3.71) Intermsofthemassattenuationcoefficient,Eq.(3.71)maybewrittenas (3.72) Thequantity pX inEq. (3.72)hasunits ofg/cm2 andis equal tothenumberofgramscontainedinanareaof1cm2ofthetarget.Thicknessesofmaterialsareoftengiveninunitsofg/cm2incalculationsofy-rayattenuation.Itmustbeemphasized,asitwasintheearlierdiscussionofneutrons,thattheintensityI giveninEqs.(3.71)and(3.72)refersonlytothoseoftenveryfewy-raysthatdonotinteractinthetarget.Thesearebynemeanstheonlyphotonsthatappearonthefarsideofatargetorshield.PhotonsthathaveundergonemultipleComptonscattering, photonsfromannihilationradiationfollowingpairproduction,andthex-raysthatfollowthephotoelectriceffectmayalsopenetratethetarget.Alloftheseradiationsmustbetakenintoaccountinshieldingcalculations.MethodsfordoingsoaregiveninChapter10.
  • 114. Sec. 3.8 y-Ray Interactions with Matter 99 Example 3.15 Calculate the mass attenuation coefficient of D02 for 1 Me V y-rays. What is their mean free path? The density of V02 is about 10 g/cm3 Solution. The molecular weight of V02 is 238 + 2 x 16 = 270. The percent by weight that is uranium is then 238/270 = 88. 1 %; the remaining 1 1.9% is oxygen. From Table 11.4, IJ.,/P = 0.0757 cm2/g for uranium and 0.0636 cm2/g for oxygen. Thus, for V02 � = 0.881 x 0.0757 + 0. 1 19 x 0.0636 = 0.0743 cm2/g. [Ans.] p The value of J1, is 10 times this number, that is, J1, = 0.743 cm-1 since p = 10. The mean free path is 1 1 A = - = -- = 1.35 cm. [Ans.] J1, 0.743 Energy Deposition Incalculationsofradiationprotectiontobeconsid­eredinChapter9, itisnecessarytocomputetherateatwhichenergyisdepositedby a y-raybeamasitpasses throughamedium. By analogy withEq. (3.8), thetotalcollisiondensityatapointwherethey-rayintensityisI isgivenby F = I/L, (3.73) where /L is the total attenuation coefficient. Ifthe y-rays were absorbedateachcollision,thentherateatwhichenergyisdepositedperunitvolumeinthemediumwouldbesimplyEF = EI/L, whereE istheenergyofthey-rays.Withboththephotoelectriceffectandpairproduction,theincidentphotonisinfactabsorbed.Unlessthemediumisverythin,mostofthesecondaryradiationemitted subsequentto these interactions-the x-rays, electrons, and annihilationradiation-isalsoabsorbedinthemedium. Thus, thetotalenergyoftheincidenty-raycanbeassumedtobedepositedintheseprocesses. InComptonscattering,however,theonlyenergydepositedisthekineticenergyoftherecoilingelectron.Let T be theaverageenergy ofthis electron. The averageenergy deposited byCompton scattering is then TI/Lc, where /Lc is the Comptonattenuationcoeffi­cient. ItisnowconvenienttodefinetheCompton absorption cross-section aCa bytherelation Eaca = Tac. (3.74) ThecorrespondingCompton absorption coefficient /LCa isthengivenby E/Lca =T/Lc. (3.75) Intermsofthiscoefficient,theenergydepositionrateperunitvolumebyComptonscatteringissimplyEI/LCa.
  • 115. 1 00 Interaction of Radiation with Matter Chap. 3 ThetotalenergydepositionrateW perunitvolumefromphotoelectriceffect,pairproduction,andComptonscatteringcannowbewrittenas where W = EI(/.Lpe +/.Lpp + /.Lea) = EI/.La, /.La = /.Lpe +/.Lpp + /.Lea (3.76) (3.77) is calledthelinear absorption coefficient.6 Itcanbe seen that, by defining Mea accordingtoEq. (3.75), itispossibletotreatallthreemodesofy-rayinteractiononthesamefooting.Thequantity/.La/p iscalledthemass absorption coefficient, andrepresenta­tivevaluesaregiveninTable11.5. ItiseasytoseefromEq. (3.76) thattherateofenergydepositionperunitmassisequaltoEI/.La/p. Itshouldalsobementioned,inconcludingthissection,thattheproductEI thatappearsinEq. (3.76) iscalledtheenergy intensity orenergyflux. Thishastheunitsofenergypercm2/secandisequaltotherateatwhichenergyinay-raybeampassesintoamediumpercm2 Example 3.16 It is proposed to store liquid radioactive waste in a steel container. If the intensity of y-rays incident on the interior surface of the tank is estimated to be 3 x 101 1 y-rays/cm2-sec and the average y-ray energy is 0.8 Me V, at what rate is energy deposited at the surface ofthe container? Solution. Steel is a mixture ofmostly iron and elements such as nickel and chromium thathave about the same atomic number as iron. Therefore, as faras y-ray absorption is concerned, steel is essentially all iron. From Table U.5, J.lalP for iron at 0.8 Me V is 0.0274 cm2/g. The rate of energy deposition is then 0.8 x 3 x lOl l x 0.0274 = 6.58 x 109 Me V/g-sec. [Ans.] In SI units, this is equivalent to 1.05 J/kg-sec. 3.9 CHARGED PARTICLES Ordinarily,thereareonlythreevarietiesofchargedparticlesthatmustbedealtwithinnuclearengineeringproblems-namely,a-rays, fJ-rays, andfissionfragments.Beforeconsideringtheseparticularradiations,itisofinteresttodiscussthewaysinwhichchargedparticlesinteractwithmatter. 6/.Lais called the energy absorption coefficient by some authors.
  • 116. Sec. 3.9 Charged Particles 101 Considerachargedparticlethatisincidentonanatomlocatedatsomepointinbulkmatter. Anumberofdifferenteventsmayoccurasthisparticlenearstheatominquestion.Firstofall,becausetheparticleexertselectrical(Coulomb)forcesontheatomicelectrons,oneormoreoftheseelectronsmaybeplacedintoexcitedstatesoftheatom,oranelectronmaybeejectedfromtheatomaltogether,leavingtheatomionized. However,thechargedparticlemaypenetratethroughthecloudofatomicelectronsandbeelasticallyscatteredfromthenucleus.Sincemomentumandenergyareconservedinsuchacollision,thenucleusnecessarilyrecoils.Iftheincidentparticleissufficientlymassiveandenergetic,therecoilingnucleusmaybeejectedfromitsownelectroncloudandmoveintothemediumasanotherchargedparticle. Also, undercertain circumstances, the incident particle, especially ifitisana-particle,mayundergosomesortofnuclearreactionwhenitcollideswiththenucleus. Finally,theparticlemaybeacceleratedbytheCoulombfieldoftheelectronsorthenucleus;asaconsequence,aphotonmaybeemitted.Thistypeofradiation,whichisemittedwheneverachargedparticleundergoesacceleration,iscalledbremsstrahlung. Itisclearfromtheforegoingthattheinteractionofachargedparticlewithmatterisacomplicatedaffair.Inanycase,itisevidentthatinitspassagethroughmattersuchaparticleleavesatrailofexcitationandionizationalongitspath.Inthiscontext,becausetheyaredirectlyresponsibleforproducingthisionization,charged particlesarereferredtoasdirectly ionizing radiation. Bycontrast,unchargedpar­ticles,suchasy-raysandneutrons,leadtoexcitationandionizationindirectlyonlyafterfirstinteractinginthesubstanceandproducingachargedparticle.Forthisrea­son, y-raysandneutronsaresaidtobeindirectly ionizing radiation. Thisisnottoimplythaty-raysandneutronsdonotdirectlyformionswhentheyinteract-theydo.Forexample,ay-rayproducesphotoelectriceffect.Similarly,aneutroncollid­ingwithanucleusmayejectthenucleusfromitsatomasahighlychargedion.Buttheionizationarisingfromthesefirstinteractionsofy-raysandneutronsisentirelyinsignificantcomparedwiththeionizationcausedbythesubsequentinteractionofthechargedparticles.It is recalledthatthe interactions ofneutrons and y-rays are describedbycross-sections. Although it is also possible to define various interaction cross­sections forchargedparticles, it is more useful to describe the extentto whichchargedparticlesinteractwithmatterintermsofeithertheirspecific ionization ortheirstopping power. Thespecific ionizationofaparticleisdefinedasthenumberofionpairsproducedperunitpathtraveledbytheparticle. Anionpairisanion­izedatomtogetherwithitsejectedelectron.Thestoppingpoweristhetotalenergylostperpathlengthbyachargedparticle;thatis,itisthetotalrateofdecreaseintheenergyoftheparticlealong its path. Ifnuclearreactions involving theparti­cledonotoccur,thenthe stoppingpower, which isdenotedby S, canbewrittenas
  • 117. 102 Interaction of Radiation with Matter s _ (dE ) + (dE ) dx col dx Tad Chap. 3 (3.78) wherethefirsttennistheenergylossperunitduetocollisions,whichgiverisetoexcitationandionization,andthesecondtenngivestheenergylossbyradiation.ThefirsttenninEq.(3.78)iscalledthelinear energy transfer (LET),anditisofspecialinterestinconnectionwiththebiologicaleffectsofradiation.AsshowninChapter9, theseeffectsdependontheextenttowhich energy is depositedbyradiationasexcitationandionizationwithinbiologicalsystems.ThemagnitudeofLET increases rapidly withthe mass andcharge ofamovingparticle. Thus, theLETofa-particlesisconsiderablylargerthanforelectronsofthesameenergy.Forexample,theLETofaI-MeVa-particleinwaterisabout90keV/JLm,whereasitisonly0.19keV/JLmforaI-MeVelectron.Forthisreason,a-particlesandotherheavilychargedparticlesarereferredtoashigh LET radiation; electronsarecalledlow LET radiation. Althoughlinearenergytransferisduetotheinteractionsofchargedparticlesinmatter,itis alsopossible torefertotheLETofuncharged,indirectly ionizingradiation-thatis, y-raysandneutrons-since,asnotedearlier,thebulkofthelo­caldepositionofenergybytheseradiationsisduetotheionizationandexcitationthatoccurssubsequenttotheirfirstinteractioninthematerial.Becausey-rayspro­ducelowLETsecondaryelectrons, theyarecalledlowLETradiation. However,sinceneutroninteractionsleadtotheheavy,highLETchargedparticles,neutronsareknownashighLETradiation.ThedistinctionbetweenhighandlowLETradi­ationhasimportantbiologicalconsequencesthataredescribedinChapter9. Alpha Particles Becausea-particlesaresomassive,theyareonlyslightlydeflectedwhentheyinteractwithatomicelectrons.Therefore,theymoveinmoreorlessstraightlinesastheytravel inamedium. However, asana-particleslowsdown, it becomes increasingly probablethatit will capture anelectron to fonnanHe+ ionandthen capture asecondelectrontobecomeaneutralheliumatom.When, ultimately, this atom is fonned, the specific ionization abruptly drops tozero.ThissituationisillustratedinFig.3.22,wherethespecificionizationisshownforanenergetica-particleasafunctionofdistancefromtheendofitstrackindryair at 15°C and 1 atm pressure. The maximum value ofthe specific ionizationshowninthefigureis 6,600ionpairs/mm. Atthispoint, thea-particleshaveanenergyofabout0.75MeV.AcurveofthetypegiveninFig. 3.22isknownasaBragg curve. Thepoint atwhich theionizationfalls tozero is called the range ofthe(t­particle.1Aswouldbeintuitivelyexpected,therangeisamonotonicallyincreasing 7Because of the statistical nature of the processes involved, all a-particle tracks do not tenni­ nate at precisely the same point. The range described earlier is the most probable endpoint.
  • 118. Sec. 3.9 Charged Particles E E ..... 8000 & 6000 � "@ 0.. s::: 4000 .s 2000 6600 o �------�------�--------�--�--� 3 2 Distance from end of track, cm Figure 3.22 Specific ionization of an a-particle in air. o 103 functionoftheinitial energyoftheparticle.Therangeasafunctionofenergyisshown inFig. 3.23 fora-particlesinair. Ranges ofa-particlesinothermaterialscanbefoundfromtherangeinairbyusingtheBragg-Kleeman rule: � -4 ,JM- = 3.2 x 10 -- Ra. Ma P (3.79) Inthis formula, R istherangein a substance ofphysical density p andatomicweight M, andRa, Pa, andMa aretherange, density, andaverageatomicweightofair, respectively. Thenumerical constant in Eq. (3.79) is computedforairat15°C and 1 atm.Forcompoundsormixtures, ,JM inEq. (3.79) istobereplacedby ,JM =Yl JAi; + Y2jM; + (3.80) where Yl , Y2, andsoonarethefractionsofatoms present having atomicweightsMI, M2, andsoon.Therelative stopping power ofamaterialisdefinedastheratiooftherangeofa-particlesinairtotherangeofa-particlesinthesubstanceinquestion.FromEq.(3.79), thisisgivenby Relativestoppingpower= Ra = 3100 �.R v M (3.81) It should be noted that the relative stopping powerisindependent ofthe initialenergyoftheparticle.
  • 119. 1 04 G 14 0 � ::s U '-' 13 1 2 Interaction of Radiation with Matter 8 > 7 0 � :>. e.o 0 6s::: 0 >. e d 5 4 ,-., 10 ,-., 3 Q:l � 0 0 � � ::s ::s � U 9 '-' 2 8 7 2 3 4 5 6 (Curve A) 6 8 9 10 1 2 (Curve B) 1 2 1 3 1 4 15 1 6 1 8 (Curve C) Mean range air-em 05°C; 760 mm Hg) Figure 3.23 Range ofa-particles in air as a function of energy. Chap. 3 7 1 3 1 9 The stopping powers ofmost materials are quite high and the ranges ofa-particlesareconsequently very short. Forexample, therangeofa5-MeVa­particleinaluminumisonly0.0022 em-aboutthethicknessofthinaluminumfoil.Mosta-particlesarestoppedbyanordinarysheetofpaper;theyarealsostoppedintheoutennostlayersoflivingtissue.Thus,theshieldingofa-particlesdoesnotordinarilyposeadifficultproblem.However,thepresenceofa-decayingnuclidescannotbeignoredinmanyengineeringproblemssince,asisshowninChapter9, thesenuclidescanleadtoserIoushealthhazardswheningestedorinhaled. ,a-Rays Theattenuationoff3-raysinmatterinsomewaysismorecompli­catedthanfora-particles.Tobeginwith,f3-raysareemittedinacontinuousenergy
  • 120. Sec. 3.9 Charged Particles 105 spectrum. Furthennore, although theyinteractwith atomsinthesamemannerasa-particles,{3-rays,beinglessmassiveparticles,aremorestronglydeflectedineachencounterwithanatom.Asaresult,f3-raysmoveincomplicated,zigzagpathsandnotinstraightlinesasdoa-particles.Nevertheless,ithasbeenfoundexperimentallythatthespecificionizationofabeamoff3-raysvariesapproximatelyexponentiallywithdistanceintoanabsorber.This phenomenon appears to be an accident ofnature, due in part to the shapeofthef3-rayspectrum. Ifi(x) isthespecific ionization atthedistancex intotheabsorber,then (3.82) where io isthe specificionizationatx = 0 andP is thedensityofthemedium.Theapparentmassattenuationcoefficient/LIP isalmostindependentoftheatomicweightofthemediumandincreasesonly slowly withtheatomicnumber.Anap­proximate,empiricalfonnulafor/LaIp, basedonmeasurementsinaluminum,is /L 17 - = -- , p E�l: where/LIp isincm2/gandEmax isthemaximumf3-rayenergyinMeV. (3.83) Amoreusefulparameterrelatedtotheattenuationoff3-raysisthemaximumrangeRmax. Thisisdefinedasthethicknessofabsorberrequiredtostopthemostenergeticoftheelectrons.TheproductRmaxP, whichistherangeexpressedinunitsofg/cm2,isroughlyindependentofthenatureoftheabsorbingmedium.ValuesofRmaxP areshowninFig.3.24 asafunctionofthemaximumelectronenergyEmax. Thesedatacanberepresentedbythefollowingempiricalfonnulas: and R P = 0 412E(1.265-0,0954 In Emax) max . max ' RmaxP = 0.530Emax - 0. 106, Emax < 2.5 MeV Emax > 2.5MeV Intheseequations,RmaxP ising/cm2andEmax isinMeV. (3.84) (3.85) Equations(3.84) and(3.85) giveslightlylowervaluesofRmaxP forairthanareactuallyobserved. Figure3.25 gives the measured ranges off3-raysforairat15°C and 1 atmpressure.ItisevidentfromacomparisonofFigs. 3.23 and3.25, andalsobycomparingcalculatedranges,thatf3-rayspenetrateconsiderablyfurtherintomaterialsthana-raysofcomparableenergy.Forinstance,f3-rayswithEmax = 3MeVhavearangeinairof13 m, whereas therangeof3-MeVa-particlesisonly 1.7 cm.However,f3-raysdonotpenetratefarintonongaseousmaterialsand,asaresult,theyarenotdifficulttoshield.
  • 121. 106 N 8 � c:i. � 8 t:a:: Interaction of Radiation with Matter 1 .6 1 .4 1 .2 1 .0 / 0.8 0.6 0.4 0.2 / / V V / /o o 0.5 1 .0 1 .5 2.0 / / V V V 2.5 3.0 3.5 Figure 3.24 Maximum range of f3-rays as a function of maximum en­ ergy; notvalid for air. Chap. 3 Example 3.17 Sodium-24 (TI/2 = 15 hr) is often used in medicine as a radioactive tracer. It emits f3-rays with a maximum energy of 1.39 MeV. What is the maximum range of the f3-rays in animal tissue? Solution. The density ofmost animal tissue is approximatelyunity. From Eq. (3.84), Rmax is then Rmax = 0.412 X 1.39(1.265-0.0954In 1.39) = 0.412 X 1.391.234 = 0.618 cm [Ans.] The range ofan a-particle of the same energy in tissue is about 9 x 10-4 cm. Fission Fragments InSection3.7,itwaspointedoutthatfissioningnu­ clei almost always split into two fragments ofunequal mass. Since momentummustbe conserved in fission, the lighter group offragments receives somewhatmoreenergythantheheaviergroup.Therefore,thedistributionoffissionfragment
  • 122. Sec. 3.9 Charged Particles 100. � :.-- �./10.0 /' /' /' lL '" ....Q.) d) E / ;, E / /I Q:::; Q.) QO :::: C<$ � 1 .0 I -f- I I I 0. 1 o 2 3 4 Figure 3.25 Maximumrange off3-raysin airat 15 C and 1 atmpressure. 1 07 kineticenergiesexhibitstwopeaks,as showninFig. 3.26, withthehigherenergypeakcorrespondingtothelighter-fissionfragments.Asseeninthefigure,thelightergroupwithA:::::::: 95hasanenergyofapproximately 100 MeV,whereastheheaviergroupwithA:::::::: 140 receivesabout68 MeV.Oncefonned,thefissionfragmentsripthroughtheelectroncloudoftheorginalfissioning nucleus as they pass into the surrounding medium. In so doing, theyusually pickup anumberofelectrons, although neverenough tofonnaneutralatom.The virgin fission fragment thus appears as a highly energetic andhighlyionizedatom.Theaveragechargeofthelightergroupisapproximately+20e; thatoftheheaviergroupisabout+22e.
  • 123. 108 Interaction of Radiation with Matter Lighter group 40 60 80 1 00 E, MeV Figure 3.26 Energy distribution of fission fragments as a function of energy in Me V. Chap. 3 Because oftheir large charge, fission fragments specific ionization is veryhigh,andtheirrangeiscorrespondinglyshort.Table3.7 gives approximaterangesofthe lighter, more energetic, and hence morepenetrating fragments in variousmaterials.Therangeoffissionfragmentsisanimportantfactorinthedesignofthefuelrodsforpowerreactors.Sincepower(fromthefissioningfuel)isproducedintheserods,theymustbecooledbythepassageofasuitablecoolantmaterialalongtheirsurfaces,asshowninFig. 3.27. However,itisimportantthatnoneoftheradioac­tivefissionfragmentsentersandtherebycontaminatesthecoolant.Topreventthe escapeofsuchfragments,thefuelis eitherwrappedinalayerofnonfuel-bearingmaterial,suchasstainlesssteeloranalloyofzirconium,orplacedinhollowtubesfabricatedfromthesematerials.InviewoftheshortrangesindicatedinTable3.7, thisfuelelement, cladding asitiscalled,canbequitethin-oftennomorethan0.05 cmthick. TABLE 3.7 RANGES OF FISSION FRAG MENTS Medium Range, 10-3 em Aluminum Copper Silver Uranium Uranium oxide (U30S) 1 .4 0.59 0.53 0.66 1.4
  • 124. References REFERENCES General Figure 3.27 A fuel rod with cladding. 109 Beiser, A., Concepts ofModern Physics, 5th ed. New York: McGraw-Hill, 1995, Chapters 3, 4, 5, and 1 1. Burcham, W. E., Nuclear Physics: An Introduction. reprint, Ann Arbor. Evans, R. D., The Atomic Nucleus. New York: McGraw-Hill, 1955, Chapters 12-14, 18- 25. Foderaro, A., The Elements ofNeutron Interaction Theory. Cambridge, Mass.: MIT Press, 1971. Foster, A. R., and R. L. Wright, Jr., Basic Nuclear Engineering, 4th ed. Paramus, Prentice­ Hall, 1982, Chapters 4 and 8. Glasstone, S., and A. Sesonske, NuclearReactorEngineering, 4th ed. New York: Chapman & Hall, 1994, Chapters 1 and 2. Kaplan, I., Nuclear Physics, 2nd ed. Reading, Mass.: Addison-Wesley, 1963, Chapters 15, 16, 18, and 19. Lamarsh, J. R., Introduction toNuclearReactor Theory. Reading, Mass.: Addison-Wesley, 1966, Chapters 2 and 3. Lapp, R. E., and H. L. Andrews, Nuclear Radiation Physics, 4th ed. Englewood Cliffs, N.J.: Prentice-Hall, 1972, Chapters 1 1-13. Meyerhof, W. E., Elements ofNuclear Physics. New York: McGraw-Hill, 1967, Chapters 3 and 5. Oldenberg, 0., and N. C. Rasmussen, Modern Physics for Engineers. reprint, Marietta, Technical Books, 1992, Chapters 16 and 17.
  • 125. 1 10 Interaction of Radiation with Matter Chap. 3 Semat, H., and 1. R. Albright, Introduction to Atomic and Nuclear Physics, 5th ed. New York: Holt, Rinehart & Winston, 1972. Serway, R. A., Moses, C. J., and C. A. Moyer, Modern Physics, 2nd ed. Philadelphia, Saunders, 1997. Weidner, R. T., and R. L. Sells, ElementaryPhysics: ClassicalandModern, 3rd ed. Boston, Mass.: Allyn & Bacon, 1980. Cross-Section Data Neutronandcharged-particlecross-sectiondataarecollected,evaluated,anddis­tributedbytheNationalNuclearDataCenteratBrookhavenNationalLaboratory,Upton, New York. Neutron cross-section data are published in the reportNeu­ tron Cross-Sections, BNL-325,andotherdocuments.Thesecross-sectionsarealsoavailablefromtheCenterondiskinstandardfonnatforreactorandshieldingcom­ andanumberofothersites.Forlatestupdateofavailability, PROBLEMS 1. Two beams of 1-eV neutrons intersect at an angle of90° The density of neutrons in both beams is 2 x 108 neutrons/cm3 (a) Calculate the intensity of each beam. (b) What is the neutron flux wherethe two beams intersect? 2. Two monoenergetic neutron beams of intensities It = 2 X 10lO neutrons/cm2-sec and h = 1 X 1010 neutrons/cm2-sec intersect at an angle of30° Calculate the neutron flux and current in the region where they intersect. 3. A monoenergetic beam of neutrons, cp = 4 X 1010 neutrons/cm2-sec, impinges on a target 1 cm2 in area and 0.1 cm thick. There are 0.048 x 1024 atoms per cm3 in the target, and the total cross-section at the energy of the beam is 4.5 b. (a) What is the macroscopic total cross-section? (b) How many neutron interactions per second occur in the target? (c) What is the collision density? 4. The {3- -emitter 28Al (half-life 2.30 min) can be produced by the radiative capture of neutrons by 27AI. The 0.0253-eV cross-section for this reaction is 0.23 b. Suppose that a small, O.Ol-g aluminum target is placed in a beam of 0.0253-eV neutrons, cp = 3 X 108 neutrons/cm2-sec, which -strikes the entire target. Calculate (a) the neutron density in the beam; (b) the rate at which 28Al is produced; (c) the maximum activity (in curies) that can be produced in this experiment.
  • 126. Problems 1 1 1 5. Calculate the mean free path of 1-eV neutrons in graphite. The total cross-section of carbon at this energy is 4.8 b. 6. A beam of 2-Me V neutrons is incident on a slab of heavy water (D20). The total cross-sections ofdeuterium and oxygen at this energy are 2.6 b and 1 .6 b, respectively. (a) What is the macroscopic total cross-section of D20 at 2 Me V? (b) How thick must the slab be to reduce the intensity of the uncollided beam by a factor of 10? (c) If an incident neutron has a collision in the slab, what is the relative probability that it collides with deuterium? 7. A beam of neutrons is incident from the left on a target that extends from x = 0 to x = a. Derive an expression for the probability that a neutron in the beam will have its first collision in the second half of the targetthat is in the region a12 < x < a. 8. Boral is a commercial shielding material consisting of approximately equal parts by weight of boron carbide (B4C) and aluminum compressed to about 95% theoretical density (2.608 g/cm3) and clad with thin sheets of aluminum 0.25 cm thick. The man­ ufacturer specifies that thereare 0.333 g ofboron per cm2 of a boral sheet 0.457 cm in overall thickness. What is the probability that a 0.0253-eV neutron incident normally on such a sheet will succeed in penetrating it? 9. Whatis the probability thataneutron can move one mean free path without interacting in a medium? 10. A wide beam of neutrons of intensity ¢o is incident on a thick target consisting of material for which aa » as . The target area is .A and its thickness is X. Derive an expression for the rate at which neutrons are absorbed in this target. 11. Stainless steel, type 304 having a density of 7.86 g/cm3, has been used in some re­ actors. The nominal composition by weight of this material is as follows: carbon, 0.08%; chromium, 19%; nickel, 10%; iron, the remainder. Calculate the macroscopic absorption cross-section of SS-304 at 0.0253 eV. 12. Calculate at 0.0253 eV the macroscopic absorption cross-section of uranium dioxide (U02), in which the uranium has been enriched to 3 wlo in 235U. The density ofU02 is approximately 10.5 g/cm3 13. The compositions of nuclear reactors are often stated in volumefractions, that is, the fractions of the volume of some region that are composed of particular materials. Show that the macroscopic absorption cross-section for the equivalent homogeneous mixture ofmaterials is given by La = fI La l + h La2 + where Ji and Lai are, respectively, the volume fraction and macroscopic absorption cross-section of the ith constituent at its normal density. 14. Calculate La and Lf (at 0.0253 eV) for the fuel pellets described in Problem 2.63. The pellet density is about 10.6 g/cm3 15. Using the fact that the scattering cross-section of 209Bi is approximately 9 b from 0.01 eV to 200 eV, estimate the radius of the 209Bi nucleus and compare with Eq. (2.3).
  • 127. 1 12 Interaction of Radiation with Matter Chap. 3 16. Using the Breit-Wigner formula, compute and plot ay in the vicinity of the first reso­ nance in 238U, which occurs at an energy of 6.67 eY. The parameters of this resonance are: rn = 1.52 meV (meV = millielectron volts), ry = 26meV, and g = 1 . [Note: Por reasons given in Section 7.3 (see in particular Pig. 7.12), the values of ay computed from the Breit-Wigner formula do not coincide with those measured with targets at room temperature.] 17. Demonstrate using the Breit-Wigner formula thatthe width of a resonance at half its height is equal to r 18. There are no resonances in the total cross-section of 12C from 0.01 eV to cover 1 Me Y. If the radiative capture cross-section of this nuclide at 0.0253 eV is 3.4 mb, what is the value of ay at 1 eV? 19. The first resonance in the cross-section ofaluminum, which is due entirely to scatter­ ing, occurs at 5.8 keY. The absorption cross-section at 0.0253 eV is 0.23 b. Calculate for 100 eV: (a) aa, (b) as, (c) at . 20. Calculate I:a for (a) water of unit density at 0.0253 eV; (b) water of density 0.7 g/cm3 at 0.0253 eV; (c) water of density 0.7 g/cm3 at 1 eY. 21. The first resonance in the scattering cross-section of the nuclide AZ occurs at 1.24 Me Y. The separation energies of nuclides A-1 Z,A Z, and A+1 Z are 7.00, 7.50, and 8.00 Me V, respectively. Which nucleus and at what energy above the ground state is the level that gives rise to this resonance? 22. There is aprominentresonance in the total cross-section of56Pe at 646.4 keV. At what energy, measuredfromthe ground state, is the energy level in 57Pe that corresponds to this resonance? [Hint: Use the masses ofneutral 56Pe and 57Pe to compute the binding energy ofthe last neutron in 57Pe.] 23. The excited states of 170 occur at the following energies (in Me V) measured from the ground state: 0.871, 3.06, 3.85, 4.55, 5.08, 5.38, 5.70, 5.94, etc. At roughly what energies would resonances be expected to appear in the neutron cross-section of 160? 24. Using Eq. (3.28), compute and plot E'/E as a function of angle from 0 to Jr for A = 1, 12, and 238. 25. A 2-Me V neutron traveling in water has a head-on collision with an 160 nucleus. (a) What are the energies ofthe neutron and nucleus after the collision? (b) Would Y0U expeet the water molecule involved in the collision to remain intact after the event? 26. A I-Me V neutron strikes a 12C nucleus initially at rest. If the neutron is elastic scat­ tered through an angle of 90Q: (a) What is the energy of the scattered neutron? (b) What is the energy ofthe recoiling nucleus? (c) At what angle does the recoiling nucleus appear?
  • 128. Problems 1 13 27. Compute and plot the average fractional energy loss in elastic scattering as a function ofthe mass number of the target nucleus. 28. Show that the average fractional energy loss in% in elastic scattering for large A is given approximately by �E 200 E A 29. A 1.5-Me V neutron in a heavy water reactor collides with an 2H nucleus. Calculate the maximum and average changes in lethargy in such a collision. 30. Suppose that a fission neutron, emitted with an energy of 2 Me V, slows down to an energy of 1 eV as theresult ofsuccessive collisions in a moderator. If, on the average, the neutron gains in lethargy the amount � in each collision, how many collisions are required if the moderator is (a) hydrogen, (b) graphite? 31. The 2,200 meters-per-second flux in an ordinary waterreactor is 1 .5 x 1013 neutronsl cm2-sec. At what rate are the thermal neutrons absorbed by the water? 32. At one point in a reactor, the density of thermal neutrons is 1.5 x 108 neutrons/cm3 If the temperature is 450DC, what is the 2,200 meters-per-second flux? 33. A tiny beryllium target located at the center of a three-dimensional Cartesian coordi­ nate system is bombarded by six beams of 0.0253-eV neutrons of intensity 3 x 108 neutrons/cm2-sec, each incident along a different axis. (a) What is the 2,200 meters-per-second flux at the target? (b) How many neutrons are absorbed in the target per cm3/sec? 34. When thermal neutrons interact with 14N, what is the probability that absorption leads to radiative capture? 35. The control rods for a certain reactor are made of an alloy of cadmium (5 wj0), indium (15 wjo) and silver (80 wjo). Calculate the rate atwhich thermal neutrons are absorbed per gram of this material at a temperature of 400 DC in a 2,200 meters-per­ second flux of 5 x 1013 neutrons/cm2-sec. [Note: Silver is a lIv absorber.] 36. From the data in Table 3.3, would you expect 232Th to be fissile? Ifnot, at what neutron energy would you expect fission to be possible? 37. Two hypothetical nuclei, A Z and A+l Z, of atomic weights M(A Z) = 241.0600 and M(A+l Z) = 242.0621 have critical fission energies of 5.5 Me V and 6.5 Me V, re­ spectively. Is the nuclide A Z fissile? 38. Fission can be induced when y-rays are absorbed by a heavy nucleus. What energy y-rays are necessary to induce fission in (a) 235U, (b) 238U, (c) 239pu? 39. Cross-sections of 235U at 1 Me V are as follows: aa = 4.0 b, ai = 1 .4 b, af = 1.2 b, and aa = 1.3 b. The cross-sections for neutron-producing and charged-particle reactions are all negligible. Compute at this energy
  • 129. 1 14 Interaction of Radiation with Matter Chap. 3 (a) the total cross-section, (b) the capture-to-fission ratio a. 40. The fission-product 131I has a half-life of 8.05 days and is produced in fission with a yield of 2.9%-that is, 0.029 atoms of 131 I are produced per fission. Calculate the equilibrium activity of this radionuclide in a reactor operating at 3,300 MW. 41. Fission-product activity measured at the time to following the burst of a nucleus weapon is found to be ao. Show that the activity at the time t = 7nto is given ap­ proximately by a = ao/IOn This is known as the 7-10 rule in civil defense. 42. Suppose that radioactive fallout from a nuclear burst arrives in a locality I hour after detonation. Use the result of Problem 3.41 to estimate the activity 2 weeks later. 43. The yields of nuclear weapons are measured in kilotons (KT), where 1 KT = 2.6 x 1025 Me V. With this in mind, (a) How much 235U is fissioned when a IOO-KT bomb is exploded? (b) What is the total fission-product activity due to this bomb 1 min, 1 hr, and 1 day after detonation? [Note: Assume a thermal energy release of 200 Me V per fission.] 44. A research reactor is operated ata powerof250kilowatts 8 hours a day, 5 days a week, for 2 years. A fuel element, 1 of 24 in the reactor, is then removed for examination. Compute and plot the activity of the fuel element as a function of time up to 2 years after removal. 45. The spontaneous fission rate of 238U is 1 fission per gram per 100 sec. Show that this is equivalent to a half-life for fission of 5.5 x 1015 years. 46. Compute and plot the parameter 1J at 0.0253 eV for uranium enriched in 23SU as a function of its enrichment in weight percent 235U. 47. Suppose that 1 kg of23sU undergoes fission by thermal neutrons. Compute the masses (or mass equivalents) in grams forthe following, which are produced: (a) neutrons, (b) fJ-rays, (c) y-rays, (d) neutrinos, (e) kinetic energy, (0 fission products. 48. The reactor on the nuclear ship Savannah operated at a power of 69 MW. (a) How much 235U was consumed on a IO,OOO-nautical-mile voyage at an average speed of 20 knots? (b) This is equivalent to how many barrels of 6.5-million-Btulbarrel bunker-C oil? 49. Consolidated Edison's Indian Point No. 2 reactor is designed to operate at a power of 2,758 MW. Assuming that all fissions occur in 235U, calculate in grams per day the rate at which 235U is (a) fissioned, (b) consumed. 50. Referring to the preceding problem, what is the total accumulated activity of the fis­ sion products in the Indian Point No. 2 reactor I day after shutdown following 1 year ofoperation?
  • 130. Problems 1 15 51. The photoelectric cross-section of lead at 0.6 Me V is approximately 18 b. Estimate ape at this energy for uranium. 52. A 2-Me V photon is Compton scattered through an angle of 30° (a) What is its energy after scattering? (b) What is the recoil energy of the struck electron? (e) At what angle does the electron appear? 53. Show that Eq. (3.62) follows from Eq. (3.61). 54. Show that the minimum energy ofthe scattered photon in Compton scattering is given by and thatfor E » Ee, , EEe (E )min = -2-E-+ - E - e ' (E')min = Eel2 = 0.255 Me V. 55. What is the minimum energy of a Compton-scattered photon if its original energy is (a) 0.1 Me V, (b) I Me V, (e) 10 Me V? 56. Calculate the mass attenuation coefficient of silica glass (Si02, p = 2.21 g/cm3) for 3-Me V y-rays. 57. Derive Eq. (3.69). 58. The mass attenuation coefficient of lead at 0.15 Me V is 1.84 cm2/g. At this energy, the principal mode of interaction is by the photoelectric effect. What thickness of lead is required to reduce the intensity of a 0.15-Me V y-ray beam by a factor of 1,000? 59. The density ofair atstandardtemperature and pressure (O°C and 1 atm) is 1 .293 x l0-3 g/cm3 Compute the mean free paths of photons in air under these conditions and compare with the corresponding mean freepaths in unit-density water at the following energies: (a) 0.1 Me V, (b) 1 Me V, (e) 10 Me V. 60. At 1 Me V, the Compton cross-section per electron is 0.21 12 b, and the Compton energy absorption cross-section per electron is 0.0929 b. (a) What is the average energy of the recoiling electron in a Compton interaction at this energy? (b) Compute the Compton mass attenuation and mass absorption coefficients at 1 Me V for (i) aluminum, (ii) water. 61. A beam of O.I-Me V y-rays with an intensity of 5 x 106 y-rays/cm2-sec is incident on thin foils of (i) aluminum, (ii) water. At this energy, the Compton cross-section per electron is 0.4929 b, and the Compton energy absorption cross-section per electron is 0.0685 b. Calculate the energy extracted from the beam per unit volume of the foils due to
  • 131. 1 16 (a) Compton scattering, (b) the photoelectric effect. Interaction of Radiation with Matter Chap. 3 62. The absorption ofradiation is often measured in units called rads, where 1 rad is equal to the absorption of 100 ergs per gram. What intensity of 1 Me V y-rays incident on a thin slab ofwater is required to give an absorption rate of 1 rad per second? 63. Detennine the range of 5-Me V a-particles in the following media: (a) air at 15°C, 1 atm; (b) aluminum; (c) lead; (d) unit-density water; (e) air at 300°C, 10 atm. 64. Detennine the relative stopping powers of the media in the preceding problem. 65. Compare the apparent mass attenuation coefficient of 2-Me V (maximum energy) f3-rays with the mass attenuation coefficient of 2-Me V y-rays in aluminum. 66. Compare the maximum ranges of 3-Me V a-rays and f3-rays in air at standard tem­ perature and pressure. 67. Near the surface of a flat fuel element in an operating reactor, fissions are occurring at the constant rate of S fissions/cm3-sec. Given that the average range of the fission fragments is R, show that the rate at which such fragments would escape per cm2/sec from the surface of the fuel if it were not clad is equal to SRj2.
  • 132. 4 Nuclear Reactors and Nuclear Power Having reviewed in the preceding Chapters the atomic and nuclear physics thatform the foundation for nuclear engineering, it is now possible to consider themannerbywhichnuclearenergyisutilizedforpracticalpurposes. 4.1 THE FISSION CHAIN REACTION Nuclearenergyisreleasedbywayofafissionchainreaction.Inthisprocess,whichisdepictedinFig.4.1, neutronsemittedbyfissioningnucleiinducefissionsinotherfissileorfissionablenuclei;theneutronsfromthesefissionsinducefissionsinstillotherfissileorfissionablenuclei;andsoon.Suchachainreactioncanbedescribedquantitativelyintennsofthemultiplication/actor, whichisdenotedbythesymbolk. Thisisdefinedastheratioofthenumberoffissions(orfissionneutrons)inonegenerationdividedbythenumberoffissions(orfissionneutrons)intheprecedinggeneration.Inequationfonn,thisis numberoffissionsinonegenerationk = --------------------------------- numberoffissionsinprecedinggeneration (4.1) 117
  • 133. 1 18 Nuclear Reactors and Nuclear Power Figure 4.1 Schematic representation of fission chain reaction. Chap. 4 Ifk is greaterthan 1, thenfromEq. (4.1) the number offissions increasesfromgenerationtogeneration.Inthiscase,theenergyreleasedbythechainreac­tionincreaseswithtime,andthechainreaction,orthesysteminwhichitistakingplace,issaidtobesupercritical. However,ifk islessthan1,thenumberoffissionsdecreaseswithtimeandthechainreactioniscalledsubcritical. Finally,inthespe­cialsituationwherek isequalto 1,thechainreactionproceedsataconstantrate,energyisreleasedatasteadylevel,andthesystemissaidtobecritical. Devicesthataredesigned sothatthefissionchainreactioncanproceedina controlled mannerarecallednuclear reactors. Inareactor,thiscontrol isaccom­plishedbyvaryingthevalueofk, whichcanbedonebythepersonoperatingthesystem.Toincreasethepowerbeingproducedbyareactor,theoperatorincreasesk toavaluegreaterthanunitysothatthereactorbecomessupercritical. Whenthedesiredpowerlevelhasbeenreached,hereturnsthereactortocriticalbyadjustingthevalueofk tobeunity,andthereactorthenmaintainsthespecifiedpowerlevel.Toreducepowerorshutthereactordown,theoperatormerelyreducesk, makingthereactorsubcritical.Asaresult,thepoweroutputofthesystemdecreases.Nu­clearbombsandexplosivescannotbecontrolledinthisway,andthesedevicesare notreferredtoasreactors. SeveraltypesofnuclearreactorsaredescribedlaterinthisChapter.
  • 134. Sec. 4.2 Nuclear Reactor Fuels 1 19 Tomakeareactorcritical, orotherwisetoadjustthevalueofk, itisneces­sarytobalancetherateatwhichneutronsareproducedwithinthereactorwiththerateatwhichtheydisappear. Neutronscandisappearintwowaysastheresultofabsorptioninsometypeofnuclearreaction,orbyescapingfromthesurfaceofthereactor.Whenthesumoftheneutronabsorptionandleakageratesisexactlyequaltotheneutronproductionrate,thenthereactoriscritical.Iftheproductionrateisgreaterthanthesumoftheabsorptionandleakagerates,thereactoris supercrit­ical;conversely, ifitis smaller, thereactoris subcritical. As wouldbeexpected,theproduction,absorption,andleakageratesdependonthesizeandcompositionofareactor.Itis showninChapters5 and6 thatitis possible toderiveaccurateanalyticalrelationshipsbetweenthesethreeratesontheonehandandthesizeandcompositionontheother.Theserelationshipsdeterminethedimensionsandmate­rialpropertiesnecessaryforareactortobecritical. 4.2 NUCLEAR REACTOR FUELS Conversion and Breeding ItisrecalledfromSection3.7 thatTJ fissionneutronsareemitted,ontheaverage,perneutronabsorbedbyafissileorfissionablenucleus.Tobecomecritical,areac­torobviouslymustbefueledwithanuclidehavingavalueofTJ greaterthan 1. IfTJ werelessthan 1, afissioninonegenerationwouldnecessarilyleadtolessthanonefissioninsucceedinggenerations,andthereactorcould neverachieveacrit­icalstate.However, TJ cannotbeexactly 1 since, asalreadynoted, someneutronsinevitablyarelosteitherinnonfissionabsorptionreactionsorbyescapingfromthesystemaltogether,andcriticalitycouldnotbereached.AcheckoffissiondatashowsthatTJ is, infact, substantially greaterthan 1 forthefissilenuclidesatall incidentneutronenergies,andforthefissionablebutnonfissilenucleisuchas232Thand238Uwellabovetheirfissionthresholds.How­ever, with thenonfissile nuclides, only thoseneutronsin areactorwithenergiesabovethefissionthresholdareabletoinducefission.Forareactorfueledonlywithfissionablebut nonfissile nuclides, the effective value ofTJ is equal to the actualvalueofTJ, multipliedbythefractionoftheneutronsabsorbedinthereactorwithenergiesabovethefissionthreshold.Regrettably,thisnumberisalwayslessthan1, anditfollowsthatreactorscannotbemadecriticalwithnonfissilematerialalone.Fissilenuclidesarethereforeessentialingredientsinallreactorfuels.Onlyonefissilenuclide,235U,isfoundinnature,whereitoccurswithaniso­topicabundanceof0.72 a/o. Theremainderofnaturaluranium,exceptforatraceof234U,is238U.Thus, only 1 uranium atomoutof139 is235U. Despitethis lowconcentrationoffissileisotopeitispossibletofuelcertaintypesofcriticalreactorswithnaturaluranium,andalloftheearlyreactorswereofthistype.However,most
  • 135. 120 Nuclear Reactors and Nuclear Power Chap. 4 modemreactorsrequireenriched uranium -thatis,uraniuminwhichtheconcen­tration of235U has beenincreasedoveritsnaturalvalue. MethodsforenrichinguraniumarediscussedinSection4.7.Asisthecaseforanynaturalresource,thesupplyof235Uisfinite.ItisshowninSection4.6thattheworld'sresourcesof235Ufornuclearfuelarelimited.Indeed,ifnuclearpowerwasbasedonthefissionof235Ualone,theeraofnuclearenergywouldbeacomparatively shortone-probably notmorethanacentury indura­tion. Fortunately, itispossibletomanufacturecertainfissileisotopesfromabun­dantnonfissilematerial,aprocessknownasconversion. Thetwomostimportantfissileisotopesthatcanbeproducedbyconversionare233Uand239Pu.The233Uisobtainedfromthoriumbytheabsorptionofneutrons.Thereactionsinvolvedareasfollows: (4.2) AstheresultofthereactionsinEq.(4.2),thenonfissileisotope232Thisconvertedinto fissile233U. Isotopes like232Th, which arenotfissilebutfromwhichfissileisotopescanbeproducedbyneutronabsorption,aresaidtobefertile. Itisarelatively simplemattertobringaboutthereactionsgiveninEq.(4.2).Naturally occurring thorium is entirely 232Th. Therefore, it is merely necessaryto introduce thorium, in one form or another, into a critical reactor where it isexposedtoneutrons.Afterasuitableirradiationtime,whenthe233Uhasbuiltuptoadesiredlevel,thethoriumiswithdrawnfromthereactorandthe233Uisextractedfromthethorium.Thiscanbedonebychemicalmeanssincethoriumanduraniumaretwoentirelydifferentchemicalelements.Withoneexception (seeSection4.5),however,comparativelylittle233Uhasbeenproducedfromthoriumsincethereisalmostno demand for233U as areactorfuel. Nevertheless, thoriumremains animportantpotentialsourceofnuclearfuel.239puisobtainedfromreactionssimilartoEq.(4.2)-namely, 238U(n,y)239U � 239Np � 239PU. (4.3) Thefertileisotopeinthiscaseis the238U;239Npistheintermediatenucleus. Torealizethesereactionsinpractice,238Umustbeirradiatedinareactor.This,how­ever,occursautomaticallyinmostpowerreactorsofcurrentdesign.Thus,asnotedearlier,mostpresent-dayreactorsarefueledwithuraniumthatisonlyslightlyen­richedin235U.Practicaliyallofthefuelinthesereactorsis238U,andtheconversionof238Uto239pU takesplaceasamatterofcourseduringthenormaloperationofthesereactors.Theplutoniumislaterextractedchemicallyfromthefuel,uraniumandplutoniumbeingdifferentelements.Thechemicalextractionoffissilematerial fromfertilematerialiscalledjUeZreprocessing andisdescribedinSection4.8.After239Pu hasbeen formedinareactor, itmay absorb aneutron andun­dergofissionorbetransformedinto240pU. The240pU, whichisnotfissile,mayin
  • 136. Sec. 4.2 Nuclear Reactor Fuels 121 tumcaptureanotherneutrontoproducethefissileisotope241Pu.Finally,the241Pumay undergo fissionorbetransformedinto 242Pu. Thus,theplutoniumextractedfrom reactorfuel by reprocessing contains, in decreasing amounts, the isotopes239pu, 240pu, 241Pu,and242pU. Thefractionalcontentofeachisotopedependsonthebumupofthefuel,definedlaterinthissection,atthetimeofreprocessing.Theconversionprocessis describedquantitativelyintermsoftheparameterC, which is calledthe conversion ratio orsometimes the breeding ratio. This isdefinedastheaveragenumberoffissileatomsproducedinareactorperfissilefuelatomconsumed. Thus,whenN atomsoffuelareconsumed, NC atomsoffertilematerialareconvertedintonewfissileatoms.However,ifthenewlyproducedfis­sileisotopeisthesameastheisotope thatfuelsthereactor,thenewatoms maylaterbeconsumedtoconvertanotherNC x C = NC2 atomsoffertilematerial;thesemaybeconsumedtoconvertNC3 fertileatoms;andsoon. Inthisway,itiseasytoseethattheconsumptionofN fuelatomsresultsintheconversionofatotalof NC 1 - C fertileatoms,providedC islessthan 1 . WhenC = 1 , aninfiniteamountoffertilematerialcanbeconvertedstartingwithagivenamountoffuel. Example 4.1 In a critical reactor fueled with natural uranium, it is observed that, for every neu­ tron absorbed in 235U, 0.254 neutrons are absorbed in resonances of 238U and 0.640 neutrons areabsorbed by 238U at thennal energies. There is essentially no leakage of neutrons from the reactor. (a) What is the conversion ratio for the reactor? (b) How much 239 Pu in kilograms is produced when I kg of 235U is consumed? Solution. 1. Each absorption of a neutron by 238U, whether at resonance or thermal en­ ergies, produces an atom of 239pu via reaction Eq. (4.3). Furthennore, each time a neutron is absorbed by a 235U nucleus, that nucleus is consumed. The compound nucleus either undergoes fission or emits a y-ray to become 236U. Thus, the total number of 239pU atoms produced per 235U atom consumed is 0.254 + 0.640 = 0.894. This, by definition, is the value of the conversion factor, so that C = 0.894. [Ans.] 2. The consumption of I kg of 235U involves the disappearance by neutron ab­ sorption of 1,000NA/235 atoms of 235U, where NA is Avogadro's number. From the definition of C, this is accompanied by the production of C x 1000NA/235 atoms of239pu, which have a total mass of C x 1000 x 239/235 g or e x 239/235 kg. Thus, the consumption of I kg of 235U leads in this
  • 137. 122 Nuclear Reactors and Nuclear Power Chap. 4 example to the production of 239 0.894 x 235 = 0.909 kg of 239Pu.[Ans.] A most important situation occurs when C is greater than 1. In this case, morethanonefissileatomisproducedforeveryfissileatomconsumed,whichisaprocessdescribedasbreeding. Reactorsthataredesignedsothatbreedingwilltakeplacearecalledbreeder reactors orsimplybreeders. Reactorsthatconvertbutdonotbreedarecalledconverters; reactorsthatneitherconvertnorbreedbutsimplyconsumefuelarecalledburners. Breedersareremarkabledevicesfor,inadditionto providing powerthroughthe energy released in fission,they actually producemorefissilefuelthantheyconsume.Needlesstosay,itismoredifficulttodesignareactorthatwillbreedthanonethatmerelyconverts.Foronething,while11 mustbegreaterthan 1forconversion,itmustbegreaterthan2 forbreeding.Thisisbecause,inanyreactor,onefissionneutron musteventually be absorbedin fueljust tokeepthereactorcritical andmaintainthechainreaction.Ifthereactoristobreed,morethanoneneutronmustbeabsorbedin fertile material toproducethe new fissile isotope. In actualfact,11 mustbe substantiallygreaterthan 2 for, as notedearlier,in anyreactorsomeneutronsinevitablyareabsorbedbynonfuelatomsorlostbyleakage.Inthis connection, itis importanttorecognizethat11 is notaconstant,butdependsontheenergyoftheneutronthatinducesthefission.Thisvariationof11 withneutronenergyis shownin Fig. 4.2 forthethreefissileisotopes233U, 235U, and239Pu. Atthermal neutron energies (i.e., E � 0.025 eV),thevalueof11 for233Uisabout2.29, whichis sufficientlyinexcessof2 forbreedingtobepossible.Thus, aproperlydesignedreactorin which most ofthe fissions areinducedbythermalneutrons-thatis, athermal reactor---could breedifitwerefueledwith233U. By contrast, the thermal neutron values of11 for235U and239pu, 2.07 and2.14, respectively,arenotsufficientlygreaterthan2 topermitbreeding.Returning to Fig. 4.2, it is observedthatin the intermediateenergy range,fromabout 1 eVto 100 keYfor235Uandfromabout 10 eVto20 keYfor239pu,thevalueof11 fallsbelow2. Breedingcannotbeachievedwhentheseisotopesareusedtofuelareactorinwhichmostofthefissionsoccurattheseenergies. Ithasnowbeendemonstratedthatsuchanintermediatereactorcanbemadetobreed,atleastinalimitedway, whenfueledwith233U.Thevalueof11 for233U isgreaterthan2 by asufficientmarginatintermediateenergies tomakebreedingpossible(seetheLWBRdiscussioninSection4.5). Aboveabout100 keV,asindicatedinFig.4.2, 11 risestovaluessubstantiallyabove 2 forall threeofthe fissile fuels. As far as the value of11 is concerned,it should therefore be possible tobreedwith these fuels provided thereactorisdesignedinsuchawaythatthebulkofthefissionsareinducedbyneutronswith
  • 138. Sec. 4.2 Nuclear Reactor Fuels 123 4.0 r-------------------------------------------------------� 3.0 <"l � 2.0 I � � V"l <"l N I � 1.0 O.O �--�----�----�----�----�----�--�----�----�----� 10-3 10-2 10-1 101 102 103 104 105 106 107 Energy, eV (a) 4.0 �------------------------------------------------------� 3.0 2.0 1.0 O.O �--�----�----�----�----�----�--��--�----�----� 10-3 10-2 10-1 1 102 Energy, eV (b) 103 Figure 4.2 Variation of 11 with energy for (a) 233U and (b) 235U. (Plot­ ted by machine from data on tape at the National Neutron Cross Section Center, Brookhaven National Laboratory.)
  • 139. 124 ::s p., 0'> ,...N I � Nuclear Reactors and N uclear Power Chap. 4 4.0 .-------------------------------n 3.0 2.0 1.0 0.0 '--_---'-__---'-__--1....-__.l.....-_---'-__---'-__--'--__.l.....-_----L__---' to-3 to-2 to-I 1 lOZ Energy, eV (c) lOS Figure 4.2 Variation of 71 with energy for (c) 239pu. (Plotted by ma­ chine from data on tape at the National Neutron Cross Section Center, Brookhaven National Laboratory.) sufficiently high energies. Reactors ofthis typeare calledfast reactors or, sincetheyareusuallydesignedtobreed,fast breeders. Itmustbenotedthat, althoughthevalueof1] for235U ishighenoughforbreeding,unfortunatelythereisnofertilematerial innaturethatcanbe converted into 235U. Thus, although afastreactorfueledwith 235U may produce more239Pu from238U thanthe235U itconsumes, the235U isusedupforever.Somefastbreeders,intheabsenceofaninitialsupplyofplutonium,havebeenstartedwith235U fuel,buttheyareultimatelyfueledwithplutonium.Thisfuelisalsopreferredover233U becauseofitshighervalueof1]. Thus,fastbreedersarealmostalwaysfueledwithplutonium.Theextenttowhichbreedingoccursinareactorisdescribedbythebreeding gain-a parameter denoted by the symbol G. This is defined asthe net increase inthenumberoffissileatomsinareactorperfuel atomconsumed. SinceC,thebreedingratio,isthetotalnumberoffuelatomsproducedperfuelatomconsumed,itfollowsthatGandCaresimplyrelatedby G =C- l. SupposethatC= 1 .2. Thismeansthat1 .2 newatomsoffuel are produced foreach atomconsumed.Thenetincreaseinthenumberoffuelatomsperatomconsumedisthenclearly0.2, andthebreedinggainis0.2.
  • 140. Sec. 4.2 Nuclear Reactor Fuels 125 Breeding is also described intermsofthedoubling time. This is definedasthehypotheticaltime interval duringwhich the amountoffissile material in (orassociatedwith)areactordoubles.Tocomputethedoublingtime, supposethatareactorisoperatedataconstantthermalpowerlevelofPo megawatts. Thisreac­torconsumesfissilematerialattheuniformrateofwPo gramsperday,wherew isthefuelconsumptionrateperunitpower. (Inthecaseofa235U-fueledreactoraccordingtoSection3.7 w = 1.23 gperdayperthermalmegawatt.)Thisisequiv­alenttotheconsumption ofwPoNAIMf atomsoffuel, where NA isAvogadro'snumberandMf istheatomicweightofthefuel.FromtheearlierdefinitionofG, theconsumptionofwPoNAIMf fuelatomsproducesGwPoNA1Mf atomsoffueloverandabovethoseconsumed,whichmeansthatthereisanetproductionrateofGwPo gramsoffuelperday.1 Ifnoneofthisnewfuelisremovedfromthereactor,thetotalamountoffuelinthesystemwillincreaselinearlywithtimefromtheini­tialfuelinventorymo. Thelineardoubling time isdefinedasthetimetDl requiredforthetotalamountoffuelinthereactortoreachthevalue2mo. Clearlythen, and GwPo x tDl = mo mo tDl = -- . GwPo (4.4) Itisnotdifficulttoseethatpermittingallofthenewlyproducedfueltoac­cumulateinthereactoris awastefulprocedure.Theextrafuel isnotrequiredtokeep thereactoroperating andcouldbetter be removedand, togetherwith fuelfromotherbreeders,beusedtofuelanotherbreedingreactor.Inthisway,thetotalpowerproducedfromall thefuel canbe increased asthe fuel mass increases inboththeoriginaland secondbreeder.Thismodeofoperationalsocoincidesmorecloselywithactualpracticesincebreederreactorsarerefueledatregularintervals(normallyaboutonceortwiceevery2 years)overtheirlifetime.Itcanbeshown(seeProblem6.8) thatthereactorpowerwhichcanbepro­ducedfromagivenfuelmassisproportionaltothemass-thatis, P = 13m, (4.5) where{3 is aconstant. However, as shownearlier,therateofincreaseinmassisgivenby dm - = GwP dt ' IThis assumes that the newly produced fuel and the fuel consumed have the same atomic weight. If they are different, the production rate must be multiplied by the ratio of their atomic weights.
  • 141. :: ;>:. .... 0 � � .. .S Q) ;::3 (..L., 126 Nuclear Reactors and Nuclear Power Chap. 4 anditfollowsthat Thisequationhasthesolution dm - = Gwf3m. dt wheremo isagaintheinitialfuelinventory. (4.6) Theexponentialdoublingtime tDe, sometimescalledthecompounddoubling time, is now defined asthetimeinwhich m reaches 2mo accordingtoEg. (4.6). Thisiseasilyseentobe However,fromEg.(4.5), In2 tDe = --. Gwf3 Po f3 = -,mo (4.7) wherePo istheinitialpower.Substitutingf3 intoEg.(4.7) thengives mo ln 2 tDe = --- GwPo (4.8) ComparingEgs.(4.4) and(4.8) showsthat tDe = tD/ ln 2 = O.693tD/. (4.9) Thegrowthintotalfuelinventoryandthetwodoublingtimes,tD/ andtDe, areshowninFig.4.3. Inpractice,bothtD/ andtDe havebeenused.However,because 2rno rno Figure 4.3 Growth in the inventory of flIel by breeding "meier two conditions: (a) new fuel continually extracted and used for further breeding, (b) new fuel left in reactor.
  • 142. Sec. 4.2 N uclear Reactor Fuels 127 tDe reflects amore usual modeoffuelmanagment,itis frequentlyreferredtoasthedoubling time. Itshouldbeemphasizedthattheactualtimerequiredbyareactorownertodoublehisinventoryoffissilematerialisgreaterthaneitherofthedoublingtimescomputedpreviously,sincethetimerequiredtoremovethenewlyproducedfuelfromthereactorandhaveitchemicallyseparatedandfabricatedintoafonnsuit­ableforfuelingthereactorwasomittedinthederivation. Nevertheless, thedou­blingtimeisareasonablemeasureofmeritfortheperfonnanceofabreeder.Thus,allotherthingsbeingequal(whichtheyneverare),thebestbreederisclearlytheonewiththelowestvalueoftDe. Example 4.2 A hypothetical fast breeder reactor is fueled with a mixture of239pu and 238U. When operating at full power, the plutonium is consumed at a rate of approximately 1 kg per day. The reactor contains 500 kg of 239pu at its initial startup, and the breeding gain is 0.15. (a) At what rate is 239Pu being produced? (b) Calculate the linear and exponential doubling times of this reactor. Solution. 1. From the definition ofbreeding gain, the reactor produces 0.15 extra atoms of 239Pu for every atom of 239Pu consumed. On a per kilogram basis, this means that the reactor produces 0.15 additional kilograms of 239Pu for each kilogram consumed. Since the consumption rate of 239pU is 1 kg/day, it follows that the instantaneous rate of plutonium production is 0.15 kg/day or about 55 kg/yr. [Ans.J 2. Thefactor wPo in Eq. (4.4) is the total fuel consumption rate, which is 1 kg/day in this problem. From that equation, it follows that 500kg 500 kg tDl = = = 9.1 yr. [Ans.] 0.15 kg/day 55 kg/yr Then from Eq. (4.9), tDe = 0.693 x 9.1 = 6.3 yr. [Ans.J Nuclear Fuel Performance-Burnup and Specific Burnup The total energy released in fission by agiven amountofnuclearfuel is calledthe fuel bumup and is measured in megawatt days (MWd). The fission energyreleasedperunit massofthefuelis tennedthespecific bumup ofthefuelandisusuallyexpressedinmegawattdayspermetrictonorperkilogram(thatis,MWdltorMWd/kg)oftheheavy metaloriginallycontainedinthefuel.AccordingtothediscussioninSection3.7, thefissioningof1.05 gof235U yields 1 MWd. Thus,it
  • 143. 128 Nuclear Reactors and Nuclear Power Chap. 4 followsthatthemaximumtheoreticalbumupforthisfuelis or950 MWd/kg. 1 MWd 106g--- X -- = 950,000MWdlt1.05g t Fuelperformanceis also describedintermsoffractional burnup (oftenex­pressedin %), denotedasf3 anddefinedastheratioofthenumberoffissionsinaspecifiedmassoffueltothetotalnumberofheavyatomsoriginallyinthefuel.Thatis, numberoffissionsfj = initialnumberofheavyatoms' (4.10) Sincethefissionofallfuelatoms(fj = 1) yields950,000MWdltandthespecificbum