HIiUduCIIUH IUPuCCui LHIHCCiIHJHIIG GI!IOHJOHH . dHdISHLate Professor with the New York Polytechnic Institute /HIHOHy J. dIdIIdPennsylvania State University !IPHIICP!3!!

iOnIicO allLppOi 5addlOiVOi, POv JOi8Oy U43Libray of Congress Cataloging-in-Publication Data is on fle. 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: Carle Anson ml 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 of Frmatome Technologies 20 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 eforts include the development, research, and testing of the theories and programs to determine their efectiveness. 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 JN ^c"cH"Prentice-Hall Interational (UK) Limited, London Prentice-Hall of Australia Pty. Limited, Sydney Prentice-Hall of Canada Inc., Tornto Prentice-Hall Hispanoamericana, S.A., Mexico Prentice-Hall of India Prvate Limited, New Delhi Prentice-Hall of Japan, Inc., Toko Pearson Education Asia Pte. Ltd., Singapore Editora Prentice-Hall do Brasil, Ltda., Rio de Janeir /|/BC |D !G||O LO|||DGThis 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 development of advanced plant designs, the signifcant 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, !ec/eakeacosaa!ec/eaIoeThe chapter is revised to include a discussion of Soviet-design reactors and technology. The use, projection, and cost of nuclear power worldwide is updated to the latest available information. In Chapter |:keaco!|ceas|agaaa]e(,the Chemobyl accident is discussed along with the latest reactor safety study, NUREG :::aA section i s also included that describes non-power nuclear accidents such as Tokai-Mura. iii iv Preface to Third Edition The basic material in Chapters .is updated to include newer references and to refect the author' s experience i n 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 ofen 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 i n 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! /|/BC |D cCDGO LO|||DGAt his untimely death in July 1 98 1 , John R. Lamarsh had almost completed a revision of the frst edition of Introduction to Nuclear Engineering. The major part of his effort went into considerable expansion of Chapters 4, 9, and .:and into the addition of numerous examples and problems in many of the chapters. However, the original structure of that edition has been unchanged. Chapter 4, Nuclear Reactors and Nuclear Power, has been completely restructured 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 efects of radiation, and there is a new section on the calculation of radiation efect. The section on the sources of radiation, both artifcial and natural, 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 effuents and radiation doses from nuclear facilities have been considerable expanded to cover new concepts and situations. Included in this chapter i s v vi Preface to Second Edition a discussion of the accident at Three Mile Island. The structure of this chapter has been kept as it was in the frst edition in spite of the earlier suggestion that it be broken up into two chapters treating environmental efects separately from safety and licensing. Several errors that were still present in the last printing of the frst edition have been corrected, including those in Example and 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 Nonan C. Rasmussen, Raphael Aronson, Marvin M. Miller, and Edward Melkonian for their assistance in the fnal stages of this revision. Finally, we ae grateful for comments and suggestions received from users of the earlier edition of thi s book. Although all their suggestions could not be incorporated, the book is greatly improved as a result of their comments. !o:eooe1982 Addison-Wesley Publishing Company Reading, Massachusetts /|/BC |D /||S| LO|||DG.................,,......,..,.......,..,.......,...,..............,...,.......,..,..,. .........,...,.9 .. ............,..,.........., ........,...........,... ...,...,,....,...,,...............,......,.....,..,...,..,.....,.....,..............,.....,..,.......,.,.....,,.......,.,...........,.....,......,,......,,.... .....,...,..,...,............,..,,.....,.,.............,.,..,..,,.,..,......,....,,...........,.vii viii Preface to First Edition ......,.....,.....,..,......,.................,4 ,..........,....,....,........,........,.,.....,.................,....,. ........,..,,.,.,................,......,.........,....,....,.,......,...........,,.,.....,.......,...............,.,.,.....,.........,...8 ...,......,.....,.......,.......,....,.,.....,.,......,..................,...,...,.9 ..:a....,....,....,.....,.,...,...,......,::...,..,.....,.....,.....,...................,.......,.................................................,......,.....,.....,...... . .,.................,...,.........,..,.,.....................,....,........................,...,,.. ....,....,.........,.,....,......,...,,..........................,,................,...... .,..,.............. . ......,....,....................................,,.....:a..,.............,........,::........,..........,............,.....,........,.,.,.......... F .... ... .., J. ....... .,...F ......,,...........................,...........................,.,.,... Preface to Fi rst Editi on ix .,.............,...........,...........,....,...,,...,.........,...,.........,,.........,............., .. .Acknowledgement ................,........,,......,....,.............................,..,3011302, .....F .......... ... . .......... ........................ .... ...... ...F ....F .....E. ..,. ...F .. .....,..... ......Dedication .............,,,...,.,..,..,LDG|G|S1 NUCLEAR ENGINEERING 1 2 ATOMIC AND NUCLEAR PHYSICS 5 2.1 ........ 5 2.2 ........ 7 2.3 ......,. 8 2.4 ...... 11 2.5 .......,, 11 2.6 .....,.. 14 2.7 .......... 15 2.8 ....,......., 18 2.9 ...,..... 22 2.10 ..... 26 2.11 ...,..,, 29 2.12 ... 33 2.13 ....,...... 37 2.14 ...., 40 xi xii .. +.. 45 3 INTERACTION OF RADIATION WITH MATER 3.1 ...... 52 3.2 ..... 54 3.3 ...... 57 3.4 .... 60 3.5 ......... 62 3.6 ..,,.......,... 68 3.7 ... 74 3.8 ..,....... 90 3.9 ..,.. 100 .. 109 .. 110 4 NUCLEAR REACTORS AND NUCLEAR POWER 4.1 .......... 117 4.2 ..... 119 4.3 ....,......... 129 4.4 .,....... 133 4.5 ............,,,.,... 136 4.6 ..,. 185 4.7 .,.,... 201 4.8 .,..., 217 4.9 ........,.. 219 .. 223 .. 224 5 NEUTRON DIFFUSION AND MODERATION 5.1 .... 230 5.2 . .... 231 5.3 ...,......, 235 5.4 ........,... 237 5.5 ....,... 238 5.6 ...........,... 240 5.7 .........,. 246 5.8 ...,...... 248 5.9 ........... 252 5.10 ...,........ 257 Contents 52 117 230 Contents xiii .. 260 .. 260 6 NUCLEAR REACTOR THEORY 266 6.1 ..,..,... 266 6.2 ..... 271 6.3 .....,. 274 6.4 ....,..,... 282 6.5 ...... 286 6.6 .. 297 6.7 .,.,..... 308 6.8 .,..... 309 .. 320 .. 321 7 THE TIME-DEPENDENT REACTOR 327 7. 1 ......... 328 7.2 .... 330 7.3 .......... 348 7.4 ..,........, 365 7.5 ......., 376 7. 6 ,...,.. 389 .. 397 .. 398 8 HEAT REMOVAL FROM NUCLEAR REACTORS 403 8.1 .....,........ 404 8.2 ........ 408 8. 3 ...,... 417 8. 4 ......... 428 8.5 ..,...... 441 8. 6 ......,... 450 .. 457 .. 459 9 RADIATION PROTECTION 466 9.1 ..,...... 467 9.2 ..... 468 9.3 ......,.,, 476 9.4 ...,....... 479 xiv Contents 9.5 ...................,. 485 9.6 .......... 495 9.7 .............. 499 9.8 ......... 506 9.9 .,......,...... 511 9.10 ............ 526 9.11 ..,.....,... 535 ..., 539 .. 542 .. 544 10 RADIATION SHIELDING 548 10.1 .....,...,...,.. 549 10.2 ............ 559 10.3 ...... 566 10.4 ...... 571 10. 5 ..,... 573 10.6 ......,.,....., 576 10.7 ........ 578 10.8 .......,............. 584 10.9 ........... 588 10.10 ..... 590 10.11 ...,..,. 595 10.12 ..... 599 10.13 ....... 604 .. 605 .. 606 11 REACTOR LICENSING, SAFETY AND THE ENVIRONMENT 612 11.1 .......,...,.., 613 11.2 ....., 614 11.3 .,........, 623 11.4 ..,........... 631 11.5 ............ 650 11.6 ..., 669 11.7 ... 681 11.8 .....,.. 701 11.9 ............. 710 .. 721 .. 723 Contents APPENDIXES .......... 731 ................ 737 .,.....,......... 745 . ...,......,..,. 751 . ....... 757 INDEX xv 761 1/UC|B| cGQ|G||GQ...,..,...................................,......,.......,..,....,..,..,,...,...,....,..........,.........,...........,........,,,.......,.......,...,..............,,...,.........,,..,.............,,..,.................,,.,,....,........,,..,......,......,.......,......,....,....... ..,................,.............,...,....,.............,..,..,,.............,..,....,,........,..,,...,..,...............,..,...........,.,.....,..,..,,,.........,..,......,........,..... ............1 2 Nucl ear Engi neeri ng Chap. 1 ....,..................,......,...,........,,..............,.,.....,.. ..,.......,...,,..,,... . ..............................................,,...,...,. ,......................,............,,..,...,.....,....,.....,..,...,,,. .,.......,,....,..,........ ....,.......,........,.....,.,...,......,........,.........,......... ......,.,.....,............. ......

......,,.....,....,........,,............,.,.......................,....,........,........,,..,.............,......,......,,..........................................,,...,.,..........,,........,,.................,,......,............... ...,.......,.....,.. . ..,a:aaaaa, ...,.,... ...., .........,........,..,................,..,.....,..................,.........,,,..........oouaaa,,......,,........,..........Lea|a.................,,.....,.........................,...,,..,........,.....,..,,.... . ....,,............,..,,...,...,....,..,....,,....,...,...,......................,..,...................,...........,....,.....,..,.....,,.........,.........,..,...........Chap. 1 Nucl ear Engi neeri ng 3 ........................,,....,,..........,..,.......,.......,.,,.,.........,.,,.........,.,...,....,......,......,,...............,.,..,,.,.............,,...........,...,...,. ..,............. ...,........,......,..,,.......,...............,.........,..........................,...,...........,......,.,..........,...........................,.................,,...,.........,,...,. .,........................,..,.............,...........,,.....,,....................,........,..........,..........,.,...,,... ..,....,. .............,...,....,....,,..........,...,.....,..,...........,.... .....,............,...........,.,...............,..........,..............,.................,.................,.....,.,...,..,...,......,......,...,...,.........,............,....,..................,.................,........,..................,.....,...,.......,......,................,,,....,................. ................,........,.,...............,,......,.....,.............,.......,...................,.....,...,.......ac|:a|oaaaa/ys|s, .........,.,.......................4 Nucl ear Engi neeri ng Chap. 1 ..,.............,...,...............,..................,.,.,..,....,..........,,.. .... . ,......,.............,.......,..1012), .......,.....,..............,......................,.,...,..c|D||C BGO /UC|B| /GyS|CS...,.......,.,............,........,..................,..,,...........,.........,.,..........,.....2. 1 FUNDAMENTAL PARTICLES ..,.,.....,....................,.. .........,..............,..........,.........,,......,.,,.......,,........,...,..,...........,........,..........,..........,...........oaoas...,.....,...................,...,....,..........,..............,.,............,,.......,........,.....,... 5 6 Atomi c and Nucl ear Physics lhoton8lholonFigure 2.1 The annihilation of a negatron and positron with the release of two photons. Chap. 2 .......,..............,....,.,..........,......,.......,...,..,.......,..,1 Electron ............

H = 9.10954 X 10-31 .,3 .......,e= 1.60219 10-19 .......,,... ..,.,..,....,e, ...,.,.,....,+e. ..,......,....,...,........,....aegaoas...,.........................jos|oas,...,.......,,.......,.......,........,,..........,.,...,....,............., 2.1. ...,.........e/ecoaaaa|a|/a|oa,...,......,,...aaa|a|/a|oaa|a|oa.Proton ...,..........H_ = 1.67265 10

.,.....,....,,.....,.....,........,....,...........,.....,.......,..,'A discussion of quark theory may be found in several of the particle physics references at the end of this chapter. According to the theory of relativity, the mass of a paricle is a function of its speed relative to the observer. In giving the masses of the fundamental paricles, it is necessary to specify that the particle is at rest with respect to the observer-hence, the ter rest mass. A 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. Sec. 2. 2 Atomi c and Nucl ear Structure 7 Neutron ............,.,.,.........,....,o, = ..

.,....,.............,..,...................,..,..........,... .,.... ..........,............,.. .....Photon .........,..............,...................,...........,..,....,.,...........,,..........,....,........,..........,,..,.........,........jaooa.....,..............,...............,..,...,..,,.c= . X

..Neutri no .......,...............,...,,.....,............,,......,.....e/ecoaaee|aos..e/ecoaaa|aee|aos)..,......,............,..,..,.,........,..............,...,,....... 2.2 ATOMIC AND NUCLEAR STRUCTURE .................,..,........................................,...,,...,.,............,.,........ ......,...........aoo|caeooe......,....,........,........ ..............,....,.....,....,.............,...............,..............,.,,..................................................aeeoaaeooe....,! ......aec/eoas-tbat.,...............,...+! = A, ..A ...aoo|coassaeooeaec/eoaaeooe........,............,.......,..........aec/|es........,.....,.......,..................,.,8 Atomi c and Nucl ear Physi cs Chap. 2 .......= A z). ......,.......,,.,z = .....,,.......

....,,...............,............eee|eoaea:yayogea),

......,z= ...............,.............,..,z.............,....H,;............. ..

................,................z......A), .......|soojes. .,,............,.

(Z = = ........... a|oac|:e).,.

..

,z= = . . ....,............................,..,...........,.....,..........,..,..................., .....,..,.,,......., ....,

.. ....

......,...,..........,...,,.........,.......,.....,.... .,..,............,....,..,........a}o. Example 2.1 A glass of water is known to contain 6. 6 1 024 atoms of hydrogen. How many atoms of deuterium eH) are present? Solution. According to Table .2 in Appendix II, the isotopic abundance of 2H is 0. 01 5 a/o. The fraction of the hydrogen, which is 2H, is therefore 1 .5 1 0-4 The total number of2H atoms in the glass is then 1 . 5 1 0-4 6. 6 7 1 024 9. 9 1 020 [Ans. ] 2. 3 ATOMIC AND MOLECULAR WEI GHT ..aoo|c e|ga..............................

..............,.

...,...,., ...,..o,z)............,az..

........

.......,.

az, M,az) , .,..,ao,z)M, z) = .X

. o, . Sec. 2. 3 Atomi c and Mol ecul ar Wei ght 9 ..,,..........,.,......,..

.....,.,. .............,. . .= ........ . ...................,......,...........a:e:age...,.........Yi ...,........,.. ..,...,.Mi, ......,...... ......................

.....oo/ece/a:e|ga......,........,...............,,.,.........

.......,.... . = Example 2.2 Using the data in the following table, compute the atomic weight of naturally occurring oxygen. Isotope Abundance (a/o) 99. 759 0. 037 0. 20 Solution. From Eq. (2. 2), it follows that Atomic weight 1 5. 99492 1 6. 9991 3 1 7.9991 6 M(O) 0. 01 [ye60)M(170) + y(I70)M(170) + yesO)MesO)] - 1 5.99938. [Ans ] ....,..............,............,...............,....g:aoaoo|c e|ga ..g:ao oo/ece/a: e|ga ...................,......,...,.........,......................oo/e......,.....,...

..., .,...,..

,... ,.........,.....................,..,......................,............,..................,,.....,,............,.........................,..........

10 Atomi c and Nucl ear Physi cs Chap. 2 ...................,....... ., ................A:ogao's/a,.................A:ogaosaeooe.........,!a...,..!a= ....X ..,.,. .....,...,.........,.......,...,...

........ .,......!a................ .

= .= .x

, ...X ... ...................................,.,.......aoo|coassea|,......................... ........,..,...,.....,o

..,.,,..,.,.,.....= x .X

,= l }!a,= .X

, ..E. . .............,...........,,......,.....,...40rdinarily a number of this type would be written as 6.0222045 X However, in nuclear engineering problems, for reasons given in Chap. 3 (Example 3. 1 ), Avogadro's number should always be written as the numerical factor times 1 024 Sec. 2. 5 Mass and Energy 1 1 2.4 ATOMIC AND NUCLEAR RADI I ........ .........................,.....,....,................,,......................,..,....,.....................,...,..........,...,,...,...........,..2 x . .....................,..............,...,....................................,....,...........,........................................ .........,,...........,....,.......,..,...,....= 1. 25fr x A

. ...........A................ ...........,..,,............,. ......V ......,,..A. .............A}\...........,......................,.......,,..........,.,...............,....,..,......,.,.............,.,.,..,,...2.5 MASS AND ENERGY .....,....... ..,..,..........,,.,............,.....,.......,..,.....ro ..........,,L,,,....,..,................C ...,,....,.......,........L = x . X

= .x

,.= .x ,..... . ........... .,,... ........ . ,....25 ........12 Atomi c and Nucl ear Physi cs Chap. 2 .....,,...........,..,..e/ecoa:o/, .,.................,,..........,....,...........,.....,....,,.,..,... 1 .= 1.60219 x 10-19 ..x 1 .= 1. 60219 x 10-19 ,...,,...,..,......(106 ..... .(103 .Example 2.3 Calculate the rest-mass energy of the electron in MeV Solution. From Eq. (2.4), the rest-mass energy of the electron is mec2 -9. 1 095 1 0-28 (2.9979 1 01 0)2 -8. 1 871 1 0-7 ergs -8. 1 87 1 1 0-14 JOUle. Expressed in Me V this is 8. 1 87 1 1 0-14 joule - 1 . 6022 1 0-13 joule/MeV -0. 5 1 1 0 MeV. [Ans. ] Example 2.4 Compute the energy equivalent of the atomic mass unit. Solution. This can most easily be computed using the result of the previous example. Thus, since according to Section 2. 3, 1 amu - 1 . 6606 1 0-24g, it follows that 1 amu is equivalent to 1 . 6606 1 0-24 g/amu -- 0. 5 1 1 0 MeV/electron -93 1 . 5 MeV. [Ans. ] 9. 1 095 1 0- g/electron ....,.......................,....(2.5) ..Hg .........o...,..,(2.5), .....H ..Hg ..o,.......o.,,...c,H ..........oa/eaegy.,...........,,,......,,.,..,(2.6) Sec. 2. 5 Mass and Energy 13 ..o ...,....,:.:...,./|ae|ceaegy t.........,,.........,,....|o,c |,| c}c:.:,:.s.........., :. s...,...,..,c}c)...,.........c c,.....,............,.,...t.:.-..............,,.................,:.-..,......,:. s,.,.......,,.,...,:.-.....,..........,,.....,: -...,.:. | a..,.....,:.-.....,.....,...,.,..,.c :: a.:cta a:t

,,, :. | | .,....,:. .......,,....a.: | | ...,:. | | ............,:. s............,.,......a.a:x a. : | | .a.a| a.| a.V .....,...... .....,..,......,.,............,...,:. s,...,.................| ,aaa...a.a:.= :a..,......,......,.....:a..,..........,..,,.........,,......,:.-.................,: -,....,...,......c= |. sx | a. :. | :..c......t.....,,........,..,.....,. :. s..:.-....,,............ ...,.,..,,,,... ........,.............,,....,,...14 Atomi c and Nucl ear Physi cs Chap. 2 ./|ae|ceaegy..,...,.. ....,.....,,......,,.,..,,..........,i=hv, . ..h...........v ..,..,...,...........,.................,,x .i..,....h.,... x ..2. 6 PARTICLE WAVELENGTHS ...,...... ....,...................,.......,. .......,.....,.....j.h = , . .|..h..,..........,.......... j.,..,j = oo. . ...o.......,...o...,......,. j......j =,.og ig..i.....,,......,........,. . .,.....,...h = ..ogL. ......................,..,...,............,.....,.,...........,.. X

= .. .........i.....,,..... .............,.j ,,...,....,......,.,.....,..

2 2 |=

i..,.- i,.

c. Sec. 2. 7 ...Excited States and Radi ati on ac=

L,,L,1 5 (2.19) ........,.........,..,.,(2.15), ...,..,...L}=c(2.20) ....L...,,.,.....,(2.20) .....,(2.14), ....ac

~ L (2.21) ..,........ ..c...,,,....,....,1.240 x 10-6 = L(2.22) ........L .. ..,...(2.22) ...,.......,........ 2.7 EXCITED STATES AND RADIATION ..Z ....................................,.,..............,., ..,.............(Z = 82), ........,........k-e/ecoa...,.....,..........|oa|:a|oa,....,. .............|a|:a|oaeaeg|es....,...........,.........,...... .......,,...............,.........goeasae.......,.......,,....,.....,,......exc|esae..eaegy/e:e/...,.....................,,,..eaegy/e:e/|agao,........,2.2 .,,....,...,,...,..................,,...............................,....,.,............,,.......,.......,....,..................,....,16 Atomi c and Nuclear Physi cs ..ss12.07 10.19 Figure 2.2 The energy levels of the hydrogen atom (not to scale). Chap. 2 ........,,,........,...... ...,.....,,..........i0. i -.., z. z).,..,.....,.......,,i 0. i -.....,(2.22), ..,..........,.). = i .z+0 X i 0

-} i 0. i - = i .zi x i 0..........,.......,....,..,..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 ke V, and so the atom minus this electron is actually in an excited state 88 ke V above the ground state. When the outer electron drops into the K position, the resulting atom still lacks an electron, but now this is outer, weaky bound electron. In its fnal 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 7 1 0-6/8. 8 1 04 - 1 . 409 1 0-"m. [Ans. ] Such a photon is i n 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. .......................,......,...........,.............Sec. 2. 7 Exci ted States and Radi ati on 17 ............,..........,,.,.....,..,,.............................,z. 1

.,...,. z.z..z. 1........,..........,..........,,...........,...........,..24 etc 22 20 18 16 14 O>12 cCU10 8 6 4 2 Figure 2.3 The energy levels of 0 carbon 1 2. 18 Atomi c and Nucl ear Physi cs Chap. 2 ....,,...

.....,..................,............,.........,........... . .......,.,..,.,....... ,..,.,.......,,,.........,........... ...,.,.........,...........,...,......,.,...,...,............,.....,-ays.....................,,,..........,.......,,.............,,................,.........,,,....................,,.............,.....,.......,....... ......., ......,..............,....... ..........,...,.........,.,.......,...............,...,.....,..Aegee/ecoas.2. 8 NUCLEAR STABI LITY AND RADIOACTIVE DECAY ,.!. +.....,.............................,........,.,..........,!.+........egecaa.caao]aeaec/|es...,.,.............,........Z ,......!0,... ...,......,...............,...,.........................,...,.....,.,.,....,.,........,..,..,,.. ...,!. +...,.............,.........,.,..,..................................,..........,......,.....,......,............,.a|oac|:eecay...........!. !, ..,..,,.,A= ).., -,........,...= . , ..i !..........,...= ...,.....,.......,......,... ..i !......,.... ...... ......., .......,--ecay....,....,.....................Sec. 2. 8 Nucl ear Stabi l ity and Radi oacti ve Decay 19 1 10 r--".".r---."..-"-"".".- 1 00 ~~ ~ , I _ ___ _

_

60

. 50

g! 40 f- -+-----+-:Fp::

5abInucIct30

~QcmtllcrsQ`cmttlcrs (orcIcclroncaplurc)20 __. Q+, Q cmtltcrs (orcIcclroncaplurc) _ o-cmtltcrs(wtlh , `, orcIcclroncaplurc)10 O o-cmtttcrs (purcor-slabIc)_ .

' ' ' ' ' ' ' ! ' 10 20 30 40 50 60 70 80 90 10 1 1 0 120 1 30 140 150 1 60 Pculron numbcr(NFigure 2.4 The chat of nuclides showing stable and unstable nuclei. (Based on S. E. Liverhant, Elementar Introduction to Nuclear Reactor Physics. New York: Wiley, 1 960. ) ,...............,...............,.... ....,.,. ... ................. I, ...,....,..... ...........! .......,...... ........,.....,, .,..,..,.............

..V ......................,...,............,.............,...,.......................,.....,,..........,,.,..........,z. . ......,.x,i) , .,.........,...,,..........,,i 20 Atomi c and Nucl ear Physi cs E Figure 2.5 A typical energy spectrum of electrons emited in beta decay. Chap. 2 ...................,...i..iL .x,i)L. ......,..............,,i,,,, .........................,.,,...i..,,...,,..c +i,,,.....,..,i~c. +i,,, ,..,.aegaeaec/ees,........,...........,..,.....ecaycaa|as....,...........,...................,..,....,..............,...............................,.........,...................,.........,.......,...........,....,......,.,...........k........,....k-cajee......,.....,....,................,....,,.....,..........,,...........,....,.,.......aja|c/e....,....,.,........,

.....,.,......... .........,.........,...........,............,

........

.........,.,...u

..

. Sec. 2. 8 Nuclear Stabi lity and Radi oactive Decay TABLE 2. 1 ALPHA-PARTI CLE SPECTRUM OF 226Ra a-particle energy 4. 782 4.599 4.340 4. 1 94 Relative number of paricles (%) 94. 6 5.4 0.005 1 7 x 1 0-4 21 ..,,.,..... ..,...,.....,........................,.,.........,,.,.....,....,......... ......... .....,....,.,..,.......,

...................,,+ ..,...,..........,............,.........,.,.,.........,........,...... ....,...............,. .,

....,.......,..,.,...........,.........ecayscaeoe. ....,.,......,..,...,...,..........

...,..,.,..,.............,.,......

.........,,.......,..........,................,,.............,.........,.,.............,,.........,......|sooe|csaes ....,......,,..,..............|sooe|caas||oa................................,....,.,,............. .....,....

...........,..,..,............. .,.......,........ .....,.....,. ..,....,.,.......

....,.....,...,.(> --).....,,,.......S Strictly speaking, the ten 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 refered to as nuclear dependent-excitation, not decay. However, the use of the ten decay is well established in the literature. 22 0.0586 IT ( 10.5 m) 99 + % 0.0 99+ % 0.013 % 0. 12 % 0.09 % 0.24 % Atomi c and Nucl ear Physi cs 2.506 2. 158 1. 332 O O ` PtFigure 2.6 Decay scheme of cobalt 60, showing the known radiation emitted. The numbers on the side of the excited states are the energies of these states in Me V above the ground state. The relative occurrence of competing decays is indicated by the various percentages. Chap. 2 ......,.............,....,..........,..,,.......,..,...,..,.......,..,..,....,........,. ....,.............,..,,..,....,.. 2. 9 RADIOACTIVITY CALCULATIONS .......,........,..,.....,.......,.........,...,..,,... ...........,.,,............, .......,.............ecay coasaa .. ..,.-....,....,........ ..a ().......,.....,a, ) ....,.....,...... .. .. +. ...........,.....,..a,) ..,...,.......,6By tradition, the same symbol, /, is used for both decay constant and the wavelength defned earlier, as well as for mean free path, defned later. Because of their very diferent uses, no confusion should arise. Sec. 2. 9 Radi oacti vity Cal cul ati ons 23 ....ac|v|(....,....,..,.a........,...,..,a( ) = a, ) . ,!. !1).,......,.........ce|es,.............,1. x i 0'..,...,.............o|//|ce|e, i 0........o|coce|e, i0-..,...j|coce|e, i 0

'-........., .......,..oeceee/,.,....,....,...,..,....,= !. 01x i 0

' ' !, ..a,)....,......................,.......,.....a,)= a, ) . ...,......,.,.a,)= a,e

' ,!. !+)..a,......... c. .,,.,....,,!. !+),,.....,....,.....,,!. !)..a,....,.= c. ....,.......,...,........,......,..,...........aa(f-/(fe...,....,.I'/-...,......,,!. !),........,.,........,........,I,/-,..ia ! 0. -1I,/-

...,.,,!. !).......,..,,!. !),..a()a,e.-,/;./-,!. !),!.!)...,...........,........,....,,!. !), .,.,......,...........,.24 Atomi c and Nucl ear Physi cs Chap. 2 ..,,.............,.......,......,...,z z).......,.. ... .........,.,..,oeaa-/e, t, ............,....,....t l } ...,,z z),..............,.. }e..........,,zzc),..........,

I,/-c c+ i . +I./-. ,z z)...,....,.......,......, z , ................,..,....,..,.............,..........,...............,,......,..............................,.,.....,...............,..,...,.,.........,...=.,..,C:E aJ2 aJe ..,.. ....!i t_Jtmc(mblra untls)Figure 2.7 Te decay of a radioactive sample. Sec. 2. 9 Radi oacti vity Cal cul ati ons 25 a}-a. ...,......,.......aa,e

- C-)t ,

,z.z)..a,..,............,... c .,,.,..,.....,.,,.....,...,z. +c)a, c,....,....a ......,......- >,a .,,.............amax ....,a.,,...........nmax ,..,. a, = c,.....,..,...,..,,.......,...,,...........,.,,.......amax ... = >. Example 2.6 Gold- 1 98 (T1 /2 - 64. 8 h) can be produced by bombarding stable 197 Au with neutrons in a nuclear reactor. Suppose that a 197 Au foil weighing 0. 1 g is placed in a certain reactor for 1 2 hs and that its activity is 0. 90 Ci when removed. (a) What is the theoretical maximum activity due to 198 Au in the foil ? (b) How long does it take for the activity to reach 80 percent of the maximum? Solution 1. The value of R in Eq. (2. 30) can be found from the data at 1 2 hrs. From Eq. (2.26), A -0. 693/64. 8 - 1 . 07 1 0-2 hr-1 Then substituting into Eq. (2. 30) gives or from Eq. (2. 26) 0. 90 - R1

e-O. 693 xI2/64. 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 1 010. ) According to the previous discussion, the theoretical maximum activity is also 7. 5 Ci. [Ans. ] 2. The time to reach 80% of Omax can also be found from Eq. (2. 30) : 0. 8R - R( 1 - eAt) . Solving for t gives t - 1 50 hrs. [Ans. ] 26 Atomi c and Nucl ear Physics Chap. 2 ..,.................,..........,...A - 8 - c - ................,...........,.,..............,...,... ...,. ..,a,.......,..... .......,..,a,.....,.....,a,. ......,.a,}.a,

,a,+,a,......,.,,!. !+)a,,.....,...,...a, .a,-t ,a,+,a,.e..a.......... c .,..,.,,!. 1 l ) ,..-gt+a,,,_ -a, a,.e e, ,.....,..,.....,....-gt+a,,,_ ., -gt a, a,.e e e ., ,,!. 1 l ) ,!. 1!),!. 11)..a...a,........ ... .,.,.......,,!. 11).,..,...,........,.,.......................,2. 1 0 NUCLEAR REACTI ONS aec/eaeac|oa.........,........,..................,......,....,. .......,a..o,...,.......,...,,...........,...,c.., .....,..,.,...a + o - c + . ,!. 1+)Sec. 2. 1 0 Nucl ear Reacti ons 27 ...............,...,...,..,.,...................,...,.....1. coase:a|oao]aec/eoas......................2. coasera|oao]caage.........,....,.............3. coasea|oao]oooeaeo.............,,.............4. coase:a|oao]eaegy...,,...,......,,........... ..,...........................,..,,.....,.........,..,..,.,.......,,....,..........,.,,.......,....,,,....,,z ++) ....,,............,..,..a..o,.........,,..,....,..,,.............,.,..c..,.........,. .,......,,....,z +)..L, , L,, .... .....,.,..a, o, .,...,z +).....,...,z +c)...,..........,.......,,.,...........,.....,..,.........,.............,..,..........,......,,z +c)........g:a/ee......,z +)...,...........g....,..,......,.......,,...,+ l......g.....,z +)28 Atomi c and Nucl ear Physi cs Chap. 2 ..,,!.1),.......,.,......|acease.....,..,..........exoaeo|c....,..,...........,..,..........eaoaeo|c. ..................................,,......................,,........ .,...,!. 1),..,............a, o, .... ....,...........................,...................,,!. 1-)..z, . z,,............a,o, ......., ,!. 1)..,....,| ,M,+z,,)+,M,+z,o,)J- ,M,+z,o,)+,M,+z,o, ) | -1 l. ,z+c)..o,..............M,+z,o,.,.............a, M,+z,o, .........o, .........,,!. 1).......,..M, , M,, ......,....................,......,...............,...........,...,...............................,.......,.......,.......,....,....,a,........,...,.o.,,.,.....,......,,!. 1+)..........a,o.c)

a,o.)c.........,,,....,....,,....,.,..............1 60 +

--1 6 +

. ........Sec. 2. 1 1 Bi ndi ng Energy 29 ....,... ..}.........,.,..,.,Example 2.7 Complete the following reaction: 1 4N + n --? + I H. Solution. The atomic number of 1 4N 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 must also 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 lef is the sum of the atomic mass numbers-namely, 14 + 1 - 1 5. Since the mass number of IH is 1 , the carbon isotope foned in this reaction must be 1 4C. 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. 01 6049 MeH) -2. 01 41 02 M(4He) -4. 002604 M(n) - 1 . 008665 M(4He) + M(n) -5. 01 1 269 Thus, from Eq. (2. 37), the Q value in amu is Q -5. 0301 51 -5. 01 1 269 ~ 0. 01 8882 amu. Since 1 amu -93 1 . 502 MeV (see Ex. 2. 4), Q ~ 0. 01 8882 x 93 1 . 502 = 1 7. 588 Me V, which is positive and so this reaction is exothenc. This means, for instance, that when stationary 3H atoms are bombarded by I -MeV deuterons, the sum of the kinetic energies of the emergent a-particle (4He) and neutron is 1 7. 588 + 1 - 1 8. 588 MeV. 2. 1 1 BINDING ENERGY ......,,......,............

... ....,.........,.,.....,,.. . ......,....30 Atomi c and Nucl ear Physi cs Chap. 2 j + a + ,...........

. . . H + , .....,..,..........,...............,,............,,.....,,...,. .......................,..............................oasse]ec..........,......................................,...................,......... . ..M . ...........,.... . .............o,...........,...,M _ +o,.,..........1 ...M+zo,.,.......M.................................,..................,...... ... ...,.,,......,.,.......,.......,....,.,.......,....,,....,....,,.......,.....................,,........o|a|ageaegy..,....,.....,,............,.............,..A.....,....,,....,.. ..............,.,,.. ........,,.................,,,..,..........,......,.,,.........,.............A.............................,,,.,....,..,.,,,... A,....A, .......,. ....................A, ....A= c............,....A .........,.,,....,......,,.........,,Sec. 2. 1 1 Bi ndi ng Energy 9 8Z*' . ' =

' Z 4 8 I I Z ! 4 I l 8 Z ZZ Z4Alomtc mass numbcrFigure 2.8 Binding energy per nucleon as a function of atomic mass number. 31 ..... .......,.,,,.....,...,.,..,.,.......,.,.....,,.....,,...,........,...........................,.,.....,,....,..........,.,,,........,..,...,.....,,......,............,........,.,....,,.........,..,.,. ...,.. ...,.,,.............= z, M(H)+!, M, M, ...................,32 Atomi c and Nucl ear Physi cs Chap. 2 .........,...,....M, , M,,......,. ....,..,..z,+z, z,+z,!,+!,!,+!,. ,8!,c)+8!,) J

8!,a)- 8!,o)J ...c....,,...,.....,.,.........,,...,.........,.,...................,.,,.,....,.......,.,,...........,.....,..,......,...,...,........,,.....,.. .......,.....,....,,.......................,.,,. .........,....,.,,. ...,.,.....

.+. .......,......,.,,..,.. . .x . .. .........,,.,,.......,,.,......

. ........!,. ... ...................,...,..............]s|oa eac-|oas......,,..,..........,,..,,.......,...,,...,..........,...,...,A.,. ..........,............,.....,...,......,.,,,....

....... ..............,..A ..= ..................,........................,......,.,,..,...,,...,c .,..............,,.....x . .....,....aec/ea]ss|oa,.......,,..........,.........,.,,,........,. ......,....... ............, ........,.....,......,.,,.,............... ......,,,..................,.,,.....sejaa|oaeaegy....,...,........,,.........Sec. 2. 1 2 Nucl ear Model s 33 ...,....,,L, ............../as.........z. .....................................z..............................

z,........,,.,..L, ...,.....L, = M,+M,' z) M,z)J-1 l.... ......,,L,.,.....................,..,....,...,,....,................,,...,.....

z,..,,L,.....,..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 binding energy or separation energy is then computed from Eq. (2.45) as follows: Met2C) ~ 1 2. 00000 Mn 1 . 00866 Mn Met2C) ~ 1 3. 00866 - M(l3C) - 1 3. 00335 Es 0. 0053 1 amu 93 1 Mev 4. 95 MeV [Ans. ] ....,...........,.,,............,.l+, .. .. .....,....,.,..........oag|c,......................oag|c aeooes. .....,..........,.....,.............................,..................,.........,....,........,..,...........,.............,..,.............,,...........,.........,...............,..............,....................2. 12 NUCLEAR MODELS ...............,...,.....,...........,.,.........,.,...,.34 Atomi c and Nucl ear Physi cs Chap. 2 .........,,.,...........,,.......,.................,..............,.Shel l Model .......,..,..........,.....,.......................,...,... ...........,.........,....,....,................,................,....... .........,,.........,............,,..,...............................,.,,.......................,............,...,..............,........,.......,....,............,.......,.....,.......,.,...,...,..............,+,...............,......., ......,.......,..............,........,...........,.,.......,.. .........,.....,.,....,H_ ......,.,...,,.....,.............,.... ....,....,......,...,.,......,,...........,........,.....................,............................,.....,......,.........Liquid Drop Model .... ...,.,,.......,....,,.....,.,..........,....................,................,......,.. ........,..,....,...,. ........,.,.........,.,.......,....,,.............,.,.........,..........,.........,...........,.........,.......,..Sec. 2. 1 2 Nucl ear Model s 35 .......,,.................,.........,..,,..., M = NMn .ZMp aA ...,..................................................,....... .................I......................,,..A /,.......M = NMn +ZMp aA.A/. .....,..........,,............. ..,.,...,,.......,.....,..............,...................,..........,......,.,............,.,

.............,....,.,........,,.,...............,....,...........................,........,.......,.,..........,N Z ..................N # Z. ...................,.... ...,................,.................,....,.....,...,..........,.....,.............,..................,...........,,..,.....Z !...........,...,.....................,..,..,8 ....,...7For a discussion of the Pauli exclusion principle, see references on modem physics. 36 Atomi c and Nucl ear Physi cs Chap. 2 ...8 ..!z.......,.......,.......,.... . ......,............,......,.,..,.....................,.....,..................,,.,..............,.a r, 8 .. .. . .. .. .. ... . ...........,,............,.....z > ....,,.........,,.............,.,.,................... ..............,,... ...,..........,..............Example 2.10 Calculate the mass and binding energy of !7 Ag 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 fve terms represents the binding energy of the constituent nucleons. The atomic mass of the !7 Ag is frst obtained by using the mass formula and noting that N is even and Z is odd. The term involving is thus taken as zero. N mZ m A +f 7 A2/3 y 7 Z2 / A 1 /3 (A - 2 Z)2 / A Mass (MeV) Mass (u) 60 939 . 573 MeV 47 938. 791 MeV -1 5. 56 1 07 MeV 1 7. 23 1 072/3 MeV 0. 697 472/ 1 071 /3 MeV 23. 285 ( 1 07 - 2 47)2/ 1 07 MeV 99548. 1 1 73 MeV 1 06. 8684 u The measured mass of !7 Ag is 1 06.905092 u L within 0. 034% of the calculated value. Summing the last four terms gives a total binding energy of 949.44 MeV or g.9 MeV! nucleon=a value slightly higher than the me(ured value of approximately 8. 6 MeV Sec. 2. 1 3 Gases, Li qui ds, and Sol i ds 37 2. 13 GASES, LIQUI DS, AND SOLIDS ....,...........,.,....,,,.......,..,.,..................,..,......,..........,..,......,.,............. Gases ...,.........,..,,................,......... .,..,....,...,.,........,..,......,..,,...,..,,...............,.............,...Sol ids ...........,......,.........,............,..,....o|cocsa/s,...................,.... ....,...................... ...,..,.....,.................,........,.....,.........,..,................so/|s....,.,.......,..,....,.........,...,......,............,.....Liquids ....,..,......,..,.........,.................,.....,,...........,....,...,.........,.........,.........,....,.........,...........,The Maxwel l i an Distribution ..,....,.............,............!,L)....,,..,...,,..!,L)L.....,..,.........,.,...L..L+L. .,.........!,L).,..,..... ..38 Atomi c and Nucl ear Physi cs Chap. 2 ..,. ..t......,..,........ .,...,/....... ...............,,,,.../ X ,.. x

.. ........,...,...,.........t,L)., .,. ....,....,,..........,......,....,. .. .................,,....t, L) ....,......,..,.,. ....,................,.................,........ ...,..............,. ...,,...,....,.....oosjooao/eeaegy............,....,. ........,,.,..,..................,,..,....t,L) ,.......,..,,LP, ........,,........,..L/....aveageeaegy,L, ..,..,.

L= -t,L) LL. t

ZL|k!J 4Figure 2.9 The Maxwellian distribution function. . ....Sec. 2. 1 3 Gases, Li qui ds, and Sol i ds .....,., ,!. !)..,,!.+)...,.,...,..,...3L/I39 ,!. )......,..../I..,.,!. 1)..,!. )..,,....,........,..,.........,..,........,,...,..I, !-1. l k,/I......./I,= 0. 0!1.` _. ,!. )Example 2.11 Wat are the most probable and average energies of air molecules in a New York City subway in summertime at, say, 38C (about 1 00F)? Solution. It is frst necessary to compute the temperature in degrees Kelvin. From the formula OK = C + 273. 1 5, i t follows that the temperature of the air i s 3 1 1 . 1 5 K. Then using Eqs. (2. 53) and (2. 55) gives 1 3 1 1 . 1 5 Ep - _ x 0. 0253 293. 61 -0. 01 34 eY. [Ans. ] Finally, E ~ 3Ep, so that E -0. 0402 eY. [Ans. ] The Gas Law ....,,....,...,....ea/gas/a,!. )..P ..,..,...\ .....n g .......,......\, k..,........I......,...,...,!. )..........tA ..,. ........t_ ...,.......n gt _ ....\ ......,..t ...........,............k}t_..../, ...... ........,.........,........P t/I. ,!. )40 Atomi c and Nucl ear Physi cs Chap. 2 ..,,!. ).....,..,...........,,.....,.......,,,.....2. 14 ATOM DENSITY .....,..,,........,.................l ...........................,..,.,,.....r..,.,....,.,...M..,.....,.......rM,...........l .....,.......x .....x ..,. .........,x,....,. ...,,Example 2.12 rxx -MThe 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 0. 6022 1 024 24 N -22. 990 - 0. 0254 10 [Ans. ] (It is usual to express the atom densities as a factor times 1 024. ) ,!. -).,...,!. -)...,,........,.........,..x ......,...,. ..M ..,......,...........,...,,,. ..,....,..,,......,,....ni , .....,......,!. 0)...,.......,,.............,...,......,....,................,.......,...,........,..........,......,..................................,,.........,.,,,............. .............,...........,,,.Sec. 2. 1 4 Atom Density 41 ...... ...,...,.....,...,......,,! -),..,....x,........,................,...............,.........,...,Example 2.13 The density of a NaCI crystal is 2. 1 7 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 of NaCI is therefore 58.443. Using Eq. (2.59) gives 2. 1 7 0. 6022 1 024 N ~ 0. 0224 1 024 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 of each atom. [Ans.] ,..,.,..,.........,....,,.....,.....!.!,.......,.....,......,......,...,.,........,.........,,..,......,................,x,. ..,.x

,,rx i 00M',!. l ) ..,, . ..,......aoojecea,...a}o. .....,..................,......,,......,.,.,...........r..,.,....,..........,..,...,...,! !)..a, ..egajecea, ...a}o, ..,....,,!.-),.........,...,...,!. 1)....,.....,.42 Atomi c and Nucl ear Physi cs Chap. 2 ............,....,,........,.,.,..,......,........,....,........,...,..........,..Xm Yn , .....,..oM,+aM,..M,..M.....,..X ..Y, .,.,...,.,.,....X .oM,wjo(X) = x l 00. oM,+ aM,! +).......,,.....,.,............,,.........................,.............,

.....,..........,+)..........,..,,......,.,......,............,..... ..............,..,..,N =LNi , ..Ni ......,. ..,...,N ..,,!. -)..Ni ..,,!. 1),..l lWi -=-LM l 00 M,,!. )...,,.........,........,................,!. l .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 1 . 00797 + 1 5.9994 molecular density is therefore, 1 8. 01 53. The 1 x 0 6022 X 1 024 N(H20) =1 8. 01 53 = 0. 03343 1 024 molecules/cm3. [Ans. ] Sec. 2. 1 4 Atom Density 4 2. There are two atoms of hydrogen and one atom of oxygen per H20 molecule. Thus, the atom density of hydrogen N(R) = 2 x 0. 03343 X 1 024 = 0. 06686 X 1 024 atoms/cm3, and N(O) = 0. 03343 X 1 024 atoms/cm3 [Ans.] 3. The relative abundance of 2H is 0.01 5 alo so that NeH) = 1 . 5 7 1 0-4 7N(H) = 1 . 0029 1 0-5 x 1 024 atoms/cm3 [Ans.] Example 2.15 A certain nuclear reactor is fueled with 1 ,500 kg of uranium rods enriched to 20 wi L235V. The remainder is 238V. The density of the uranium is 1 9. 1 g/cm3 (a) How much 235V is in the reactor? (b) What are the atom densities of 235V and 238 V in the rods? Solution 1. Enrichment to 20 wi L means that 20% of the total uranium mass is 235 V. The amount of 235V is therefore 0. 20 x 1 500 ~ 300kg. [Ans. ] 2. The atomic weights of 235V and 238V are 235.0439 and 238. 0508, respectively. From Eq. (2.63), 235 20 x 19. 1 0. 6022 7 1 024 N( V) -1 00 x 235. 0439 - 9. 79 x 1 0-3 x 1 024atoms/cm3 [Ans. ] The 238V is present t o the extent of 80 wi L and so Example 2.16 238 80 x 19. 1 x 0. 6022 x 1 024 N( V) -1 00 x 238.0508 = 3. 86 x 1 0-2 x 1 024atoms/cm3 [Ans. ] The fuel for a reactor consists of pellets of uranium dioxide (V02), which have a density of 1 0. 5 g/cm3 If the uranium is enriched to 30 wlo in 235V, what is the atom density of the 235V in the fuel? Solution. It is frst necessary to compute the atomc weight of the uranium. From Eq. (2.65), 1 1 30 70 M -1 00 235. 0439 238.0508 ' 4 Atomi c and Nucl ear Physi cs Chap. 2 which gives M = 237. 1 41 . The molecular weight of the V02 is then 237. 1 41 2 1 5.999 = 269. 1 39. In view of Eq. (2.64), the percent by weight of uranium in the V02 is (237. 1 41 /269. 1 39) 1 00 = 88. 1 w/o. The average density of the uranium is therefore 0. 88 1 1 0. 5 = 9. 25 g/cm3, and the density of 235V is 0. 30 9. 25 = 2. 78 g/cm3 The atom density of 235V is fnally REFERENCES General 235 2. 78 0. 6022 1 024 N( U) 235. 0439 = 7. 1 1 1 0-3 1 024atoms/cm3 [Ans. ] Arya, A. P., Elementar Moder Physics, Reading, Mass. : Addison-Wesley, 1 974. Beiser, A., Concepts of Modern Physics, 5th ed. New York: McGraw-Hill, 1 994, Chapters 3, 4, 5, 1 1 , 1 2, 1 3, and 1 4. Burcham, W. E. , Nuclear Physics: An Intrduction, 2nd ed. Reprint Ann Arbor. Foster, A. R. , and R. L. Wright, Jr., Basic Nuclear Engineering, 4th ed. Paramus: PrenticeHall, 1 982, Chapter 2 and 3. Goble, A. , and K. K. Baker, Elements of Moder Physics, 2nd ed. New York: Ronald Press, 1 971 , Chapters 8-1 0. Kaplan, I. , Nuclear Physics, 2nd ed. Reading, Mass. : Addison-Wesley Longman, 1 962. Krane, K. S., Nuclear Physics, 3rd ed. New York: John Wiley, 1 987. Krane, K. S. , Moder Physics, 2nd ed. New York: John Wiley, 1 995. Lapp, R. E. , and H. L. Andrews, Nuclear Radiation Physics, 4th ed. Englewood Clifs, N.J. : Prentie-Hall, 1 972, Chapters 1 -7. Liverhant, S. E. , Elementar Intrduction to Nuclear Reactor Physics. New York: Wiley, 1 960. Meyerhof, W. E., Elements of Nuclear Physics, New York: McGraw-Hill, 1 967, Chapters 2 and 4. Oldenberg, 0. , and N. C. Rasmussen, Moder Physics for Engineers, reprint, Maetta, Technical Books, 1 992, Chapters 1 2, 1 4, and 1 5. Semat, H" and J. R. Albright, Intrduction to Atomic and Nuclear Physics, 5th ed. New York: Holt, Rinehart & Winston, 1 972. Serway, R. A. , Moses, C. J. , and Moyer, C. A. , Modern Physics, 3rd ed. Philadelphia: Saunders, 1 990. Tipler, P. A., Moder Physics, 2nd ed. New York: Worth, 1 977. Wehr, M. R. , Richards, J. A. , and Adair, T. W. , Physics of the Atom, 4th ed. Reading, Mass. : Addison-Wesley, 1 984. Probl ems 45 Weidner, R. T., and R. L. Sells, Elementar Modern Physics, 3rd ed. Boston, Mass. : Allyn & Bacon, 1 980. Williams, W. S. C. , Nuclear and Particle Physics, Oxford, Eng. : Clarendon Press, 1 99 1 . Nuclear Data Iaecaao]aetec/es...............,....,.................... ............. .. ................................,aj//. ae. oa/.go:.. ... ..... Iao/eso]Isoojes,....., ..........,...............,...,..........,.................... ... .....,,...,................................,,...................,......,..............,............aj//. aac. oa/. go:/aacsc.PROBLEMS 1. How many neutrons and protons are there in the nuclei of the following atoms: (a) 7Li, (b) 24Mg, (c) 1 35Xe, (d) 209Bi, (e) 222Rn? 2. The atomic weight of 59CO is 58. 933 1 9. How many times heavier is 59Co than 1 2C? 3. How many atoms are there in 1 0 9 of 1 2C? 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 0.01 5 Atomic weight 1 . 007825 2.01 41 0 5. When H2 gas is formed from naturally occurring hydrogen, what percentages of the molecules have molecular weights of approximately 2, 3, and 4? 46 Atomi c and Nucl ear Physi cs Chap. 2 6. Natural uranium is composed of three isotopes: 234U, 235U, and 238U. Their abundances 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 7 1 0-24 g. 14. The planet earth has a mass of approximately 6 7 1 024 kg. If the density of the eath 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 7 1 07 J in heat energy. The conversion of 1 g of mass into energy is equivalent to the buring of how much coal? 16. The fssion of the nucleus of 235U releases approximately 200 MeV How much energy (in klowatt-hours and megawatt-days) is released when 1 g of 235U undergoes fssion? 17. Compute the neutron-proton mass diference in Me V. 18. An electron starting from rest is accelerated across a potential diference of 5 million volts. (a) What is its fnal kinetic energy? (b) What is its total energy? (c) What is its fnal mass? 19. Derive Eq. (2. 1 8) . [Hint: Square both sides of Eq. (2.5) and solve for mv. J 20. Show that the speed of any particle, relativistic or nonrelativistic, is given by the following fonula: Probl ems v = c _

Eest 2 ' Etotal 47 where Erest and Etotal are its rest-mass energy and total energy, respectively, and c is the speed of light. 21. Using the result derived in Problem 2.20, calculate the speed of a I -MeV electron, one with a kinetic energy of _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 1 0-10 cm is called the Compton wavelength. 2. 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 of me ; (c) speed in units of C, (d) wavelength in units of the 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 of the emitted photon, show that Ey E I - ' 2Mc where M is the mass of the atom or nucleus. (b) Compute E - Ey for the transitions from the frst excited state of atomic hydrogen at 1 0. 1 9 e V to ground and the frst excited state of I 2C at 4.43 Me V to ground (see Figs. 2.2 and 2. 3). 27. The frst three excited states of the nucleus of 1 99Hg are at 0. 1 58 MeV, 0.208 MeV, and 0.403 Me V above the ground state. If all transitions between these states and ground occurred, what energy y-rays would be observed? 28. Using the chart of the nuclides, complete the following reactions. If a daughter nucleus is also radioactive, indicate the complete decay chain. 4 (a) 1 8N (b) 83y (c) 1 35Sb (d) 219Rn Atomi c and Nucl ear Physi cs Chap. 2 29. Tritium eH) decays by negative beta decay with a half-life of 1 2. 26 years. The atomc weight of 3H is 3.01 6. (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 1 0 millicuries per millimole ( 1 0 mCilmM). What fraction of the carbon atoms i s 14C? 32. Tritiated water (ordinary water containing some 1 H3HO) for biological applications can be purchased in 1 -cm3 ampoules having activity of 5 mCi per cm3 What fraction of the water molecules contains an 3H atom? 33. Afer the initial cleanup efort at Three Mile Island, approximately 400,000 gallons of radioactive water remained in the basement of the containment building of the Thee Mile Island Unit 2 nuclear plant. The principal sources of this radioactivity were 1 37Cs at 1 56 JCicm3 and 1 34Cs at 26 JCicm3 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 of the a-particles are neutralized and retained in the free volume of the capsule? (b) What is the pressure 1 0 years afer the capsule is sealed? 35. Polonium-21 0 decays to the ground state of 206Pb by the emission of a 5. 305-MeV a-particle with a half-life of 1 38 days. What mass of 2 1Opo is required to produce 1 MW of thenal 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 1 2. 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 thenal to electrical efciency of the system is 5.4%. Calculate (a) the fuel efciency in curies per watt (thenal); (b) the specifc power i n watts (thenal) per gram of fuel ; (c) the power density in watts (thenal) per cm3; (d) the total electrical power of the generator. 37. Since the half-life of 235U (7. 1 3 1 08 years) is less than that of 238U (4. 5 1 1 09 years), the isotopic abundance of 235U has been steadily decreasing since the earh was foned 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 bombardment of X according to the reaction 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 of A and B are equal. (b) Show that, for t tm, the activity of B is less than that of A, whereas the reverse is the case for t tm. 40. Show that if the half-life of B is much shorter than the half-life of A, then the activities of A 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 originates 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 of nuclides), may be present in uranium mines in dangerous concentrations if the mines are not properly ventilated. Calculate the activity of 222Rn in Bq per metric ton of natural uranium. 43. According to U. S. Nuclear Regulatory Commission regulations, the maximum penissible concentration of radon-222 in air in equilibrium with its short-lived daughters is 3 pCiIiter for nonoccupational exposure. This coresponds 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 of R atoms/sec. Derive an expression for the activity of B as a function of time. 45. Complete the following reactions and calculate their Q values. [Note: The atomic weight of 1 4C is 1 4.003242. ] (a) 4He(p, d) (b) 9Be(a, n) (c) 14N(n, p) (d) 11 5In(d, p) (e) 207Pb(y, n) 46. (a) Compute the recoil energy of the residual, daughter nucleus following the emission of a 4. 782-MeV a-particle by 226Ra. (b) What is the total disintegration energy for this decay process? 50 Atomi c and Nucl ear Physi cs 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 diference = 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 of MeV. 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) = 1 4. 95 MeV, eH) = 1 3. 1 4 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 deuterium. The reaction is 2H + n -3H + 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 for 2 H of 2. 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 of the reaction using the results of part (a). 52. Using atomic mass data, compute the average binding energy per nucleon of the following nuclei: (a) 2H (b) 4He (c) 12C (d) 5 lV (e) 1 38Ba (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. Probl ems 54. Compute the separation energies of the last neutron in the following nuclei : (a) 4He (b) 7Li (c) 170 (d) 5 ly (e) 208Pb (f 235U 51 55. Derive Eq. (2. 53). [Hint: Try takng the logathm of Eq. (2. 52) before diferentiating. ] 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 of graphite 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 lo in this isotope if the physical density of the uranium is 1 9. 0 g/cm3 ? 60. Plutonium-239 undergoes a-decay with a half-life of 24,000 years. Compute the activity of 1 gram of plutonium dioxide, 239PU02. Express the activity in terms of Ci and Bq. 61. It has been proposed to use uranium carbide (UC) for the initial fuel in certain types of breeder reactors, with the uranium enriched to 25 wlo. The density of UC is 1 3. 6 g/cm3 (a) What is the atomic weight of the uranium? (b) What is the atom density of the 235U? 62. Compute the atom densities of 235U and 238U in U02 of physical density 1 0. 8 g/cm3 if the uranium is enriched to 3. 5 wi L in 235 U. 63. The fuel for a certain breeder reactor consists of pellets composed of mixed oxides, U02 and PU02, with the PU02 comprising approximately 30 wi L of the 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 . c/G||BC||DG D/ /BO|B||DGA||G BD|...,......,............,,.......,........,....,......................,.................,.........,,.....,..,.............,...........,,,.3. 1 NEUTRON I NTERACTIONS ..,..,................,....,....,........,.,....,.........,......,.....,..........,........................ .....,............,..,. El astic Scateri ng ...,........................,...,........ .....,,.....52 Sec. 3. 1 Neutron I nteracti ons 53 .......,.....................e/asca//yscaee,..... ............... .........,..,...I nelastic Scateri ng ...,..........,.,...................,,...,.........,........ ......,..,..,... ........,. ...,...... ,.......,............,.,.........,.,..ae/asc,ays.Radiative Capture .......,.,...........,..cajee,ays-e.................,.. ...,..........,........,.............aosojoaeacoas.Charged-Paricle Reactions .....,....,,........,.....,,.o) ...,........,.......Neutron-Producing Reactions ....,,...........,............,..................................................,.,.,.........,...,..,....

...

.....,...........,,Fission ....,........,.........,.,...,]ssoa............,...,.,.......,,,...,,.....,.......,.......,,......,......,.........,....,.............,..,....,..............,..,..........................,..............., ..,..,...,,........,..........,..,......,........,........,...........,.,,...,..........,...,..........,....,.. ......... ..........,.....,.....,..............,....,....54 I nteracti on of Radi ati on wi th Matter Chap. 3 3. 2 CROSS-SECTIONS ......................,..........cosssecoas......,...,,,.,....,,.........,..,.,,.....,.,......,.....X ....A ......., 1. l ..a....,.......o...,........,...,1 = ao ,1. l ) ...aeas(................o..l ............oA ....,....,......nvA = I A .........,,........I A }A = I .,.............,..,,.

.............,..................,..,....,..........,1 . l ..,,.....,...,......,....,.............,,........,.....,N ..,.............,................,.,........,.....,= a IN AX, ,1.!)..a, .,,..,.......cosssecoa....N AX ..,,1. !)........ ...,........,.......,......,..a I . ....a .,.........,.........,.....,...........,....,.......,.I A ........,,....a I ......,,..........,....aI a AI A Neutrons in beam r I\V\o l'|X\\!I|/7

Figure 3.1 Neutron beam striking a target. ,1. 1)-- Area A Thickness XSec. 3. 2 Cross-Secti ons 55 .,...,.,.......................,,1. 1)..O ................O ....,........................cosssecoa..........,......oaas,.......,..

.

.............o//oaa,....,..,........................,.......,.,..........,...........,......,.......... ...............,......................,....... ,...............,..................,..,......,....O , ......,,.......,....O , ..,) ......,.,..,.....Oy; ...,........O f ; ................,..............oa/coss-secoa....,..,.O , ...,1.|).............,.,.........,,,.............,..............,............aosojoacoss-secoa....,O . ....,1. )..O_ ..O_ ......... ,... a) ... .....,,1. ), ...,............,.,.....,.....,....................,..........Example 3.1 A beam of I -Me V neutrons of intensity 5 7 1 08 neutrons/cm2-sec strikes a thin 1 2C target. The area of the target is 0. 5 cm2 and is 0. 05 cm thick. The beam has a crosssectional area of 0. 1 cm2 At 1 Me V, the total cross-section of 1 2C 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? 56 I nteracti on of Radi ati on wi th Matter Chap. 3 Solution 1. According to Table I . 3 in Appendix II, N ~ 0. 080 X 1 024 for carbon. Then from Eq. (3. 2), the total interaction rate is atI N AX = 2. 6 X 1 0-24 7 5 X 1 08 7 0.080 X 1 024 7 0. 1 7 0. 05 = 5. 2 7 1 05 interactions/sec. [Ans. ] It should be noted that the 1 0-24 in the cross-section cancels with the 1 024 in atom density. This is the reason for writing atom densities in the form of a number 7 1 024 2. In 1 sec, I A = 5 X 1 08 7 0. 1 ~ 5 7 1 07 neutrons strike the target. Of these, 5. 2 7 1 05 interact. The probability that a neutron interacts in the target is therefore 5. 2 7 1 05/5 X 1 07 = 1 . 04 X 1 0-2 Thus, only about 1 neutron in 1 00 has a collision while traversing the target. Example 3.2 There are only two absorption reactions-namely, radiative capture and fssion-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 fssion will occur? Solution. Since ay and a f are proportional to the probabilities of radiative capture and fssion, it follows that the probability of fssion is a f / (a _ + a f) = a f / a _ =582/681 ~ 85. 5%. [Ans. ] ....,,1. !), .............,.....,INat x AX, ,1. )..at .....................,.,.......,,,..,...AX ........,....,,1. ).........,.....,......co//soaeas(i,.,..,,1. )..,......,N ..........., ,1. ), ..,..,..,........,..,.,....,..,.L.....acoscoj|ccosssecoa.II ,....,.Nat = L, ...oacoscojcoa/cosssecoa,N as bs ...oacoscojcscaeagcoss-secoa,.... ..N ..a ..........

For reasons explained in Section 3. 6, neutron cross-sections are tabulated at an energy of 0.0253 C NSec. 3. 3 Neutron Attenuati on 57 .,.,L ....... .......,........., .. Example 3.3 Referring to Example 3. 1 , calculate the (a) macroscopic total cross-section of 1 2C at Me V; (b) collision density in the target. Solution 1. From the defnition given previously, bt = 0. 08 X 1 024 7 2. 6 X 1 0-24 = 0. 21 cm-l [Ans. ] 2. Using Eq. (3. 8) gives I = 5 X 1 08 7 0. 21 = 1 . 05 7 1 08 collisions/cm3-sec. [Ans. ] These collisions occur, of course, only in the region of the target that is struck by the beam. 3.3 NEUTRON ATENUATION ..,.,....,.......,...,...,,.....,.....X .,............,

..................,......., . ...........,.....................,...,.............,,.....,....,..,.,..............,.........,.............Incident neutrons Scattered Uncollided neutrons Detector Figure 3.2 Measurement of neutrons that have not collided in a target. 58 I nteracti on of Radi ati on wi th Matter Chap. 3 .I,x)....,..........ao.,...,....x...,........,........x,....,........,....................,....,....l .

........x...,,1. !), ........,.,..,I,x) !o,I,x)xL,I,x)x. ,1. -)...,......,......I,x) i,e ' ,1. l 0).....,..............,..,..........,.....,..........,.,...,...,1. l l ).........,....,...,..,......................................,.......,.,..........,....,.....,..........,.,........,,1. l l ),..,.........................,.....,...,....,.......,...........,,1.-)..,I,x) , ....I ,x) --L,x.I ,x),1. l !)...,...,I,x) .,.............I,x)...x,....I,x)}I,x)..,.,...............,x.............x.....,,1. l !), L,x .,...,.,..........x, ....,....L,..,.,,..,...,..........,............................I,x)..............,...,....x, ..I ,x) }I, e '.,...,.,............,.............,.....,...,j,x) x.,.,...............x...,..x. ......,,...,.,............,x.........,.,............x. ...L, ..,.,...,,...,.j,x)x.,..,Sec. 3. 3 Neutron Attenuati on j,x) x e 'x x e 'x.59 .....,....................oeaa]eejaa....,...,.....,..,..,.,.,......,...x, .......,....................,.,...j,x),...,....,........ xj,x) xxe- 'x,1. l 1)Example 3.4 Calculate the mean free path of 1 00-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 1 024 The macroscopic cross-section is then bt = 0. 0254 X 1 024 3 4 X 1 0-24 = 0. 0864 cm-l The mean free path is therefore A = 1 /0. 0864 = 1 1 . 6 cm [Ans. J .......,...........,.X ..Y, ....,,..!,......,,..O_ ..............,........,.,,..,..................,,., ! _O_ ............,,., !,o,....,.,,..,....................,1. l+).............,......,,1. l +).......,..................,,,..,....,........,.......,........,..!...XmYn ,. ..!_ o!.!,a!, ....,,1. l +).........O - HO_ +ao, . !,1. l ).,....,1. l +) ..,1. l ) ..........,......X ..Y ..,..,........,...........60 I nteracti on of Radi ati on wi th Matter Chap. 3 ..........,.....,,..........,,...,.....................,..Example 3.5 The absorption cross-section of 235U and 238U at 0. 0253 e V are 680. 8 b and 2.70 b, respectively. Calculate Ia for natural uranium at this energy. Solution. By use of the methods of Section 2. 1 4, the atom densities of 235U and 238U in natural uranium are found to be 3. 48 7 10-4 7 1 024 and 0. 0483 7 1 024, respectively. Then from Eq. (3. 1 4) Ia - 3. 48 X 1 0-4 7 680. 8 + 0. 0483 7 2. 70 =0. 367 cm-1 [Ans. ] Example 3.6 The scattering cross-sections (in bas) of hydrogen and oxygen at 1 Me V and 0.0253 e V are given in the following table. What are the values of as for the water molecule at these energies? Solution. Equation (3. 1 5) applies at 1 Me V, so that as (H20) ~ 2as CH) +as (O) = 2 7 3 + 8 = 14 b. [Ans.] Equation (3. 1 5) does not apply at 0.0253, and as (H20) = 2 7 21 + 4 = 46 b. The experimental value of as (H20) at 0. 0253 eV is 1 03 b. [Ans. ] 3. 4 NEUTRON FLUX H o I Me V 3 8 .Z e V 21 4 ..........1. !...............,I ........,.......,...,..,I= L, I. ,1. l )..L,.....,..........,...,,.....,1. 1, .........,..,........,..................................................,,.....................,.....,......................,I= L, ,I+Ie+Ic+. . ) . ,1 l )Sec. 3. 4 Neutron Fl ux

Figure 3.3 Neutron beams striking a target. 61 .................., ..., ,1. l ),....,......) c. ,1. l )..aA, ag, ..........................c......,..aA ag +a+ .,..a, ....,.......,..,.,, 1. l )..,1. l -).........,,......,........,.............,.a//.......,,1. l -).......a.......,..,...I..,...,...,ac..,,1. l -)...aeaoa]ax,........,.......,....,......= ac. ,1. !0)..............................,...,.....

............,.,1. !l ) ........,.............,.....,...,Example 3.7 A certain research reactor has a fux of 1 1 01 3 neutrons/cm2-sec and a volume of 64,000 cm3 If the fssion cross-section, 2, in the reactor is 0. 1 cm-1, what is the power of the reactor? Solution. The power may be obtained from the fssion rate using the relationship between the energy released per fssion (200 Me V) and the rate at which fssions are occuring: 62 Power I nteracti on of Radi ati on wi th Mater Chap. 3 1 MW 1 W-sec 1 . 60 1 0-1 3 joule 200 Me V f' SSlon rate 1 06 watt joules Me V fssion I MW 6 7 3. 2 X 1 0-11 watts/fssion/sec fssion rate 1 0 watt = 3. 2 1 0-17 MW/fssionsec fssion rate From Eq. (3. 21 ), the fssion rate is Fission rate = br< = 0. 1 cm-I \ 1 013 neutrons/cm2-sec. The reactor power/cm3 is then Power/cm3 = 3. 2 1 0-17 MW/fssion/sec 1 1 01 2 fssion/sec-cm3 The total power is the power/cm3 times the volume of the active core. Power = 3. 2 1 0-5 MW/cm3 64, 000 cm3 = 2 MW [Ans. ] 3. 5 NEUTRON CROSSSECTION DATA ...............,,...............,...............................................,.,..................,,..........................,....,...................,.............Compound Nucleus Formation ........,.2 .,............,..,.............,........,.....A z, ..,.......A+l z. ...,........,...,.......,. ...,... ..........,..,.......

..........,.,,..,.............,.......,....,....

+......,

+. ......,+. .

+.....,.+.......Sec. 3. 5 Neutron Cross-Secti on Data 63 .....,.........,,..,.,..................................,...........esoaaaces........,..,.....!. ................,..,...,.............. ...............,.......,.......,........,,............,.......,........ ...........,..............,...,............,.....!. l l ......,,...,.....,....,,................,....,,.,,..........................................,..................,...,,,......,,......,....,....,,..,.,,.......,......

El astic Scateri ng .......,............,,.............,.......,,,.re ..,,...,.........,...,....,.,...........,......,..,......,....,...........,.....,.,..,re ,....., +r.

,1. !!)............,..,.....,,....,...........,...........,..,........,........................,.....,.re .........,..,.,....,,,.1. +.......,....,............,,.............,............,. ...,.....,.

,....,.......,.,l..Example 3.8 Using experimental elastic scattering data, estimate the radius of the C nucleus. 2This discussion is somewhat simplified, center-of-mass efects having been ignored. For a more complete discussion, see Intrduction to Nuclear Reactor Theor, noted in the references. 6 C =COVooc\I nteracti on of Radi ati on wi th Matter Chap. 3 Elastic ad Total Cross Sections C MT 1 ZConstant to 0.01 5 eV Potential scattering 1 0 Resonance region Smooth region 10-1 10

100 10 I 1 02 1 03 1 04 1 05 1 06 1 07 1 08 Neutron Energy (eV) Figure 3.4 The elastic scattering and total cross-section of cabon. (Plotted from data received over the Interet from the Korean Atomic Energy Research Institute using ENDFPLOT and ENDFIB 6. 1 . ) Solution. From Fig. 3.4, it is observed that _ 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), 4R2 - 4.8 X 1 0-24 and R = 6 2 X 1 0-1 3 cm. [Ans. ] Inelastic Scateri ng ...,.....................,,,...,.......... ....O ...,......,,..,.,...,..,,.....................,...........,..O .......,.,,,...........,..............,.+. 0. ......,+..

.,.......O ..,.,,..O .Radiative Capture ..........,...........,.......,......,,3Because of center-of-mass efects, the threshold energy for inelastic scattering is actually somewhat higher than the energy of the frst excited state. Except for the very light nuclei, however, this can be ignored. Sec. 3. 5 104 1 03 CC1 02 `COVUUc1 01 \1 0 Neutron Cross-Section Data Radiative Capture Cross Section Au- 1 97 MT 1 02 65 1 0-1 10-3 1 0-2 1 0-1 1 00 1 01 Neutron Energy (eV) Figure 3.5 The radiative capture cross-section of Au- 1 97 at low energy. (From ENDFI 6 plotted over the Interet using ENDFPLOT from the Korean Atomic Energy Research Institute.) 1 0 ,.....a y ....l},, ..L......,,.......,c.,,...,........ay .....l/c. ....,,,.ay .........i /c eg|oa..........,..,,............l/c...,,.......,......,l }!. ........,1. , .....l /c,..............1 97 ....,....cy .......l /c......,,.........aoa- l }caosooes. ..l /c,....,................,..........as . ............,,L,ay .,..,...,...... r-g rn rg a ~ ay 4J , L L)+r2/4 . ,1. !1)....,...r......,........,,L .g.............sa|s|ca/]aco, r ..r g ........,.,.aeaoa|a...a|a|oa|a,..r = r n +r _ ...oa/|a....,.....a y ..................,.66 I nteracti on of Radi ati on with Matter Chap. 3 L r }!. ...r ...............,.......,...|a........,.........l .. .....,.........,,.,..,..,...Oy ,..,,.....,.,.......Charged-Paricle Reactions .....,.......,,...................,,.....................,.,..............,.........,..........

...

...............,1 . .....Oa ..,.,...,,.....

.........,,..........., 1. ..Oa .l/c.......,....,,...................,l /c ....

. . .. .........,............,,......,.........,........,.........,. ., cCC.OVUOC+\1 03 1 02 1 01 n- alpha Cross Section B- l O MT 1 07 1 00 1 0-2 1 0-1 1 00 1 01 1 02 1 03 104 1 05 Neutron Energy (eV) Figure 3.6 The cross-section for B- I0, n-alpha reaction from O. Ol eV to 1 0,000 eY. (From ENDFIB 6 using ENDFPLOT from the Korean Atomic Energy Research Institute over the Interet.) Sec. 3. 5 Neutron Cross-Secti on Data 67 ....,.,.......,....

..,..,......,,...,.. ......,..,.....- ...,..,....,.,...................,,,.............Total Cross-Section ..O ..................O ...,,...........,....... .,......,,O ......

cO =...+ _, ....c ..............,..............,....,...........,............,......,,....., .......,........O .........,,........O . ....,,,.......,.O ............O ..O , ...........,............,..,........,.O .........,....,,.......,. Hydrogen and Deuteri um ..... ..

.....,....,.........,.................................,....... ..

...........,........,........,...........O , ......,l 0....O/....,. ...............(.....,...,,..........,...Example 3.9 The value of Lfor 1 H at . .. eV is . b. What is Lat 1 eV? Solution. Since Lis 1 /v, it can be written as L(E) -L(Eo)

where Eo is any energy. In this problem, Lis known as ... eV and so it is reasonable to take this to be the value of Eo . Then . .. L( I e V) = . = ... b. [Ans. ] 68 L.P'Incident neutron I nteracti on of Radi ati on with Matter

_ '

Scattered neutron

nucleus ' " " Recoiling nucleus _A,

Figure 3.7 Elastic scattering of a neutron by a nucleus. 3. 6 ENERGY LOSS IN SCATERING COLLISIONS Chap. 3 ...........,...........................,,................,,......,......,....,,.,.,..,.........,......,,..........,......,,..........,...,,.........,...............,,........i,p ..L' , p' ....,,...................,.,..LA ..r..,,.........,................,i; ..........,( .,1. ). .................Li+La. ,1. !)...............,p = p' + r,..,,...,.......,1. . ...,......I j+,j ) !jj.i. P

Figure 3.8 Vector diagram for conservation of momentum. ,1. !)Sec. 3. 6 Energy Loss i n Scateri ng Col l i si ons 69 ..........

= !MLa,j-= !oL, ..' = !oL', ..M..o..................,.,.,...,1. !)........MLa= oL+oL' !o_LL'.i. ,1 !)..M}o..,,...,,..A, ................,,1.!).,....ALa= L+L' !_LL'.i. ....,La..,,1. !),........,,A+i ) L' !_LL'.i ,A l ) L ...,....,....-.........L'L

.i +A-~..

i]

,A+i ) ,1 !)..........,....,,1. !). ..,...,..i ..,,...,,......,,1. !),..L'= L.....,....,,...........,..,....,,..........,..,....,,1. !)...........L' ., L' )_,, .....i = R. ..........,............,.,......,,i = R, .,,1. !),....A i -, L'),,-= L= DL.A +l

'...co//|s|oajaaoee.....D .,.....1 . l . ,1. !-),1. 10).....,....,.,,....,.................,,.....,......,,.................................,....,...,...............,.........,.....,,....-0 ........,,........,,.......,,..,i = R}!..,,1. !), ....,.., L'),,-= 70 I nteracti on of Radi ati on wi th Matter TABLE 3. 1 COLLI SI ON PARAMETERS Nucleus Mass No. O; Hydrogen 0 1 .00 H2O * 0.920t Deuterium 2 0. 1 1 1 0.725 020 * 0.509t Beryllium 9 0.640 0.209 Carbon 1 2 0. 71 6 0. 1 58 Oxygen 1 6 0.779 0. 1 20 Sodium 23 0. 840 0.0825 Iron 56 0. 93 1 0.0357 Uranium 238 0.983 0.00838 *Not defned. t An appropriate average value. ...............,,..,A,1. !-),..,...., L') _,.L...a//.....,.,,.Chap. 3 | ..,,1. 1 l ) ............,.,,.....,.........,.........................,.....,.................,,,......,.,,...........,.............,.,,.......,...,,...,,L'= , l . t.....,.,,.. L..L= L L'= | - .) Lg......,....,,...L1 - , | - . L,1. 1!)...,.................,..,,.... ..238U, ...,.,,1. 1!).............,,| aa. ..,..,...,,......................,,.,,1. 1!) . Sec. 3. 6 Energy Loss i n Scateri ng Col l i si ons 71 .., .... .....D ...,,...........,..,......,...., ........,....,,...... .....,,.........,...........,...........,,.......,,......,....D .... ....,,.............D ..............,,...............,,......,..... ......,...........,........ooea|oa. ..,.,............................,.,...............,,.............,,..,.......................,...........,,......,...,..,.......,.............,,. ...,......,....,.....,.....,....................................,...,.,.................,.......,.,................................../eaagy....,...,..,..,.e......e

. ., . ...,.....,.,,....,....,...,,......,...., ........,..,,.......,,..................,,.... .....,..., ...,, .......Ae, .....,....,,....,,1. 1!)}, ..,....,,...... ..,...,Ae.,,......,....,..,.........,..,. .,........,.,,............,..,

. ..............,.... ..,....... ...,,...,...,...

....., ...,.,... ........,..... 72 I nteracti on of Radi ati on wi th Matter Chap. 3 Example 3.10 A I -Me V neutron is scattered through an angle of 45 in a collision with a 2H nucleus. (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 0 = 45 into Eq. (3. 28) gives immediately 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 Du = u' - u = In(E / E') -In( 1 /0. 738) = 0. 304 (a unitless number). [Ans. ] Polyenergetic Neutrons ....1 !..1. 1.............,.....,.............,..............,.......,,..............,..........,,..,.,.a(I)L........,....,...L ..L +L ..............,.....,.......I,L)= a,L) c,L) L. ,1. 1)..c(I)...,.,..,..,,L. .,.,,1. )...........,...I, L)a,L) c,L) Z, ,L)L...,....Z, ,L) .....,...........,...I,a,L) c,L) Z, ,L)L. ,1. 1)......,.......,.......,.......,...,,.....,....,,.Z,,L)..,,1. 1),..,,,.......,.,.,........,...................,......,.............!. l 1. .............,,.......aema/eacos,...........,.+.......,...Sec. 3. 6 Energy Loss i n Scateri ng Col l i si ons 73 .......... .I, ,a,L) c, L) L, ,L)L. ,1. 1)..L,,L).....,..,.........,...........,..,..0. l....1. , ...,.......,.......l } c..,...........,......,......,.L,, L) ......,1. 1-)..L,.....,.,,..o,...,..,.,...,,1. 1-).....,,1. 1), .c(L)......I, L, , L,) c,,a,L) L. ,1.+0)......,.,..,......,......a,...,,1. +0)..,1 +l ).,...,1.+i ) .......i/c .....,....,....,,.............,..........,.,,.....,....,. .....,..............,....,.. . ............,..........,.,,L, .....,ao,.................,..,,...,......i/c.....,.,,L, 0. 0!1 .....,..,.,.o,!, !00.............0.0!1...,..aema/coss-sec|oas.............,.....,.........,.....,,.. ...,...,ac,..,,1.+i ) ..!,!00..,........,...ac,

,1.+!)....,.......,,,1.+1)74 I nteracti on of Radi ati on wi th Matter Chap. 3 ..,..,..,...,...... ............,.,.......,...............,.....,.........,..,..,.,,1. 1). .....,......,..........,...I ..,.......a(L) ....., , L) . ...,......,..a(L).......... . ...,.I, ....,...,.... ........,...I, .......,............,.....,1++)..g, , I) , ......aoa- i }c ]aco,..........,L,)..,.....,......0.0!1...... ...,.....1. !. ..,..,...................,.,.,,,,..,....,...........,....,...