THE UNIVERSITY OF MICHIGAN INDUSTRY PROGRAM OF THE COLLEGE OF ENGINEERING IRRADIATION INDUCED ATOMIC DISPLACEMENTS IN METALS Jean Michel Planeix A dissertation submitted in partial fulfillment of the requirements for the degree of Doctor of Philosophy in the University of Michigan 1958 September, 1958 IP-315

Doctoral Committee: Professor Ho Jo Gomberg, Chairman Assistant Professor B. A. Galler Professor E. H. Rothe Professor Mo J. Sinnott Assistant Professor G. Lo West, Jr. Professor Eo F. Westrum, Jr. ii

ACKNOWLEDGEMENTS The author wishes to express his gratitude to the members of his Doctoral Committee for their interest in this work and their understanding, and particularly-to Professors H. J. Gomberg, M. J. Westrumn, and M. J. Sinnott for frequent and fruitful discussions. Many thanks are due to Mr. R. White for his very competent work in building the components of the cryostat under construction and to.Mr. J. Mannlein for his accurate job on an unusual liquid helium transfer. Finally, the author is indebted to the Industry Program of the College of Engineering for having accepted this dissertation for distri' bution. iii

TABLE OF CONTENTS Page DEDICATION......... o......... O................ e o o e ii ACKNOWLEDGMENTS.o...... o. o.. o......***....*.*o..o.a iii LIST OF TABLE.................... vi LIST OF FIGURES......... o a...........b o......,........ O..... vii LIST OF DIAGRAS. o o... ix LIST OF APPENDICES... O......,..... o,....... X CHAPTER I. INTRODUCTION......O o... O............ O...0. 1. CHAPTER II. BACKGROUND.....o oo 0 0o o0 0 0 0 ooo*0oooo. 0 0o 1.2 CHAPTERE III. COLLISION BETWEEN TWO IDENTICAL PARTICILES INTERACTING THROUGH A MUTUAL POTENTIAL ENERGY AND ACTED UPON BY AN EXTERNAL FIELD (Application to the Collision of a Knock-On and a Stationary Atom in Copper).......... 18 1e Generalitieso 6oooooo o.v***oooooooovooeoao 18 2. Choice of an Interaction Potential Energy.o.. 22 3. Classical Treatment of Scattering by a Center of Force Giving Rise to an Interaction Potential Energy Depending on the Distance On.ly...............-.......... 29 4~ Application to the Selected Potential Energy in the Case of Coppero.................. 34 5. Numerical Calculation of Knock-On Displacement Cross Section and Mean Free Path in Copper......,...... a................. 66 CHAPTER IV. DISCUSSION OF MODEL. COMPARISON WITH BRINKMAN THEORY. o.......o. a....o..e.o.oo..o......o.....0 0... 0 oo 71 1. Brinkman Theory,.......O...........O.... 71 2. Main Features of the Proposed Model. Application to Deuteron Irradiation of Copper. o..... a. ~. o. o.o o..... o o. o 0000 82 3. A Possible Explanation of the Phenomenon of Radiation Anneal. Comparison of Charged Particle and Neutron Irradiation.. *.O.fi..... 95 4. Extension of the Model to Other Metals than Copper. Expected Effects of Charged Particle and Neutron Irradiation as a Function of Atomic Number and Mass Number,... 109 iv

TABLE OF CONTENTS (CONT'.D) Page CHAPTER V. CALCULATTON OF TriE NUMBER OF ATOMS DISPLACED rER PRIMARY KNOCKON a o o o oa o o o o o o o o o 12 1i Generalities-Obtention of the Primary Integral EquatiQo on oO aO:.0 0 09 0 ~~,~~ Q0 000' 122 20 Obtention of an AsymIptotic SQoLution for the Model of Interaction Usedo Comparison with the Asympttic Solution of the Snyder and Neufeld- Equation. COnfrontation of Calculati.onal and Experimental Resultso... o o..... 128 CONCLUSIONo o. 6.. o, oO,,o.,., oO.O.O.... o, O Oo o0 or0o o a O...o...a00 O- 0 134 APPENDICESo OOO.O.O O O0000 0000a O O O O O O O O O0 0 0 OO 0 0 0 0 OOOOO 135 BIBLIOGRAPHYC 0 O. 0 O O O O 00 000 00. O O. O..000000000000 o o o.........o O..... 245 NOTE: The following signs are used throughout this dissertation., meaning proportional to _ meaning approximately equal. to r\3 —~~

LIST OF TABLES Table Page I Values of V(r)..................... 26 II R, R, Ru for E = 10-1 Mev, p = a/100............... 52 III R, Re, Ru for E = 10-1 Mev, p = a.................... 54 IV, R. R, Ru for E = 10-2 Mev, p =a/100. 0...o....... 56 V R, Rg, Ru for E = 5 x 10-3 Mev, p = a/100......... 58 VI R, Rp Ru for E = 10o3 Mev, p = a/100............. 60 VII R, R, R Ru for E = 10-4 Mev, p = a/100............. 62 VIII R., R, Ru for E = 25 ev, p = a/100................ 0. 64 IX au(E) and -ce(E) for Various Eo e....00,......v.0oO 66 x (Pd)u' (Pd).' (Xd)u, (Xd)2 for Various E.........67 vi

LIST -OF FIGURES Figure Page 31 Diagram for the Collision of TwO Partiles o...O.. o 21 2 Bri nkman Potential Energy0. di 0 0 v o' * 0 O J.0. 24 3 Distane of Closet Approache,...0 * o o 0 * o 0 o o o o 0 o. o o 28 4 Polar Diagram for Kepler Probxlewm..?o E O'..O O O oo o.. O O.0 30 5 Branches of t'. Trajectory.*..o.oo.. aoooo a..o 32. Synmetry of the Trajectory...oo...... o..... *.. o o 32 7 Bondary Conditions at r/2..**.....oo.0.o 0 o 5. 32 8 Preferential Interaction..... o e o.-....e... o... O o * 36 9 Beginning f InteractiOn,. 0.e. 0.o.0..... 0 o.0.0. o. o 0o 356 10 Possible Confi.gurations When the First Particle Reaches Distance r2 from the Center -of Collision... 38 U1 Diagram of Angles and Speeds in Laboratory Frame,... 38 12 End of Colli sion.'* 0*0 0 0 o 0 0ooO0 0o 0 0o0o 0 0 0 a oo0 38 13 Two Body Interactiono.,o...............,............ a *O.0 a O. 41 14 Equivalent Condition at Infinity*..:.. 0 o o 00...o0 o 0 e o 0 41 15 Origin of Angles..0...0,o.0, 0. 0 D e 000.I4,0..0..:o o.....o... o 41 16 Prolongation of Trajectory to Infinityo.o.0 o 0 oo.o 45 17 PMaximum Value of p Due to Approximation,0 *. O o o oo.. 48 18 Upper and Lower Approximations.. *..*o. 0 o.0 o 0 o. o0, o o oo- O. 48 19 Angle of Asymptotes.0..... o. o_,: 0....O O 48 2.0 The Function F(b/a.) * o o..o,*.D o..,.:o o, ooo...... 81 21 Track of an Energetic Primary Knock- 0Or (, Mev),..... 86 vii

LIST OF FIGURES (CONT'D) Fie Page 22 Frequency Function cr(E', E) of the Cross Section for Energy Transfer by Charged Particle...,........ 92' 23 Illustrating the Possible Interaction of Defects of Different Generations in Charged Particle Irradiation... ~.o o...... 0 9 0 0,0 0 0 e O O O 0.. O o o o ~.O O a o. O, o O ~* 94 24 Frequency Function a(E', E) of the CrOss Section for Energy Transfer by Neutrons, in Elastic Collisions, Isotropic in the Center of Mass Frame..,....0o-o 101 25 Illustrating the Possible Interaction of Defects of Different Generations in Neutron IrradiatiOn.~,,, 103 26 Possible Variation of )d for Light Metals.o........... 111 27 Effect of Radiation Annealo, *oo..o, o o,.o, o.......... i111 27a Neutron Collision... OO a OOO o.., a... o..0l0 O- 00 0 0 114 28 Angle of Scatter in the Center of Mass Frame.o.oo... 12j3 29 Variation of ~ versus p.o.aaOOaOOa, OaOaOOOa*O... 129 viii

LIST OF DIAGRAMS Diag ra Page-.1 PIlt Of V(r), V(r) V() and u2'o.o.o....eo a,.o. 51 2 R, R., Rua for E = 10-1 Mev, p = a/10.000o0o. 0.eo.o0o 53 3 R, Rg, Ru, fr E = 10J- Mev, p = ao.*,.. o~... o. o o o 55 4 R, R., RRu for E =.10-2 Mev p - a/l.. o.0 o o....O 57 5 R, R. Ru, f'or E = 5 x 1'o:3 Mevr, p = a/000 0.o.,0.. o 59 6 R, R, Ru, for E = 10-3 Mev,. p a/t100.0O0.,0 oo oo.O 61 7 R Ry, R..,Ru fQr E = 104 Mev, p =:/l o.o..... 63 8 R, R Ru for E = 25 ev, p = /l o... o0.O..o. o...o 65 9 Limits of p and Possible Region of many Body Collision.,.a. o.. o o o. o 0 o ooo'. 0 o: o, o o. o o, o o. 68 10 Limits of Xd, Possible Region of many Bordy Collision, and Brinkman Estimate of O.....o f... o.oe 69 11 Change po in Electrical Resistivity versus Integrat:ed Flux. for Cyclotron Irradiation of Cu with 12 Mev Deuterons at Liquid He. o o oooO..oO O 96 12- Near Liquid Helium. Temperature React Irradiation of Varis~ Met.als,.and.A!Iy S a ao Ox a0.,. o... o..o, 1o o o 1.06

LIST OF APPENDICES Appendix Page I Corresp:Qndence Between a Born-Mayer Interaction and the Potential Energy used in the Paper, at Large Separation........... 135 II Reduction of the Schrodinger Equation for a System of Two Particles.....*.. o... o.......... 138 III Wavelength of a Copper Knock-On "Reduced" Particle at Various Energies............................. 140 IV Constant of the Law of Areas and Angular Momentum 4......... 0.0.... o. O.... o o o 141 v Sign of dV/dou.....,,... e.........,...... 142 VI First Integral of the Equation of Motion. Boundary Conditions at r = ro/20.....00,.0..... 0 143 VII Obtention of Equation of Motion...o.......a......... 144 VIII General Relations for Elastic Scattering............ 145 IX Discussion of Plot of V(r). e*............. o o ooo 151 X Maximum Value of -c~ S n-\ 1C t (E)t 2- - p p- 152 XI Calculation of Qo, Angle of Asymptote OB with the Velocity vl at AO (r = ro/2). * O e 154 XII Calculation of (Pd)u, l, (ad)u,, (Zd)u,, (Xd)u, for Various Energies.,........................ OO 156 XIII lMutual Potential Energy of Two Rigid Charge Distributions with Screened Potential.,........,.. o o,... 159 XIV Energy Transferred in the Impulse Approximation, for the Interaction Energy Used by Brinkman........o 162 XV Average Number of Atoms Disp aced Per Primary in the Range 1.5 Mev - 2.3 x 10 ev,........,... 166 XVI Number of Defect Pairs Per Primary in Displacement Spikes................. o..... o.. oo....... o... 167 x

LIST OF APPENDICES (CONT. D) Appendix Page, IXVII Average Energy Transfer in a Displacing Collision ftr the Interaction Cross-Secti=on Adoptedo..... o 168 XVIII Fraction.f'Primaries, Average Number of Displacements Per Primary, and Fraction of Atoms Displaced in the, Irradiatiot of Copper by 12 Met Deuterons a0 0 o 000:o 000 000 00000..0.0 0.. oo 171 XIX'Calculation'of the Cefficient B of Radiation A nea.,O * O~- O o o o o.. O O a o. O. O 0 O e o - o o o-o o o o o0 o o o o ~ O O ~ 0 ~ 177 XX Upper- Energy-at Which Classical Approach Remains Valid in Light "Metals000 0 0r o0 0o0O 178 XXI Variation of the Fractin of Primary Knock-Ons with Z and A in Charged Particle Irradiationo. o oo 180 XXII Solution of Snyder and Neufeld Primary Equation and Asoeciated Problems..0,0o0.0 1681 XXIII Direct Obtention of the Primary Integral Equation for the Model Used in the Collision Problem..00... o 186 XXIV The Design of a Cryotat for Pile Irradiationo..... 187 Sect ion I- Int roduct i on 00o 0.. O o... O o. O O 0, o 0 a o o 188 1-Basic Idea o..... o, 0...oooooooO 0 0 0....... 0oo. 188 2-Application to. the Study of Atomic Displacements by Fast Neutron Bombardment..0o 190 3-Mass of Cryostat. Consumption of Liquid Nitrogen and Heliumo o0000000.. 0....o o 0...0oo 192 4-Construction Schedule,.a.0... o o00,.0....00 193 5-Demand on the Reactor Schedule...0o.0.. 0000. 194 Section II-Feasibility of In-Pile Measurement of Change in Electrical Resistivity Due to: Neutron Irradiation oo a.. 000.. 196 1-Fast Flux Available-Change in Electrical Resistivity Expected. 0 00 oao. 00 0.00 0 o 196 2-Feasibility of Measurement o.o 00. 4...o O. 0 b O o 197 xi

LIST OF APPENDICES (CONT' D) Appendix Page XXIV (Cont d) Section III-Design of the Cryostat................ 201 1-General Considerations........&...,O...O....O0 201 2-Calculation of Gamma Heating...... 201 3-Other Sources of Radiation Heatingo....0..... 218 4-Heat Transfers Into the SystemO..o.. -t......O 219 5-Total Volume Rate of Consumption of Liquid Helium and Nitrogen and Total Consumption,,..... 220 6-Cooling Prior to Experiment............................ 221 7-Expenditure of Liquefied Gases During Constructiono Total Expenditure of Liquid Helium and Nitrogenoo...,.,...000o.......... 222 8-Selection of Aluminum Alloys....O....00,...... 223 9-Mechanical Resistance...................... 225 10-Thermal Contraction on Coolingo..e.......o...... 227 11-Activation of Nitrogen-14..................... 0 229 Section IV-Preliminary Experiments................... 233 ilMeasurement of Electrical Resistance,......... 233 2-Transfer Tests of Liquid Nitrogen and Helium.. 238 Addendum............. O 0 0.O...O..... e 242 Bibliography to Appendix XXIV 06*......80O...o.. 243 xii

CHAPTER I INTRODUCTION Of the two broad categories of the effects of radiations on matter, namely atomic displacements and ionizations: the first one is of basic importance in solid state studies, since it is closely dependent the interaction ofl identical atoms of the irradiated sample-, Atoicdisplacements constitute the only permanent radiation damage in non fissile metals and they are only produced to an important extent by massive particles, those coming into consideration being neutrons for pile irradiation and protons, deuterons, alpha —particles for cyclotron irradiationo The present status of theoretical and experimental research in this field has been summarized in a previous paper0 (l) Some indications relative to this status will be given in Chapter II, Background., Much more experimental work has been done than theO-retical worko The two most important recent pieces of theoretical work areBrinkan theory f displacement spikes(2) and Snyder and Neufeld(5) calculation of the total number of atoms displaced per atom displaced by bombarding particles, i.e. per primary knck=on.o Brinkman assumes an interaction potential energy between a knock-on and a stationary lattice atom which is the ele-ctrostatic mutual potential energy of twGo rigid charge distributions having each a typical screened potential. This potential energy is negative deep within the atom, which stems from the fact that closed shell.repulsion between the two atoms is neglected in this treatment, and leads to a diseontinuity in the interaction whe the sep-aration of the two atoms drops below the value for

-2which the potential energy has its minimum. Brinkman uses an impulse approximation to assess energy transfer between moving and stationary atoms and obtains a mean free path between displacing collisions which is smaller than the interatomic distance when the knock-on energy falls below a certain transition value, typical of the metal. Hence he concludes to the existence of highly disturbed localized regions, which he calls "displacement spikes". This theory leads to a qualitative explanation of the phenomena observed when samples are thermally an-ealed after low temperature irradiation. Snyder and Neufeld proposed the only calculational model allowing qualitative confrontation to be made with experimental results0 This model assumes hard sphere scatter between moving and stationary atoms, at all energies. It is worth noting that a direct confrontation is impossible. It is necessary to adopt a theoretical value for the change in'electrical resistivity due to a fraction of one percent of atoms displaced, i.e. of Frenkel pairs if we regard the sample as perfect before irradiation. This value has been calculated for some monovalent metals by Jongenburger(4), Blatt(5), and Dexter.(6) The author(l) has made an independent calculation, assuming a screened potential interaction between defect and conduction electron of the same form, but opposite sign, depending whether the defect considered is an interstitial atom or a vacancy. His result, for 1% Frenkel pairs in coppe-ris about the same as IDexter's estimate and abut three, times smaller than Jongenburgers.. Dividing the observed change in

electrical resistivity by the adopted value of the inftuence of o.ne percent Frenkel pairs on resistivity (2.7, Q x cm is generally admitteds after Jongenburger) furnishes a first estimate of the ~raction of existing.defect pairso If the experiment shows radiatin anneal. the tangent t. the.curve at the origin is extrapolated to replace the.experimental curve A second estimate is obtained by calculating the fraction of lattice atoms becoming primary knock-ons, fom cross sections, and bombarding -particle fluxesand the number of atoms displaced per primary, using Snyder and Neufeld.methodo Such a confrontation made.g in the cas'e of pile neutron irradiation of copper and cyclotron deuteron irradiation of the same metal shows that the second estimate is 4 times and 6 times higher than the first one, for pile and Cyclotron irradiation respectively0 Similar treatment of other properties,. such as changes in Hall coefficient and neutron scattering cross section of graphite -duri:ng neutron irradiation and volume expansio:n f capper under deuteron bombardment, have shawn deviations in the same direction, and of about the same magnitude, between the two estimates. It seems,. therefore, that Snyder and Neufeld method over estimates the number of atoms displaced per primary knock-: on.'Measurements of electrical resis tivity have been performed during cyclotron irradiation(7), (8) and reactor irradiatin(31), in both cases:at very low temperatureo The cyclotron experiments show that, as irradiation progresses, the line representing the change in resistivity versus integrated particle flux deviates from linearity,

-4bending downward. Thus some damage is recovered in this way and this process is called "radiation anneal". Seitz and Koehler(9) have investigated the problem of recombination between defects of various generations, i.e. corresponding to various primary displacing collisions between bombarding particles and lattice atoms and conclude that the values of knock-on range required for such process are higher than the values one would reasonably expect for a moving atom. But the case they considered was that of extreme recombination, i.e, saturation, when the rate of recombination equals that of formation. The reactor experiments show that, for exposures of 150 hours in a fast flux of the order of 7 x 1011 neuts cmrl sec-l, the change in resistivity is proportional to irradiation time, in other words that no appreciable radiation anneal is taking place. The aim of this dissertation is to study in an independent fashion the problems of interaction energy and energy transfer between knock-ons and lattice stationary atoms, knock-on displacement cross section and mean free path, radiation anneal, and number of atoms displaced per primary knock-ono The results are compared to those obtained from the existing theories and to experimental data, qualitatively and quantitatively. Comparison of expected damage in light and heavy metals, both for charged particle and neutron irradiation is drawn up. Basic principles are obtained for the design of a pile neutron irradiation experiment.. Assuming, subject to check by the results, that the problem at hand is a two body problem, the collision of two identical particles

interacting through a mutual potential energy and. acted. upon by an external field which is essentially the same at the positioof both particles at any time of the interaction for the separations coming inato considleration is first reduced to a prOblem (in the eenter of mass frame) of scattering of a particle by a center of force, the potential energy in the reduced problem being the same as the interaction potential energy of the two identical particles~ The potential energy adopted for two interacting atoms is the sum of the electrostatic potential energy-of two rigid charge'-distributions with screened potential and of a term corresponding to closed shell repulsion of the two atomso It has the correct Coulombian form at small separation and takes a value with the correct positive sign (repulsion) and a magnitude consistent with compressibility data, at separations of about 5 to 7 times the screening distance. The case of copper is treated numerically in -detail. First it is seen that, for E = 25 ev, the distance of closest approach is about the atomic radiuso This tends to show that an atom receiving an energy of the order of 25 ev in a collision will not be able to pass through the nearest neighbors, but will be pushed back to its normal lattice site. At any rate, it is clear that low energy d efect pairs have a small separation. It is shown that classical treatment is valid over the whole range of knock-on energies for pile neutron irradiation and for deuteron irradiation with deut-eron energy up to at least 12 Mev. Instead of using the impulse approximation, the center of mass frame

reduced problem is treated as a Kepler problem0 A strict first integral of the motion is obtained and more proper boundary conditions than a condition at infinity are employed. This result is quite general, independent of the form of potential energyO Angle of scatter, displacement cross section and displacement mean free path at various energies are then bracketed between an upper value and a lower value by replacing the potential energy by two functions of the separation, one overestimating the interaction, the other underestimating it and such that the only remaining integration giving the equation of motion can be performed.. The lower value of the displacement mean free path is shown to be a good approximation at high knock-on energy. An advantage of this method over Brinkman's is that it yields an upper and a lower limit for the interaction parameters. The lower values of the displacement mean free path are close to the estimate of Brinkman at energies down to 10-3 Mev and are larger below. This is consistent with the conalusions of a critique of Brinkman model, namely of the form of interaction energy and the screening distance it uses, and of the impulse approximation, all of which tend to overestimate the interaction, es-, pecially at low energy. The results obtained essentially check the validity of the assumption made of a two body collision problem. In the model of this paper and the case of copper, there is an energy region, estimated to lie between 10l1 and 10-3 Mev, where the displacement mean free path is comparable to the interatomic distance and where we may consider that a knock on creates a displacement spike,

But this model "attaches a tail" to Brinkman displacement spikeso a knock-on slowed down tQ 10o3 Mev within the spike can escape from it and travel quite a distance, creating displacements, before it is trapped as interstitial, or recembines with a vacancyo However, the low -energy pairs formed toward the end of a track have a small separation0 The region of the sample disturbed'by a primary knrck-on and its progeny is seen as a oylindrical region,, along the track of the primaryy essentially linear down to energy 103ev, which we call a'damage spike"-o It comprises the displacement spike in its middle. For a I Mev primary, its lengthg or about the range of the primary knock-ony is estimated of the order of 660 r. for copper, ioe 1.68 x 10-5 cm0. This is still much less than the estimated range -4 x 1o-4 cm-of Uranium fission fragments in.Uranium; it is also com_patible with the number of secnaries for a primay of that enrgy0 This shows another superiority of the method, naely the obtention of an analytical expression approximating thedisplacement cross section, manageable enough to permit the calEaulation of the average energy transc fer in a c:llision,: hence that of the approximate range0 Using results obtained later in the paper, the number of atoms displaced ger primary is obtained for a copper knock-n having the aver age energy of a primary produced by a 1 Mev neutron and for a copper knock-on having the average energy off primary produced by a 12 Mev deuteron; these average energies are 3.1 X 104 ev and 275 evg respectively, the numbers of displacements per primary, about 600 and 6 respectively,. Hence displacement spikes will oc.cur in the neutron irradiation, not rin the charged particle itrradiation0

-8Consideration of the distance, from the point of birth of a primary knock-on, at which small separation defect pairs are formed, at the end of the track of the primary, of the number of defect pairs per primary, and of the separation of the points of birth of the various primaries, shows that, for irradiation of copper by 12 Mev deuterons(8), at one fourth the full irradiation used in the experiments interaction of defects newly formed with defects previously formed can be appreciable. For neutron irradiation, a similar reasoning shows that such interaction would not be appreciable with the exposures coming into consideration in experiments. Hence, for charged particle irradiation, there is the possibility that the thermal or electron spikes of the knock-ons of a generation (i.eo corresponding to a primary collision) will cause appreciable recombination of low separation defects of a previous generation. This would explain the phenomenon of radiation anneal which, on this basis, is not expected in pile neutron irradiation, for the exposures and fast fluxes coming in consideration in experimentso As pointed out before, this last prediction is confirmed by reactor experiments (31) The type of recombination proposed explains quite well the differential equation which Cooper(l0) found to fit closely the curve of change in resistivity versus integrated flux in the deuteron experiment. Classical treatment as used in the case of copper is still valid for pile neutron irradiation of beryllium and even cyclotron irradiation of beryllium by deuterons of more than 20 Mev. But, for low atomic number, the model is inadequate at high energy, where ionization is important. The potential energy of the form used, where a

screening distance depending on the atomic number is involved, is shown to decrease motonically with the atoic number Z at all separations coming in consideration, so that energy transfer and displacement cross section decrease with Z and the displacement mean f'ree path increases when Z decreases, For Z.ow enough, it is possible that no displacement spikes are formed at any knock-on energyo For charged particle irradiati'on there will be less disturbed regions in light metals than in heavy %etals, but the mean free paths will be larger, so. that the chances of recombination will not be appreciably changed.. Hence we expect radiation anneal to hav approximately the same effect in various metals. This conclusion is borne out by the experimental results of Marx C er, and.enderson. (8) For neutron irradiation,. the n.riber of disturbed re gions mainly depends on the scattering cross section. The number-of defects in a region varies -on the average as the reciprocal.of the mass number, so: that, although radiation anneal is not expected.for "reasonable" irradiati.Ons, chances for its appearance are larger for light metals with highneutrn.scattering cross section. As said before it it.possible that, for an atomic number low -enQugh displacement spikes wi11 not formo These conclusions show the interest presented by in pile measurements. fr neutron irradiation If knock-on collisions with stationary atoms can be described at all. energies by differential cross sections analytically known, an integral equation, replacing that of Snyder and Neufeld. but more general,

can be studied in the asymptotic case of a primary with high energy compared to the energy needed to displace an atom from a normal lattice site. The application of the method to the approximate interaction cross section in copper found earlier in this paper shows that a linear asymptotic solution is a possible approximation, io.e, number of atoms displaced per primary varying linearly with energy of the primary, the coefficient of the variable- energy being almost equal to that found by Snyder and Neufeld (2 x 14 in this paper, against 2.24 x 104from th.enyder and Neufeld. equation, E in Mev). Since the approximate cross section used overestimates interaction at low energy and most primary knock-ons in charged particle irradiation and secondary knock-ons in neutron irradiation have low energy (say, below 103 ev), the estimate of the number of atoms displaced per primary is too high, hence, also theSryderardl Neu-feld estimate is too higho This agrees with experimental results, as shown before in this Introduction. We also expect the estimate to be better for neutron irradiation, where an appreciable fraction of primaries have high energy, than for charged particle irradiation (at reasonable particle energies, say 20 Mev at most for deuterons) where most of the primaries have low energy. This trend seems to be followed by the results of experiment and of calculations for the 12 Mev deuteron experiment(8) already mentioned and a pile experiment(ll) performed in the same conditions of temperature (liquid helium). In summary, the interaction potential energy used yields collision and damage parameters qualitatively and quantatively compatible with ex-' perimental results and the method of investigation throws light into the

~main processes of atomic displacements both in the case of neutron and czharged particle irradiation, It allows for useful comparisons between these two modes of irradiation and shows the' importa.nce of the in pile measurement during neutron irradiation~

CHAPTER II BACKGROUND In -this brief review, the most important pieces of theoretical and experimental work in the field will be described, with the exception of that by Brinkman and by Snyder and Neufeld, already described in the Introduction, and on which we shall spend some time, later in this dissertation. a. Determination of the energy Ed needed to displace an atom permanently Ed has been calculated theoretically by EHuntington. (12) He employs an interaction potential energy between atoms, of the form\'" ('' i.; where r is the separation of the interacting atoms, r'f the normal spacing of the lattice, A and:!, constants. Depending on the direction in -which the atom is displaced, it is found that, for copper, Ed should be bracketed between the two values 18 ev and 40 evo The above formula is of the Born-Mayer type. The constants A and p are chosen to fit compressibility data. In Appendix I it is shown that it is comparable to and yields values of the same order as the potential energy used in this dissertation for the interaction of two copper atoms, at large separation. Eggen and Laubenstein( 13) have measured Ed experimentally in copper by observing the threshold electron energy for which atomic displacements are evidenced by X-ray inspection. They found the value 25 ev.

-13Denny [unpublished work, quoted by Seitz and Koehler(9)] has studied the effect of electrons on precipitates of iron in a CuFe alloy with 2.4% iron and found that Ed should be 27+ 1:ev in iron. b. Irradiation of Cu3Au by alpha-particles Dixon and Bowen (14) irradiated Cu3Au with 36 Mev alphaparticles. They found that the initial disordering rate, as measured by change in electrical resistivity)was proportional to the bombarding flux. c. Irradiation of iron, nickel, and cobalt'wires by 10 Mev deuterons Wruck and Wert(l5) irradiated iron, nickel, and cobalt wires by 10Mev deuterons, at -1500C. For an integrated flux of 1017 deuts cm'2, the relative change of electrical resistivity, measured at -150~C, was 0.5 for irbn, body centered cubic) and 0.1 for nickel and cobalt, hexagonal close packed and face centered cubic. d. Irradiation of/.copper wires by 20 Mev deuterons Dieckamp and Crittenden(16) bombarded high purity (99.999%) polycrystalline copper wires by 20 Mev deuterons at -1750~C The recovery of the change in shear modulus was observed at close temperature intervals between -1960C and +3000C, with an allowed annealing time of 15 minutes at each temperature step. The shear modulus decreased by 1.5% upon irradiation (84t A hr cm-2). One third of this change annealed at -125~C, further very little recovery took place at -1000C and -750C and practically the final two thirds were recovered continuously between -500C and +1000C.

e. Irradiation of copper, silver, and gold by 12 Mev deuterons Cooper, Koehler, and Marx(827 ) irradiated pure thin wires of copper, silver, and gold at 10'K, using 12 Mev deuterons. Measurement of electrical resistivity during irradiation showed the occurrence of radiation anneal. After irradiation, the samples were left to warm up. It was found that, for copper and silver, a very rapid recovery takes place near 430K (40-50% anneal) and 300K (13-24% anneal) for each of the two metals, respectively. Recovery continued gradually from 50 to 220~K, becoming more rapid above 2200K. At 3000K, the remaining changes were 8% of initial change for copper 10% of initial change for silver and gold. f. Irradiation of copper by 19 Mev deuterons McDonnell and Kierstead(l7) irradiated a bent tube of commercial copper by 19 Mev deuterons, at -180~C. The volume expansion of the sample was measured by the Change in bendingo A relative volume change of 0.068% was found for 1.15 x 1017 deuts cm-2. g. Irradiation of copper, silver, gold, nickel, and tantalum Marx, Cooper, and Henderson(7) irradiated thin foils of copper, silver, gold, nickel, and tantalum by 12 Mev deuterons at liquid nitrogen temperature. This experiment, performed, chronologically, before the experiment quoted in (e) above, gave results similar to those obtained in that experiment. The comparison of the two shows the influence of thermal annealing of the defects below liquid nitrogen temperature.

-15~h. Isothermal annealing of irradiated copper Overhauser(18) has followed the annealing of the damage induced in copper by 12 Mev dMuterons at -145~Co He found that the activation energy for amnnealing.varies linearly with temperatures at low temperature, and that.there possibly exists also a single iso'lated recovery, at -3Q~C,, of activation energy 0.68 ev. i. Pile irradiation of copper and gold Redman, Noggle, Coltman, and Blewitt(ll) irradiated copper and gold in.the Oak Ridge reactor, at 170K, for 154 hours. The theoretical change in resistivity, obtained from.Snyder and Neufeld method(3) for the fraction of defects formed, from Jongenburger value(4) of an increase of resistivity of 2.74 aL cm per one per cent;Frenkel defects, and from relations by Holmes [unpublished. but quoted by Seitz.and Koehler(9)] expressing the neutron flux. +(E) in the experimental hole used, has been calculated in the Preliminary Study.(l) It is found that the theoretical value is about four.times higher than.the experimental one. J. Pile irradiation of UCr Tucker and Senio(l9) used fission thermal spikes.by bombarding uranium containing.-2% —hromium in the Brookhaven reactor. X-ray observation after irradia#Lon failed to show the presence of betauranium, which should be retained by dhromium if nucleation after melting took place in a,thermal or displacement spike. k. Pile irradiation of copper and aluminum McReynolds, Augustyniak, McKeown and Rosenblatt(20) have irradiated copper and aluminum in the Brookhaven reactor, at liquid nitrogen temperature. Electrical resistivtity and critical shear

stress changes were measured after irradiation, and their recovery was followed during thermal anneal. For copper, it was found that there is recovery of electrical resistivity in a lower temperature process, between -80 and +200C, and in a higher temperature process, between 300 and 3500~C, this last process being accompanied by the recovery of the critical shear stress. For aluminum, recovery of electrical resistivity and of critical shear stress takes place in a single process around.600C. 1. Evidence of melted regions in the spikes. Denney(21) has irradiated ferromagnetic samples of a FeCu alloy with 2.4% copper in a cyclotron. Such an alloy is metastable, iron precipate s, the precipitate being paramagnetic, but going over to a ferromagnetic form under the influence of cold work or particle irradiation. The ferromagnetic precipitate is stable, except when the sample is heated above the two phase region. In the experiment, it was found that irradiation decreased the ferromagnetism of the sample, from what one can induce that melting,has taken place in some regions. m. Low temperature pile irradiation of various metals and alloys with measure of electrical resistivity during irradiation Blewitt, Coltman, Holmes, and Noggle(31) have bombarded various metals and alloys, including copper, aluminum, nickel, iron, gold, cobalt, Cu3Au, brass, around 220K. The interesting result of these experiments, for the purpose of this dissertation, is that the increase of electrical resistivity varies, for all metals and alloys investigated, proportionally to the time of irradiation, i.e. that no radiation anneal is apparent.

This list is very incomplete but, nevertheless, contains more background than will be used in the study. It is believed it gives a fair cross section of the status of atomic displacement studies.

CHAPTER III COLLISION BETWEEN TWO IDENTICAL PARTICLES INTERACTING THROUGH A MUTUAL POTENTIAL ENERGY AND ACTED UPON BY AN EXTERNAL FIELD (Application to the collision of a knock on and a stationary atom in copper) 1. Generalities Consider 2 particles of masses M, M2, subjected to an external potential li such that, if E and E2 are the coordinates of the particles in the laboratory frame, the potential energy of the system of the 2 particles in the external field is U (l, R2) Assume a potential energy of the system of the 2 particles, isolated from the external field, of the form X' (r) where The Schroedinger equation defining the wave function' (R,, 2 t) representative of the system of 2 particles is: _ =FL~7' 24 a&V \ I-V(r)-\(Q, RD. (1) Where the dot denotes derivation with respect to time and the Laplacians are taken with respect to the coordinates of each particle. Defining,& = 1 R_1 + 1-Q,' where i is the reduced mass of the system, n~2 ml equal to mlmn,L-.PRis the coordinate of the center of mass in the ml+m2 18

-19laboratory frame, and Equation (1) can be thrown into the form (see Appendix II), ~* 1c, L,, + A V2r}.-iV(r) +U CRi)&S1SV where M = m1 + m2 and the Laplacians are taken with respect to the componenets of ~ and r. We now consider the interaction between two idenetical atoms, one a knock-on moving through the lattice, the other one stationary before collision. The potential energy U is due to the other atoms of the lattice, i.e. the nearest neighbors of the stationary atom: U (R R2) U (R) + U(2) For a typical screened potential interaction between the two atoms, it will be shown later that for r approximately equal to r/2, half the interatomic distance, the interaction is weak in copper. Hence we may assume that, during a "collision", r is small compared to the distance of the center of mass to any neighboring atom-except the struck one and approximate. the. U's thusly, if both atoms are on the same side with respect to the minimum of the potential trough: u () = U (XA) + (RU a) 7k U U (2) =U (+ -(R 2 d)A U Since ml = m2 = m, (R -A) + (E2 -)= o If the two atoms are not on the same side with respect to the minimum, they must be close to the minimum during the collision, and

-2-0In both cases, U (.,.2)? 2U (~) Hence the variablesk, r of the spatial part'4is of k can be separated and a solution is s (r i t ()r () with 1 and V2 satisfying F V.- F 2F U V(l) - (3).~S -r -Y(r)X ulz cr) =-z\Y(r) e (4) Equation (3) is the equation of motion of the center of mass in the external field. (4) is the equation of relative motion and may be considered as the equation of motion of the reduced mass, about the center of mass, i.e. in the center of mass frame, at a distance r = R - R2 from the center of mass. Substantially, this shows that a two body treatment is permissible. The velocities U> a and ["c 9, in the center of mass frame, of the knock-on Al and the stationary atom A2, after collision, are colinear and their support passes through the center of mass G (Figure 1). Since r = ERl R 2 = A2A1, the angle of scatter P of the knock-on in the center of mass frame is equal to the angle of scatter of the particle P with reduced mass At in its motion about G. Hence the problem of finding h, which furnishes the interaction cross section, is reduced to that of studying the motion of the mass -t about the center of force G, in a field giving the potential energy V(r) to the particle P of mass i.

-21/ "Pafter // Al / _G A2 before A:2CA2 Figure 1. Diagram for the Co Figure 1. Diagram for the Collisc= of Two Partieles.

-22 - If the velocity of the incoming knock-on, in the laboratory frame, is CT before collision, the initial velocity of P (in the center of mass frame) is, since the atom A2 is stationary and hence R = 0 initially:. 0 r _ For elastic collisions, P, in its motion about G, keeps a constant total energy equal to where E is the absolute energy of the knock-on before collision. It will be noticed that this treatment neglects ionization and atomic and conduction electron excitation. The second one is in effect, a case of inelastic scattering. It has been shown by Seitz(9) that the third one is negligible in all cases and by Cottrell,(22) on the basis of a classical criterion, that the first two are only significant in light metals (beryllium and aluminum). The release of an atom from a normal site is envisioned as a two stage process: 1. The incoming knock-on transf rs energy to the stationary atom by elastic process. 2. If the energy transfer has been large enough, the initially stationary atom, by losing an energy Ed to the field of the neighboring atoms, escapes from its site. In this paper, Ed will be taken equal to 25 ev.(12' 13) 2. Choice of an Interaction Potential Energy Brinkman(2) has studied the interaction of two similar atoms, considering a rigid atomic charge distribution corresponding to a

-23potential, at distance r from the atom, V (r) =ZE exp(- r) r a where Ze is the cha3rge of the nucleus and a the screening distance. This leads(see Appendix XIII) to a potential energy of interaction between two identical atoms V(r) = r (1 -)exp (-) (5) r 2a a The force F, counted positively from O (center of force) to P (particle), is, for such a potential energy (Figure 2), F = dV = 2 exp(- r) 2a2 + 2ar - r2 dr a 2a'r2 toi.e. F is > o for o < r < a (1+ i3) F is < o for r > a (1 +3) Hence V(r) is repulsive for o < r < a (1 + T3) and attractive for r > a (1 +:T3) The screening distance a is much smaller than ro. An accepted value for a is ah z-l/3, where ah is Bohr radius 0 for hydrogen. For copper, this gives a = 0.172 A. Hence such an interaction potentiall energy corresponds to no physical reality. * V(r) should he repulsive up to r = r and attractive only for values of r 0 exceeding ro. However, for r << a, Equation (5) gives the correct Coulomb unscreened form of interaction and, for r >> a, the magnitude of V(r), i.eo 22 V(r) = Z.E exp(- ~) 2a a when employed for a repulsive interaction, leads to values agreeing.well with compressibility data, after Huntington(l2) and Brinkman(2), and as

V(r) k F>O r - F<O 0 Center of Force o r Figure 2. Bri n Potential Energy.

-2:5further shown in Appendix I. Hence it appears that an..interaction of the form 2.2 V(r) Z (1+ r) exp(-r) (6) r 2a a is more realistic~ It is derived from Equation (5) by adding to the interaction obtained from.two.rigid charge distributions with typical screened potential, a term Z2e2.a- exp(-.a) which. would account for closed shell repulsion, admittedly neglected by Brinkman. For copper, this potential energy is V(r) = 1.21 x 102 (+ i) exp(. r ) (7) r.54.4 0.172 where V is in Mev and r in A. Table I gives numerical values of V for r = 2R (R nuclear radius), a/100, a/10, a/5, a/2 through 15a.by increments equal to a/2, and 20a. The problem at hand may be treated by the methods of classical mechanics provided the "dimension of the scatterer" is large compared to the wavelength of the incoming particle, i.e. b where b is the smallest value of the radius vector r in the motion.of the particle with mass i.equal to the reduced mass around the center of mass GI a is the reduced wavelength of P when it has the speed of the incoming knock -on.

- 26 - 22, TABLE I. V(r) = Z (1+L) exp( —) r 2a a for Cu, V(r) = 1.21xlO'2 (1 + 1 e)' a 7I7 Mev; r in A r 0.344 r u r r r u + 2.9 rQ u2 WU) V-V(?) V-V(E) - WO 2 0.172 a eeO.72(u + 2.9) U 2. e 0.i72(u+ 2.9) ~ e~ A A -2 Mev 2M U v-Mev 2R 1.192x10'4 8.37x103 b.92x10'4 1.000 8.37x103 8.37x103 l.01xlO2 7.00x107 1.44x10'6 a/100 1.72 xlO'3 5.80x102 1.00x102 9.9oxo-1 5.83x1lO2 5.78x102 7.00 3.35x105 2.09x10'5 a/10 1.72 x10'2 5.80x1O 1. OxO' 9.05xlO'1 6.09x10 5.50x10 6i. 65x10'1 3.35x103 1.99x10'4 6.65x10'1 1.99x10-4 a/5 3.44 x10'2 2.90x10 2.00x10'1 8.19x10O1 3.19x10 2.60x10. 3.14x10-l 8.40x102 3.74x10'4 3.14x10'1 3.74x10'4 a/2 8.60 x1O'2 1.16xO10 5.00x10'1 6.07x10'1 1.45x10 8.80 1.07x10-1 1.34x102 8.00xO-4 1.07xlO'1l 8.00x10'4 ~a 1.72 x10'1 5.80 1.00 3.68x10' 1 8.80 3.20 3.86x102 3.35x10 1.15xlO3 3.86x10'2 1.15x10-3 35/2 2.58 x10'1 3.88 1.50 2.23xlO-101 6.78 1.51 1.62x10' 2 1.50x10 1.o08xo10- 1.62x10'2 1.08x!0'3 2a 3.44 xlo'l 2.90 2.00 1.35x10'l1 5.80 7.82x10'1 9.50xlO' 3 8.40 1.13x10O' 9.50xlO-3 1.13xlO3 5a/2 4.50 xlO1'l 2.32 2.50 8.21x10'2 5.22 4.28x10'1 5.18x10S' 5.39 9.60x10'4 5.16x10'5 9.60x10-4 3a 5.16 x10-1 1.94 3.00 4.98x10-2 4.84 2.40x10'1 2.90xlO3 3.75 1.75x104 2.88x10' 7.75x10-4 7a/2 6.02 x10'1 1.66 3.50 3.02xl0w2 4.56 1. 38x10'1 1.67x10' 5 2.75 6.08x10'4 1.65x10' 5 6.08x10'4 4a 6.88 x10'1 1.45 4.00 1.83x1O' 2, 4.35 7.95x10'2 9.65x10'4 2.10 4.60x10-4 9.40x10-4 4.48x10'4 9a/2 7.74 x10'1 1.29 4.50 1.11x10-2, 4.19 4.65x102 5.64x104. 1.66 3.40x10'4 5.49x10'4 l.26x10o4 5a 8.60 x10'1 1.16 5.00 6.74x10'3 4.06 2.72x10'2 3.30x10O4 1.34 2.46x10'4 3.05x1O-4 2.28x10-4 lla/2 9.46 x101'l 1.06 5.50 4.09x10'3 3.96 1.62x10'2 1.96x10-4 1.12 1.75x104 1.71x10-4 1.53x10-4 6a 1.052 9.70x10'1 6.00 2.48x10'5 3.87 9.60x103 1.16x10'4 9.40x10'1 1.23x10-4 9.10x10-5 9.70x10-5 13a/2 1.118 8.95x10'1 6.50 1.50x10O,3.795 5.68x10'3 6.88x10'5 8.00x10'1 8.60x1O5 4.42x105 5.53x105 7a 1.204 8.30x10'1 7.00 9.12x10-4 3.730 3.40x10' 4.12x10-5 6.90x1i-1 6.Ox1O'5 1.66x10'5 2.41x10.-5 15a/2 1.290 7.75x101 7.50 5.53x104 3.675 2.0o3x10o 2.46x10'5 6.OxlO'1 4.10x10'5 0 0 8a 1.376 7.26x10'1 8.00 34 35xlO10'..620' 1.21x10-3 1.46x10'5 5.30x10'1 2.76x10-5 17a/2 1.462 6.85x10'1 8.50 2.03xlO'4 3.585 7.30x10'4 8.85x106 4.70x101 1.88x10' 5 v(1 ) V 2 2 9a 1.548 6.45x10'l 9.00 1.23x10'4 3.545 4.37x10-4 5.30x106 4.15x10'1 1.28xo105 = 2.46x10'5 Mev 9lga/2 1.634 6.1x10'1 9.50 7.48x10o'5 3.511 2.62x104 3.17x10' 3.74x10'1 8.50x106 loa 1.720 5.80xlO'l1 1.OOxlO 4.54x10'5 3.480 1.57x104 1.90xlO6 3'.35xlO'l 5.70x10-6 21a/2 1.806 5.55x10'1 1.09Iyl 2.75x10'5 3.455 9.50x10 11510 -6 3.07x101 3.75x10'6 lla 1.892 5.30x10'1 1.lOx10 1.67x10'5 3.430 5.72x105 6.95x10'7 2.80x10~1 2.48x10-6 23a/2 1.978 5.05x10'1 1.1SxlO 1. OlxlO'5 3.405 3.44x10'5 4.16x10'7 2.55x10'1 1.64x106 12a 2.064 4.85x10'1 1.20x10 6.14x106 3.385 2.08x10'5 2.52x10'7 2.35x10-1 1.07x10-6 25a/2 2.150 4.65x10'1 l.25x10 3.72x106 3.365 i.25x10'5 1.52x10o-7 2.16x10'1 7.05x10'7 13a 2.236 4.48x10'1.L. 30x10 2.26x10'6 3.348 7.57x06 9.1 -8 2.OOxlO1 4.57x107 27a/2 2.322 4.20x10'1 1.35x10 1.37x10'6 3.320 4.55x106 5.50xlO8 1.76x101 3.13x10o-7 14a 2.408 4.15xO1-1 1.40xO10 8.31x10O7 3.315 2.76x10'6 3.34x10 -8 1.72x10'1 1.95x10'7 29a/2 2.494 4.Ox10'1 1.45x10 5.04x10'7 3.300 1.66x106 2.01xlO-8 1.60x10'1 1.26x10'7 r 2.557 3.90xl0-1 ib1.50xlO 3.06x10o-7 3.290 1.00ooxlO-6 1.22x10-8 1.52x10'1 8.15x10'8 15a 2.580 3.88x10'1 1.50x10 3.06x10'7 3.288 1.oOxlo6 1.22x108 1.50x10'1 8.15x108 20a 3.440 2.90x10'1 2.00xlO 2.06x10'9 ]5. 190 6.60x10'9 8.00x10'11 8.40x10'2 9.50x10'10

-27If bo is the "distance of closest approach", i.e. the minimum value of r for a particle shot directly at the center of force (p = o) with the same speed v, it is clear that (Figure 3), b > bo for V(bo) E and V(b) = E'-.w2(b) 2 2 2 E being the initial energy of the incoming knock-on, w(b) the speed of P at distance b from G. Hence V(b) < V(bo) and b> bo. Naturally, for p = o, b = bo. For pile neutron irradiation of metals, the energy of a primary knock-on is practically always smaller than 0.72 Mev, which is the maximum energy transfered by a 2 Mev neutron to a Be 9 atom in an elastic collision. For copper, with the interaction potential energy (7), Table I shows that forE/2 = 0.72 Mev, bo = a/10 = 1.72 x 102 A for E/2 = 25 ev, bo = lOa = 7.5a = 1.29 X In Appendix III, the following reduced wavelengths are obtained: for E/2 = 0O72 Mev, ~ = 1o35 x 10'13cm < = 1.72 x 10'10cm for E/2 = 25 ev, = 3.0 x 1011cm << = 1.29 x 10-8cm For smaller energies, the same inequality will hold, even more so. It also holds for E = 1.5 Mev, maximum energy transferred by a 12 Mev deuteron to a copper atom. Hence, classical treatment is applicable over the whole range of energy of the knock-ons, for a potential energy such as (7) between two copper atoms and in the case of pile neutron irradiation. It remains

-28/,bo Figure 3. Distance of Closest Approach.

applicable, in the case of deuteron irradiation, for a deuteron energy of 12 Mevo In Table I, it is worth noting that V(r) = 2.46 x 10-5 Mev for r = 15 a/2'i rJ2, so that an atom receiving in a collision an energy smaller than E= 25 ev will not approach another atom closer than ro/2. It is then plausible that this atom will be pushed back to its site. It is clear that an atom receiving an energy slightly in excess of 25 ev will at most become interstitial at the interstitial position closest to the site from which it has been ejected. Hence a knock-on becoming inter~ stitial at low energy will have a small separation from its vacancy and should recombine easily with it. 3. Classical Treatment of Scattering by a Center of F6rce Lving Rise to an I.nteraction Potential 2aergy DEpending on the 71stancetqly Use polar coordinates (Fi'gure 4) with pole at the center of force G and arbitrary axis G X. for the origin of angles, The equations of conservation of energy and momentum for the motion of the particle P of mass i about Pd are: 1=1 E (8) r pWw2+ V(r) = E iw = where t is the speed of P at -the point considered, r the distance GP, V(r) the potential energy - of P at distance r from P, assuming V(oo) = 0,. v the speed of P at,infinite separation and E the initial energy of the incoming atomn in the laboratory frame, and r2 = K = pv dt p being the impact parameter.

d50 dt P dt r ~~/ ~~~~~~ ~/ ~~~~~8 Figure 4. Polar Diagram for the Kepler Problem.

We have W, = r _ (id dr dC3 Cied a r =(d- tat +r d Putting into (8), E (t&1 r4 i\t v +\1r) {CLt\= t rl < -I ~ 1r Let 1 u, then the above relation becomes E) E For a repulsive potential, with origin of i selected so that Q is counted positive cloc1kwise, and for a trajectory such as AB (Figure 5), Q increases from u = o to u = ump (maximum, vertex 0 of the trajectory). Then: it is clear that the + sign is to be taken in (10) for the branch AO. For a trajectory such as CP, the + sign, applies to the branch O'D. A minus sign would correspond to the branches OB and CO', with a same positive direction for the angles. We shall remove any uncertainty in sign by taking the + sign and the positive direction for the angles such that, initially, Q increases with u and considering only the first half branch.. It is clear then, that only a configuration such as AB is to be considered and (10) is to be taken with the + sign.

B B (v)O 8,, A —Figure 5. Branches of the Trajectory. Figure 6. Symmetry of the Trajectory. v / Figure 7. Boundary Ccoditiona at ro2, or for r = r/2.

-33There is obviously symmetry with respect to GO. For, consider two points, Al on AO and B2 on OB (Figure 6), both with the same u and respective angles 91 and ~23 call Om the angle corresponding to point o. Let then el i L I CIL j e LA. 13 Lk Ct _ __- + _ d 2 UG e~t2(1-~) i.e. Q1 + 2 = 2@9m hence the symmetry claimed. Pbint 0, i.e., um; is determined by the condition s =t OL ke ( ad,ue = 00) in other words, um is a physically acceptable root of R m O. It is easy to see that V(u) is a monotanically increasing function of u (Appendix V), since F = dV is always positive for repulsive potential. Hence dr the curve Z1 (u) = V(u) + u2 intersects only at one point with the parallelZ2 (u) 1 to the u axis. Hence R = o has only one root, u It is acceptable, since R is positive for u < umn, as may be seen as follow s: V(u) = E (1-p2u2) = maxi. of V hence u <u j 1 - p2U2 >1 p2Um; V(u) <E (1 p2u) <E (l -p2u2),2~~~ p~

and 2 22 < 1 v(u) -p u hence R >o It is interesting to treat now the case where the initial conditions cannot be taken as w = v for u = o (r =-). For example, consider the case where it is necessary to consider the interaction beginning at r = ro/2, ro interatomic distance. Use the notations of Figure 7. Equation (8) now becomes w2 + V(r)=- 1 v2 + V(ro/2), where v is the initial speed. Equation (9) remains unchanged, but p is now the distance of G to the support of velocity v at the initial point considered. Equation (8) being only altered by changing V(r) into V(r)- V(rJ2), Q is obtained from (10) by replacing V(u) by V(u) - V(2/ro). Hence, du for the initial branch of the trajectory: de 2 4. Application to the Selected Potential Energy V(r) Z2E21 +) exp(.fr) )( 2a a In the case of copper: V(r) = 1.21 x 10 2 (l/r + 2o9) exp(-r/a) Mev, r A We have assumed that the problem of collisions between knockons and stationary atoms is a two body problem, i.e. that the knock-on is scattered by only one stationary atom at a time. We shall show later

35that the results of the calculation are reasonably consistent with this assumption. This is equivalent to saying the V(r) varies sharply enough around r = ro/2 = 15a/2 so that, say V( - a) is appreciably higher than V(r2~) From Table I is is seen that V(ro/2) = 2.5 x 105 Mev, while V(' EQ a) = 7 x 10.5 Mev, i.e. V( - a) is 298 time greater than V(Q) Therefore, we consider that a knock-on 1, being at a point A (Figure 8), with velocity vy will collide with the stationary atom 2 if the impact parameter p relative to 2 is smaller than ro/2. 1 would collide with 2*, nearest neighbor of 2, if its velocity v* were such that p* were smaller than ro/2. It is clear that there is only one close stationary atom for which p is smaller than ro/2. In our assumption, this is the only atom with which the knock-on 1 will experience its next collisiono Then, we can assume that collision of 1 with 2 begins at a point such as B, on a,sphere of radius to r0/2 centered on 2 and that the interaction of 1 with any nearest neighbor 2* is negligible compared to that with 2 when 1 is at point B or closer to 2 (Figure 9). In the equivalent problem of scattering of a particle.of mass 4 (reduced mass) by a center of force G, this means that we consider the interaction beginning when P is at distance ro/2 from G. It remains tow to select an appropriate criterion for the end of the interaction. When atom 1, after.,collision, reaches the surface of the sphere of radius rJ/2 centered on the site of atom 2, it is ready to enter a region where it is at a distance of less than roJ2 from a nearest neighbor of 2, that is where the predominant interaction with 1 will be that of the nearesft neighbors provided, at that time, the distance between 1 and 2 is greater than ro/2 (2 is in motion!). Hence, it is satisfactory

Figure 8. Preferential Interaction. Fge9 BenigfItatn 22 Figure 9. Beginning of Interaction.

-37to consider that collison between atoms 1 and 2 ends at distance rJ2 from the lattice site of atom 2 when this corresponds to a distance between 1 and 2 greater than ro/2. -The two diagrams of Figure 10 indicate the two conf.igurations possible after'collision. It is shown in Appendix VIII that, for elastic collision of two particles of the same mass, the two particles fly off at 90 degrees from each other, in the laboratory frame, after collision (Figure 10). Generally, the particles interacting are considered in zero field before and after collision. This leads, for two particles of identical mass M to one energy balance and two momentum balance equations, namely (Figure 11), in Mthe2 hrato iy M(r12) Men Swn = M Li > (12) M U, = MU-) CvSS + C M -a Ws* J in the laboratory frame. If we consider that the collision of 1 and 2 ends when the distance between the two is equal to rJ2, and since we have assumed that collision begins when 1 comes at distance rJ2 of 2, we can write the energy balance equation 2 ( z ) Z l +' 1.,.V(ro and two momentum balance equations identical to those in (12), i and U2 being taken when the separation of the 2 colliding atoms is ro/2, i.e. at the the "end" of the collision (Figure 12)* Hence, this choice will preserve the usual results for two identical particles (See Appendix VIII) namely:.. +y =T/Z j T= E san-' 4Q/a,

-38/v; 2 V, > VV2 > V, Figure 10. Possible Configurations When the First Particle Reaches Distance rJ2 From the Center of Collision. VIXJ Figure 11. Diagram of Angles and Speeds in Laboratory Frame. Fl r Figure 12. Erd of Collision.

-39where C) and T are the angles to the initial velocity _l of the incoming knock-on 1 at which this knock-onad the initailly stationary atom 2 fly off at the end of collision (i.e. when their separation is r/2) in the laboratory frame, T is the kinetic energy transferred to 2, E the kinetic energy of I before collision, both in the laboratory frame, and ( is the angle of scattering of 1, in the center of mass frame, corresponding to O). In the equivalent problem of the scattering of a particle P of mass p4 equal to the reduced mass by a center of force G, this means that (P is the angle with vl for which P is at distance ra/2 (after collision). From Figure 12, it is seen that 1 and 2 have to travel some distance, after their separation has reached ro/2, to reach the surface of the sphere of radius.rJ2 centered.n the lattice site of 20 This they do practically free of interaction between each other or with any nearest neighbor. Hence, it is essentially equivalent to consider the end of the interaction when 1 or 2 has reached the surface of this sphere., at which time the separation of the two atoms is larger than rJ2 and the sum of the angles CJ and J is still aboutT /2, or when their separation is rJ2 In summary, for the problem in the center of mass frame, we shall consider that interaction begins and ends when P is at distance rZ/2 from G. We therefore adopt formula (11). It will be recalled that Rutherford scattering formula(235) was established considering the beginning and the -end of the collision when the moving atom is at infinity from the scattering atom 2, this corresponding to single scattering of the alpha-particles when they pass through the foil. The assumption was justifiable since 1) the foil was assumed

very thin (practically, Geiger and Marsden(28) used a gold foil only 2.1 x 10-5 cm thick in their experiments to check Rutherford formula); 2) the incoming particles fell perpendicularly on the foil; 3) the incoming particles were energetic~ On the contrary, in the problem being studied, the knock-ons move in all directions and their energy can be quite small, so that they experience multiple collisions in the sample. For energetic knock-ons, say E=0.75 Mev, whichis practically too high for neutron irradiation of copper, r is always greater than a/10, so that r/(ro/2) is always greater than a/10 = 1/75. Hence r /2 cannot be considered as infinite. 7.5a However, it will be shown shortly that taking the boundary conditions at infinity yields a justifiable approximation. It will also be noted that Rutherford treatment of the scattering of alpha-particles assumes a two body situation: it is implicitly postulated that, when 1 crosses the foil (Figure 13) through two nearest neighbors 2 and 2*, only one of the two scatters 1; otherwise the simultaneous deviation of a particle by two centers of force would have to be studied. At r = r/2, the Coulomb energy between two copper nuclei, Vc(r) = 1.21 x 10=2/r (Mev, A), is about 10-2 Mev. The interaction potential energy used here is 2.46 x 10-5 Mev. Hence, we should be justified in using a two body approximation. Calculations of the displacement cross section, using this model, give results reasonably consistent with the two body assumption, as will be seen later. Now, relation (11) is not integrable. Appropriate approximations to R = 1 [v(u) -V(2/ro)] p2u2 are looked for, at various values of E.

I m *2 Figure 13. Tro Body Interaction. C \\ A' V~ B Figure 14. Equivalent Condition at Infinity. A, r5.o'r Figure 15. Origin of Angles.

If an acceptable approximation of the form V(u) -V(2/ro) O((E) u2 can be found for each E, then (11) can be approximated by d3/du - c iE)+ (12a) which can.be written hence G is then obtained analytically. (12a) is the differential equation for the motion of a particle P of mass p, speed i.-= at infinity from the center of force G (Figure 14), the asymptote OB passing at distance p from G and the potential energy being X (E)u2. Actually, the velocity.of' P at r,/2 from G should be such that its support is at distance p from G and its magnitude is vo In the approximation (12a), if we call 1 the velocity of P at A and P1 the distance from G of the support of vly we have hence & f - with E =,v2. It will be seen later that the maximum values of (2/E) O((E) are 1.20, 6.52 x l0-1, 1.92 x l0-1 for E = 104, 3, 10-3, 2 Mev Hence, E C (E) r.) takes the values 0.517, 0.785, 0.940, respectively, for these three values of E. It follows that, down to E = 10-3 Mev, it is not a serious error to

consider that the angle of scattering is the angle CP of the asymptotes OB and OC. To gain an idea of the error of doing so when E is smaller than lOj3 Mev, the following analysis is made. Consider the correct conditions at Ao, i.e. velocity v and v passing at distance p from G. Count O clockwise, from a parallel to v in a configuration as shown in Figure 15. Then the integration of (13) between (O(, 2/ro) and (@, u), yields (See Appendix VII): for % corresponding to the asymptote of the first part of the trajectory, Om to the center of the trajectory. Let = S E)+l 7; - C. dE) o In Appendix X it is shown that c) <,j < - 1, in the range rbch p which limits p to <- 2 E

-44We write LCEIE 0 2 So | E (E)et (14) Now, the approximation V(u) - V(ro/2) -i( (.E.)u2 can only be valid for u>2/ro, i.e. for the branch Ao Al. If we use it for the branch between infinity and A0, this means that V(u) > V(~) for u < 2, i.e. that the potential corresponding to V(u) is attractive for ro 0 < u < 2/ro, has a minimum around u = - (the approximation is, of course, ~~~2 " ~r 2 wrong for u = 2 ), and is repulsive for u > Hence the trajectory of -O P resulting from (14) starts with P below the asymptote OB (attractive portion). It is shown in Figure 16. ~e have 0 -l /|E i(LE)4+| (15) Asymptote OC is defined by the angle @\= be/| E (E)+ 1 (15a) and the angle C9 is given by: TT_ 0 - T, m (16) The angle of scatter to consider is IFP I= [- |1 00- ) o)i al = I o _ > 6' (17) Appendix XI gives 90 for E = 10-1 Mev and E = 10-4 Mev, for the two values p = a/100 and p = Fr2/4 - (2/E) ( (E). The values o((E) taken are what will be called later CK, i.e. a maximum leading to a too strong interaction (too large displacement cross-section, too small

.-45 - A, I/i -*0 ~~~~~~O ~ ~ ~ O I~ """""""""""""""""""" T /2 Os A,~~~~~~~~~~~~~~~~~~~00 ~~~~~~~~ B~~~~~~~~~~~~~~~~~~~~~ Figure 16. Prolongation of Trajectory to Infinity.

-46displacement mean free path)o It is found that: for E = 101 Mev and p = a/lO00 Qo 0, while t for E = 10-1 Mev and p = r/4 - (2/E)ok = 1.270 Y 7.5 a,.o 0 - 0.019 rad, while f 2-2.20 x 10O2 rad, i.e., although -29o is larger than k ), 0'= 0O240 rad, remains small. For E = 10-4 Mev and p = a/100, g0 o 2o72 x 10-4 rad while TT for E -= 0Mev and p = J..u= 0o655 = 38 a, go = ~ 06265 rad while (= 1.54 rad. In summary, the approximation I f ~ is worse for large p ( C y P ), the displacement cross section so calculated will be slightly too small.'On the other hand, using the upper value O(t overestimates scatter, sothat, for cross sections calculated-with 0(Q, there should be some compensation of the two errors and we expect the result to come about right, somewhat too large, probably, for E < 10-1 Mev, sinde the approximation V - V(ro/2)' 0( u2 appears worse than the approximation Ua'h.for such energies. On the contrary, cross sections calculated using a lower value OQ, as will be explained shortly, should come out too small, since both errors are underestimating the interaction. Then, we accept, for scattering angle, in the center of mass frame, an angle ( given by (16). In other words this discussion has shown that it is a permissible approximation to consider the boundary condition at r = -. Equation (16) has the general required form, since it gives

no interaction (9= 0) for p =~O or O.(E) = 0, i.eo V(u) = V(2/ro.), and backscatter ( T =T). for p = O, i.e. head on collision. From (12a), it is seen that the maximum value of p for which it is possible to have the trajectory pass through a point at distance ro/2 from G is (Figure 17): ko - E i (E) This is entirely due to the mathematical approximation made. Actually, P is not scattered by G, but by another center of force, corresponding to a nearest neighbor.,of the atom co.nsidered, if its initial velocity (at A0o);, is tangent to the sphere centered about G of radius ro/2. Hence, the maximum value of p for'scattering is pib = ro/2 and the scattering cross section 6 = ps 2 == (ro/2)2 For copper, ps = 1.278 A ps2 = 1.628,2 6 s =5.11 A In order to find what value of O((E) to select for the approximation V(u) - V(ro/2)' m(.E)u2 two values were considered, a lower value h(and an upper value d,.,is such that R,=i 2 d ~ts Ul- u b(ei)-\g(ef -' ze= Rb for all u < u1, u1 being defined by Ko _/e,, uE _

Figure 17. Maximum Value of p due to \Po Approximation.,v Ao "~o/ i, i FurUe 3..Ageo smp eOe Figure 19. Angle of Asymptotes.

-49Ot is such that e- E 2 ) - E C ( ) t 2 )- R for all u < u2, u2 being defined by -t d -2 LEH) U1 --. It follows that u2S, is greater than um. Figure 18 illustrates the situation. ( u and 0(~ are chosen by inspecton of tables giving R as a function of u for various values of E. Strictly, Q is not chosen so that R, > R for all u < u2, which would lead to very small values of ode but so that R > R in the portion of the R(u) line which has an appreciable curvature-i..e. the range of u where scattering takes place —and that R, = R in the portion where there is no',appreciable curvature. Disregard the approximation on CP for the time being and con,sider Cp' defined by the asymptotes. Call e4),an. d LOTj)O the two values of eT obtained by selecting the approximation corresponding to o, and o(,. respectively (Figure 19). ( T)u, corresponds to scattering by a stronger potential than the actual one. Since the initial conditions are the same in the three cases, it follows.that U- T < (E~)..hence, We assume elastic collisions, hence the kinetic energy T transferred to the stationary atom by the incoming knock-on of energy E is (Appendix VIII), T= E S: ~o/ 2 This and"Equation (16) give pd, maximum of p for displacing collision, when T is equal to Ed. Thus, = z-d- t./.'/, /T-Q- z, (20)

-50SL~e~ d /2. E d ~/E,~ g(21) Call (Pd)u and (Pd)e the values of Pd obtained by selecting At h and i( for i. In the same way, call C d)u and (Wd), (Zd)u and (Zd)~, (Xd)u and (Xd the corresponding microscopic and macroscopic cross sections and mean free path for displacing collisions, assuming t'i-,. Ca116d, o. the true values of these parameters. From Equation (20), since 2d is fixed by Equation (21), and recalling that <>d =a (pd)2, it follows that (6 d)u' (Zd)u are proportional toQXu (Xd)u is inversely proportional too<u (6 d)t, (Zd) are proportional to c/ (Xd), is inversely proportional to Q. Hence, (6d)z < 6d' (_a)q <a.d Xd < ()Xd However, since, for 0u, the two approximations V V(ro/2)'~'o(uu2 and ~J=' partly compensate each other, we shall write (6d)Q <4.u (d) < 6a()u () < a (d)u, X (22) (Xd)u < xad< (Xd) Diagram 1 shows the variations of V, V-V(r0/2), and u2. This diagram is discussed in Appendix IX. Tables II through VIII give the result of the numerical calculations made for E = 10-1 Mev, 10-2 Mev, 5 x 10-3 Mev, 10-3 Mev, 10-4 Mev, 25 ev and for p = a/100 (in addition, also p = a for E = 10-1 Mev), in order to select Ku(E) and & (E).

t GOU~4Sra S UTUGGIos JO SUO Th4Id UT passajdxa ax * SA En puP'(O )A - (1.)A'((T)A *1T G eTTIa /o5gl oL Z/D~1 09 I /Dfl 09 /0ll 6 ob r Z/DZ 0~ Z/og DZ Z/0 0 /,ol I: -' 1'" i -: f"' -t i... -I - i "..i..gl PI \\ Li 1I 01'~ ~\ f-I ZEOI e01 i01,p2 9,.

TABLE II. E = 10-1 Mv; p = a/100 p2 = 2.98 x 10o6 -; =3.35 x 105 2 2; _= 20 E a. = 1.15 x 10-3; =a 2.30 x 10-2; = 6.00 x 10-4; E a~6= 1.20 x 10-2;r'b v-v E2 2Y:r - u2 u v( ] 1 ] 22 2u2 R. 1 (-2 i~ _ - -.,..,. u a/2 1.34x102 1.07xlO-1 8.00x-104 2.14 1.14 o 1.l14 + + + a 3.35x10 3.86x10-2 1.15x10-3 7.72x10-3 2.28x10'1 2.28xl0-1 7.71x10-1 2.29x10-1 2.29x10-l-4.01xlO-1 5.99x10-1 5.99x10-1 4.78x10-1 4.78x10o1 3.60x10-3 3.60x10'3a/2 1.50xlO 1.62x10-2 1.08x10-5 3.24x10-1 6.76x10-1 6.76x10-1 3.45x10-1 6.55x10-1 6.55x10-1 1.80x10-1 8.20X10-1 8.20x10-1 8.20x10-1 8.10xlo-1 2.10x10-5 2.12x105 2a 8.40 9.50x10-3 1.1x10-3 1. OxlO-l 8.10xO1'1 8.1xlo-1 1. 3xO1- 8.07x101 8.07x101 1.01x10-1 8.99x10-1 8.99clO-1 9.0oo10- 8.98x10-1 1.91x10 1.9x10 5a/2 5.39 5.16x10-5 9.6ox10-4 1.12xlO-1 8.80x10-1 8.80x0o-1 1.24x-101 8.76xlO-1 8.76xlO-1 6.45x10-2 9.36x10'1 9.56xl0-1 9.40xlO1 9.5x10-1 1.8x10-5 1.84x10-5 3a 3.75 2.88x10-5 7.75x10-4 5.762x10-1 9.4 2x10-1 8.61x10-2 9.1o x10-1 9.14x10 - 4.48x10-2 9.55x10-1 9.55x10-1 9.70xlO-1 9.57x10-1 1.77x10 1.80x10-5 \ 7a/2 2.75 1.65x10-5 6.08x10-4 3.3010-2 9.67x10-1 9.67xlO-1 6.32x10-2 9.37xI-1 9.37x10-1 3.3010-2 9.67x10-1 9.67x10-1 9.84xO1-1 9.68x10-1 1.75x10-3 1.78x104a 2.10 9.40x10-4 4.48x10-4 1.88x10-2 9.81x10-1 9.81xlO-1 4.83x10'2 9.52x1l-1 9.52cl10-1 2.52x10-2 9.75x10l1 9.75xlO-1 9.90x10-1 9.76x10-1 1.74x10-5 1.76x10'5 9a/2 1.66 5.40x10-4 5.26xi0-4 1.08x10-2 9.89x10-1 9.89x10-1 3.82x10-2 9.62x10-1 9.62x10-1 1.99x10-2 9.80xlo-1 9.80c10'1 9.95x10-1 9.80o10-1 1.73x10' 1.76x105a 1.34 3.05x10-4 2.28x10'4 6.ox10xo-3 1.00.00 8x10-2 9.6910 9.69x10-1 1.61x10-2 9.84x10-1 9.84x10-1 1.00 9.85x10-1 1.72x10-3 1.75x10'5 lla/2 1.12 17. 55x10-4 i53x104 1.00 1.00 2.58x10-2 9.74x10-1 9.7%4r10-1 1.00 9.87xl0-1 1.72x10-5 1.74x10'5 6a 9.40x10-1 9.10x10-5 9.70x10-5 1.82xlQ-5 1. 00 1.00 2.16x10-2 9.78x10'1 9.78x10-1 1.00 9.89xo101 1.72x10-3 1.74x10'5 13a/2 8.00(xlO-1 4.42x15 5.53x105 1.00 1.00 1.84x10-2 9.8ExlO-1 9.82xlO-1 1.00 9.9gx10-l 1.72x10-3 1,73x10'5 7a 6.90x10-1 1.66x10-5 2.41x105 3.32x10-4 1.00 1.00 1.58x10'2 9.8cx10-1 9.84x10'1 1.00 9.93x10' 1.72x10- 1.73x103 15a/2 6.00x10-1 0 0 0 1.00 1.00 1.38x10r2 9.86xol-1 9.86xlrl 1.00 9.94xo-1 1.72x10o - 1.73x10-3....~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

6Tw/s = d iAaW piow= - O aHuOTST8Ea 9 SutuaxJOS JO suOTIo6d-u1T passaadxa a ( I) u'(x) )"'(.x)a e (gaZ x12s Mo91 0 I wl 09,/OII l 9 /D 6 o0 OZt c.~ Z//o DO Z/o; O Z/l rO Z'O -'0 ~'0 C I L:. _'0 I _ 9'0

TABLE III. E = 10-1 Mev; p = a p2 = 2-95 x 10-2 A2 oq = 1-15 x 10-3; 2 au = 2.50 x 10-2 rrU2 V-Vv-v) (! 2[V-V( ) ] 1- [ ] p2u2 R 2 C2 u2 1- Ru u E E E E a/2 1'34 x 102 1-07 x 10-1 800 x 10-4 2.14 1-14 3'95 5'09 a 3'35 x 10 3'86 x 10'2 1'15 x l2-3 7872 x 10 2 -28 x 1 9-85 x 10-1 7.57 x 10- 7 71 x l0'l 2- x x 10-1 7'56 x 103a/2 1'50 x 10 1-62 x 10-2 1.08 x 10-3 3.24 x 10'1 6-76 x 101 4.43 x 101 2 33x ll 3-45 x 10-1 6-55 x 10-1 2.12 x 102a 8-40 9 50 x 1- 13 x 10-3 11-90 x 1Q01 8-10 x 10-1 2.48 x 10-1 5.62 x 10-1 1~93 x 10-1 807 x 10-1 5.59 x 101 5a/2 5'39 5-16 x 10-3 9-60 x 10-4 1-12 x 10'1 8-80 x lo-1 1~59 x 10-1 7-21 x 10-1 1-24 x 101' 8-76 x 10-1 7.17 x 103a 3.75 2.88 x 10-3 7.75 x 10-4 5.76 x 10'2 9.42 x 10'1 1.11 x 10-1 8.31 x 101 8.61 x 10-2 9.14 x 10c1 8.03 x 10-1 7a/2 2.75 1.65 x 10-3 3.30 x 10-2 9.67 x 10-1 9.67 x 10-1 8.10 x 10-2 8.85 x o10-1 6.32 x 10-2 9.37 x 10-1 8.56 x 101 4a 2.10 9.40 x 448 x 10-4 1.88 x 10-2 9.81 x 10-1 6.20 x 10-2 9.19 x 10-1 4.83 x 10-2 9.52 x 101 8.90 x 10 9a/2 1.66 5.40 x l0-4 3.26 x 10-4 1.08 x l0-2 9.89 x 1 0-1 4.9 x l0-2 9.40 x 10o 3.82 x 10-2 9.62 x 10-1 9.13 x 10-1 5a 1.34 3.05 x 10-4 2.28 x 10-4 6.10 x 10-3 - 1.00 3.95 x 10-2 9.61 x 10-1 3.o8 x 10-2 9.68 x 10-1 9.30 x 10lla/2 1.12 1.71 x 10-4 1.53 x 10 -4 1.00 3.30 x 10-2 9.67 x 10'l 2.58 x 10-2 9.74 x 101 9.41 x 101.6a 9.40 x 10-1 9.10 x 10-5 9.70 x 10-5 1.82 x 10'3 1.00 2.77 x 10-2 9.72 x 10-1 2.16 x 10'2 9.78 x 10ol 9.51 x 101 13a/2 8.oo x 10-1 4.42 x 10-5 5.53 x 10-5 1.00 2.36 x 10-2 9.76 x 10-1 1.84 x l0-2 9.82 x 10-1 9.58 x 101 7a 6.91 x 10-1 1.66 x 10-5 2.41 x 10-5 3.32 x 10-4 Ov 1.00 2.09 x 10-2 9.79 x 101 1.58 x 10-2 9.84 x 10-1 9.63 x 10-1 15a/2 6.00 x 10'1 0 0 0 1.00 1.77 x 1O2 9.82 x 10 10-2 9.86 x 10-1 9.69 x 10-1

CT*d (a W yT _ ( A a0 e Ouss9T PuTuagaoS Jo SUt4o0vied UT passajdxaI J (J)nu pus (J)a OC m* J i96(' 91 wL v-~ o w ZJ;l iF -66....o ~.... i....... D./o I'0 ~0 go 90 // 6 / O' 60 01

TABLE IV. E = 10-2 Mev; p = a/100 p2 = 2.98 x 10-6 = 9.60 x 10-4; 2 2 au0 =1.92 x 10-1; = 2.28 x 10-; 2o = 4.56 x 10-2 E E~~~~~~~~ U2 V-V () v-v' -xR[v 1-V[] p2u2 R 2 R2. 2 2 ro V-V 2 2 E u 2~~~~~ ] pV- u2 R] Eu2 ~ ~ - ~. 5a/2 5.39 5.16x10-3 9.6ox1o0-4 1.03 3.2x10-2 0 <0 1.05 3.00xlO-2 <0 + + 3a 3.75 2.88x10'3 7.75x10O-4 5.76x10-1 4.24x10-1 4.24x10-1 7.21x10-1 2.79x10-1' 2.79x10-1 1.70x10-1 8.30x10-1 8.30x101 %n 7a/2 2.75 1.65x10-3 6.o8x1i-4 3.3oxl0-l 6.70x10-1 6.70x10-' 5.28x10-1 4.72x10-1 4.72x10-1 1.25x10-1 8.75x10-1 8.75x10-1, 4a 2.10 9.40x10-4 4.48x10-4 1.88x10-1 8.12x10-1 8.12x10-1 4.04x10-1 5.96x10-1 5.96x10-1 9.60x10-2 9.04x10-1 9.04x10-1 9a/2 1.66 5.4oxli-4 3.26x10-4 1.08x10-1 8.92x10-1 8.92x10-1 3.20x10-1 6.80x10-1 6.80x10-1 7.57x10-2 9.24x10'1 9.24x10-1 5a 1.34 3.05x1o10-4 2.28x10i-4 6.10xlcO-2 9.39x10-1 9.39x10-1 2.58x10-1 7.42x10-1 7.42x10-1 6.11x10-2 9.39x10-1 9.39x10-1 11a/2 1.1' 1.11-4 1x0 1.53x10-4 2.70x102 9.73x10-1 9.73x10-1 2.16x10-1 7.84x10-1 7.84x10-1 5.11x10-2 9.49x10-1 9.49x101 6a 9.40x10-1 9.10x10-5 9.70x105 1.82x102 9.82x10-1 9.82x10' 1.81x10-1 8,19x10-1 8.19x101 4.29x10-2 9.57x10'1 9.57x101-1 13a/2 8.00x10-1 4.42x10-5 553x10-5 4.84x10i3 9.95x10-1 9.95x10-1 1.54x10-1 8.46x10-1 8.46x10-1 5.66x10-2 9.63x10-1 9.63x101 7a 6.90x10-1 1.66x10-5 2.41x10-5 3.32x103 9.97x10-1 9.97x10-1 1.33x10-1 8.67x10-1 8.67x10-1 3.14x10-2 9.69x10-1 9.69x10-1 15a/2 6.00xlO-1 0 0 0 1.00 1.00 1.15x1l-1 8.85x10-1 8.85x10-1 2.74ix10-2 9.73x10-1 9.73x101

1.0 oe:. 0.9 / 0.8 /_ I / I / I_, / / c- 0.4. _ 0.32_ / 0.2., 0.1 ~/2 4 3a/2 2. 5a/2 3. 7T2 4. 9~/2 50 i./2 Go 13a/2 7 150/ Diagram 4. R(r), R(r), R.(r). r Expressed in Fractions of Screening Distance a E 10'2ev, p = a/LO0.

TABIE V. E = 5 x lO0-3 Mev; p = a/100 p2 = 2.98 x o1-6 X2; au = 7.75 x 10-4; Xu = 3.1 x lo-1; a, =1.53 x 10-4; 2 ae = 6.12 x 10-2 r u2 -V() V-V( 2[VV( ) 1 2 p2u2 R 2a 1 2 Ru 2 atu2 2 RI -ur2 E - E" _._ - + + 3a 3.75 2.88x10-3 7.75x10-4 1.15 1.52x10-1 -O < 0 1.16 1.60xlO-1 1.60o10-1 2.29x10-1 7. 7.71x10 - 7.7x1 7a/2 2.75 1.65xio-3 6.o1xio-4 1 + -I-+ 7a/2 2.75 1.65x10-5 6.08x10-4 6.60x10-1 3.40x10-1 3.40x10-1 8.51x10-1 1.49xlO-1 1.49x10-1 1.68x10-1 8.32x10-1 8.32x10-1 4a 2.10 9.40x10-4 4.48x10-4 3.76x10-1 6.24x1o-1 6.24x1o-1 6.51x10-1 3.49x10-1 3.49x10-1 1.28x10-1 8.72x10-1 8.72x10-1 O 9a/2 1.66 5.40x10-4 3.26x10-4 2.16x10-1 7.84x10o1 7.84x10-1 5.15x10-1 4.85x10-1 4.85x10-1 1.02xlO-1 8.98x10-1 8.98x101 5a 1.34 3.05x10-4 2.28x10-4 1.22x10-1 8.78x10-1 8.78x10-1 4.15x10-1 5.85xlO-1 5.85x10-1 8.20x10-2 9.18xlO-1 9.18x101 lla/2 1.12 1.17xL0-4 1.53x10-4 6.12x10-2 9.39x10'l 9.39x10-1 3.48x10-1 6.52x10-1 6.52x10-1 6.85x10-2 9.32x10-1 9.32x10-1 6a 9.40x101 9.10x10-5 9.70x10-5 3.64x10-2 9.64x10-1 9.64x10-1 2.92x10-1 7.08x10-1 7.08x10-1 5.75x10-2 9.43x10-1 9.43x10-1 13a/2 8.00xlO-1 4.42x10-5 5.53x10-5 2.21x10-2 9.78x10-1 2.48x10-1 7.52x10-1 7.52x10-1 4.90x10-2 9.51x10-1 9.51x101 7a 6.90x10-1 1.66x10-5 2.41x10-5 6.64x10-3 9.93x10-1 9.93x10-1 2.14xlO-1 7.86x10-1 7.86x10-1 4.22x10-2 9.58x10-1 9.58x101 15a/2 6.00xlO-1 0 0 0 1.00 1.00 1.86xlO-1 8.14xlO-1 8.14x10-1 3.68x10-2 9.63x10-1 9.63x10

'00'T/ = d'Aa X = d: O1X = SDFS J t IoTUXJOS JO suOy9vxj ut p8assa. I (J)tg {(lng ()I'5 Is9a PADz| D2 ZXDl D9 Z/D Do /D bb E/AD1ID; /6 5; DZ b /I Z/D I0 ~~I~~~~~O // ~'0 / /_60 /~~~. /~'

TABLE VI. E = 10-3 yv; p = a/-l0o p2 = 2.98 x 10-6 R2; au = 3.26 x 10-4; 2 au 6.52 x _0-1; 3.03 x 10a5; 2 ag,= 6.06 x 10-2 2 2 E E E E E E 9a/2 1.66 5.40X10-4 3.26x10o4 i.o8 8.00x102 0 < 0 1.8 <0 <0 + + 5a 1.34 3.05x10 2.28x10'4 6.loxl10'ol 3.90x10 3.910xLo- 8.75x10'1 1.25xlo10 1.25x10'1 8.12x10-2 9.lgxl0 9.19x101 la/2 1.12 1.7Jx10-4 1.53x10-4 3.2x10-1 6.58o101 6.58xi01 7.30x101 2.Y~101 2.YCx101 6.80x10x2 9.32x101 9.32x101 6a 9.4xlox-1 9.10x105 9.70x105 1.82x10o- 8.18xio-1 8.18x10o1 6.13x10-1 3.87x10-1 3.87x10-1 5.7C0c10-2 9.43x10l- 9.k3x101 13a/2 8.O0xlO-1 4.A2x10-5 5.53x10-5 8.8ix102 9.]2x101 9.12x10-1 5.22x10- 4.78x10-1 k.78xi01 I.85XiO2 9.52x10-1 9.52x10-l 7a 6.90ox1o1 1.66xio05 2.4x105 3.32x10-3 9.97x101 9.97x10-1 k.50x10-1 5.50x101 5.50x10-1 f.18x10-2 9.58x10-1 9.58x101 15a/2 6.00x10l 0 0 0 1.00 1.00 3.92-xl0-l 6.08x101 6.08x10o1 3.6)x10-2 9.6Ix101 9.6ix10-4

OQ/V _ a'Ata C-.OT = g tV aOuscSTa 2UTUa.3S JO qUOT40vad UT passajdxiT X *(a)H'(J)H'(I)1 9 U1hsaTC[ k/gI oL Z, l lo9 o/Dll, ODg Z/O6 to'/oL Do Z/Dtg oZ;/oR 0 0O / -eo /1 / / 9&0 i,a go

TABLE VII. E = 10-4 Mev; p = a/100 p2 = 2.98 x 10-6 2; oU = 6.00 x 10-5; au =1.20; 2a =a 05; 2 a= = 2.0oo x 1O-1 r u2 V Vr~ V-V( 2 [V.V( )] 1 =l _ p2U2 R 2 auu2 1 R 2 2 E2 2 1u E E E E E E 6a 9.40x10-1 9.10xl10-5 9.70x10-5 1.82 0.82 0 0.82 1.13 0.13 0.13 + + + + 13a/2 8.00xlO-1 4.42x10-5 5.53x10-5 4.84x10-1 5.16x1l-1 5.16x10-1 9.60x10-1 4.00x10-2 4.00xlO-2 1.6oxlO-1 8.40x-101 8.40x10-1 7a 6.90x10-1 1.66x10-5 2.41x10'5 3.32x10-2 9.67x10-1 9.67x10-1 8.28xl0-1 1.72xlO-1 1.72xlO-1 1.38xlO-1 8.62x10-1 8.62x10-1 15a/2 6.00xlo-l1 0 0 0 1 1.00 7.20x10-1 2.80x10-1 2.80x10-1 1.20x10-1 8.80xco-1 8.80xcl0'

1O 0.9 08 G7R 0.6 0.5 0.4 0.3 / / 0.2 / Ru0 / / I O/ a/2 a 33~/ 2a 5a/2 3 7? /2 4a 92 650 lO 0 60 130/2 7a 15%Z2 r Diagram 7. R(r), Ru(r), R2(r). r Expressed in Fractions of Screening Distance a. E = 10-4 Mev, p = a/100.

TABLE VIII. E = 25 ev; p = a/100 p2 = 2.98 x 10-6 2; a, = 1.90 x 10-5; 2 u = 1.52 r u2 v-v(-) V-Vu( 2[V-V( )1 1 -[ p2u2 R 1 u, R E E E 7a 6.90x10-1 1.66x10-5 2.41x10-5 1.33 0.33 = 0 < 0 1.05 5.00x10-2 < 0 15a/2 6.0x10-1 0 0 0 0 1 9.5xlO-1 5.00x10-2 5x10- 2

-651.0 R 09 0.8, 0.7 0.6 05 R 04 0.3 02 0.1!/2 ~ 3q2 2o 50/2 3o 70/2 4o 90/2.S 11o/2 6o 130/2 70 15a/2 Diagram 8. R and Ru for E = 25 ev; p - a/100.

-66Diagrams 2 through 8 show the variations of R(u), Ru(u), R1(U), for.the same energies and p = a/100 (also p = a for E = 10'- Mev), The approximation V - V(ro/?) ( Ju2 is very good at high energy (10 and 102 Mev), acceptable at medium energy (.l0-3 Mev), and mediocre at low energy (10'4 Mev). But, in any case, lower and upper limits for 6 and d are obtained from the relations (22)o The results for the values of cK are sunpmarized in the following Table. TABLE IX E CK u1~ (E) 1Z 1.E) c-Y(E)' (m) 25 ev 1.90 x 10-5 1.52 10 Mev 6.00 x l025 1,20 1L05 2.00 x 10-1.0~.Mev 53.26 x 104 6.52 x l0=l 5.03 x 10-5 6,06 x 102 5 x 103 Mev 7.75 x 10-4 3,10 x 10~ 1.53 x 10-4 6.12 x 10-2 i0o2 Mev 9.60 x 104 1.92 x 10-1 2.28 x 104 4.56 x 102 10lO Mev 1.15 x 103 2. 6.00 x 10-4 1.20 x 10'-2 It must be noted that, at very low energy, close to 25 ev, the moving atom cannot approach the stattonary one closer than about rJ/2, since from Table I, V(r) = 2.46 x 105 Mev for r = 15a/2' r0/2. Thenthe approximation V - V(ro/2) i ou2 is, of course, very poor and willoverestimate the interaction., whatever' the value of O( selected. 5. Numerical Calculation:of Knock-on Displacement Cross Section and Mean Free Path in Co. pper The value..~ of for displacement is obtained by with Ed = 25 evo

From this 0(Ad>>= TT a.) LI N(2d) = (6d G ) with N = 8.5 x 1022 atom cm-3 for copper. Finally This is done for E = 10, 1.012, 5 x 10-3, 10-3 and 10-4 Mevo For E 25 ev, T =1Tand naturally. The numerical details are carried out in Appendix XII. The results are as follows. TABLE X E (Pd)u (Pd)Q.((Xd)u lQ-4 Mev 0.-384 ro 0.157 ro 1.510 ro 9.060 ro 10-3 Mev 0.653 ro 0.199 ro 0.525 ro.5.620 ro 5 x 10o 5 Mev 0.691 r 0,3o8 r0 0.465 ro 2.O50 ro.102 Mev o.675 rb 0O329 ro 0o502 ro 2.110 ro 10lO1 Mev 0.415 ro 0.300 ro 1,290 ro 2.510 ro Diagrams 9 and 10 illustrate the situation. It is seen that, in the region roughly limited by the energies 2.5 x 10-4 and 5.6 x 10'2 Mev, the problem may be one of many body collision, since:in this region. (Xd)u < r and (Pd)u > ro0/2

0.7 ro 0.6 ro 024 r0 0.1 ro, I,,, I 16-4 25 10-3 5 I0-2 5.6 10 E MEV Diagram 9. Limits of pd ///// and Possible Region of m.y Body Collision XXX.

S~~~ X I4 6~ 1 4~~~~~~ I 4EV Diagram 10. Tjidtt Of Ic, Possible Region of MSMY Body Co ision:3I~, Br-Ck~ Estimate of Xcl - afll for Cu.

-70O However, the approximations made to obtain (Pd)u and (Xd)u have..been shown to be mainly good at high energies, specially at 101.Mev. We can then expect that (Pd)u is actually too large and (Xd)u actually too small at smaller energies and the the two body approximation is a meaningful one over the whole range of energy. This justifies the two body treatment. The above results will now be discussed by comparing them to those of the most recent existing theory of atomic displacement, namely Brinkman theory of displacement spikes.

CHAPTER IV DISCUSSION OF MODEL,.COMPARISON WITH BRINKMAN THEORY lo Brinkman Theory Brinkman(2) has treated the problem of atomic displacements by using an interaction potential energy between two identical atoms of the form (5). How this formula is obtained from two rigid distributions of charge with potential is shown in Appendix XIII. Brinkman uses an impulse approximation, similar to the original Bohr approximation, for energies E >> Edo In Appendix XIV, it is shown that the use of such an approximation for an interaction of the form (5) yields an analytical expression for the displacement cross section, namely,. 2.[ K- ( C EE. 2 K(23) where Co - C a Z (24.) aEh being Bohr radius of hydrogen, Ri Rydberg energy for hydrogen, C taken equal to 2,09, and Fl1. the inverse of the function Fc=:[K,() C KCT (25) K denoting the modified Bessel function of the second kind of order no

For E of the order of Ed, Brinkman notices that the absolute value of (5) for the asymptotic case where r>> a, namely, VCr)= ) (24a) gives an approximate representation of the interaction. It is of the Born-Mayer type VCr)=- Aex (-t n) ) (25a) representing closed shell repul sion at distances of the order of 7a, A and B being chosen to fit compressibility data, Numerical values from (25a) agree well with values obtained from (24a). He assumes hard scatter for E of the order of Ed, which yields with b, minimum value of r, determined by V(6)= as_ 4 -& ) =E The resulting variations of.oL versus E are shown in Diagram 10, where they are compared to (~A),. It is seen that Brinkman values almost coincide with the values of (A.d)4 for 10=3l.' E 10-1 Mevo On the contrary Brinkman value is much smaller that (~)u at 104 Mev. The two portions of the curve Ad vs E do not exactly join, due to the fact that two different potentials have been used: (5) at high energy and (5) changed sign at low energy, hence a discontinuity has been introduced Brinkman calls "transition energy", urt, the energy for which, while the knock-on slows down, its displacement mean free path becomes approximately equal to roo He proposes that, for E.> Eg., the displacement

-73cross section being comparatively small, the knock-on mainly loses its energy in setting stationary atoms into vibration, i.eo in furnishing the lattice thermal energy. Frenkel defects formed in this energy region should be enough spaced to remain stableo. He calls the, disordering effect along the corresponding path of the knock-on, a "thermal spike", For E <Etr, Brinkman sees the displacement process as an intensive, localized.one-: the displacement cross section is comparatively large and practit cally all atoms of the region affected are displaced. He calls such a region a "displacement spike"t. However, since high temperatures are attained in a displacement spike, and since they endure long enough for appreciable relaxation to take place, it is postulated that most Frenkel defects do not persist in such a region, the damage- being speculated to consist mainly, for example, of dislocation loopso. In addition, it is recognized that the relaxation of Frenkel pairs may progress only until the fraction of interstitials is decreased to the order of 10-3, the remaining defects, with large separation, becoming "frozen in" as the displacement spike cools off~ Brinkman theory does not attempt to furnish a quantative estimate of the number of stable interstitials remaining after irradiation, Furtherm.re, it postulates such defects as dislocations, whose effects on physical properties are unpredictable quantitively. Therefore, it cannot be tested against experimental results, However, although this has not beedone in the eiteratulre-, it can furnish reasonable qualitatisue explanad tion of some experimental results. This will be discussed rnow.

-74First, consider the case of the low temperature cyclotron irradiation of copper by 12 Mev deuterons (8) Since Etr is about 2,3x104 ev for copper(2) and since the maximum energy transfered by a 12 Mev deuteron to a copper atom in an elastic, center of mass isotropic collision, is about 1o5 Mev,. it follows that, according to Brinkman views, knock.ons with energies between 1.5 x 106 and 23 x 104 ev will create atomic vibra tions and Frenkel pairs and those with energy below 2.3 x 104 will induce displacement spikes. A typical value of the heat of fusion for a metal, at atmospheric pressure, is 0ol to 0.2 ev per atoms. Assuming that the melting point is raised because the displacement spike is held at high pressure by the surrounding lattice, we accept a value of the order of 0. 5 ev in our case, Finally, taking into account heat losses to the unmelted lattice, we could expect that about 1 ev is needed to melt one atom of the lattice in the spike. All the atoms of the spike are assumed melted, hence the number of atoms in a displacement spike starting at energy Etr is of the order Etr/l, which, for copper gives about 2 x 104 atoms. With the figure, advanced above, of a fraction of 10-3 of these atoms remaining in interstitial positions, we find that there would be 2 x 104 x 10o3 = 20 Frenkel pairs per displacement spike starting at Etro It has been shown in a previous paper (Preliminary Study(l)), that the differential scattering cross section for Rutherford collision between a charged particle and an atom may be placed in the form 6(E'E) dE=Th2E EE 4 EE ) = E 1e oP ot~h~rwise2

where E' is the energy of the bombarding particle, E the energy of the atom after collision, Em its maximum (i.e. a function of E'), and 0o the distance of closest approach between atom and particle. If we admit that, in the range 1.5 Mev - 2.3 x 104 ev, the Snyder - Neufeld model(3) describes sufficiently well collisions between knock-ons and stationary atoms, we are led to accept a number of displacements per primary knock-on given by 3CL-) 0 5 6 ( I +) forX>) 1, where XCl= (E-Ed) /E If we also admit that all progeny knock-ons Of a primary with energy in this range do not appreciably participate in displacement spikes, we can write for average of $ for these primaries _ CXc) 6E',E) d E xCitr 6 (E',E)dE where X..5)410, -E 0 _.3Q2 j - Ea Ed E This yields (see Appendix XV) ) = 2.34 All knock-ons with initial energy E K.Etr only create displacement spikes, the number of atoms per spike being of order E ev. Hence the total number of atoms remaining interstitial in displacement spikes, per primary knock-on, is

(E r z2 o 5I 6 (E', E) dAE 3 E 6(E',E )dE a o5,E)d i 6aE',5)dE Ed_ Ed where the first term is term is the contribution of the primaries with initial energy greater than Etr, the second one that of the primaries with initial energy smaller than Etr. Hence i..(../Etr /Em). + 0 - L.t. Etr/E. I/Ea - I/Em \/E - l/Em The numerical calculation is carried out in Appendix XVI. The result is <i = 1o9 x 10 v so that the total number of surviving Frenkel pairs is, per primary knock-on + )Q = 2,34x 0.19 = 253 Now if the Snyder and Neufeld model is assumed over the whole range of energy, the average value of ) is 1Xm G ]6 = = 5GI -O = o.6 6(E', E)d _ E IrI, dI, 0 L V \+,) 0 561 l/ C ( L+ ( +m) M i _ 1/(,1 +jCt") Xx3C

-77i.e. with X ~6x 104, ) r 0.561 Ln 6 x 104 =-6.16 Hence the fraction of Frenkel defects expected from Brinkman model would be 2053/6.:16 = 0.41 times the fraction calculated with the Snyder an Neufeld method6 Using Jongenburger(4) value of 2~.7 St SL- Cm for the increase in electrical resistivity caused by 1% point defects in copper, it has been shown by Seitz and Koehler.(9) that the Snyder and Neufeld method gives a result of change in resistivity by 12 Mev deuteron irradiation 6 times higher than the value obtained by replacing the experimental curve by its tangent at the origin, in Gorder to account for radiation anneal. Hence, Brinkman model would give a numerical result closer t. the actual change,- about 2.4 times too high. In the deuteron experiment(8) discussed, anneal was conducted after irradiation and, at 300~K, the fraction of initial change in resistivity still remaining was 8%. One may assume that this residual change is due widely spaced Frenkel pairs which have not annealed, i.oe presumably, the Frenkel pairs surviving the cooling off of displacement spikes. The remaining fraction of initial change expected from Brinkman model is then __ _ ol q 7- 5 ~/o 0+0, 2.53 which agrees well with the experimental result of 8%,: Consider now the neutron irradiation copper and aluminum at 800K. Changes in electrical resistivity and in critical shear stress were followed during anneal after irradiation. It was foundthat, for copper, th increase in electrical

-78resistivity anneals in two stages: between -80 and +200~C (2/3 of initial A and between 300 and 5500C, while the increase in critical shear stress is removed only at the stage between 300 and 350~Co In aluminum increase in electrical resistivity and increase in critical shear stress both anneal in a single stage at -60~Co Brinkman model furnishes an acceptable qualitative explanation for these results. In copper/most of the stable (at irradiation temperature) Frenkel pairs are produced by knock-ons in their high energy range (corresponding to the') calculated above in the case of deuteron irradiation), a lesser fraction are produced in displacementspikes (K1), these are assumed to be more separated than the former ones. In addition, dislocation loops are presumably formed in displacement spikes. Anneal is then seen. to proceed in the following way, During the low temperature annealing stage (-80; +20CC) the high energy range Frenkel pairs ( ), less separated, anneal out. Since, conceivably, in this case also 9)> )1, it is quite understandable that 2/3 of the initial increase in electrical resistivity may be recovered in this stage. During the high temperature annealing stage (300; 350~C), the displacement spike Frenkel pairs (%v) recombine and the dislocation loops they pinned. down until then anneal out. Hence the residual increase in electrical resistivity and the whole increase in critical shear stress (presumably mainly due to the dislocations) are recovered in this stage. In aluminum, Etr is very low, about 1200 ev, so that the displacement spikes contain at the most (see above) of the order of 1200 atoms. Practically all defect pairs are high energy range ones,

-79they anneal at low temperature (-60Oc) in a single process and., with them, the small dislocation loops corresponding to the small displacement spikes created.. Despite the apparent quantitative agreement with experiment sho above in the case of the residual increase in electrical resistivity at room temperature, it is clear that too many assumptions have been made, so that only relative significance can be attached to quantitative results obtained by such interpretations of Brinkmaa's model. First, the figure of 10-3 atoms remaining interstitial in a displacement spike was adopted arbitrarily, on the assumption that pairs in which the defects are separated by 10 r er more would survive and not recombine during the cooling off period of the spike. Secondly, the treatment given above is incorrect in that knock-ons produced in the high energy range must enter the spike range while slowing down, unless we admit that, energy transfer being small for fast primaries (which is not in accord with Snyder and Neufeld model), the secondaries only produce negligible spikes. Thirdly, the model used to treat the interaction of two atoms has weaknesses in several respects. The interaction potential energy used becomes negative (attractiv potential) in a region where, physically, it must be positive (repulsion of the interacting atoms). This leads to adopting a discontinuity in energy transfer and displacement cross section. The energy transfered to the stationary atom is found to be (see Appendix XIV), T CL+ zz R )

-8owhere Z is the coamon atomic number of the two atoms, Rh Rydberg energy for hydrogen, E the energy of the. incoming knock-on, C the constant of formula (24) giving the screening distance a, and F the function defined by (25). The variations of F are shown by the full line in Figure- 20. It is seen that T = 0 for Q- 2[,4 a, meaning that there is no interaction for an Incoming knock-on with an impact parameter having the value 2.4 a. This is due to the fact that the potential used makes a transition from repulsive to attractive for r = a(l + \3) f 2,73 a, as explained earlier. Brinkman then Tbridges the gap" between the two branches of F by the broken line shown in Figure 20. The impulse approximation itself, valid when collisions with small impact parameters or collisions between a fast massive particle and a much lighter one are considered, is certainly less correct when comparatively slow particles experience with particles of the same mass encounters where the impact parameter may be large (here, of the order of ro/2) and, consequently, the path of interaction comparatively small. The use of such an approximation tends to overestimate the interaction and the displacement cross section, hence to underestimate the displacement mean free path, At high knock-on energy, this approximation is justifiable, in the.case of copper, since as has been seen in discussing the Kepler problem boundary conditions at infinity are acceptable and since, as will be seen later, the knock-on path is then closely a straight line,, Finallyg the value C = 2.09 of (24) used by Brinknkan was a value used by Ozeroff(24) for a Thom'as-Fermi atomic model0 Seitz(9) considers it too high for this application and Bohr(25) adopted. the value C = 1 to

-81F 100 10-4 _ \ 0 2\ lo b/a'2.4 Figure 20. The Function F(b/a) Full Line: F. Dotted Line: Bridging of Gap.

treat problems of atomic interactions. A value of 2d09 for C leads to a screening distance a about twice then the one used in this paper,. The. potential energy VCr) = ~iz2 j r -2-) eS (-< ) becomes larger, for all r, if a becomes larger. Hence, on this score also, it appears that Brinkmaseatment tends to.overemphasize the interaction and the displacement cross sect ion The limiting values -of displacement -mean.free path found in this paper are consistent with these observatEns,- si.nc the lower ones almost cincide or are larger than these Of Brinkrian. This is perhaps a good check for the model us-ed One of the most important assets of BrinkrARman's work is to have shown that the displacemeAnt mean free path may become comparable to ro0 and tO have. introduced the cOncept.f'displace-ment spike, be-6 of a regin of high.disturbance --'Main Features of tbh Pro-oed Model Aplication to Deuteron Irradiation of Copper The results, Of the.alculations made using the proposed model (Diagram 10) show that there is also an energy region of the knock-ons for whih the displacement mean free path is close to ro0.. This region,can be t aken for the case of.opper considered, as extending from 105 ev (Etrl) down to.!0 ev (Eta). Any k:.ck'. born with energy larger than Etrl xperienes first spaced displacing.lisins in which it produeese a) secondary knck-.ons with energy larger than Etrl- b) secondary knock-ons with energy in the range Etrl -Etr2 and. c) secondary knock-Ons with

-83eitergy smaller than Etr2o Group a) 1mock-ons have collisions of the type just under discussion until they reach energy Etrl Group b) knock.-.ons are in the displacement spike regi; they and their progeny, while their enery is larger thans Etr2, move in a confined highly disturbe-d region. When their energy becmes smaller than Etr2 (but larger than Ed), they may mo;e out.f the highly disturbd region and, like Group c) knock.-ons, they create point defects of various separations The- process through which the primary knock-on considered undergoes slowing down when its energy becomes smaller than Etrl,,, then Etr2 is, of course identical to that discussed above. In any case, defect pairs created with. energy transfer of order 25 ev have small separationr Hence, we see the displacement spike as a region in which perhaps a majority of the atoms have received substantial energy. The atos remaini.ng at their normal site may have received up to 25 ev, so the region may have melted. This region comprises vacancies corresponding to.the departed atoms, ie. those which have left the spike with energy between 1. and 25 ev. Those create defect pairs within and outside the spike and lodge themselves as interstitials or fill a vacancy at a distance fro. the spike's periphery which may be largej as will be seen shortly. The whole pattern of the damage would then be as follows. Narrow regions with holes and possibly dislocations fmed upon resolidification with also Frenkel pairs of various separations, depending on the energy transfered during their f:ormation, Around these regions,

84-v Frenkel pairs with various separations and interstitials far away from their corresponding vacancies within the spike. The process of knock-on collisions and diffusion through the lattice shoCuld perhaps be treatedby stochastic methods and a Monte Carlo approach may be a profitable one. This is out of the scope of this paper, but an attempt will be made to picture a disturbed region and to find an order of magnitude of its size. In Appendix XVII., the average energy transfered in a displacing collision is calculated approximately, the collision being described by the cross section established earlier in this paper, namely9 E I-(I-~T/) E i (1- z S ) It is found that this average transfer is approximately E, in ev, being the energy of the colliding knock-on. The approximation underestimates T for E> Ed. It is not valid for E' Ed, for which values it seriously overestimates T. 2 Hence, a knock-on with energy E = (100) 400 ev will displace little. The secondaries of a knock-on with energy 106 ev will have, on the average, an energy smaller than 5 x 103 ev Etr2 = 103 ev. Hence, most of them will not create displacement spikes, which shows that, perhaps, Brinkman's treatment gives too much importance to these chaotic regions.

Consider a fast primary knock-on with energy 106.ev Figure 21,. in copper, First, tice that the angles f scatter for transfer Of T remain small da to E = 10 ev, so that we may consider the track as straight in t.he high.eergy regiono At E = lO4 ev, T = 5 x O2 ev. To transfer this energy, the angle of scatter in the center of mass frame must be -p=2 Sb-' /E = 2'l 5~L- E0. 52 rad S26~ The angle -of scatter in the laboratory system.must be Y, = Y/ ~ 13 ~ (see Appendix VIII, where Cp is the angle of scatter in the laboratory frame,f the angle of scatterin the center of mass frame. ) At E = 103 ev, T = 5 x.316 x 10 = 158.ey y 2 -s 2 s m0 5a o. oi r4 = 4 6 0 L=23 At E = 400 eyv T = 100 ev Ip 2 o O5 ~a6; L= 3~o Let us reason on cllisins in which the avwerage energy is transfered. Between 106 and 105 ev, the energy transfered is large enough's that either defect pairs with comparatiely large separation are formed or the secondary knock-ons can. frm pairs with comparatively large s-eparation (say several ro) The displacement mean free path will be taken equal to 2 ro in this energy region,. The track is linear. An order of magnitude of its length can be gained by the following reasoning. In d,' the primary loses energy E, as a good approximation, since every other collision. is a displacing collision.

-86O VACANCIES + INTERSTITIALS KNOCK ON BENDING AT END OF TRACK /\ I / \.. W 1.-,. -. 400 ev ~~~~~~~~~~~PRIMARY ~TRAPPING OF PRIMARY AS INTERSTITIAL O ev 10ev IOev CIXI5OAd r Ad 5rO FAST REGION DISPLACEMENT SLOW REGION SPIKE A- DEFECT PAIRS OF LARGER SEPARATION B - DEFECT PAIRS OF SMALLER SEPARATION Figure 21. Track of an Energetic Primary Knock-On (1 Mev).

The loss per unit path is 9E/Ao. Hence, -dE c) = CS / Ad) A dC And the track length between E 106 ev an. E = 105 ev is, x,= (2 Aa /s)(CI- 3.16 10 )l o 5.5 x 10 ro Between 105 and 1.3 ev, the primary is in the displacement spike region, d is about equal to. Displacing c:llisions are closeo The region is intensely disturbed around the track~ Defect pairs formed by the primary or the secondaries may still have an appreciable separation. The track may still be considered linear. Its length in this region may be approximated to C 2 = (s r./5)(3.16 x l& - 316 x 10) = 114 ro From 105 ev down to 400 ev, ~d will be taken of order 5 r. (Diagram 10). Displacing collisions are spaced0 Secondaries have small energies. Defect pairs formed have small separation, of the order of rq, They should recombine easily. The track is bent appreciably~. From 400 ev to 25 ev, the primary ppractically does not displace any more It may travel appreciably, before it is trapped when its energy reaches 25 ev. To obtainr an estimate of the total crow flight of the primary, we take the total track length for the "linear" portion of the track) ioe., Z+,IX = 6.o.6 x le ro With ra = 2.556 A, this gives, in copper, a range of 1.68 x 10-5 cm* This may appear high, but can be compared to an estimated range(2):of 4 x 10-4 cm fr fission fragments of uranium in uranium and a range tf 1.85 x 103 cm(26) for RaC' alpha-particles in copper. It is true that

-88fission fragments have an energy of about 160 Mev, but uranium has a high atomic numbers so that the interaction should be strong, and, in addition, for fission fragments, ionization is important, due to the high value of the speed. RaCl alpha-particles have four energies between 7.6 and 10 Mev, about. Alphasparticles of this energy mainly ionize and lose energy fast. It may also be observed that the order of magnitude found for the range is compatible with the number of secondary knock-ons per primary knock-on, This may be seen as follows. The number of secondaries directly formed by the primary while it has energy between 1 Mev and 103 ev, would be Xl4 + ~i = 545/2 + 114= 330 The total number of secondaries produced as a result of collisions of the primary between 1 Mev and. 10 ev is, using the Snyder and Neufeld method, 0.56l (106/25 - 103/25) = 2.24 x 104 So that%, on the average, each secondary produced by a collision of the primary, would, in turn, produce about 70 secondaries, which is reasonable. Finally, it is interesting to remark that the range of fission fragments has been experimentally found(2 ) to vary as the square root of the energy. The same energy dependence obtained- at high energy, for the range of a copper knock-on thus appears plausible. It should be noted that non-displacing collisions have not been taken into account, but this does not introduce large errors, since the average energy transfer is very small for such collisions.

89If-we use the approximation, found later in this paper, of a total number of displacements per primary given by Qtjc1) O 5 (1+"t,) where DCI = E/Ed, which should still have some validity at E = 103 ev, as will be seen later, we see that the number of defect pairs formed along the portion of the track of the primary corresponding to E 103:ev should be about 0.5 (1-+ 40) 20 Hence, we see a region disturbed by a fast primary knock-on (1 Mev) and its progeny as some sort of cylindrical portion of the sample, around the track of the primary. Pairs with large separation, but close to each other, are formed in the first and longer part of the disturbed region. Pairs with still appreciable separation, but very close to each other, are formed in the middle part, i.e. the displacement spike. Pairs with low separation and at appreciable distance (say, 5 ro) away from each other are formed in the last part, at:maximum-distance from the point of birth of the primary.-There should be about 20 pairs with small separation, located at distance of the order of 660 ro from the point- of birth -of the primary. We shall call this cylindrical disturbed region a.."damage spike". It comprises the displacement spike in its middle. In Figure 21, it will be noted that the tracks of the secondaries have been drawn perpendicular to that of the primary. This is intentional, since, as observed in Chapter III, Section 4, and shown in Appendix VIII, two

particles of equal mass should leave from an elastic collision at 90 degrees from each other, in the laboratory frame. Now, consider the case of the bombardment of copper by 12 Mev deuterons.(8) It is shown in Appendix XVIII that on the basis of the Snyder and Neufeld model(3), for an integrated flux of 4 = 7 x 1016 deuts x cm'2, the fraction of atoms becoming primary knock-ons being h Wrp = 4.90o x 10-4, the fraction of atoms displaced is C' 53.04 x 10-3 and the average number of displacements per primary =)' 6.16 These values agree rather well with experimental results, giving a change in resistivity about six times that found experimentally, if the value 2.7 ~L tc Oym of Jongenburger(4) is adopted for the change in resistivity due to C = 0.01. On the grounds that a deviation of the same order is found between calculation and experiment for neutron irradiation and for changes in other properties under cyclotron irradiation, it may be prep sumed that thedeviation is due an overestimate of c*) by the Snyder and Neufeld model. It will be shown in Chapter V, Section 2 that the overestimate is a fact, but its magnitude will not be found. We must aries in the case of irradiation under discussion.

It has been already mentioned that the energy transfer cross section between a charged particle and an atom may be placed in the formb E2. ) where E' is the charged (bombarding) particle energy, E the energy transferred to the atom hit. The frequency function, Figure 22, has therefore small values for E large, ioe. small energy transfers are favored. The average energy of a primary knock-on is _ _ E6_LE',E)dE _ Ln(E,/EA) ~ E L Em Em being the maximum energy transferred to an atom (in an elastic collision). For EB= 1.5 x 10 ev, eorresponding:to 12 Mev deuterons in copper, this gives E = 275 ev We recall now the remark, made earlier, that in most instances, a knockmon with energy smaller than 400 ev will not displace atoms. This is in keeping with the -observation that most primaries may not produce secondaries in the case at hand.. However, some primaries will have enough energy to displace atoms, but the secondary pairs formed will have small separation.. The number of such pairs is impossible to obtain~ In the low energy region, we may take A = 5 ro, so that the small separation pairs will be formed, on the average, at distance 5 ro from the point of birth of the primary.

o (E',E) EI _\E EE_ nE Figure 22. Frequency FuactiCon (E' E) of the Cross Section for Energy Transfer by Charged Particles.

-93Let us divide the sample in equal spherical cells centered at the point of birth of each primary. At one fourth full irradiation, i.e. Yv 1.75 x 1016 deuts x cm2, there are Nnprm = (4.90/4) x l0o4 N such cells, N being the number of atoms in the sampleo The atomic radius is ro/2. The radius a of each cell is given by (1/4) x 4.90 x 4 lO4 Na = N(ro/2)3 i.e., a l10 r, Imagine, Figure 23, that in cell 1, corresponding to primary 1, or generation 1, there is the primary vacancy, point of birth of the primary, the primary as interstitial, and a small separation pair A, at distance 5 ro from the point of birth. In cell 2, of generation 2, take only a primary pair. The knock-on B, before becoming interstitial, causes a thermal perturbation, or "thermal spike" which may be felt by the pair A with enough strength to cause its recombination, Assume a spherical thermal spike, originated at the center of cell 2, with the release of 275 ev. Take 20 r0 for radius of the spike, which then would comprise the pair A. There would be (20 ro)3/(ro/2)3 = 6.4 x 104 atoms in the spike. If thermal equilibrium were attained, each atom would receive an energy 275/6.4 x 104 4 x 10-3 ev = 6.4 x 10-15 ergs, corresponding to a temperature increase T = 6.4 x 10-15/1.4 x 10-16 = 46~K.

.94 — CELL I + \ +,,-I0 ro /+ 5r0 B+ CELL 2 CELL Figure 23. Illustrating the Possible Interaction of Defects of Different Generations in Charged Particle -Irradiation.

-95It is found, experimentally, that an important thermal annealing occurs at 50~K, se thgt although the treatment is by no means rigorous, recombination of close pairs by thermal spikes due to knock-ons of later generations appears as a possibilityo It is.lesar'that no displacement spikes will be formed in this particular case of irradiation, or is any expected, in general, for charged particle irradiation with the usual energies (say, smaller than 20 Mev). Note, finally,.that electrons displaced in cell 2, for example, by ionization may well transfer enough energy to the interstitial of the small separation pair A to force recombination of the pair. 3.e A Possible Explanation of the Phenomenon of Radiation Anneal. Comparison of Charged Particle and Neutron Irradi tion First we describe the phenomenon of radiation anneal. For two experiments(718) made by bombarding thin foils and thin wires, respectively, by 12 Mev deuterons, the temperature was that of liquid nitrogen and liquid helium, respectively. The change -AP in electrical resistivity was measured during irradiation. It was found (see Diagram 11) that the line 4 fversus, integrated deuteron flux, is not straight, but possesses a downward curvature. Hence, some damage is recovered while irradiation progresses. This process is called "radiation anneal". The recovery is apparently not due to thermal anneal at the temperature of irradiation, at least for the lower temperature experiment, since the workers who performed this experiment report that the damage is stable when irradiation is turned off, but the low temperature maintained.

2.5 2.0_'. 5 E I0 1.0 0.5 0 0 10 20 30 40 50 60 70 80 ~ IN I0'5deut. cm-2 Diagram 11. Change p in Electrical Resistivity Versus Integrated Flux for Cyclotron Irradiation of Cu with 12 Mev Deuterons at Liquid He.

97On the basis of what has been said at the end of the last Section, we see a possible explanaation to the phenomenon, namely the interaction of new defects, in the form of thermal spikes or -electrons, with defects already formed~ Although, no strict proof can be given, the root of the argument is the smallness of the separation of the (few) secondary pairs -and the closeness of the "disturbed regions"d If this is correct, radiation annealing should be noticeable only after a minimum.integrated flux.has been attained. Although the experiment has been performed with great care by Cooper, Koehler and Marx(8) and eight experimental points have been obtained between -= 0 and - = 20 x 1015 deuts x cm-2, it is doubtful that the slight curvature shown by the curve (given also by Seitz arid Koehler(9)), for << 20 x 1015 is meaningful, so that the reasoning made above remains valid~ Diagram 11 is a reproduction of the curve obtained by Cooper, Koehler and Marx for copper. Cooper(lO) observed that such a curve fitted very exactly into the appropriate integral curve:of the equation ~dC _ ia +-SAC d (26) with deuteron integrated flux, C fraction of atoms displaced for the value, l and constants. C and -will now be determined approximatelyo Integration of (26) with the boundary condition C = 0 for @ = O yields -C = ^/B )(l-e- $), (26a) Hence (d C/ d _h~ O

-98On Diagram 11, we see that, for ~ = 2 x 1016, = 5 x 10-2 jS) urm Since, for C = 1%, we accept a value ae= 2.7 gLq m it follows that c = (5 x 10-2)/(2.7 x 100) for ~ = 2 x 1016 Hence' ~= (5/2.7) x 10-4/(2 x 1016)' 10^-20 And (26a) yields A = 2.7 x 100 x (10-20/B )(1 - e-I) For ~ = 70 x 1015, i.e. A = 1.32 x 10-1, this gives (see Appendix XIX),'o0.5 x 10-17 In (26), f C is the fraction of defects (number of defects per lattice atom) which recombine, per unit, at time corresponding to the value o0f for which the value C( ) obtains. To obtain the value of the ratio defects recombined defects produced about the value ~, we observe that N c, where N is the number of lattice atoms, is the number of defects produced per unit H. Hence above ratio is r= N C(bVNlk - -B For = 7 x 1016, i.e. about maximum irradiation of experiment, then te- ~ =) e-7 x 0.5 x 10-1 00.7; r = 0.3 For = 2 x 10162 e j _ e_ -2 x 0.5 x 10-1 0.9; r = 0.1

-99These two values, one for extreme irradiation, the other one at about one quarter of the total irradiation, are reasonable. From what has been said before, it is quite conceivable that one third of the defects in a disturbed region produced after long irradiation will recombine with defects of another region, and that one tenth of the defects of a disturbed region produced..after light irradiation will do so, on the average. The qualificatives "long" and "light" apply, of course, only to the experiment discussed. It is alsoseen thatsaturation (r = 1) can only be -attained for = o, so that, if one attempts(9) to determine the required order of magnitude of the range of a knock-on so that saturation be attained for a finite ~pthe range obtained appears too large. Not it is interesting.to see what the situation is for neutron irradiation. In this case, the differential cross section for energy transfer from a neutron to at atom of the sample is, for elastic, center -of mass isotropic cClli.ionb (conditions wholly justifiable), C(E',E) dE 6S E') d o E A E =0 otke/Cse, where E' is the energy of the neutron, E the energy transferred to the atom, 65(E') the scattering cross section of the atom for neutrons of energy E and.Em the maximum. of E, namely Em= L[A /(A+l] E' where A is the mass number of the atoms considered- Hence the frequency

-100function6(E',E) is independent of E; all energy transfers are equally likely, in the meaningful range (0; Em), Figure 24. The average energy transferred in displacing collisions by neutrons of energy E is E E. dE /i dE E d i.e., for the same Em, it is much larger than in the case of bombardment by charged particles. For comparable energies of the bombarding particles, the cross section 6d is higher for charged particles than for neutrons Another difference is that pile neutrons have a spectrum of energies and that a flux of very energetic neutrons, say, E larger than 1 Mev, I M e of order 1011 neuts cm-2 sec-1 is difficultto obtain, while it is easily obtainable with a cyclotron. This value corresponds to an integrated flux of i= 5.5 x 1016 particles cm-2 for one week irradiation (- 5.5 x 105 sec), which was exceded in the deuteron experiment previously discussed. For E -1 Mev, in copper, Em = (4 x 63)/(64)2 Mev = 6.25 X 104 ev. The average energy of primary knock-ons produced by a 1 Mev neutron is, therefore, E =3.1 x 10 ev. For this energy, a knock-on is in the range Etrl; Etr2) and will produce a displacement spike. But as seen in Chapter IV, Section 2, only few of its secondaries will create displacement spikes.

I o'1 (E') E', E) =E) Em I I I I I I I I I I Ed Em E Figure 24. Frequency function a(E', E) of the Cross Secton for Energy Transfer by Neutrons, in Elastic Collisions, Isotropic in the Center of Mass Frame.

-102-'Assume now, for the purpose of comparison, a monoenergetic neutron flux:of energy 1 Mev and magnitude 1 = l0ll neut cm-2 sect1 and an irradiation of one week, i.eo about 5.5 x 105 sec. The cross section of copper for 1 Mev neutrons is 65 = 3 barns. Since, practically, all collisions of 1 Mev neutrons displace copper atoms, the number of primaries produced during the whole irradiation, per lattice atomL is about Mprim = 3 x 1024x 1011 x 5.5 x 105 = 1o65 x l107 If as before, we diviLde the sample in spherical cells round the point of birth of each primary, the radius a of these cells is then a = (10/165)1/3 x 102 x (ro/2) = 91 r0 Repeating the reasoning made in Chapter IV, Section 2, but for a primary with energy 351 x 104 ev, this time, we find that an -order of magnitude of the length of the damage spike is (2 ro/5)(1076 x 102 3.16 x 10) = 58 ro If we consider the damage spikes as cylindrical, Figure 25, of radius 5 ro, which is probably too large (see Figure 21), and length 58 ro0, they occupy, at full irradiation, a volume n x (5 r)2 x-58 r x:N x 1.65 x 10-7 N being the total number'of atoms in the sampleo Compare this to the volume of the -sample, i.e., N x (4 /3) x (ro/2)3. This is comparing.2.4 x l10-4 to 1.7 x lo0l

-10358 r 90ro5sr DAMAGE SPIKE DISTURBED \ REPRESENTATION REGIONS (CELLS) } OF A DAMAGE SPIKE Figure 25. Illustrating the Possible Interaction of Defects of Different Generations in Neutron Irradiation.

-1o4This shows that the total volume of the damage spikes, at full irradiation, is less than 2/1000 of the total volume of the sample. Add the fact that, in the first part of the spike, the defect pairs have relatively large separation, the pairs in the displacement spike (center of the damage spike) still have appreciable separation, and only the pairs at the end. of the damage spike have small separation and..are highly susceptible to recombination, there being.20 such pairs, compared to 0o5 x (5.1 x 104/25 + 1)' 620 pairs in the whole damage spike, then it can be concluded that there should be little interaction between defects of different generations, for the case considered (see Figure 25) o Hence, for a typical metal like copper, except for extremely long irradiations, of'the order of one year, which generally do not come into consideration in experiments, with fluxes of the order 1011 - 1012 (radius of cells 90 ro, 30 ro, respectively) above 1 Mev, we do not expect radiation anneal, Displacement spikes will form and contain many ~atom-s o. The number of atoms displaced for a knock-on energy 35l x 104 ev will be 620, as just seen. Hence, the distrubed regions will contain many displaced atoms. The typical damage spike considered covers more than 8 x l03 lattice sites. As it has been noted earlier, the conclusion just reached concerning the absence of radiation annealing in pile neutron irradiation, for reasonable exposures, is completely confirmed by the results

-105of the only reactor.experiment(31) (to the best knowledge -of this author) for which the temperature was near that of liquid helium and which, at the same time, incorporated the measurement of electrical resistivity during -irradiation. In this experiment, conducted at Oak Ridge, several metals and alloys were irradiated.in-a cryostat maintaining a temperature of 14 5~K within-+ 1.5~K. Diagram 12, reproduced from the report on the experiment, shows that no radiation anneal took place, for'any of the.materials irradiated, for an exposure of 150 hrs, corresponding to a fast neutron integrated flux of 3.78 x 1017 neuts cm 2, The accuracy and reproducibility of the measurements were good. The experimenters concluded, in line with the numerical results obtained in this dissertation, that it can be inferred that there is only slight, if any, interaction between adjacent damaged regions. It is interesting, in connection with this experiment, to make an approximate calculation of the predicted change in electrical resistivity, in the case of copper. For this we use the result, which we shall establish in the next section, that the fraction-of atoms displaced is approximately given by c E E) ckE E (A+| A,)- Ea/-4 A where t is the time of irradiation, in seconds, 6s the scattering cross section >of the irradiated raterial and A its mass number.

- 106 - (x o10-8) 5 o POLYCRYSTAL (COPPER) * ZINC V Cu3Au (DISORDERED) A ALUMINUM 4 0 NICKEL_ * COPPER SINGLE CRYSTAL; COLD-WORKED ol, V IRON' oGOLDO' x Cu3Au (ORDERED) /,IRRADIATION TIME (hr) 20 40 60 80 TEMPERATURE (OK) Diagram 12. Near Liquid Helium Temperature Reactor Irradiation of Various Metals and Alloys.(31)

-107For copper, A = 63 and we can take 6C = 3 barns. In the experiment discussed, t = 150 hrs = 504 x 105 sec An analytical expression for the neutron flux-in hole 12 of the Oak Ridge reactor, which is the hole in which the cryostat is installed has been given by Seitz and Koehler(9) from unpublished data furnished by Holmes. It is the following: ( ) 0414 x loll4 cp (E) 0.414x1011 for 0.025 < E<25 x 10 ev, E &/(E) = 1.508 x loll(16,7e-2e5E + 1.02e"l~51E) for 2.5 x 10-2 Mev E, with E in ev in the first formula and in Mev in the second oneo With this, C _ - 5.4 x 10 5 x 10-24 l 104 x10-6+ 25 x 10-6 x 64 400 1o508 \[16o7 x E e=20o5E Ee 1E0.10'2 ddE + 11.02 2.5 x 10-2 25 x C. 1-V104 44 2 o + 1e 1508 x 167. ec30.51 5 x o-12 - 1 - 20o5 2025 +1,508 x 1l02 e-378 x 1-2 x 102 1 + 1.e L2.5 x.o2+ 2 l 51 C 10o4 (lo2 + 1.23 x 0.61 x 7.4 x o 10-2 + 2.5 x 10-2 + o.66) C- 104 (10-2 + 5.56 x 10-2 + 205 x l0'2 + 0.66) > C = 7.5 x 10 5

With Jongenburger estimate of [ 7= 2.7 Lcv\ for C = 1% in copper, we obtain for estimate of increase in electrical resistivity after 150 has irradiation, 7.5 x 2.7 x 10-3 = 2.02 x 10-2 c. r~ On the line for copper in Diagram 12, we read, (8e)e>,= 3.75 x 10-3 Si r Cat Hence =_= _ 20.2 4 ~ e)p y 3575 Thus, we see that the theoretical estimate is about 5.4 times the observed change. This is consistent with the remark which will be made in Chapter V that the model used to calculate the number of secondary displacements per primary knock-on reads to an overestimate. This is also true of estimates made by the Snyder and Neufeld method. If anything, this may tend to prove that Jongenburger value of the increase in resistivity due to one percent displaced atoms has the correct order of magnitude. The comparison between charged particle and neutron irradiation may be summarized as follows. Charged Particle Many primaries. The disturbed regions are small and contain few defect's, Even for a reasonable irradiation, their distance apart is not large, compared to their size, Radiation anneal is expected to take place for such irradiation. Displacement spikes are not likely.'"Dislocations" are not expected. Neutron - Comparatively few primaries. The damage spikes are far apart for a reasonable irradiation. They contain many defects, Radiation anneal is not expected, Displacement spikes are formed and "dislocations" are expected,

l1094, Extension -of the Model to Other Metals Than Copper Expected. Effects of Charged Particle and Neutron Irradiation as a Function of Atomic Number and Mass Number. It is out the scope of this paper to make for other metals the same analysis as made for copper of the scattering by a potential leading to an interaction potential energy of the form (6)o However, we shall now attempt to extend to other metals the results just obtained for copper and to see to what measure they must be modified. In Appendix a it is shown that classical treatment is valid for pile irradiation of beryllium and even for cyclotron irradiation of beryllium with deuterons up to more than 20 Mev. However, the model is limited to probably Z 13, because of the importance of ionization for lower atomic numbers~ In the interaction energy. (6) V(r) = (Z / r)(i - +r/cL ) X C- r/ c)) with a = ah z'1/3, we see that, the smaller Z, the larger the screening distance a (the potential remains Coulombian at larger distances)o It is clear that (a) = C(+ r/2c)ex. -r/&) is a,mon-otonically increasing function of a, for dOF/dct = - r/2a- + r/ao) xL C- r/. o) is always greater than zero. Hence, F increases when Z decreases,.monotonically, for- all. values of ro Since V(r) =(gE/r) Z2F, it is possible for V(r) not to vary monotonically with ZO.... Let us study the function G(Z), such that i(er) = (e)r ) G C/) GCz)Z) = r iaZ C+r2&'13leX/ S (rlagAz'1'

-110or, with Z1/3 =; (') = [E 6* (i Haag) (g i [- (r/ ) %r / dG/d =- 16'5 +(7 r/2 4z q - / C, ~ r 1 ex[-(r/ia ) For the metals susceptible of being studied under neutron irradiation the smallest value of Z is 4, hence that of' is 1.59, for which 5 = 10.12. All the values of r considered are smaller than ro, which, in turn, is smaller than 4 ", for all metals. Hence 6 5 > 6 x 10.12 r/a < 4/0.5 = 8 ioe., in the range of r coming into play, we have ~g~5 > r/a4 for all r, and dG /dc > o In other words, G and V are monotonically increasing functions of'S and Z. With an interaction energy..of the form (6), for all r in consideration, the interaction will be weaker for Z small than for Z.large. In lighter metals, energy transfer and displacement cross section between knock-on and stationary atom will be smaller, displacement mean free path larger, at the same energy of the knock-on. For Z low enough, Diagram 10 may be altered to give a variation of (d)4 such as shown in Figure 26, i.eo such that it would be possible to conclude that no displacement spikes are formed, This is in accordance with Brinkman views, which are that, the transition energy being small for light metals, their displacement spikes will be small.,

Xd Figure 26. Possible Variation of kd for Light Metals. AP I A 70 x 1015 70 Effect o Radiati Figure 27. Effect of Radiation Anneal.

Consider first charged particle irradiationo In Appendix XXI, it is shownr that, for a given type -f particle, energy and integrated.flux, the fraction of primary knock-ns Pprim varies like Z2/A where Z is the atomic number and A the mass number. of the metal irradiated. In Appendix XVIII it is shown that the average number o) f atoms displaced per primary varies like Lh (I c, j where ( (Ed )i zs Ed (M 4A EA M being the mass number of the bombarding particle, E' its energyo Hence, s) varies little from metal to metal, for the same conditions of irradiation, Take M = 2,' = 1 Mevo For Be 9, DIv= 7x 12 x 106 29 x 104; Li (1 + C,) = 12.6 121 25' For Au 197, X _> = 1972 x 48 x 104 = 95 x l0; Ii (l +,X ) = 9.85 (2 + 197)2. 85 Hence, for a light metal, we- have less disturbed regions than for a heavy -metal since nPrim is smaller, but each region does not contain appreciably. more -defects than a region in a heavy metal~ The overall damage is greater in a heavy metal. This is borne out by experiment, in particular by the results of Cooper et al, HOwever, the knocklon mean free path will be larger for the same energy and the small separation defect pairs will be farther away from the

1135 point of birth of the primary. Hence, there is comnpensation on the change eof- the two paraeters which influuence recombination in the form -of "'radiation.anneal"' when,one goes from irradiation of a heavy metal to that of a light metal and we expect that the effect of radiation annealing will not change appreciably from metal to-metal, in charged particle irradiation. oThis checks with the, few.,experimental results available. In the helium temperature deuteron irradiation(8), samples of gold (Z = 79), silver (47),, and copper (29), were irradiated~ For I)= 70 x 1015 deut cm2, the ratio OA/OB (see Figure 27), where A is on the Ae vs curve and B on its tangent at the origin, was 49o36/63.6 = o078 for -Au 34/45 5 = 0 75 for Ag 5o.5/38o? = 0779 for Cu (the numbers, such as 49.6, o.,, were measured in mm on the curves given by Seitz and Koehler (9)).Consider now neutron irradiation. Denote by r (a vector) the position of-a point A of the sample, Figure 27ao Define.n(r,E',S) dEt dSLU the volumetric density, at r, of neutrons with energy in dEl about E, going -within a small solid.angle dAS about the direction Sa. Call v" the velocity corresponding to EB and' and.v the modulus of v' The number of such neutrons which cross unit area perpendicular to $L " at A, in the directional, per second, is'-n Cr)C E' _ E~'d = SI'_ CC)' E, E' )) sL'L ) by usual definition -of the angular neutron flux. Here " is the flux in the sample, not the original flux before the introduction of the samples.

- % --—,, o/ V \ I re 27a. Netron Colisi Figure 27as Neutron Collisioa.

-115The number of neutrons, from those in dEt, dSi before collision, deflected by collision in da about S, per second, per atom present at A is, from definition of the microscopic differential cross section, Cr, El', 9/)dE'dS 6 CEG 5l _ 6 (EB, T,) d. may be written in the form of an -energy transfer cross section, by the condition of conservation of momentum and energy (elastic scattering assumed here), and the above number written as Cr,E', ) A)EdAS' 6 CEK,Ed) E) (27) where E is the energy transmitted to the atom hit. This number is then seen to be the fraction-Qf atoms present at A which receive energy in dE about E, per second,, from neutrons with energy in dE' about El, moving -in directions in dS'.about L'o Coherent scatter (crystal effect) is assumed not to take place and the lattice nuclei are considered at rest before the neutron.colli — sion. These two assumptions are perfectly valid for displacing colli= sions, where Ed:= 25 ev, at least, is transmitted to the atom. Hence, 6 is independent of QL t, Integrate (27) over'all. S calling Ctr ~E)'-,,, ~ ( r| E' jL) a A' iQeo,, the "normal" flux, to obtain + jr e $E') (E' E cL (28) as the fraction of atoms present at A which receive energy in dE about E,

per second, from neutrons with energy in dE' about Eo. The samples used are small (for example, 00005 inch d1iameter metal wires, 1 to 3 inches long), hence we may take ( as constant over the sample, for the purpose of calculating the fraction of atoms displaced. Then CE') )E' 6 (E', EWE (29) is now the fraction of atoms in the sample which receive energy in dE about E, per second, from neutrons with energy in dE' about Eg' If we add. the assumption of center of mass isotropic scattering to that of elastic collision, which is also justified here at least for small and medium mass numbers, and a good first approximation for heavy metals (Preliminary Study(l)), (29) can be written S CE') )dE (E') tE 0, os E o) E E yv~(30) O ) otherwise, where Em is the maximum of E, The fraction of primaries formed per second is obtained by integrating (30) between E = Ed and E Em, and between E' = Elhr and E' = O, where Elthr, threshold energy for atomic displacement, is defined by Ed = 4A E' t After an irradiation time t sec, the fraction of primaries is -____ ~ A+ (M (A- 1dE E. ___ EaE'

-117It is not seriously wrong:to take ~6 (E') 6S =. ct, neglecting,resonance scattering, Hence we may write 00 h Ed m NtLs,% [1- (A+1)E Now, the frequency function is not peaked around-Eth, which is far above thermal. -Hence, since 4A) Ed I, s +' E' S E( the term of the integral corresponding to ~ (E').dE' will predominate over that corresponding to (A E ) 5Se' and another L4A E'.justifiable approximation. is ~ ht;M C sL 5 LE')d E', (31) 4 is the disturbed flux, ioe... after introduction of the experimental device'and. the sampleso / +A) - _/4- LE' decreases when A increases. QA4+ I) E:d'A [ takes the value 25 = 5605 ev for Be 9 and the value =19 x. 1230 ev for Au 197. If, for the purpose of -estimate, we assume ( (E') r 1/Et, and.adopt a cut off energy of'2 Mev, the integral is proportional to Log (2 x 106/560.5) =- 2.3 Ln 3 554 x l04 for Be 9 and to Log (2 x 106/1250). - 2.3 Ln 1.63 x 103 for Au 197, so that the

118ratio of the values for beryllium and gold is about 4/3. At any rate, it appears clear that nprim is fairly insensitive to A, for irradiation in a thermal reactor. If we compare two metals irradiated in the same conditions, i.e same reactor core configuration and neutron level, same experimental device, and presumably to a lesser degree of importance, since the samples will be small and very thin, same flux distortion, we can say, from (31), that nprim varies like GS However, in fact, about same size of different samples is adopted and flux distortion by the samples will depend on 6 -. BHence, it must be emphasized that, only if the flux distortion by the samples (not by the experimental device) is small, and this may be said to be true for such dimensions as mentioned above, can we say that nprim r 6, approximatively. The scattering cross section for neutrons of energy above, say 1 Kev, does not vary drastically from metal to metal. Hence, even for lighter metals, in which Ad will be larger, no radiation anneal is expected. It will be seen later that an approximation (overemphasizing displacements, however) for the number of knock-ons (including primary) per primary knock-on formed with energy E is C) o) ~ os C\ tX,) where Ci= El/Ed and E1 E- Ed, E being the same energy as used above. Hence, A)(E) - 0. E/Ed

oor. and the, fbaction of at mdsaisL -ed a fer irr idiation t sec is qE_4_ A EE A P Ck'+'~' (A +' Ed s a( t )E | Af E dE d jA+SEE'/ d A Ed Following ithe teasoning,iade before we can neglect, in -the. integrand Ei in eomparison to.Hence, we have 00 -,, ( ( +i )~ E E' (32) For a..E' fast flux,, and neglecting the lower limit, the integral is a tonstantt so that, fo i rradiation in a thermal tectorE and since A/ (A F 1)2 = 1/A, we. expect C to be approximately proportional to 6s/A. Since nprim isap tely proportional to S., ) is approximately inversely proportional to A.,,Hence, the.. a-ge spikes. will contain more point defects for A small than A large and, since d will be larger, there will be more chances for recombinationof def.ects by radiation anneal forA.small. We shall now.su..arize the comparison of'ligt metals and -heay tals for both charged particle and neutron irradiation, Ch.ged Pa.rticle Smaller number o primar.ies (i..e..of disturbed.regions) forlight metal than for a heavy. metal (nprm 2/A). About the -s me-number pri.m 2A u h enme

of point defects per region (s = ct). The regions are larger for the light metal, but the displacement'mean free path, A, is larger. Hence, radiation anneal is expected to be of the same order for light and heavy metals. Displacement spikes and'"dislocations" are not expected. Neutron The number of primaries depends mainly on 6s, to which it is roughly proportional~ The number of point defects per region is approximately inversely proportional to A, hence is higher for light metals0 Radiation is not expected to take place for "reasonable" irradiations (order of a few weeks with a flux of order 1011 l1012 above 1 Mev). For heavy irradiations, chances for radiation anneal are higher for metals with high 6 and low A (nprim, A )and..all large)~ For Z (io.e also A) low enough, displacement spikes may not.form. The above conclusions. are borne out by experiments in which property changes have been measured during irradiation, insofar as radiation annealing is concerned, both in the case of charged particle and neutron irradiation. (7, 831) Mechanical properties such shear'strength, which should be sensitive to dislocations, hence, presumably, to the formation of displacement spikes, have.not been measured for charged particle irradiation. At least one reactor.experiment (Reynolds et alo (20)) checks well with the picture of no displacement spikes produced in a light

metal. The results of this experiment have been discussed in connection with the critique Of Br in.'s modelo Hence,. the interest- of measurements of change in properties..then, practically, the only property which.is m'nageable is. electrical resistivity - -during irradiation in a reactor -is -obvious We rnote ialso that such&an -experiment must be made at or near liqid helium temperature, so that posible radiation anneal is not ma.ked by thermal:ann;al, incipient already below liquid nitrogen temp'.W erature6 We remark that, -for purposes of checking the effect of irradia tion, several metals should be. rradiated in the same disturbed flux ~.conditionso Finally, the- success of'the experiment -demands a compaiatively. high fast flux if the duatian is to remain reasonable, Now, an attempt will b made to calculate the umber of seiondies per primary, using a displacement cross section given by (20)o

CHAPTER V CALCULATION'OF THE NUMBER OF ATOMS DISPLACED PER PRIMARY KNOCK-ON l. Generalities =Obtention of the Primary Integral Equation We consider a collision between a knock-on, called primary-, and a lattice atom, called secondary, as happening in the following way6 First the primary, of energy E, transfers energy T to the secondary Then the secondary escapes from its lattice site if T >Ed, by losing energy Ed to the lattice, therefore retaining energy T - Ed as a free secondary knock-on. After the collision, the primary has energy E, Since the collision is assumed elastic,. we have E = El + To Call P(E,T)dT the probability that, in a collision, i.e., given that a collision occurs, the primary transfers energy in dT about T to the secondary. Then P(E,T)dT = probability of having a collision wtransf.in dT ab.T probability of having a collision Hen.ce, P(E,T)dT = 6 (E, (p. )dSLT' 6s(E), where G (E, (PT )dTLris the differential scattering cross section at angle YT in the center of mass frame, such that YrT corresponds to T; YT is the angle of scatter of the primary in the center of mass frame, Figure 28; dST is an element of solid angle of the center of mass frame, limited by the angles QT and &T + d cT a i.eo d T = 2TT sin QT d 9.o

-123SECONDARY PRIMARY PRIMARY Figure 28. Angle of Scatter in the Center of Ma'ss Frame.

-124We replace 4T-by its value in terms of T and E, from the relation (see Appendix VIII) expressing elastic scatter, namely T = E sin2 CPT/2. (33) from which we see that T increases withPT T Hence,, we can -write P(E,T)dT = 6 (E,T)dT/6s(E) (34) T varying from 0 to E. In the same way, if we call P(EE' )dE' the probability that, in a collision, the primary will emerge with energy in dE' about El, we can write P(E,E' )dE' = (EE' )dE'/6S(E) (35) where 6 (E,E' )dE' is obtained by replacing in 6 (E, wY )d SLT, %T by its value from E' = E - T = E(1 - sin2 Y /2). (36) E' decreases when LT and SLT increase, and will be made to vary from E to 0. Now, the following reasoning, generalized from Snyder and Neufeld(3) to various forms of cross section which may be considered, is made. Every knock-on will suffer at least one collision in the sample, since, in the model used, the collision cross section is TT (ro/2)2 Note that this has been assumed by Snyder and Neufeld, for a model based on hard sphere scatter, center of mass isotropic, arbi-'.. trarily, We have seen before that the slowing down collisions of knock-ons in copper can be described by an approximate formula for the cross section 2 o'E) d /.1 (37)

-125The approximation is much better at high energy than at low energy. Similar fo.rmulae would. hold f or -other metals than coppero The important point, for the time being, is that we can describe s1owing diown and energy transfer by appropriate differential crOss sections 6 (E9Ej )dE and 6 (E,T)dT, respectivelyo Call d (E) the total number of atoms displaced by an knock-on (including the knockson itself) of energy E after release, i.e. E is the quantity by which the energy transferred tto it in the releasing collision exceeds E.d Then, when the secondary receives energy T in the collision, the total number of atoms displaced as a result of its being released is (T Ed), and, when the primary -emerges from the collision with energy E., the total number of atoms displaed as. a- result of this emergence is 4 (E~t)o We can clearly write 0 )(E)= C)T-Ed) t ET) T(dT (38)_ the limits of the integrals corresponding to values of T and E' which define the range of displacing collision~ ioeo, Ed T E and. O ~ Elt <E respectively, and E going from E to 0, while T goes from 0 to E and w from 0 to T YT

_126We first apply (38) to elastic, center of mass:fraie isotropic scatter, Then, E), YT)& d T = GS E) C LT/4T Ffm V and. 7 T - T S;' eT T Tna AT =E/2) S QT FIT ) we have 6 CE,,)dQT - 6S CE) AT/IE 6 CE,T) dr. In the same way, aEj: _ (E/z) S~ ~. d Qr and G(Ei,Y) QT = - 6S CE) cLE'/:E = 6CE,')aE# (38) becomes E) S (T-E) _ T I. CE')a E/ Let E/Ed= I, t\-Ed)/E d =/ Tz E'/EI -x?, e The equation becomes finally, xl~ d,)= 4X,),) d:Y,. (39) This is Snyder and Neufeld p'.rimary equation,. whose solution, and various associ.ated. roblems,- like treatment by Laplace transfor, consideration of replacement and recombination of defects, are cinsidered in Appdenix XXIIO.. Return now to Equation(38). Call 61 (E,T) the primitive of 6(ET) with respect to T, and 6s(E,Et ) the primitive of 6(EE' ) with respect to El, defined by T, E, 7') - J 6 C/E,T') dT 6 E, E) F- 6 EIEf-), d

-127We. then have the following relations: CE, ( T)dT = 0 6d TE) 6S, E,E) = o E,E) - 6C E')s CE) 0 where 6d is the displacement cross section, 6 the scattering cross secetion. Integrate Equation (38) by parts. This gives, s E)..E) LtE) E +6 CE) (0) G R E) E 1- (40) 6 E) 6S CE E/ For continuous satter, ife. for a differential cross section of the form pdp, with p going from zero to infinity, (40) is identically true, since 06So0.

-1282. Obtention of an Asyptotic Solution for the Model of Interaction Used,06 Comparison with the AsymtOptic Solution of the Snyder and Neufeld Equat ion, Confrontation of Calculational and Ex peimental Resultso For the model used in this paper, 6(E, i Y-r) CC EL TTlT) C 0 QT =L < 1|| O:otherwise. Henae, 6s is finite and (40) can be use-d. By efinition, (E) = 1 for E = 0, sinrce, when the primary emerges from the particle cillision with energy 0, the total number of atoms displaced is reduced to that primary knock-on. Using this, (40) can be written, Gde(tE) c& 3CT cT-E. d) (ET)T'(E')'teV,E')E, (41) Ed TE Now, assume possible a solution.of the form 3CCE) = A +BE A,B constant. Then, we should have.GoE) = | 61CET) dT + G a (EE') El G, CEITdT Since p decreases (Figure 29) from 00 to 0 wlhen (9T goes from 0 to n1 (whi.le T goes from 0 to E and E' from E to 0.), w~e have 6'(E,pYT)do =6 CE2T)dT= GCEE')FE' = -C Ed().E From (20),

d129AN\10T p+.dp Figure 29* Variation of QT versBU p.

-130with (T connected to T and E' by (33) and (36). Hence, tET+) d ~TF 7 as a { (l S' -'S~z/[E ( VA E-F.*( E 1'[E From which, 6,(5,T)= E c or (I-Tr rs~I T7 Ptutting these expressions in (42), we obtain E 6CCE) _( - 2_ SiT_ (l T iE ) E AinS JEa) Sl s~L4 V~.,'(43) Let E - E' = T in the integral at the right (dE' - - dT, limits 0 to E)o Call f(T) dT the integrand, now common to both integrals. The right hand side of the equation becomes, - +Eds + 1L T) (r dT -T N1i ~i ~ i )dLb:~ (jT 0 E4 o'E o Now, assume E~ Ed. We can make the approximation EadsE) _ l-4NfE dr _ -Ed) LrZ P 4 ir -(./,, 1 7 L~~TTYE /) J Ed d, LE 61 CEEL) =rr ___ E E

Hence, the following must hold: 41e 4 B 4 E5 d -I ta) and, since E >Ed is assumed, I 8 Ed FT 4TgE c which gives B3 _ /2 Ed (44) Since B, so determined, is effectively independent of E., an asymptotic solution, i.e, for E > Ed, of the form ) (E)' A + BE, with E in Mev, is possible, With Ed = 25 ev, B = 106/(2 x 25)=2 x 104 Snyder and Neufeld Equation (39) has an asymptotic solution SC(XI= 0.561 (t4X) Since xl = E/Ed, we can write Z (E) = (0.561/Ed)(E + Ed) = 0.561 + (56.1 x 104/25)E; E Mev. We see that the coefficient of E is practically identical with that found in this paperO Such. solution leads to fractions of atoms displaced 4 to 6 times that shown by experiment on copper and gold, if Jongenburger(4) value of the change in electrical resistivity due to a fraction of 1% atoms displaced is eployed This is consistent with the observation that the model used in this paper overestimates displacements at low energy, from which we expected that sOt tlculated would yield too large fractions of atoms displated.

-132The apprOximation E> Ed in the last calculation is good even for E =103 Mev, since,. then, Ed/E 2.5 x 10-2 (4/T j 0.201. Neglecting (4/tr )f]7 in comparison to 1 is not quite correct, but remains commensurate with other approximations made in the model used. At any rate, the estimate of (E) for this value of E is certainly too large. We would expect the approximation on 4 to be worse in the case of charged particle irradiation, where the primaries have low average energy, than in the case of neutron irradiation. However, we must remember that, in neutron irradiation, there is a large number of secondari-es, most of them having low energy, so that most displacements must happen at low energy Two experiments only allow comparison, the helium temperature cyclotron irradiation of copper by 12 Mev deuterons(8), and the helium temperature reactor irradiation of copper, (ll) The deuteron experiment gives a change in electrical resistivity 6 tinmes smaller and the reactor experiment a change 4 times smaller than the change calculated using the < (E) furnished by the Snyder and Neufeld model. Now, return to the approximation on 0 (E). We cannot determine the constant A, Physically, if the solution were good over the whole range of E, we would have A = 1 for E = 0, A must be negligible in the,ange where the approximation is good, so that its value is of little importance. Since E + Ed is the energy transferred to the

-133knock-on (primary; fr example) before its release from its site, it is convenient to adopt ) (E) -B(E + Ed), EEd in Mev, that is A BE&d We can also write (Xl) = BEd (+ Xl (45) with B = 2 x 4, BEd = 05. Note that the study of ) (E) has proceeded on quite general lines, applicablJe t:. various types of cross sectional models, up to Equations (40) ad (41). A direct treatment of the case of the:odel used in this paper is given in Appendix XXtII, In summary, it will be said that the model of interaction between knock-ons and stationary atoms us-ed: in thit paper leads to a number of secoadaries per primary, at high primary energy, practically identical to that obtained by Snyder and NeUfeld on the assumption of hard sphere scatter~ The values so obtained, for the number of secondaries, are certainly tao large, since the model overestimates interaction at low energy. This is conform to confrontation of experimental and calculational results

CONCLUSION The form of interaction potential energy used and the method of approach adopted in this paper have proved successful. The values of displacement mean free path and fraction of atoms displaced obtained are quite c mpatible with experimental resultso They show that recombinat n of defects by interaction with them of defects of a later generation can reasonably be expected in charged particle irradiation, but, not in reactor neutron irradiation, for exposures cwming into considerationn iLexperimentso The two extremely useful models of Brinkman and Snyder and Neufeld have been investigated critically, by comparison with the model used in this paper. The direction of the error in the parameters they arrive at has been defined. The method used has afforded useful com~parisons between charged particle and neutro irradiation and shown the great interest of in pile measurements during low temperature neutron irradiation. The design of a helium temperature cryostat which could be used, among various purposes, for experiments on neutrondinduced atomic displacements in metals, has been made. Construction has begun and some development experiments have been carried out -Appendix XXIV gives a description of this work, Low temperature neutron irradiation experiments help clarifying the problem of radiation annealingo It may be added that the study of the recovery of mechanical properties durig themal anneal after a low temperature charged particle irradiation should be helpful in the study of the problem of displacement spikeso 1354

APPENDIX I Correspondence between a Born-Mayer interaction and the potential energy used in the paper, at large separation Born-Mayer interaction (Huntington(l2)): V, (r) -A rex (, of- ro with two sets of constants, for Cu, P = 13, A = 0.053 er - = 17, A = 0.038 ec which constitute a bracket for and A. At r~Za., (6) of the text becomes V (r) 2 z ) In fact, this will be a good approximation even for r7c-. At -=ro, V, (r) A V(w) = 1.22 x 10-8 Mev. from Table I, so that the potential energy V(r) used checks with the value of A, within a factor of at most 4. Write V1 (r) A eP exp (-~ C) l-P= 13; logloeP 13 5.65; eP= 4.47 x 105 2.3 AeP- 5.3 x 10-2 x 4.47 x 105= 2.36 x 104er,,ZeZ 1.21.2., 0' o 344 x 10-2 Mev = 3.52 x 104ev, roa 2.556 13 = 0.197 -135

-136We can write VL Cr) = 2.36 x 104 exp (- r) ru compared to V\/r) = 3.52 x 104 exp (- ) l r0.172 For r - - i) r 2.556-6.5; -&ot 1e6.5 6.5 = 2.83 0.197 00394 2.3 e 6.5 = 6.77 x 102; V, (. - ) = 3 _ 34.8 6.77 x 1 Oii) _ 2.556 = 745; log10e745 = 7.45 = 324 0.172 0.344 2.3 e 7~45 = 10.74 xV ( ro. = 3.52 x 1o4 = 205 1.74 x 103 At r= ro 15, it is seen that V(r) is sufficiently close to V\ (r) 2p-.=17; 1og10 e= -17 = 7.40; e = 2o52 x 107; Aef= 3.8 x 10-2 x 2.52 x 107 = 9.55 x 105 =- 2556 = 0.150 17 / 00150 For r-ro 9 r 2.556 = 8 52 log oe8o52 82.52 = 3e7; e8o52 = 5.02 x 103 10 - 23190 V(~) = 955 x 105 = 190 ~bk7

1l37Actually, from Table- I, V(m. = 246 x 10-5 Mev- Hence, V r0) is rather close to V (r) for the lower value F = 13, but smaller, so that V(r) is somewhat tQo small at large r, but yet takes accetable values.

APPENDIX II Tl Reduction of the Schrbdinger equation for a system of two particles 7Z.'i V, - =; V. - Czar_ —3. Let( be the coordinate of the center of mass, i e. (R = & 1R, +k R ) = 2 XY / ) I Hence X = ) (\ t fo z 3 ) with r =3 t, X z Z,X a3 X L +x - 2 X" aX -i38 -

In the same way, a$c. p MX L a,n-'I ) a lVR 1 1 ='_ t~ZR -a- r - _ b)(2X. Z x2t;0i,27g+5;7r+2\p>> a~tO x,w, K(R + I -VGE, Vr) where 3'L | + A\; _ t_ 1M M. k+ h) + ml Hene th rii nal eat. ce.to (2)of he text Hence, the or~iginal equation reduices to (2) 0~ the text.

APPENDIX III Wave length of a copper krick on "reduced" particle at various energies 1) E = 7.2 x 10-1 Mev 7,2 x j0-1 x 1.6 x 10-6 =' 63 x 1.67 x 10-24U-2 in ca x sec1 l 2 - 1.44 x 1.6 x 10-6 _ 2.305 x 10-6 = 2 19 x 1016 6.3 x 1.67 x 10-23 1.05 x 10o22 f = 1.48 x 108 cm x see-1 2 X = 1, 05 x 1027 = 1.35 x 1013 cm 63 x 1.67 x 10 x 1,48 x 10 2) E =25 e,-r 25 x 1.6 x 02 = 63 x 1i67 x 10-24r2 2 _ 50 x 1.6 x -102 1.X 24. 3x c. 63 x 1.67 x 10-4'1 N lo05 x 10 7 r3 X 10 ( 2 63 1.0567 x 10-24 x 16'-"......3 x O-11cm

APPENDIX IV Constant of the law of areas and angular momentum J = dO/dcLt 3 u I/r 3 r7d/& = l u~ --- c) Hence = rr dr r very large, dreases when in dt onofangdroment, i ct arameter. Hence, ) Recall t hat~r ~ i~s the-c....Omp.'.onent o...f momentum ~perp~end~icul~ar.t~o...the rad.ius r eer CerG dd-,r c..t'.'r r oo dIC) sine For r very large, r derreases when t increases, since =| — RZ | i.e. the searation of the moving atom from the stationary one aereasesi We have spee d at nftminity antd F ro cons-ervaFtion-of angular Eaom~entum, S rld @ = Ole > impact parametero. Ienee,...., Recall that tr E is the enent af entutm perpendicular to the radius vect~3t t since i41

APPENDIX V Sign of V/.r_.A 1k. (()19) d\/ = t:ati e- I i I ( + 2 d~l _ A 2.Se r | &u~~~~~~~~g isL~ 3 a aeAc

APPENDIX VI First integral of the equation of motion. Boundary condition at r = ro /2 X- _ __ -> +(, V _. 3 1L~t r

APPENDIX VII Obtention of equation of motion — ~I-~Z eU~. _1 _ Anr~'~iL..T-M.-~~ -l~Y5 I'_h ~r-.r..,U.[>. ).-.. l o 1 44 =

APPENDIX VIII General relations for elastic scattering _I_ C' _ Incident particle: mass M1,. velocity U2 1 in laboratory frame, before collision; Y 1 Lr, - c~ % in lab frame,, after collision - _J}1C in center of mass frame, after collision. -= velocity of center of mass in lab frame. Stationary particle: mass M2, velocity 2 in lab frame, after collision, e = angle of scatter of incident particle in the center of mass frame, in the lab frame. jC - reduced mass = M M,+M We have IMl+M24 t - = U Vj where

-146For elastic scatter, there is no change of speed (the magnitude of the velocity vector) in the center of mass frame, hence 6b}~M 2 -. M and t4'4 Cr-~l r~ I) LOW 5:<?8 _ M2 Sl -(jt V co e ( 1) _ being the angle of scatter @ith _-'~I' —' ~' —J 4 respect to"~l) of the stationary a -. particle, in the lab frame, we also have, M M;; 2 M<&I Z CI2 o S@ + Mt (see.GsZ L';" 1i~'M + M 2.) (see f. ex., Glasstone & Edlund, p_ 149) _ M~ ~ 2- kM1 IL'' M tj v7 = j t= _ M Ii t M 24 -- L tM M N A Mi~~ Mr 1M )r O ~~~~M If Mr) L~~~~~~

~~2 _Md lM~a)~ MQ 2t Q, 8 o. 2 2M +l,MM2M.COS(+Mtl tz., M,2 1 - cos 0 M,1+ M, M2 Cse+F l~ _____ LML, M CwaS e 4-sV' I Z Ml' 81 - oo Mz) uQu u+j Hence, sinecet and' are both ~, it followb Q:au y = aTr _E) (2) 1) M >... mass of moving particle greater than mass of statiary one then (p goes from 0 to CtrE < n when goes froem O to0; ~tta 94 ( =v5 = ML' \ <t; (-t4 (v) {3 =5 3; hence there is a ad.~ = ose(~ c l OS.+Ml) + M 0,SLAe M.+ MI M2 cos e d Ct2 Oo; 582l (M31258Cs f or.Cos G (2) LA ia M a 1de - 0 ha+h

d.8 W has a zero for Maximum of t9 T6 q 9,_..__.... _ Tr/ _.2) Mi< M, mass of moving particle smaller than mass of stationary one 30 the corresponding value of. For 0> O = O- -— _ ) ta ~ O ando (T=. Hence 7 goes from O toTT. f aries monotically, since ed -'9 St M t - has no zero. It increases with ho

-149iT use 2 3) 1 MI M22it= SVV - hence - /aZ (3) and., from (2) = -(4) The final velocities of two particles of __ -Qt~ —-.eqgual mass, in the lab system, are perpen.dicular, for an elastic collision. Return now to t the stationary particle after collision. - 2/ _ __._.,_,_ (M _+ M +2$......... S and hence

-150 The kinetic energy transferred to the stationary particle is T= S, M 1'7,gi The maximum of T is T7. = E. Hence r7 Ten 7 If M- M;T = E and %A= N. L

APPENDIX IX Disussion of plot of V r) r -2.) __lLl I (r +1e7 =r Il <lo ~ +-, ) e C o i.e. Vc) is represented by a stratight line o slope -/a in sedm-log paper (10gV r ) I fact., since r +/. a - U+.- 9 waries slowly with fo..r Y> a, a straight line is practically obtalned for r >/ la For rca.s V(rv)Nr, hence an byperbOlic branchfor small r Notice that V(.t)-J wVpl/) yarie sharply when r approaches ro/2, butt, up to -r= I the. approximation ( 0 would be a justifiable one. 45l5

APPENDIX X Maximum value of JZ5~'r- LF+t' a G i ~ Let /- -- - with 3l,._ t _) _ __ (-J9'" - ro _ _, - | d>Ti+ a~, _ 11 | j t E ~7 Fo 0 ForenCe ad _ wh ~,d, For 3Ct=I />3 d=3 +p-.~ _. t-a 3 1 _-a =, mog-,4

Since P/t&/-,c /c/dc -d/d s > oforall, it follows that A has its highest value for 2/ r = /XIt 4 ilw/Et2 2)PL. which tis. AB = $r/ -) = E +E sY% (7 |11i/ ) Hence the highest value ofJ3 is Since We have Hnc <i a2 r - tha T _ Hence is always smaller than_.. Xence $ is lways s~sr~lJ~r t~a ~ $

APENDIX XI Calculation of' angle of asymptoteO with the velocity t at r r at A Qrz~%/zir o=. w 8 _ S C-, (,:/ -o. f _ -........ t b / = 10 201 Me2 2Di, 2,30 x 10-2; % - 5556 _ - i) 1 = /lOJ1, = 2.98 x lo6) 2 C. / cT(9 X- 2. 30 x1 ( i'.5 hence ii ) C -- " / = 6,51/4L - o30 x 10- 1.605 = 1.270.' = 1; = sin- 1 1 2.556 sin-1 Z5 5+ -7, =a;, 1 1.270 2.S 0a = 2.556/2 = (1 6- 2705 - =299 x 1O" radoA

-155Although, here, (9 = 2.199 x 10-2 + 0.o21i8 = 0.240 rad, compared to Cp = 2.199 x 102 rad, it is probably not a very serious error to take instead of cq, since (.'remains a small angle. However, this treatment will slightly underestimate the interaction. 2) E lo-0-4 mv; 2e, = 1.20 i) = /loO =172x 10-3 2Ot/E +$ 7 1 20>; Q2 w E +1|= 1.05 J V2dlE-2 2, =_ 1. 095 x 2 = 0857 lo a2.556 sin-l 0O.857 1..030 rad, 1.030 - (1/1.72 x 103) 1.o95 x 2 x 1,7~ x 10-3/2.556 = 1,030 1- 90 = 0.73 rad. 2.556 = -.172 x 10,L 0.173 = 2,72 x 10-4 rad. 1.095 Since -=T1 (1 1.72 x Lo-3/1.095) 1 TV, the approximation = _(' is here fully justified.., ii) t = (T /4_2 ft, / ),'t-' 6=. 1.120o= o.4288 =: o.65 - =sinl 1 1 2556 sin-1 2 x 0.655.0.518 rad. o0655 2 2,556 8e _ 0...65.5 0.518 = - 0265 rad =7T (1 - 0.655 ) = 154 rd. ~.556/. Henee the errr ade by making the approximation ( is not negligible, but still not large

APPENDIX XII Calculation of ) )C'). o.. for various energies 1) E - 10-4 Mev E,/=E =25/102; ~(E (E)= 1/2; % = 2 x 0.5236 = 1.0472 rad; Ledal/ = 1.o472/3.1416 = 0.333; (1 - a/1T) = 0.667; (1 c~/rr )2 -.4 45 1- (1 -_y/TT )2 0 555; 2 4/E = 1.20; ()2 = - 1.20 x 0.445/0.555 = 0.965 A2 (t4d)t = 0.980 5.70 0= 7 3o384 o r (6~)u. = TT x 0.965 = 04 2 3.04 x lo016 cm2 (2)cx = 8,5 x 1022 x 3.04 x 10'16 - 2 58 x 107 cm-1 (Pd), = 1/0.258 A= 3.87 A = 22.5 C 1.51 rQ ( Nat2 /(Qd)/2 = 105/6 x 10-5 = 1/6 (=d)e (5.7/2.45) a = 2.33 a- 0.157 ro (~d). = 6 ( A), = 9.06 ro 2) E = 10-3 Mev 4E/.E= 25/103; T(E/-E) = 1.58 x 101; d ^ 2 x 0.1571 = 0.3142; t lTT = 0.1; 1 - cl /T =T 0.9; (1 -d )2 - 0.81; 1 (1 -Pl/1T)2 = 0. 19; )2 6.52 x 101 x 0.81 = 2e78A2; (j )u = 1.67 = 9.7 a= o.65 ro L1. 0.19 677_6 _ (6A)M = IT x 2.78 = 8.74 2 = 8.7,4 x 10-16 crm2 l 14. 1o'6 =,~'4 x 10 7o- C (' )L, = 1/0.744 = 1 342 = 7.8 = 0. 525 ro ()422/(b) 2 = 3.03 x 10-5/3.26 x 10-4 = 9.3 x 10-2 (Fgcl)e 5 05 x io1 x 0.653 %o = 0.199 C (\)~Q = (~d)> /9.3 x lo2 = 10,7 (Ad) = 5,62rc -156

-1573) E = 5x 10-3 Mev Ea /E = 25/5 x 1o3 = 50 x 10-4; f(CdlE)= 7.07 x Lo2 1'.2 x 0.0707 = 0.1414; (Qa/rT = 0o045; 1 -_XIlT = 0.955; (1 -d/T )2 = 0.910; 1 - (1 - dTT )2 = 009; (~),2 = 351 x 10o x 0.91/q x 10-2 = 3-14. 2; (vLd). = 1o77 1 = 0o,3 c = 0.691ro ( 6) - =7T x 3,14 = q 4 x 1o-'16 c; (o)~. = 8,5 x 1022 x',j x 10-'16 = 8.41 x 107 cm-1; ( 1/.81 =.=1i88 = 6A - 6 - 0. 465 ro ()l) 1.53 x 10-4/7.75 x 10-4 = 1/506; (r6) = 0o691 t- /2.245 = -0.308 Cc ( =d) - 5.06 (Xd = 2.35 rp 4) E = LO-2 Mev /E L= 25/104; fS(E4l= 5 x 10-2 d 2 x 5x 10-2 = lo 1 cdl / T = 0.0318; 1 -Ya/t = 0.9682; (1 -d/l t )2 - 00 938;.1 - (1 -Y/r)2 = 0.062; (9)2 1.92 x 10Q x 0.938 = 2.91x2; o.062 (k) I = 172 X = 10.= 0.675 r0 (6od) =TT x 2.91 = 9.15 A 2; (~)u. = 8.5 x ].22 x 9.15 x 10-16 = 7~78 x o17 cm-l; (Ld) = 1/0V 778 a= 1.282 - 7.45 CL = 0502 ro (Ld)22/() = = 2 28 x 10-4/9.60 x 10-4 = 0.237 (a). -= o0.488 x o.675 ro= 0,329 ro (Ad)Z: = (AS)Jo.237 = 2.11Yro

-1585) E = lo-' Mev Ed/E = 25/105; (EdI/E)= 1658 x o10-2; t 2 x 1.58 x 10-a = O0e316; Cdl/T =oo; 1 -(Cd/ TT= 0.99; (1rT )2 =o 098; 1i (1-'/1T)2 =.02 (~L)J2 — = 2.3 x 10.2 x 0.98/0.0 = 1..3.2 ().). = 1. 2 j = 6 18 0 =2 O. 415 ro (6Sj)L =1Tx 1-.13 = 356X2 (EO,). = 8.5 x 022 x.56 x -16 =. x cm-l ( d) /= 1/0.302?a = i. = 19.2 0 = 1.29 ro (d)g/(()c2 = 6x 10`4/1.15 x 103 = 0.522 (la ='0.724 x 0.4 o15 ro = 0,30rLo (Ad)L= (Ad/)~o.522 = 1.94 ( O)LC. 2.51 rB

APPENDIX XIII Mutual potential energy of two rigid charge distributions with screened potential. Consider two non-identical atoms separated by a distance R, one at point R1, the second at point R2, with respective potentials at r: [=r- ] X The charge density is) o X r_- _ * Q O - L -- j e J- e, ~f~ Oz ) 41T where vz 2 7 since there is spherical symmetry. For r = R, we can write 5 being the 3-dimensional Dirae functiono This gives, correctly, edIr=4 LV4 3r +-E ST- c3 r =r - S V + S + ~ = 4rd r a t -159

Hence +lr-R-, % " l r-, I_-fL1 + U The il.nteraction energy is ~LtP. s &SV'. Jca tI _____. ~,. 1) ~_~ — 4g:(R T'tr./ Ir- Pt ] From (1), we also have Hence.S-,- U2( R- v] L4rr a 2-r ct )

So that QR LI) a _ 2..\/ = -L 1z al - () Gl (o,,Cl (2) 4i a $( 4)_a>hh( 1 Nowmae e diff-eren; e; 2-C) then Fin4ally Relaig a aio J ON - L) a-) t a ~ - 0= 4- AL(C ) C ( caII 2 the differencea~2-, then StV -( Z2LA ~/Q:) -)0N aLC /t C K E) a. L, CL 6L ~;.L Q a-, a.l J, -a,, = a1 - ct) - __ a ); aQa_-C aI> C a Finally, =(Rt (l- ) - Replacing iby r givres Equation (5) f the text0o

APPENDIX XIV Energy transfer in the impulse approxcimation, for the interaction ener;gy used by Brinkman, FL Fr I I I_(,::F- %d. =~2 ME FzAc Pt~vY M mass, E kinetic energy of disturbing particle F~r= )w ); Fr = - ~ 0)( L ar ) a_ F1=FS =r F j Rrr_4j d(=rTrl) d4Jr Hence (V2E G e (J- +J, +J3 ~, _ ( e ~ r - (oo a ( (, ror' +;'Lr - K>3 - ~r 33 _..(,.. r ~-'~f-.~) /

163Denote by L a LaPlace transform~ Then 1-=-, LfICF)i s with transform variable l/c, where GCr) =o o0rc & = I/~r_-6)/ / C tZoo From Churchill, Appendix III, po 301, g Cs) = lv,, tht s) = L! F.(t) 0 F(t)- o o itttl =t=- lt)-/~ kc t Hence <0o modified Bessel function, 2nd kind, zeroth order. NOW, let l/L=oX. Then - 3C i e r; ~)lllr; r% =r2>-4 eCI-) = d (w24.\2 We have z cI;hence J5k- x o|3 t0 (3c d x, where the limit0 is taken because Jl = O t - =- oO From this, We also see that 4I. ---- -t) - cC -

For; YC=o0 3 =- O, hence J:3 r i 0 t(o (Qx)dix But, it can be shown that I Cn3a i OU A L3L) d CU c -S- C'-X) L) Hence 0 3 iQL viz ( a -) k< ( C 3C)c X Finally,, +0L Uwhere = g- (;E, which is Equaton (25) of the textL With Q4 C & Bohr radius for hydrogen, ie. ~z T energy T trasfeed to the statiRydbergnary atom isenergy O,=_.., t F(. — for hydrOgen. we can write ( ~- )- ( Ca _/)( Ct

-165Hence T=4ZI~ia~FI~) (I) Cc- E E The displacement cross section is with such that Tj given by (1)) equals E,.e. F-1 = inverse of F. Hence whih is tion () of the text which is Equatio (23) of the text.

APPENDIX XV Average number of atoms displaced per primary,, for primary energy in the range 1.5 Mev 2'03 x 104 ev, Xa =.1.5 x 106 i y6 x 104 = 2.3 x 104 902X 25 T1 ~ i04 _ ni~X dx" 0_. U Js ( L+X1+) the denominator being proportional to the total probability of displacing interaction0 6 x o1 + 1')=0 561L 92 9 xl+ 1=00561x4o175=2 o34 I - I 6 x 1'+ =166

APPE.NDIX XVI Number of defect pairs per primary in displacement spikes, 1 20 E -Ey /Ev-Ed Eeat, Em / EdEm + o-3 EE E [L (E /E])/(E,-E 7=201.5 x 1O -.3x 104.25 + 1.5 x 106 25 2.3 x 10 10-3' x 25 x 1.5 x 106 x LOg. 2,3 x 10425 l 5 x 510 - 25,1 =2 x 10-2 + 2.5 x 10-2 Ig 9.2 x 102 = o1-2 (2- + 17) = 1.9 X 1l-1

APPENDIX XVII AVerage energy transfer in a displacing collision, for the. interaction cross section adopted The.ross section adopted is ES - 7)T-TT) for transfer of energy in dT about T; rL goes froml Wto o, T from o to E (see Chapter IV of text)o It is clear that the frequency function 6 (E6T) is infinite for T = o, i.e. for no, energy transfered. Low energy transfers are strongly favored, Actually, we recall that our mathematical model does not allow L to go to infinity, The average energy transfered in a displacing collision is i Ed~ T d -,r./E)Z ETA 4. TE rrE E c (l -,~,. 7 _ It is perhaps more convenient to place the- differential element in the form CT)d.T I- __\ QJ ) I T ~_~_ ~ ~~-,~~,]~'

Values of f (T) T/t e 10o 6 10-4 10-2 4x10o2.16xlO-1 4, 9xl-1 8,bx1co 9.9x10,l 1 1 -2 sin-1 1 0.999 0.994 0.934 0.87 0.75 0.51 0.38.6 0 (1 —. sin-1; )2 1:0.998 0.99.8y7 0.76:0.56 0.26 0.14 3.6x10-93' L(1 - snjf) ~ 1& 10 4.4 2.04 1 1 o~64 2 q-' Mr f (T) co 5x108 1o6 6x102 76 10 2 1.4 0.64 2 2. tT

-170 For E = 1 Mev for T = 10-6 Mev, the integrand of the upper integral takes the value 1-6 x 5 x 108 = 5 x 102 for T = 10`4 Mev, 104 x 06 1 022 fer T = 102,Mev, 10-2 x 6 x 1&2 =6 for T-= 1.6 x 10-1 Mev, 1o6 x 10-l x 10 = 1.6 So that most of the contribution to the integrals is due t, the range T <EO Then we can approximate: 5 2 T d = oT TIE_ 4_ 7 dT | T/, ) EA cE T E 2z (.1/~ -\/-') E- 75',' ~.~-, It is seen that the energy transfer is small. which justifies the approximation made. In fact, the denominator is better approximated then the numnerator, so that T is somewhat underestimated for E ~> E. For EB' EB, the approximation is not valid. If employed, it certainly overestimates T seriously, since T = 25 ev requires E = 25 evo

APBMTIX XVIIl Fraction of primaries average number of displacements per primary, and fraction of atoms displaced in the irradiation of c-oper by 12 Mev deuterons ~ Differential cross sectin.for energy transfer f.m charged particles to lattic atoms: 6(, E)) E= TT t EE -E(1) 13- - E where E is the charged particle energy, E the energy transfered, Em the EtaximIm of E, b the distance of closest approach; b is given by ~= reduced mass = MA M mass of the charged particle.; A mass of the lattice atom; - atomic number of the: lattice atoms;atomtic number of the charged particle; intial velocity of the- charged particle, i.e. related to El by E /=, M'' Since elastic cllisions are assum:ed, E 4M E: Integrating (1) between Ed, (minimum value of E for an atomic displacement) and Em, one obtains the atom cross section for production -171.

=172' of primaries by charged particles, namely - ~ - t E Ej E vv\ 4 ked &[Lz 4t 4. I 4 _A E_' We have E-/Zc4 = 4,, P j B4 Bohr radius for hydrogen, Rh Rydberg energy for hydrogen. Hence we can write' M +-AL ) / E h- _' E' ER M and A can be taken as mass numbers, since they only enter through their ratio. Write with. - M - _ __R.M9,A, mass nimbers of particle and atom, respectively. Now, admit, after Seitz and Koehler(9), the following formula for the range of the charged particles in the sa;mple X (E') C E' ~ with'= 1,63 and C another constant. If N is the atomic density, N 66[2)is the probability, per unit path, for small paths, at X, that a bombarding particle will form a primary. If ( is the integrated charged particle flux (i.e. in particles x cm'2'), the number of primaries formed per unitvolume at X is; N 6j ),

-173S being an area perpendicular to the beam, the number of' primaries f ormed...., — _Id in the volume defined by S and a length equal to the- dop X of range is rX-X -NS X The correspnding number per atom in the volume is obtained by dividing byNSAY, so that -X - X-~ where X' is a range (hence dX' = -dy)o Using the formula for the range) i)( "X.A (s QI) S- (2) Let X be the thickness of the sample and assume it is smaller than the range X of the particles at energy E'. C X[ Q Zt I X + ( X ) ---- % 1 ~- c %v~~~Cr~~ kb ~

,, J 2l E' { ~ &i ) ( )( -, ( X ) 6 ( W )( z )( )( X ) =@ t r X A XlEXsI l AX\X tirch I X) L 6 Reoall that this is correct to the third power of (u), and for a thin target, i.e. X X; AX ais the thickness of th sample, X -the initial. ange of the b.ambarding charged particles. Note an error,'relative to the "order which (3) is valid. in Sei tz - and Koehler' s review article o (9)

-175Admit Snyder and Neufeld(3) result that, for. EEd, the number of atoms displaced per primary of energy -E before release is Q ) = -0.561(1 + 1) where XK= (E-Ed)/E The probability to get a primary with energy in dE about E in a tollision of the bombarding -particles is measured by 6(E/, 6) dL = - E. h E So that the average of' Q is = 0.561 I dX, /d I Ea' E EA where Qr r is the maximum of YC, i eo EdK Ea This yields. =.0o561 I+ Lot g (1 +XlC = 0.561 Log X3Cm Note also an error in Seitz and Koehlerts (9) review article, on this result The fraction of atoms.displaced is For the low temperature experiment (8), thin wires of 5.mils 0.13 mm diameter,t of Cu placed perpendicularly to the beam, were irradi-2 ated by 12 Mev deuterons. The flux was = 7 x 1016 deuts x cm72 at

-176maximum exposure. The range of 12 Mev deuterons in Cu is 0.2 mm, so that LX =0.13 mm X =0.2mm and approximation (3) is certainly very god. = 4T (. 5351 x 10-8)2 2 1(29)2(15.54)2 = o.68 x 10-9 63, 25 Mev x cm2 The bracket has for value 1 o.13. 2.63 0-13. 2 1 + 1 0 1 2.65 (0 15) = 1.25 2 x''63 x0.2 6 (1.63)2 0.2 T prim. = 7 x 1016 x.68 x 10-19 x 1025 = 4.9 x -4 12 Zll= 1 i4 x 63 x 2x 12 x 106 = 5 7 x 104 25 (63 + 2)2 cH = 6.16 Hence C= 3.04 x 10-3

APPENDIX XIX Calculation of the coefficient B of radiation anneal is given by 10o2 x 101 2o7 x 10o18- (31 -e-x 7 x 1016) or e 7 x 1016 =1- 49 x 1016 j, ~i~ For. =.0, the two sides equal 1. The right sideis zero for3 =! x 10-!6 12 x I0-17 For this value 4-7 \ QfAe-7 X 16x-6 =e-lo4 7 02470 Try 3 = 0.5 x 1017; - -7 x 1O6J - e -~.35 0.7 1 - 4,9 x 10163 = 1- 49 x o. x 0.5 -l = 100755 Hence we can accept'' 0.5 x 1017 -177

APPENDIX XX Upper energy at which classical approach remains valid in light metals The condition of validity for classical approach is where bo is the distance of elosest approach. Hence, since, at high energy, the potential must be practically Coulombian at closest approach, The condition is k/2fA x 1.67 x 10-24 E) C< t2E'/E A.mass number, Z atomic number of metal, E energy of the knock-on (in ergs). The maximum value of E is 4-Pt EM Lm ( ) X M mass number of bombarding particle, E' its energy (ergs), assuming elastic collision, Hence, we must have or 43. -3LEA~T~~ I~ —-Z~Y~~ FvAA M E 2-~) oE < (M+tA) 354 x 1o-24 ( 4MrR For 4 Be 9 and deuterons (M = 2) E'C (li)2 34 x lO24 0 2 8 l.l x 105 46 x 1032 x 13.7 x 10-36 ergs = 6 x 12 ergs = 6.3 x 102 = 3.94 x 104 Mev 1,6 x 10 -178

-179Hren;e. classical treatment is valid. for -deuteron irradiation be Be,. even for deuteron energy of the order of, say, 20 Mevm It is clear that it is also valid for pile neutron irradiation of Be.

APFENDIX XXI Variation of the fraction f primary knock-ons with Z and A in charged particle irradiation. Return to Appendix XTIII, Equation (i) We see that, for a given irradiation i.e. the same and E', A M and Zo mass and atomic numbers of bombarding particle, A and Z,mass and atomic numbers of the sample irradiated.. Hence, for the same bombarding particles and. different samples, f-t which is the proportionality claimed in the text. This result was first obtained by Seitz. (30)

APPENDIX XXII Solution of Snyder and Neufeld primary equation and associated problems. Snyder and Neufeld primary equation is.C Cx F:)CI JC~ 4X,- I 0 0 By differentiation one obtains the e-quivalent differ'ntial equation with bo -ndary condition.: XIC9'CX,) = 3 CX,-I ) (2) where pri-me means derivativeo Take l. 1 42, hence O'X- 1 < 1, then, in this range,. and 3CC )= LogX1 + C 2 CI()0= C = 1 O(y,) = 1 + LogK* If the boundary condition (3) is disregarded, (2) admits a s olution which must be the asymptotic solution f~r ZC>) L, since, then, (2) reduces to O'LCx,)4Ctc,) = -/X, which has the solution 0 Q3C) = C C \) ) -181l

-182 - We can app ioximately determine the constant a in the following way 3 (2) = 1 + Log 2 = 3a, hence a = 0.564 A correct numerical treatment(3) yields a - 05610 The curve of s6XI) obtained numerically is reproduced in the diagram next page. Note that (2) can be treated by LaPlace transform Using the rel ations L t tY it )c ~ i:h ah)W ( S)c between the function'(t) and its transform'(s), callingL(() the transform of ) (Yl), (2) transforms into - cCs) - s3YCS) = e Jf LCs) The exponential term is obtained using the fact that From this _ S ~ c + i S S and, by inverse transformation, 3U, C~c~) SCE S The expansion of the exponential integral around the pole yields a term in e = 0.561, where T is Euler constanto

-183. 3 2/ I-!,, f/ THE FUNCTION N/(XI) 5 | // SOLUTION OF SNYDER a& NEUFELD EQUATION (I) l/ 1 V=- NO. OF ATOMS DISPLACED PER PRIMARY. XI: E ~~~~~/ ~Ed // E:PRIMARY ENERGY. Ed= ENERGY TO DISPLACE PERMANENTLY AN ATOM. 0 I 2 3 4 5 6 XI, a -

Equation (1) assumes that any atom receiving 25 ev is released from its site and that the incoming kxock-on always emerges from the collision. Now, assume that,, if the incoming knock.-on has energy smaller than 25 ev after collision, it replaces the secondary dislodgedo The integral Equation (1) becomes Xt )C_ I Y-I 4The differential Equation (2) remains unchanged, but the boundary condition (3) must be replaced by Cnas) XI) I, o42 An asymptotic solution l) = aC + X) also exists. For 1 j, sXXrI) = 1, s that C(y)= C + LOgX1,; (2) = C + Log 2 = 1; ) (3) = 1 - Log 2 + Log 3 = 1 + Log 3 4 a; this yields a = 0351. Finally, assume that, not only the primary replaces the secondary if 06)lj4, but also that the secondary recombines with its vacancy if OL C The integral Equation (1) becomes 3-c~ I I (2.) remains unchanged, but its boundary condition (3) becomes For 4X xL 4 3 s,) - C + Lo c, ) (4)=1- Lo +Lo 4- 5a. This yielis a.= 0258,

It is teen that these two models, assuming replacement, or replacement and recombination, would reduce the overestimate inherent to the Snyder and Neufeld method. But they remain artificial ways and means to bring calculations closer to the. desired result.

APiPENDIX XXIII Direct obtention of the priPmary integral equation for the -model used in the collision problem., The differential cross sections used are CI;?I'i E (EE')E _ -- E_....'. E ) Put into (38) of the text, assume a soluti.on (E) = A + BE and integrate (38) by parts, to obtain r 5E) ) =Q- E -d) ~ ~) 7 Ed E) _I E.- I. S Z] Ed l-(I- wtr S-t>E _e;o) t4E)+ ON 7 )C )2ETs. -E-_ ~ 1-(1- -f S -' _,SCE) CE) =CLo)d CE) (EC) S CE) ET TiE 1 Finally since CO) - I) E wh ic) /s E l5 i o h which is Equation (43)'o f tA'e text~

APPENDIX XXIV THE DESIGN OF A CRYOSTAT FOR PILE IRRADIATION Note: This Appendix has been arranged as a separate. complete part of the. dissertation, with its own numbering and bibliography. -187

INTRODUCTION SECTION I 1-Basic idea The application of cryogenics to particle irradiation is of general interest. Low temperature irradiation allows for eliminating the effects of thermal motion, diffusion, for example. Because of this importance of cryogenics, in particular for pile neutron irradiation, it is considered that the design and construction of a multi-purpose low temperature cryostat for the Ford reactor is a worthwhile undertaking. In the design of such a cryostat, to be used at liquid helium temperature —i.e. the minimum temperature practically attainable in the application proposed —one is severely limited in the mass of the experit — mental set up. A large mass of the components liable of supplying heat to liquid helium by conduction or radiation means a high rate of helium evaporation from gamma-ray heating in the reactor. Within this limitation, the low temperature chamber and the tube leading to it must be of such dimensions that specimens of substantial size or chemical test tubeis can be inserted in the chamber, so that the installation can serve many purposes. Mass is reduced by the adoption of a practically all aluminum construction. Diagnan 1 shows the general dimensions and the arrangement of the cryostat. The low temperature chamber and helium in and out transfer tubes are placed in a vacuum jacket, reducing heat intakes other than from absorption of nuclear radiations to radiative and free molecule transfers between jacket and liquid helium chamber or helium transfer tubes. In -188

D itagYr.Al I CRYOSTAT Generai arranyern eivt -and dim en v Sons 63a H —- - 29 __~~~~~~~~~~~~~~~~~1 ___. t —----— * r —---. 3k —-. —._____ @3 /~~~~~~~~"R TYPICAL Q a — 2 F~~~~~~~~~~~~~~~~~~~~~~ i-~~~~~~~~~~~~~~~~~~~~~~~~~~~~~i o I'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~o H-Cnta Tue1-Ntoe otTb HC0-Hl Cbabe 0-Guide li-Roti -m L~ A-Sup~port ("'Mock. up Fuel Elemecnt") I -Sa~nler. B-J~acket Envelope, Lover Part J-Flectrical Leads C-Jacket En~velope, "Long Tube" K41~eliumP in Tube D-Jacket Envelope, Ronritotal Tube L-Nlitrogenn in Tdbe I-Central Tube X41tragen out TubeF-Witrogen Cham~ber Nl-Thernal Shields O-Reliutm Chamber O-Guides 11-Rod P-KOVara

-190order to save on helium,which is an expensive commodity, radiative losses are minimized by interposition of thermal shields between-liquid helium chamber and jacket and, also, between -helium tubes and jacket, the -latter at positions which do not make construction too difficult. The shield around the liquid helium chamber is attached to a liquid nitrogen chamber, fed by in and out.nitrogen transfer tubes. The shields around the tubes are welded to the nitrogen out tube. They shield the nitrogen in tube, as well as the-helium in tube. Liquid helium and nitrogen are transCered to their respective chambers by applying a light pressure of helium gas to their containers. Continuous transfer is planned and no automatic devices are provided. The liquid level in the helium chamber will be monitored by continuous measurement of a resistance and that in the nitrogen chamber by a thermocouple in.contact with the nitrogen out tube. It will.be noted that this arrangement is predicated upon a continuous, constant power level operation of the reactor. This is not an additional restriction placed on the operation of the reactor, since such conditions are requisite to the knowledge.of the irradiation.neutron integrated flux in the range of energy of interest. 2- Application to the study of atomic displacement by fast neutron bombardment. The first use that is intended for the cryostat is a low tet — perature neutron irradiation of thin metallic wires, in order to study atomic displacements induced by such irradiation. It is known,(l) that, during cyclotron charged particle bombardment of metals, at liquid helium temperature, the variations of the increase in.electrical resistivity induced by irradiation versus

integrated flux are represented, not by a straight line, but by a curve bending down. Since the increase in resistivity is found to be stable at helium temperature when irradiation, is ceased, the recombination of defects evidenced by the curving down of' the irradiation curve cannot be ascribed to thermal annealing. Hence the process is called "radiation anneal". Considerations3 developed in the text, relative to the number of atoms displaced. by secondary collisions of moving atoms, per atom displaced by a bombarding particle (primary knock-on), and to the moving atom mean free path between displacing collisions, show that radlation anneal can be explained, for charged particle bombardment;, in terms of interaction of defects of various generationso They show that no such explanation seem valid for neutron bombardment, where, for the fast fluxes and irradiation times practicable in experiments, the total number of primary knock ons is much smaller than th.e corresponding number for charged particle irradliatlion with Integrated fluxes remaining reasonable. A low temperature, in-pi.le measurement experiment performed at Oak Ridge, to studyg not radiation Zanneal,, but thermal anneal after irradiation., shows as a by-result, that no radiation anneal takes place, and, thus checks the theoretical. deduction recalled above. The experiment pLanned will. serve as a further check. Also, since samples of several metals (four different metals, as a first step) wil.l be irradiated in the same fl.ux conditions, indications on the dependence of tthe change in el.ectrical resistivity —i.e. of the fraction of aitoms displaced —on atomic number or mass number will be obtained. In addition tpo sheddi.ng light on these two important points of the theory of atomic displacements, the experiment wil.l provide valuable training in cryoginic tecaicju.es.

KL92Since heat intakes have mainly for source the absorption of prompt fission gammas and uranium capture gammas, it is important to place the experiment in a regi on of comparatively small thermal neutron flux. On the other hand, a maximum damage is inflicted to the sample for a displacing neutron flux (energy greater than 400 Mev for copper) as high as possible. Irradiation out of a beam hole, or within the thermal columns using a converter plate, leads to a simple design of cryostat, but only allowsvery small displacing fluxes, hence exigl long exposures. In addition, such set ups demand an extensive shielding, in order to prevent the escape of fast neutrons into the experimental space around the reactor. A high fast flux is best attained in the core of the reactor and the maximum ratio fast flux-thermal flux is obtained right in the center of the core. This position of the irradiation low temperature chamber has been selected. It will be noted that such a positioning of the experimental set up can only be obtained with a swimming pool reactor, hence the Ford reactor is ideally suited for the kind of experiment planned. In order to maintain a reasonable rate of consumption of helium, a reactor power level of 0.1 MW has been selected. It will be shown later that meaningful results should then be obtained for 50 hr exposure. 3-Mass of cryostat-Consumption of liquid nitrogen and helium The total mass of the cryostat is 7.35 kg, not including the "fock-up fuel element" to which the cryostat is attached at its lowest part, most of which is due to the outer vacuum Jacket. Only a small part of this tbtal mass is placed in a region of appreciable gamma fluKx It

-1935= is estimated that about 38 gr absorb gammas with resulting heat flow to. liquid helium and about 200 gr absorb gammas with a resulting-heat flow to liquid nitrogen. A mass of 16 gr liquid helium.and 30 gr liquid. nitrogen, about, should be present at all times in the respective chambers. Liquefied gases expenditure is estimated as follows: (see Section III, paragraph 5) TABLE I Use Liquid He, Liquid N2, ~~~~~~~~~~~~~~~~~~ liters liters J....., j.' i,.,_1 Transfer tests, during construction. 25.25 Pre -experiment cooling. 15 2 Experiment. 280 26 Totals 320 53 With a figure, of about $3.00 per liter of liquid helium, the cost of preparing and running the first experiment amounts, for this commodity, to 320 x 3 =.$960. This includes investment costs consisting in developing equipment and techniques, which will not be incurred for further experiments o 4-Construction schedule The design was completed on -November 22, 1957. Aluminum sheets and tubing were ordered at the same date and received a week later. Construction began in the first days of December (lower part of vacuum jacket). Welding of comparatively thin aluminum tubes commenced on February 21, 1958, when an appropriate Heliarc "torch" was received.

-194An aluminum transfer tube of 20 feet length, consisting of an inner tube of 1/16 in. inside diameter, i/8 in. outside diameter (Alcoa "Utilitube") and of a peripheric outer vacuum jacket of 1/2 in. outside diameter, 0.028 in. thickness (Alcoa 3003 H 14) was completed at the end of March. This transfer tube is represented in Figure 1. It proved possible to Heliarc weld the 3003 H 14 tube, but attempts to weld the Utilitube failed and threaded compression fittings were used for the passage of the inner tube at the ends of the vacuum jacket. The transfer tube has been used to check the possibility of long transfers with tubes of small diameter and to attempt to measure liquefied gas evaporation during such a transfer. From the beginning of April to mid-May, a vacuum line was built to evacuate the vacuum jacket of the test transfer tube. The system was tested for leaks, which resulted in the replacement of the oil diffusion pump. A thermocouple vacuum gauge (RCA 1946) was installed and calibrated. In April the shop welded the flanges on the sections of the long tube which constitutes the main part of the vacuum jacket of the cryostat After the transfer tube proved to be of too small a diameter (inner tube) to transfer nitrogen, two transfer tubes with larger diameter were ordered from the shop, on May 20. It is worth noting that a low temperature cryostat, manufactured at Hanford, has required 1 1/2 years to be completed. 5-Demand on the reactor schedule It is re-emphasized that 50 hours at constant power level operation —0l1 1W —will be demanded by the experiment.

JACKET r O.D. TO VAC. PUMPS LQUIDS GAS TUBE I I.D 0/ -1 D.| FROM UQUIFIED. GAS CONTAINER I TO DEWAR RECEIVER. TOTAL LENGTH OF JACKET FROM END TO END 20 feet. Figure l.. Test Transfer Tube.

FEASIBILITY OF IN PILE:MEASUREMENT OF CHANGE IN ELECTRICAL RESISTIVITY DUE TO NEUTRON IRRADIATION SECTION II 1-Fast flqx available-Change in electrical resistivity expected Even with the experimental device in place, we can expect to obtain in the samples, at a reactor power of 0.1 NW, a fast flux ET 12neut cm2 - of the order of 10 neut cm sec H. ere, E is the cut off energy for the thermal region. This estimate is made from information contained in the document "Research Reactors", (3) which. gives data obtained of the Bulk Shielding Reactor. It is probably sufficient to admit, for the purpose of estimate, a fast flux of the form XCE)4E = kE-{dE/EL Assume, to be on the safe side, that E, is taken equal to 1/40 ev, and take a neutron cut off energy of 2 Mevo Then, K is determined by, K SoE/E =\) Hence K- 5,5 x 10~ From Equation (32) of'the text, we have, Cal ts/ E s)t/f\5.5 s*OtOX us o6 where C is the fraction of atoms displaced after an exposure t sec. For copper, 6$ = 3b, A = 64, E = 25 ev. -196

-197Since we know that the theory of displacements overestimates the fraction displaced by a factor of about 5, we shall take for estimate of:C, after 50 hr irradiation, C'' _50_3__xlo_ _3 0_ x xl 2 O10 = 7 4S xIo- =7.4 x I10 AI/oO Taking Jongenburger value(4) of e = 2.7 Slcm for C = 1% in copper, the estimate for the increase in electrical resistivity of copper after 50 hr irradiation is Ae= 2.s7 %7.45 txlO 42 l 2-Feasibility of measurement Take, again, the typical example of copper. First calculate the electrical resistivity at the temperature of the experiment. 0 denoting the Debye temperature, T the temperature considered, lT the thermal component of the resistivity at temperature T, pf2 the thermal component of the resistivity at temperature0, 9T / varies practically linearly with T/0( for T/,>O 0.3 and. is expected to vary like T5 for T<~)(Gruneisen function, see Figure 2). For copper, 0 = 3330K and, from the curve PT/46 versus given by Kittel,(4) it is found that i 9r 1.9JS cm, and that R/ / ~ i 5 x 10-3 for r/D = 0.1. The coefficient K of PT =:KT5 is determined by 5 x 10-3 x 1.9 = K x (33.3)5, i.e. K = 2.24 x 10-1O. The temperature of the helium chamber will be maintained at about 5~K while liquid helium will fill the chamber. At this temperature, the thermal component of p should be about, 5~K = 2.24 x 10-10 x 55 7 x 10-7& 4,

T.,. 0.3 T Figure 2, Gruneisen Function,

d199Should the supply of liquid helium fail to maintain the level and the temperature of the samples rise to 15~K, the thermal component of the resistivity of the samples would become 150K = 2.24 x 10-10 x (15)5 C' 1.72 x 10-4glQ m\\ It is quite feasible to obtain commercially metallic samples whose residual resistivity —i.e. non thermal component, due to impurities and crystalline imperfections —is 5 x 10-4 times the total resistivity at O~C. For copper this residual resistivity would then be about 1.69 x 5 x 1'-4 C 8.5 x 10-4 jSe LCV Hence the total resistivity of the copper samples may be taken as Pot 5~K' 7 x 10-7 + 8.5 x 10-4 - 8.5 x 10-4 S~l at 5~K2 and eT0t) 150K - 1.72 x 10-4 + 8.5 x 10-4 1.02 x lo-3JQ.xtYat 150K. At one tenth of full irradiation, Ae would be about 2 x 10'4 a As we have seen before, F should be of the order 8.5 x 10-4 at 50 K, so that there is no difficulty in obtaining a meaningful measurement of Ae provided the temperature is maintained reasonably constant. For example, if the temperature in the helium chamber should jump from 50K to 15~K between two measurements, the thermal component.T would pass from 7 x 10\-7j. cm to 1.72 x 10-4 J2 cm and the two measurements could not be compared (i.e. plotting their representative points on the same curve would be meaningless). However, a change of 2~K, about 50K between two measurements distant of 1/20 of the total irradiation time would be permissible, since the At corresponding to that length of irradiation would be of order 10-, while the change in ~T due to the temperature

,200 - change would -be of order 7 x 10-7[(7/5)5 -61" 3.08 x 10 6, i.e. only 1/30 of the increase in lp. If the level of liquid helium is maintained above the samples.at all times, the temperature of the samples can only change if the pressure changes, At the pressure envisaged (slightly above atmospheric pressure), the temperature of saturated liquid helium is fairly insensitive to changes in pressure, hence constancy of temperature is not considered a difficulty, provided the samples are alway immersed in liquid helium.

DESIGN OF THE CRYOSTAT SECTION III 1 -General considerations The design of the cryostat has proceeded along.the following lines. The general feature of the cryostat were first selected, starting from the basic ideas expressed in Section I, Paragraph l. Diagram 1 shows the layout. The material selected was aluminum, at least for the parts in or near the core, for reason of small mass, hence comparatively low gamma absorption and also because, with this metal, there is no activation problem. The criterion used to select the dimensions was that the mass of the portion.of the cryostat placed in high gamma flux should be as small as possible, in order to limit gamma heating and, therefore, liquefied gas consumption. The dimensions selected must meet the requirements of mechanical resistance, unrestricted flow of liquids and gases, and usability of the cryostat as a multi-purpose low temperature reactor irradiation device. The selection was therefore made by trial and error calculations. Diagram 1 shows the dimensions arrived at. 2-Calculation of gamma-heating The cryostat will be placed in the core, in such a way that the center of the helium chamber will be approximately at the center of gravity of the.core. To simplify the calculation, since the geometry of the cryostat is quite complicated, it was decided to consider as concentrated at the center of gravity of the core the whole gamma-absorbing portion of the mass of the device. This is equlivalent to assuming a space independent gamma flux around the experimental device. This is probably not a -201'

-202bad assumption within the core. Since the gamma flux should fall rather fast outside the core with distance from the core face, the gamma absorbing portion of the cryostat is assumed to extend only about the one foot above the top.of the core. The portions of the helium and nitrogen systems so considered are shown schematically in Figure 3. Since the experiment is placed in the center of the core, for gammas originated in the reactor itself we shall consider only gammas born in the core, i.e. we assume, which is reasonable, that gammas born in the reflector will be absorbed by the core and will never reach the experiment. The cross sections for capture and fission are much greater for thermal neurtrons than for fast neutrons, hence only the thermal neutron flux will be considered. Core gammas are: -thermal fission prompt gammas and uranium capture gammas. -fission product decay gammas, -aluminum capture and decay gammas, -water capture gammas. The first two sources can be lumped together, as, for example, in the work of Clairborne et al. (6) Call Ni(E) dE the number of photons with energy in dE about E produced, per interaction, by kind i of the processes mentioned above, and Zi the macroscopic thermal cross section for that process~ Call S(E)dE the number of photons with energy in dE about E produced per second, per W, by all (total) the processes above. Call 4 the thermal flux per NW, averaged over the core. Then the source corresponding to a volume dV of the core is, S(E) dE = dVZE Z Ni(E)dE (1)

-203- co_ co I~n~ I I I I I ~ He SYSTEM CENTER UNE..... OF CORE MOCK UP FUEL ELEMENT N2 CH. N2 SYSTEM THERMAL SHIELD Figure 3. Gamma Absorbing Portion of N2 and He Syatems.

-204. Consider (Figure 4) the experiment concentrated at 0, center of the core. Consider a volume dV of the core, concentrated around a point P, as a punctual source. Call r the distance OP, J c (E) the linear absorption coefficient of the core, and e (E) that of the experimental set up (homogeneized) for gammas of energy E. Call fe the density of the homogeneized experimental set up. Neglect build up in the experimental set up, since it is mainly made of thin aluminum walls. In aluminum, for 1 Mev gammas/,J'/P ~ 0.09g cm2 x gr'l (5) hence ra. 0o g x 2 7 0.24 cml. The thickness is smaller than 0.2 cm, hence XJ < 0.24 x 0.2 =..048 and the build up factor B(1 Mev, x) is about unity (B:= 2.0. for J'{= 1, E= 1 Mev, and B =1.53 forjX' = 1, E = 4 Mev.(5)) Call B(E) the build up factor in the core, admitting that it varies little with r. If the core is assimilated to an homogeneous mixture of water and aluminum, the build up factor is taken equal to B CEDE r - ue Ae aft VL u)n A (E~T)+ \/tHi0/Vo{G4e)SX lf) (2) ) where.E is the energy of the gammas considered and r the distance from the gamma source considered. Uranium will be neglected. In the Ford reactor, the following approximate ratios Cbtain, Voe he/Voe 0 e. cz vL a VO. Ha O /\40 4woe = O.5. (3) Hence, B leE) r a42 B CEr) +0O5s8 -LOEl. (4) From simple arguments (in particular, isotropy of emission of gammas is assumed)y it can be shown(6) that the energy spent in the homogeneized set up, per unit mass of its material, per MI of reactor power, by gammas w with.energy in dE about E and originated in a volume element

-205 - w. CORE EXPERIMENT Figure 4. Volume Element of Core Source of Gammarays.

-206dV at distance r from the set up (concentrated)is, in Watts, t (E)dE = la 6 x\O-l" Sh4(6E) BrtEl e P.r E S CoE (5) ) where S(E) is given by Equation (1), and the other symbols have been defined above. We write LCE)E L=- F(E, r) E' C) S dE, (6) with FCI~)=l~~~ld I3S(eLE)b cE) SE iCEr (7) Now inthe work of Clairborne et al.(6) the energy is divided into seven groups, setting an arbitrary average energy in each group, and defining an equivalent number of photons for each group and each process, such that the average energy E, in the n-th group, limited by energies and E, (n, 1, 2,...7) and the equivalent number of photons Neq,,; for the same group and for process i are related by N, 4.. E N ELE E. (8) From Equation (6) and Equation (1) we can. obtain the total energy expended in the experiment by gammas with energy in the n-th group, produced in.dV by all processes, namely, FCF)r).SE)dE, (9) th i F lE\r(Er)E+ EPrl~ Tz N;(EjdE a (10) -R-,=$dVhvjy CEar)E NtE)d E. (11)

-207If we can consider F(E,r) independent of E, i.e. equal to a function Fn(r) in each group n, then, R,= Qdv F.r)ET( j EZ EN,*(E)cE, (12) a-, = 4 dV Fh (()r El Z (> Nee l) * (13) We therefore look into the possibility of considering F(Er) constant in each group. Table II gives, in the six columns at the left, E Vcd N TABLE II n Energy F,. Ne, LI E group Mev i =1 i =2 = 3 Mev -tiO Ae Fission Al Water (prompt, capture capture 1U cap- and ture, decay fission product deca*,::) 1 O.o. 00-. 7 5 5.680 0.000 0.000 O. 1 0.171 o0. 169 o.6 o.89 0.078 2 0.75- 1.50 0 4.oo000 0- 425 0.000 1.0 0.071 0.061 3 1.50- 3.00 2.0 2.345 1.113 1. 115 3 0 0.049:0. 043 4 3.00- 5.00 4.0.441 o. 727. 000 4.0 0 -.034 0.031 5 5. 0-700 6.0 0.054 O 202 0. 000 6.o 0. 028 0.027 7.00- 09.0o 8.0o 007 0.344 0.000oo 8.0 0.024 0. 024 7. 9.00-11.00 10.0 ( 0.0009.000 0.000 10.0lo 0.022. ~.0.023:* -Note - The.decay term is variable -with.time. of operation The value adopted is.an average for the-decay- at 85% saturation. Strictly, the rate of consumption of helium should vary somewhat as irradiation progresses.

-208from Clairborne et al.(6)andj/Pe for water and aluminum from the Reactor Handbook(5) in the two columns at the right. It is seen that thejt S of water and aluminum may be taken as constant in any energy group except the first one ( n equal to one). The error made by taking the t'S constant in group 1 may be appreciable, since N eq 1 A is important for i = 1. However, to be on the safe side, i.e. overestimating rather than underestimating,.3e, hence ~, we shall take a high value forage in this group. We denote by subscript n the constant values adopted. The values selected appear in Table III below. TABLE III Group, n) (3j1) Hdo.AC ":"~ ~ [', _,;',., i'.. -. -.; I.. 1 0.150 0.150 2 0.065 0,055 3 0.055 0,045 4 0.034 00031 Since, for'E > 5 Mev, A_ has changed in the ratio 0.034/0.17 =0.2 for both aluminum and water,.and since Nq OSI /Ne, 0.05/5 = 0.01(6) for the most'important process of gamma production, we shall consider only n = 1,2, 2, 3, 4 "B(E), given by Equation (4), is very sensitive to changes in EB at energies' smaller than -2Mev. We shall adopt rather strong values of B(E) in the various intervals. Call Ba afe A.) B the build up'.factors*, considered constant, for the n-th group, in-water, aluminum, and the core. The values of s E for the different groups is calculated.in fthe following way. First an appropriate value for the linear gamma

-209absorption coefficient of the core m l.is calculated for each group n. The core -is assimilated to an.homogeneous.mixture of water and aluminum, and YJc is calculated by'c> e ) I roL P Hlo~, ( A+ where the (i) /S are for the pure component in the mixture. Uranium is neglected, For aluminum, F = 2.7. HSILOI =o 0 S j -,, X VA6 _ I 15 Hence, Y(Ch= 148 (:L0 ) 0 Table IV below gives the values of JCLr obtained. TABLE IV Group Tt' 1 0, 150 01087 0.150 O. 169 0.256 2 0o065 0o038 0 055 0.062 0.100 3.0C55 0.032 0.045 0.051 0. 083 4 0.034. 020 0.031 0.*035 0.055 Now the B's are calculated Group 1 E = 0.5 MeV; cl -= 0.256 For r = 10 cm, J4cl = 2.56, e 0.08 *For = 5 cm, )ch r = 1,28,' 0.27

-210We certainly can neglect all volume elements ar distance greater than 10cm. Hence, the maximum values of the B's are, Water- JC= 0.150, r= 1.5, BH20(0.5, 1o5) ='3.5 Aluminum- ('4 = 0.405, J.X = 4.05, BA1 (0.5, 4) 9~5. B = 6000. Group 2 E l )= 1 Mev; 3J,' _= 0.100 For r = 10 cm, JcJhr l J = 0.332 For r = 15 cm, 3tcnr = 15, e = 0.223 t We take values at 15 cm Water- hL = 0. 065, ICr= 0.975, BH2(l( Oo98) = 2.13, Aluminum- = 0.148, L r = 2.220, BAL (1, 2.22) 3.50 B = 2.70 Group 3 ~E3 = 2 Mev; hL c = 0.083 We take r maxi equal to 15 cm Water- t = 0 055. r- = o.825, BH2O (2, 0.825) = 1.60 Aluminum- & = 0.121, (Ar =.810, B 1.81) B 2.40 B 1.93 Group 4 E~,4= 4 Mev; LM = 0.055 Again, we take r maxi. equal to 15 cm. Water- = 0.034, O4r = o.510. BH0 ( 4, 0.5) 5 1.20 ~ Aluminum- = 0.084, Yt = 1.260, BA (4, 1.26)' 1.70 B = 1.415 With these adopted values, we can write Equation (7), for E h- E 4 E) _+ _ rT 1P r-lc

-211: The total energy expended in.the target (experimental set up) by all gammas.originated in dV (i.e. with all energies), per unit mass of target, per lW., is obtained by summing Equation (13) over n and using,Equation (14), thus, ~,~B = c — d V s 2 ~ ~.[}(ea N~ P (15) 47 ee I:L,!,,C - - Call M.the total mass zof the target, P the reactor operating power in in.MN, then.the.total energy expended by core gammas in the. target is, Y )<t~:fh~t,\DE6e4>,)8tV4 (16) in Watts. Now, for the sake of sinmplicity, we assimilzte the core to a sphere -whose c.eter is. the point at which the experiment is concentrated and of radius 15 cm(Figure 5), This does not introduce a large error, compared to the actuFl geometry, since for group 1, the largest contributtor, and for r 10= lOcm, eCw Sc 077. N.ow, Hence Equation (16) becomes, iA^X1[i;;( P\ ) h (h (17) H Me~ x\~6xld-IS 4e) RhE (1-'1 ZJ~13~ ~ ~ ~~.9-'',/~ av - 4,,.,,::_ [7

-212o EXPERIMENT Figure 5. Reduction to a S3pherical. (ore.

.213 Since the only parameter of Equation (17) depending.on the naturt of the target is I / le,,f and since it is known that, at a given energy, J~/9 is ~about the same for all elements, we can use for M the total mass which absorbs gammas, whatever the nature.of the materials contributing to it and replace e,K /~ by (t/~)Ae,o Equation (17) can be written, b-=89+Kl.GxO- AA Watts, (18) with, A_ Ah, (19) and,} A,,,= I>e) 8 EO, kr I- Z) N,,; (20) The macroscopic cross setis to consider are, 1 Thermal fission, U 235 2 Thermal capture, Al 3.Thermal capture, Water If'-T is the most probable speed of neutrons at temperature To, assuming a Maxwell Boltzman energy distribution and that the A, /S follow a B/v law, calling';(ro) ) the cross section for process i at speed trTo T we hAve at temperature -[T LT- 35 e2.r Troo BNL 325 gives Q (jT 0) for To 2Q 00, i.e. for 2T = 200'YA Sc'-P

-214- If we admit that Tr. is the moderator temperature and is equal to 50QC, then, Gir J3 T) |; NL = 0.845 6o:Y GA NL From BNL 325 and its supplement$ the following values are found, at 2200 m see, 6~{v,5 =581 ss; ~.a -lm~ j6, al 3 ja A schematic description of the core of the Ford reactor is given in Figure 6. The volume percentages in the core are: vol. water 57.7%; volo aluminum 42.3%. The active length of the elements (and of the core) is 60 cm. Each element contains 140 gr of U 235. Z, is calculated from the formula above where N; = A No''rY mass of constituent i, A mass number of constituent i, V total volume (core or cell), No Avogadro's number. The calculation of the zi's is made in the Addendum. The results are, 1 = 4~75 x 10-2 cm-l1 = = 4~83 x 10-3 cm-1 = 1.09 x 10-2 cm-l The value of the quanity A, defined by Equations (19) and (20), is calculated in Table V. It is found that A = 0.922. The mass M to consider in Equation (18) is obtained as follows. Two "systems" are considered. A "helium system", in which gamma absorption results in helium evaporation, and a "nitrogen system", in which gamrma absorption results in nitrogen evaporation.

-215 - LO A,B,C,D — ti t -- FUEL ELEMENTS.. I,,1 89" U+AI+ HIO.L; _ 3.189" Figure 6 Spacing of Fuel Elements.

TABLE V HEAT FROM CORE GAMMAS l l | _ l l 1 2 1 3=1x2 4 5 =3x4 6 5x6 C~e,n Cn-15.c n - 3. -159c,n Group | n | Pe 1 |B B E 1 e,n | Bn Eav,nBnEav n 15n e l-e l An Pe — c, n Pe Pe, n Pe 9c, n Pe i 1 10.256 0.150 5.85x10-1- 6 0.5 3 1. 655 3.84 0.021 1.00 1.655 2.7x 1 68 i | i Ne|n,i E Ne|,n,i 1 Fission |1 4,.75x10-2 5.68 0.270 2 Al Capture Thermal 2 4.82x.0-3 0 0 i 3 120 Capture 3 1.0 x10l2 0 0 0,270 2 0.100 0.055 5.5xlO0- 2.70 21.0 2.70 1.485 1.50 0.223 0. 777 1.154 1. 92xlo 0.222 i Ei Neq~Tn,i Ei Neq,n,'1 4.75x10-2 4.00 0.190 2 1.82x10-3 o0.425 0.02 3 l.09x10-2 0 0 0.192 3 0.083 o0.045 5.42xlO-l 1.93( 2.0 3.86 2.092 1.224 0.300 0.700 1.464 1.28xlO-11.187. I.....................1. i % | Neq, n,i | Ne]|n, 1 4.75x10-2 2.345 0.111 2 4.82x10-3 1.113 0.005 3 l.09x10-2 1.115 0.012 0.128 4 | 0.0-55 | 0.031; 5.65x10-l 1.42 4.0 5.680 3.209 0.825 0.44 0.560 1.797 2.5x10-2 0.045 i Neq, n,i l Neq,n, 4 -.75x10o-2 0.441 0.021 2 1.82x10-3 0.727 o.o 004 3 1.09x10-2 0 0 0.025 A = 0.922 = MP x 1.6 x 10-13 A W A = An

217For the helium system ( see Figure 3), the upper part of the chanber, the cryostat central tube, the electrical wires, are intensively cqoled by helium gas leaving the chamber. This gas has a specific heat of 4.99 cal/gram atom, or 4.99/4 cal/gr. Assume a helium consumption of 3 liter per hr (liquid), i.e. 3 x3 x x 0.126 = 378 gr/hr. If helium gas warms up to 100K above liquid helium temperature, then it can absorb (4.99/4) X 378 x 4.18 x 10 JoLcA kr or 5~5 Watts,. For a gamma absorption of 0.1 watt/gram of experimental set up, it is seen that the leaving gas could possibly evacuate the gamma from 55 grams of material. This is more than the gamma absorbing mass of the helium system. It was decided that 1/4 of the mass in the upper part,of the gamma-absorbing region (i.e, above the liquid helium level in the helium chamber) would Convey heat to the liquid helium by convection and the whole mass of the lower part would do so. Then, the gamma absorbing mass of the helium sysbem is of 22 gr of aluminum and samples plus 16 grams of liquid helium, or a total of 38 gr. In order to limit to an. acceptable value the rate of consumption of liquid helium, the power level of the reactor is chosen to be P = 0.1 EIW, i.e. the thermal flux of the order of 7 x 10 neut cm-2 sec1- (average). Equation (18) gives, for the gamma heat per gram, CH~ r = 7 x 1011 x 1.6 x 10-13 x 0.922 = 0.,10 watt/gr. For the gamma heat in the helium system, ( ) ie = 38 x o.103 = 3.92 watts,

-218For the nitrogen system, it was.considered'that the total ~gmma absorbing mass contributed to nitrogen evaporation. This total mass is found to be 200 gr, including 30 gr of liquid nitrogen. Hence the gamma,heat in the nitrogen system is estimated to be LH f) = 200 x 0.103 = 20.6 Watts. The latent heat of vaporization of helium is 24 cal/gr atom at - 2680C, i.e. 6 cal/gr, or 25.11 Joule/gr. The mass consumption of helium is then, (C )He, s = 3.92/25.11 gr He/see The volume consumption.: is jbXvC.3.92 x3.6 ( keJV = 2'25.11 x 0.124 lit He/hr = 4.6 lit/hr The latent heat of vaporization of nitrogen is.42.8 cal/gr near atmospheric pressure, or 4.96 x 10-2 Watt hr/gr, or 40 Watt hr/liter of liquid nitrogen. Hence the volume consumption of nitrogen due to core gamma heating is, Ct y,=-'2~0906- 0.52 liter/hr liquid N2. 3-Other sources of radiation heating Heat is also introduced into.the system by the.following.procLsses, -scattering of neutrons, particularly fast neutrons. This effect is small and usually neglected; -self-absorption of gammas originated in the experimental device by (n,~ ) reaction Al 27 (n, f) Al 28; -self-absorption.of betas produced by the decay of A1 28.

-219The calculations by Clairborne et al.(6) show that, for an aluminum test sample placed on.the face of the BSR core, while the'heat due to core gammas is 0.177 Watt/gr NW, the heat due to self-absorption of gammas is 0.0027 Watt/gr w and the heat due to self-absorption of betas is 0,023 Watt/gr AW. Since a significant part of the absorbing mass of the helium system is constituted by liquid helium and the absorbing mass of the nitrogen system-has been willingly overestimated, and since the core gamma heating has also been overestimated, it is corncleuded that selfabsorption.of gammas and betas can be neglected. 4-Heat transfers into the system The cryostat.will be jacketed by a vacuum space at a pressure of the.order of 0.1l mercury. Heat transfer from the outer wall of the vacuum jacket into the helium and nitrogen systems can take place through the following.processes, -radiative transfer; -free.molecule convection; -conduction at the upper end of the cryostat, where helium and nitrogen systems are in contact with the outer envelope through the central tube and the helium and nitrogen in and out tubes. There exist formulae to calculatej for simple geometries, heat transfer by the first two processes, (8-9) However, they employ assumptions and empirical coefficients which render doubtful the accuracy of the resutls. On the whole, it appears safer to adapt to the system under study experimental observations made on other cryostats. For a

-220large volume cryostat with short transfer lines,(lO) it was observed that, when irradiation was turned off, the helium consumption, including transfer losses, amounted to 4.7 liters per day. The volume of the helium chamber and of the liquid helium transfer tube is much smaller for the cryostat planned, although the transfer line is much longer. It seems safe to admit that the transfer losses will not exceed five times the loss of helium by heat transfer (other than due to absorption of radiations) into the chamber and to adopt a global consumption of 1 liter per hour of liquid helium for all these contributions. Paragraph 2 of Section IV indicates what liquid helium transfer experiments are being performed to check this point. For nitrogen, since the problem is less critical, no such contributions will be added, in this estimate. 5-Total volume rate of consumption of liquid helium and nitrogen and total consumption. Our estimate of the volume rate of consumption of liquid helium and nitrogen is then, C ~e, 5. 6 liters/hr.He, vol' CN2, vol 0.52 liters/hr For 50 hours exposure in the reactor, the estimate of liquefied gas consumption is, 280 liters helium..26 liters nitrogen

6-Cooling prior to experiment. First, the cryostat will be cooled by liquid nitrogen and niM troogeniexchange gas produced by evaportion, then by liquid helium and helium exchange gas. It is computed that,the mass (practically all aluminum) to cool from, say, room temperature, or about 3000K, to liquid nitrogen, or 77~K, is 850 gr and that the mass (also all aluminum) to cool from 770K to temperatures between 40K and, say, 15~K, is 636 gro For aluminum, C = 5.78 cal/gr atom = 0.214 gal/gr, at 20~C. Actually, it is much less than that at low temperature. For example,(ll) C. 0 5 x 3R for T/O' 0.25, whereeis the Debye temperature of the metal, R Mayer constant, or Boltzman constant for a gram-molecule. For aluminum, 0 = 3940K, hence, at 100~K, Ck = 0.5 x 6 = 3 cal/gr atom. For simplicity:we shall keep C u = 0.214 cal/gr atom. We can ~treat the cooling problem in the following.way. Suppose the mass M to cool, of specific heat C- is placed in the cooling liquid, whose latent heat of vaporization is Lo The gas produced by the evaporation catributes to cool the mass M. Call C h its specific heat. When a.mass dm of the liquid vaporizes, L dm is furnished to cool Mo Further, if To is the temperature of the liquid and T that of the mass M, the gas produced by evaporation brings a contribution which we may estimate as oc C dm (T - To) In view of the geometry, it is not exaggerated to take oi = 0.1.o We can write, - M C udT = dm. L +( Cy (T - T)] or, with T - To = 0, - M Cr c(iO = dm [L +dCT.,]

-222 - Call Ti the initial temperature. The mass mo of liquid needed to cool the mass M from Ti to To is Mo = (M C,/ ) C )Log [1 + (L C /L)(Ti -.)1 We first apply this for the cooling of 850 gr of aluminum from 3000K to 77~K with nitrogen. For nitrogen L is comparatively large, 42.8 cal/gr, so that for i( small, brvo ~ (M CL/L)(Ti - To) = (850 x 0.214/42.8) x 233 = 956 gr This represents a volume (965/0.808) x 10'3 = 1.195 liters liquid nitrogen. We shall consider that a volume of 2 liters liquid nitrogen will be needed for this phase of the cooling. Next, we calculate the mass mo of liquid helium needed to cool 630 gr of aluminum from 77~K to, say, 40K. For helium, Cp 1.25 cal/gr, L = 6 cal/gr. m'o = [636 x 0,214/(0l! x 1.25)] x 2.3 log [ 1 + 0.1 x 1.25 x 73/6]'l103 gr. The volume of liquid helium needed for this phase of the cooling will be, 1/0,124 = 8.1 liters As a safe estimate we shall count on spending,15 liters of liquid helium for this operation. 7-Expenditure of liquefied gases during construction. Total expenditure of liquid helium and nitrogen. Transfer tests through long transfer tubes have been made with liquid nitrogen and are being made with liquid helium. A total consumption of 25 liters of each liquefied gas is estimated for these tests.

The various consumptions are summarized in Table I ( Section I, ltragraph 3). The estimated global consumption of liquid helium, including construction tests, amounts to 320 literss that of liquid nitrogen to 53 liters. 8-Selection of aluminum alloys For most parts it is not possible to use the series 1.oo of Alcoa because of low mechanical strength and welding difficulty (practically pure aluminum). Commercial Alcoa aluminum alloys are aluminum with copper (2,..), aluminum with manganese (30..), aluminum with magnesium and silicon (6o..), aluminum with silicon~ Al 27, abundance 100%, has a thermal reaction cross section of 0.23 b. Al 28, formed 1by (n,, ~) reaction, undergoes beta decay (2.865 Mev), half life 2.27 min, with a gamma transition of 1,782 Mev. The only isotope of silicon whidh captures thermal neutrons to form a radioactive product is Si 30, abundance 3.05%, with a thermal rer-t action cross section of 0,4 b. Si 31, f'brmed by (n,' ) reaction, decays by beta emmision (1.48 Mev), half life 2,62 hr., gammas 0.17 and 0q52 Mev. The only isotope of magnesium which captures thermal neutrons to form a radioactive product is Mg 26, abundance 1129%,r giving Mg.27 by (n,-6) reaction (reaction cross section 60 mb). Mg 27 decays by betq emmision (1.8 Mev), with gamma transitions of 1.01 and 0,84 Mev, half life 9.45 min Mn 55, abundance 100%, has a thermal reaction cross section of 13,4 b. Mn 56, formed by (n, ) reaction, decays by beSa.emission (2.87 Mev), 2.576 hr half life, gamma transition 0,822, 1.77s and 2.06 Mev.

224- Cu 63, abundance 69.'4, has a cross section of 4.3 b for thermal, reaction. By (n,, ) reaction it gives Cu 64, which decays by electron capture, beta, and positron emmision,, with.half life 12o8 hr (betas 0.571 Mev, positrons 1o.65 Mev,, gammas Lo34 Mev). Cu 65, abundance 30.9%, has a thermal reactonx cross section of 2.11 bo By (n, ) reaction, it gives Cu 66, which decayrs by beta emmision. (2.63 Mev), half life 5.10 min, gamma transition, 1.05 Mev. The above data have been collected from standard texts on neutron reactions data.(5' 1,2913) Because of comparatively high thermal reaction, cross sections coupled with non-negligible half life of the radiosotope formed, it is advisable to use alloys containing copper and manganese, i.eo for example the series 2.0~ and 3... of Alcoa, only where necessary. We may note, however, that even the activation of a large sample containing 5% in volm:e of: copper would not create very serious problems. Take a hollow cylindrical sample of 5 cm.diameter thickness 0.2 cm, length 1.00 cm. Assume the whole sample placed in a thermal neutron flux 1013 neut/cm2 sec for 50 hrs. The half life of Cu 64 is about 13 hours, hence we may' consider th.at saturation is attained. For simplicity, consider that al.l the copper present is Cu 63. The activation. cross section is 4.3 b. The rough estimate of the activity of the sample at the end of th.e exposure is 1 -2 23 24 13 _7 x 0x 5 x 10 x 0.75 x 10 43103 x 43. x 10 x 314 x 5 3.7 x iL~1() x 0,2 x 102 1. 2 x 104 curies. If the sample is left in the pool, away from the core, for 20 days, i.e. about, 40 halsf lives, hle activity at the end of this period is 1.2 x io4/(2)40 = i2 x 0-8 curieo

-225The alloys selected are 6061 T 6, 5052 0 for most parts, 3003 H 14, 2024 T 3 and 2024 0 for components which may pose problems of weldability (helium chamber) or are not in the neutron flux,as well as Il00 aluminum (all Alcoa)o In addition, during construction, it was decided to build the long tube of the cryostat in three sections assembled by flanges. The lower section is made of 2024 T 3, the upper section, the horizontal section and the T for assembling the horizontal section are made of brass (see Diagram 1). These last sections are not placed in the neutron fluxo 9 -Mechanical resistance The following values are admitted(l2) for the yield strength of aluminum and various aluminum alloys, A1 (more than 99%) annealed 1.22 x 108 dyne cm'2 Al 75% rolled 10.60 x 108.dyne cm-2 Al 5 Si, cast 9.65 x 108 dyne cm'2 Al 3.8 Mg 11.00 x 108 dyne cm2 Al 12 Si 12.40 x 108 dyne cm-2 The following critical parts were calculated for mechanical ~strength. Helium chamber On the cylndrical body the stress is greater than on the spherical caps. It is given by, R =p r/2h with p internal pressure, r radius, h thickness. r =1 i nch, p < 2atm = 2 x 0.981 x 106 dyne cm'2. We try h = 0.03 cm 12 milse

-226This gives R < 1.64 x 108 dyne cm-2 We can try Alcoa 1100 in 12, 16, and 20 mils, and 2024 0 and 5052 0 in ~12 mils. Nitrogen chamber The critical parts are the two annular discs that form the bottom and the cover of the chamber. The stress is maximum along the small circumferenceof' the discs. It is given by(14) R K p r2 / h2 K is a coefficient equal to 0.9 for a.disce built in at the inner edge and prevented from rotating at the outer edge, and for a ratio outside diameter/inside diameter equal to 2.2, p is the internal pressure in.psi, r the outside diameter, h the thickness We consider using a silicon or magnesium alloy, hence we take a yield strength of 9 x 108 dyne cm'2. We have already selected r = 5.62 cm. Take p = 2 atm. The formula gives h = 2.5 mm' 0.083 In. The alloy selected is 6061 T 6. Envelope of the vacuum jacket Similar considerations as those albove lead to adopt a thickness of 5 mm = 1/4 in. for the disc closing the envelope at the bottom of the cryostat and 4 mm = 3/16 in. for the annular disc that forms the top of the lower part of the envelope (liaison with the long tube)* Another consideration,for the envelope, is that of possible collapsing of the long tube under extermal pressure. At the bottom end of the long.tube the water head is about 23 feet or 690 cm, hence the pressure 1 +-O.69 = 1.70 kg em 2. mTh erit2ical pressure(l5) is given by E 3 4 r

-227E elasticity modulus, r radius, h thickness. E = o.98 x 106 dyne cm'2 is a good value for Al Si, Al Mg Si, Al Mg. With this value one obtains, h > 1.32 mm The above results were used in addition to the requirement that the dimensions selected should permit multiple use of the cryostat and yet lead to an acceptable gamma absorption, for the trial and error selection of these dimensions lO —Thermal contraction on cooling Due to the different temperatures of helium and nitrogen systems, in their coldest parts, and the cryostat envelope, and since appreciable lengths are involved, relative positions of these elements will change appreciably from the beginning of cooling, at room temperature, to the end of cooling, when the various components have reached their operating low temperatures. Consider a linear distribution of temperature along a given element, from top to bottom, the top (top of the cryostat) being roughly at room temperature To, the bottom being at the operating low temperature T1. Call x the abscissa of the point on the element, counted vertically, from top to bottom. Let Tx be the temperature at that point. Call 1 the length of the element, from top to bottom, and o( the average linear coefficient of expansion of the material at the temperatures considered. The change Se of length of a portion dx at x is 3ZT =d i T0.-TS) = iCG TCOT r and the total contraction of the element is c k. T0-Tz a

-228Central tube This is the tube leading to the helium chamber (see Diagram 1). Say that Te= -269~C, To = 24~C. For aluminum, o( = 18 x 10-6. The length is about.696 cm. This gives a' 1883 cm This is an upward motion, since the central tube is attached at the top. The helium out tube will contract of about the same quanity. The helium in tube, colder at the upper parts will contract more. The differential contraction will be absorbed by an appropriate bend in the helium tube. Nitrogen out tube and thermal shield The bottom of the thermal shield is at distance about 707 cm from the top of the cryostat. Take the temperature at the bottom as -1900C and that at the upper part of the nitrogen out tube as.240C. Then, AQ~ 1.36 cm Hence, the clearance to provide between bottom of nitrogen chamber and top of helium chamber is 1.83 - 1.36 = 0.47 cm For safety, we shall leave o.6 cm, The nitrogen out tube is suspended from a support (lucite) at the top of its vertical portion. Its contraction will pull upward the nitrogen chamber. The nitrogen in tube will contract more than the nitrogen out tube. The differential cantraction will be absorbed by a bend in the nitrogen in tube and the nitrogen chamber will be guided in the Jacket envelope.

-229 The helium in tube is cooled roughly from 24 to -2690C. Its length is about 270 cm, The contractionl with respect to the envelope, is 1.41 cm. This will be absorbed by -the bend in the helium tube already provided to compensate the central tube contraction. The nitrogen in tube-also possesses a bend, as seen before, which will absorb its con - traction. 11l-Activation of nitrogen-14 Radioactive N 13 is formed by (T, n) or (n, 2n) reaction on N 14. C 14 is formed by (n p) reaction on N 14. The flatter does not constitute an activation problem, since its half life is 5568 years. The N 14 (~, n) N 13 reaction has a threshold of 10.5 Mev. The N 14 (n, 2n) reaction has a threshold of 1103 Mev. The total reaction cross section of N 14 at 10 Mev is smaller than 1.5 b. (6 t, n) reaction %~/cj is of the orderof 002 for 10 Mev gammas(5) We certainly have -,~ ~,) c t = 0.02 x 1.25 x 10'3 = 2.5 x 10-5 cm'l for nitrogen gas, and,'(~,n)( 0.02 x 0.81 =1.62 x 10'2 cm-1 for nitrogen liquid. We may neglect the reaction in the gas phase and consider that the reaction with liquid nitrogen only takes place in the nitrogen chamber, since the volumes of nitrogen in and out tubes are small. The liquid nitrogen chamber may be considered as a point absorber of volume 37.5 cm3, concentrated in the center of the core. It is found(6 that the equivalent number of gammas of energy greater than 10 Mev, per fission, is 9 x 10-4

-230Consider a volume element dV of the core and, at distance r, a surface element dS, normal to the line joining it to the volume element and subtending a solid angle dSL. Call the average thermal neutron flux in the core, the thermal macroscopic fission cross section of the core, ~, the linear absorption of the core for 10 Mev gammas, B(r) the build up factor of the core for 10 Mev gammas and penetration r. The number of gammas of energy greater than 10 Mev which cross dS per sec is, Zdav x 9 x 10 x ex dS:(r)/4 ir2 and the number of gammas with energy greater than 10 Mev absorbed in a colume element dV' = dS dr is, Zg,dV x 9 x 10-4 er B(r) (,n) dV' /4 r2 0 Take (.,n) = = 1.62 x 10-2 cm-!, and consider a spherical core of radius 30 cm. The number of (,n) reactions per sec in the liquid nitrogen chamber is, 3.75 x1ZF x 9 x 10-4 x 1.62 x 10'2 B(r) (1 - JC,, for a water-aluminum core, may be taken equal to (0.58 + 1.13) x 2 10-2 = 3.42 x 10-2 cm-1 An overestimated value for B(r) is 1.4jZ has been calculated previously and found equal to 4.75 x 102 cm"l. Take ~ = 7 x 1011 We obtain, as an overestimate of the number of N_ 13 atons formed per sec, N = 3.75 x 10 x 7 x 1011 x 4.75 x 102 x 9 x 10-4 x 1.62 x 10-2 x 1.4 x ( 1 -e-)/3.42 x 10-2 = 4.7 x 108, For a nitrogen consumption of 0.5 liter/hr, the time of exposure in the nitrogen chamber (volume 37.5 cm3) is 37.5/(5 x 102/3o6 x 103),

or about.250 sec. Since this is smaller than the half life, 10 minm, of N 13, we shall disregard N 13 decay. Hence, the number of N 13 atoms continuously formed is 1.18 x 1011. The 37~5 cm3 of liquid nitrogen correspond to a volume 37.5 x 0.808/(1,.25 x 10-3) = 22 x 103 cm3 of the nitrogen gas at atmospheric pressure and OOC. In a tube of 4 mm I D, this volume occupies a length, L = 22 x 103/(3o142 x 16 x 10'2/4) cm l18 x 103 m The number of N 13 atoms per meter is, disregarding decay, in this overestimate, (1.18/1.8) x 108 = 108 N 13 has a half life of 10 mmin, hence a disintegration constant of 0.692/(6 x 102) = 1*15 x 103 sec-1 The linear activity of the nitrogen out tube is much smaller than 108 x 1.15 x 10-3/3.7 x 1010 = 105 curies/meter (n, 2n) reaction Z(n, 2n) is smaller than tot (10 Mev) for all possible reactions. Again, consider reactions in the liquid nitrogen chamber alone. Then, Rt (10 Mev) = 1.5 x 10-24 x o.81 x 6.02 x 1023/14 = 5.3 x 10-2 cm1l It is difficult to obtain a reasonable estimate for the inte.grated flux at energies above 10 Mev. It has been observed(3) that, for the BSR while the thermal flux at the center of the core is 1012, the flux of neutrons with energy greater than 8.1 Mev (Al [tn,o] Na 24 threshold reaction)is about 7 x 109 at the same position. Operating at 0.1 IW,

-232i.e. with a thermal flux of order 1012 at its maximum, we expect a flux of neutrons with energy greater than 10 Mev of less than 5 x 109 at the center of the core. Hence the rate of formation of N 13 atoms is certainly much smaller than 37.5 x 5.3 x 10-2 x 5 x 109 1010 This is about 22 time what was found for a maximum estimate of the (", n) yield. Hence, we expect that the activity of the nitrogen out pipe will be much smaller than 2.3 x 10'4 curies/meter. This should give no great shielding difficulty.

PRELJ.AINARY EXPERIMENTS SECTION IV 1-Measurement of' electrical resistance The length of the samples will be about 8.cm, their diameter 1 mil, i.eo their cross sectional area 5.06 x 10-6 cm2. The measuring circuit to be used during the experiment is represented schematically in Figure 7. Actually, the potential difference will be measured across the sample and the rod which supports it in series. This is permissible, since the resistance of the rod is very small compared to that of the sampleo With f = 8.5 x 10-4 JtScm at 50K for copper, the resistance of a copper sample would be, -at that temperature, R5K = 8.5 x x 10-4 8/(5.06 x 10-6)= 1.34 x 103 JcnL After 50 hours exposure, the increase in resisti'vity.due to irradiation is expected to be 2 x L0'3 S0 cm, i.e. the total resistivity at 50K is expected to be 8.5 x 10-4 + 2 x 10-3 = 2.85 x 10-3 JScfmn and the resistance of the sample, 2.85 x 10-3 x 8/o 06 x o10)> 4 5 x 103 S.. After 5 hours exposure (1/10 total irradiation), the estimated increase in resistivity is.2 x 10-4 SL cm, i.e. the total resistance of the copper sample at 5~K is expected to be 1o05 x 0o-3 x 8/(5 o6 x 10-6) 1: 66 x 103 L @ Leeds and Northrup potentiometers reading d-own to 10-7 V are available in the Department of' Electrical.Engineering of the University. These would in principle allow to take readings every hour.. The change in resistance between readings would then be 4 x 10-5 x 8/5.06 x lo-6)= 63.1 E kC -233

MA4MP STANDARD 0 o RESISTOR ---- INDIVIDUAL N 3 LEADS L AN H AD | SAMPLE SAMPLE POTENTIOMETER /A POTENTIAL COMMON LEAD CURRENT LEAD LENGTH OF LEAD FROM A TO B OR C TO D: ABOUT 30 feet. Figure 7. Measuring Circuit. (4 samples)

-235. With a current of 0.1 A —small enough to prevent heating of the samples —this is an increase of the drop of potential across the sample of 6,31 V. The lowest range of. e.m.f that can be read with the potentiometer corresponds to 1 V per large division of the drum. To read 10-7 V, the tenth of a division.must be estimated. This is problematic. However, even an accuracy of one large.division is sufficient, since we want to.read increases of 6 divisions. However, account must be taken of.the accuracy with which the galvanometer can be read and the value of the current can be measured, A check.of. the accuracy obtained was made, using aLeeds,and-Northrup potentiometer type K, A standard resistor of O.lS..was placed in series with a resistor of O.OlS, to be measured, two dry cells of 1.5 V in parallel furnished a current through the circuit set at o.l A by a slide potentiometer. A sensitive wall galvanometer was used in.connection with the measuring potentiometer. In a first series of readings,,the eom.f, for the measuring potentiometer was furnished -by a dry cell of 1.5,V. Table VI shows the readings and the value of the resistance obtained. It is seen that the value obtained is O. olS, exact to the fifth decimal, i.e. 10O5 = lOJt&lcan be measured. This is sufficient, although with not much accuracy to spare, for precise measurements of R every hour during irradiation. A second series of readings was made with a stable power supply as e,m.f. for the.potentiometer. Table VII shows the readings and the value of the resistance obtained. This gives the same accuracy as the first series. Attempts to measure small resistances with.other galvanometers than a spot deflection galvanometer, namely using two electronic galvanometers, failed to give the accuracy of 10 -SL desired.

-236TABLE VI Measurement of resistance-e.m.f. for potentiometer:dry cell Current e.m.f. Resistance A V n"k..g09906. oo0009906 0.0100000 9903 9908 50 99045 99045 00 9902 9902 00 9902 9904 20 9900 9902 20 9900 9902 20 9900 9903 30 98985 98985 00 9897 9905 81 9899 9899 00 98975 98975 00 9894 9895 10 98835 9884 05 9884 98849 09 98849 9885 01 9885 9885 00 9885 9885 00 9886 98865 05 9885 9886 10 9886 9886 00 98835 98915 81 9885 9885 00 98855 98865 l0 9885 98855 05 98855 98855 00 98855 98865 00 9885 9887 20 9887 9888 10 98805 98825 20 9880 9882 20

-237TABLE VIIT Measurement of resistance e.m.f. for potentiometer:st&ble voltage supply Current e m. f. Resistance A V SL o.09885 o.ooo9885 0.0100000 9880 98805 05 98705 98705 oo 9864 9864 00 9855 9856 10 9849 98505 15 9842 9844 20 98375 98385 lo,9834 9835 10 9832 9832 00 9830 9830 00 9 8275 98275.00 9826 98265 05 9822 98225 05

It may be concluded that, irradiating thin samples ( 1 mil diameter, at least for copper) and using a sensitibe wall galvanometer, the measurement of change in resistance may be made with about 50 experimental points for 50 hr irradiation, per sample. However, this is using the available equipment to the limit of its possibilities. A more sensitive potenetiometer (108 V'; Rubicon, for ex.) would permit the use of a galvanometer less cunbersome than a wall.type model and the adoption of thicker samples (5mils diameter, for ex.). 2-Transfer..tests of liquid nitrogen and helium Since the loss of liquid during transfer cannot be estimated on the paper with a reasonable degree of accuracy, it was decided to measure the fraction of liquid helium which evaporates during a transfer through a long tube. Figure.8 represents the principle of the measuremento A transfer tube of 20 feet.length, diameter of the vacuum jacket 1/2 inch, inside diameter of the helium tube 1/16 inch, was built, as explained in Section I, Paragraph 4. The envelope is in Alcoa 3003 H.14, the helium tube in Alcoa "Utilitube". The vacuum.jacket is pumped down to less than 1 _.. mercuwry by an.oil diffusion pump.and a mechanical pump. Vacuum is read with a RCA 1946 thermocouple tube. The calibration.curve obtained for this thermocouple, using a McLeod mercury gauge, is.given in Figure 9. An attempt;to pass liquid nitrogen through this tube showed that the mass flow was extremely slow, even applying a pressure of 20 lbs psi nitrogen gas on the Dewar containing the liquid nitrogen. After two hours no liquid nitrogen was collected at.the outlet of. the transfer tube. The volume flow of nitrogen gas out of the transfer tube was too small

b~~~~~ a I:S % oz 1~~1~1 zU j zz LaJ. UI C~~~~~~~~~: Z ~ ~ 11..._ I

-240100 o\ a 3: a. I 10 11 12 13 14 15 16 17 s1 19 20 2 22 3 24 GAGE READINGS Figure 9. Calibration of Thermocouple.

-241to be measured. Although there is a.serious reduction in the section -of the inner tube caused by the use of threaded compression fittings, it was decided that.the diameter of 1/16 inch -was too small for the transfer of liquid nitrogen Two -transfer tubes for.nitrogen, with a.nitrogen -tube of diameter of 1/8 and 3/16 inch, respectively, were ordered from the laboratory shop. Inasmuch as their use is only a test of.dimensions, steel has been selected as the material. There remain, however, the possibility -that the aluminum tube would be adequate for the transfer of liquid helium, whose viscosity is extremely low.

AD:MDUM Calculations of the macroscopic cross sections U025 N25:!4~x!No 235 (3.189 x 3.035) x (2.54)2 x 60 = 1.6 x 10-4 N0 Zf th, 25 o.845 x 582 x 10-24 x 1.6 x o104 x 6.03 x 1(o23 = 4.75 x l0-2 cm A1 N Al, m ~ lx N~ _ 13 x 6o03 x 10 3 = 2.49 x 1022 27 27 a th, Al 0.845 x 0.23 x 10-24 x2.49 x 1022 = 4082 x 10-3 cm-1,-0 a 1H20= NH6H + N0,o = H2) + 0 = H20 m x = 058 No = 3.22 x 102 x 6003 x 1023 H20.' 18 18 1.942 x 1022 (a.H20)core, 2200 - 1942 x 1022 (2 x 0.332 + o0.002) x 10-24 1.29 x 10-2:C3 = ~a th, H1 20 0= 845 x 1.29 x 10'2 - 1.09 x 10'2 cm-1 -242

BIBLIOGRAPHY TO APPENDIX XXIV L, Cooper, II. G., Koehler, J. SW, and. Marx, J, W, t"Irradiation Effects in Copper, Silver, and Gold Near 10~K," Physo Rev,, 97, No. 3, (1955)o 2, Marx, J, Wo, Coper, Ho Go., and iHenderson, Jo Wo "Radiation Damage and Recovery in Copper, Silver, Gold, Nickel, and Tantalum," Phys. Rev., 88, No. 19 (1952) 3. Selected Reference Material, United States Atomic Energy Program, Research Reactors, (1955). 4. Kittel, C, Intioducti to Solid State Physicsy John Wiley and Sons, Inc., New Yor.k (1956 5. The Reactor Handbook, Vol. 1, Physicsdeclassified edition, published by Technical Information Service, U.oS Atomic Energy Commission, (May 1955)o 6. Clairborne2 H. Co, Copenhaver, C. M., Bertini, Ho W., and Fowler, To BW, "Calculating Gamma Heating in Reactor Structures,' Nuc.leonics 15, No. 10, (1957). 7. Blewitt, To I,, Coltman, R. R. Holmes, Do Ko, and Noggle, To S., "Mechanism of Annealing in Neutron Irradiated Metals," ORNL 2188, p. 52, (December 1956). 8. Wexler, A., "Evaporation Rate of Liquid Helium," Proceedings of the NBS Semicentennial Symposium on Low-Temperature Physics, NBS Circular 519, (1952) 9. Dushman, SO, Scientific F ti f Vacuum Techniue, John Wiley and Sons, Inc.., New York, (1949)0,. 10o Mapother Do, Eo, and Witt, F. E. Lo "Cry ostat for Cyclotron Irradiation at Liquid Helium Temperatures," Rev Sci. Instr., 26, No. 9, (1955). 11o DeLaunay, J,, The Theory Qf Specific Heats and Lattice Vibrations," Solid State Physics, Advances in R search ansd Appications, Vol..2 Academic Press Inco., New York, (1956). 12. Hughes, D, Jo., and. Harvey, J. A, Secto 8,. Nuclear Physics, Para. 8h, Neutrons, American Institute f Pf ysics Handbook, Gray, D. E,, Editor, McGraw.-Hill, New York, (1957)o. 13. Hughes, D. Jo, and Harvey, J, Ao. Neutron Cross Sections, BNL 325, (July 1, 1955), -24I3

BIBLIOGRAPHY TO APPENDIX XXIV (CONT ID) 14o Marks, Lo S., Mechanical Engineers' Handbook, McGraw-Hill, New York, (1941)o 15. Foppl, A., Resistance des Materiaux et Elements de la Theorie Math4e matique de 1' Elasticit', Gauthier-Viliars et Cie, Paris (1930

BIBLIOGRAPHY 1. Planeix, J. M., Preliminary Study, University of Michigan, (1958). 2. Brinkman, J. A., "On the Nature of Radiation Damage in Metals," J. Appl. Phys., 25, No. 8, (1954). 5. Snyder, Wo S., and Neufeld, J., "Disordering of Solids by Neutron Irradiation," Phys, Rev., 97, No0 6, (1955), Phys0 Rev., 99, No. 4, (1955). 4. Jongenburger, P., "The Extra-Resistivity Due to Vacancies in Copper, Silver, and Gold," Appl. Sci. Research B, 3, No. 4-5, (1953). 5. Blatt, F. J., "Effect of Vacancies and Interstitials on the Electrical Properties of Copper," Bulletin Am. Phys. Soc., 29, No07, (1954). 6. Dexter, D. L., "Scattering of Electrons from Point Singularities in Metals," Phys. Rev., 87, NO. 5, (1952). 7. Marx, J. W., Cooper, Ho Go, and Henderson, Jo W., "Radiation Damage and Recovery in Copper, Silver, Gold, Nickel, and Tantalum," Phys. Revo, 88, No. 1, (1952). 8. Cooper, H. G., Koehler, J. S., and Marx, Jo W., "Irradiation Effects in Copper, Silver, and Gold Near 10'K," Phys. Rev., 97, No. 3, (1955). 9, Seitz, F., and Koehler, J, S., "Displacement of Atoms during Irradiation," Solid State Physics, Advances in Research and Applications, Vol, 2, Academic Press Inco, New York, (195 10. Cooper, H. G., "Irradiation Effects in Copper, Silver, and Gold, Near 10~K," PhD Dissertation, University of Illinois, (1954). 11. Redman, J. Ko, Noggle, T. S., Coltman, R. Ro, and Blewitt, T. H., "Very Low-Temperature Irradiation of Metals: Change in Electrical Resistivity," Bul. Am. Phys. Soc., II, 1, No. 3, (1956). 12. Huntington, Ho B., "Creation of Displacements in Radiation Damage," Phys. Rev., 93, No. 6, (1954). 13o Eggen, D. T., and Laubenstein, M. J., "Displacement Energy for Radiation Damage in Copper," Phys. Rev., 91, No. 1, (1953). 14. Dixon, C. E., and Bowen, D. B., "Radiation Ordering by Cyclotron Particles," Phys. Rev., 94, No. 5, (1954). 15. Wruck, D., and Wert, C., "Crystal Structure as a Factor in Radiation Damage," Phys. Rev., 94, No. 5, (1954). -245

BIBLIOGRAPHY (CONTt D) 16. Dieckamp, H., and Crittenden, Eo C., "Shear Modulus of Irradiated Copper," Phys. Rev. 9_4, No. 5, (1954). 17. McDonnell, W. R,, and Kierstead, H. A., "Expansion of Copper Bombarded by 21-Mev Deuterons," Phys, Rev., 93, No. 1, (1954). 18. Overhauser, A, W., "Isothermal Annealing Effects in Irradiated Copper," Fbmys. Rev., 90, No. 3, (1953). 19. Tucker, C. W., and Senio, Po, "On the Nature of Thermal Spikes," J. Appl. Phys., 2_7, No. 3, (1956). 20. McReynolds, A, W., Augustyniak, W., McKeown, M.,, and Rosenblatt, D. Bo, "Neutron Irradiation Effects in Copper and Aluminum st 806K," Phys. Rev., 98, No. 2, (1955). 21. Denney, J. M., "Experimental Evidence for Melted Regions in Metal Crystals Resulting from Particle Bombardment," Phys. Rev., 2, No. 5, (195k4), 22, Cottrell, A. E., "Effects of Neutron Irradiation on Metals and Alloys," Metallurgical Reviews, Vol. I, Part 4, (1956). 23. Rutherford, E., "The Scattering of Alpha and Beta - Particles by Matter and the Structure of the Atom," Phil. Mag., 21, p. 669, (1911). Reprint of original paper in Foundations of Nuclear Physics, Beyer, R, T,, Editor, Dover, New York, (1949)..24. Ozeroff, J. D, "Atomic Displacements Produced by Fission Fragments and Fission Neutrons in Matter," KAPI20S5, AECD-2973, (1949), 25, Bohr, N., "The Penetration of Atomic Particles Through Matter," Det. Kgl. Danske Videnskab. Selskab. Mathematisk-fysiske Medd,, XVIII, 8,(19411.). 26. Kaplan, I., Nuclear Physics, Addison-Wesley, Cambridge, Mass., (1956). 27. Cooper, HE G., Koehler, J, S., and Marx, J. W., "Resistivity Changes in Oopper, Silver, and Gold Produced by Deuteron Irradiation Near 100K," Phis. Rev., 94, No. 2, (1954). 28. Geiger, H., and Marsden, E., "The Laws of Deflection of Alpha-Particles Throu h Large Angles," Phil. ag, 25, p. 604 (1913). 29. Hughes, D. J, and Harvey, J, A., Neutron Cross Sections, BNL 325, (July 1, 1955). -~246u

BIBLIOGRAPHY (CONT' D) 50. Seitz, F.o "Disordering of Solids by Action of Fast Massive Particles," Discuss. Faraday Soco, 5, P. 271 (1.949). 31o Blewitt, To H., Coltman, R. R., Holmes, D. K., and Noggle, T. S., "Mechanism of Annealing in Neutron Irradiated Metals," ORNL 2188, p. 52, (December 1956). 32. Waltson, Go N., "Fission Recoil and Its Effects," Progress in Nuclear Physics, Vol. 6, Pergamon, London, (1957).

UNIVERSITY OF MICHIGAN 3 9Il 1111111 111 11 11111 3 9015 03695 2235