Introduction to Health Physics PDF

INTRODUCTION TO Health Physics Notice Medicine is an ever-changing science. As new research and clinical experience broaden our knowledge, changes in treatment and drug therapy are required. The authors and the publisher of this work have checked with sources believed to be reliable in their efforts to provide information that is complete and gener- ally in accord with the standards accepted at the time of publication. However, in view of the possibility of human error or changes in medical sciences, nei- ther the authors nor the publisher nor any other party who has been involved in the preparation or publication of this work warrants that the information contained herein is in every respect accurate or complete, and they disclaim all responsibility for any errors or omissions or for the results obtained from use of the information contained in this work. Readers are encouraged to con- ﬁrm the information contained herein with other sources. For example and in particular, readers are advised to check the product information sheet in- cluded in the package of each drug they plan to administer to be certain that the information contained in this work is accurate and that changes have not been made in the recommended dose or in the contraindications for adminis- tration. This recommendation is of particular importance in connection with new or infrequently used drugs. INTRODUCTION TO Health Physics FOURTH EDITION Herman Cember, PhD Professor Emeritus Northwestern University Evanston, Illinois Thomas E. Johnson, PhD Assistant Professor Department of Environmental and Radiological Health Sciences Colorado State University Fort Collins, Colorado New York Chicago San Francisco Lisbon London Madrid Mexico City Milan New Delhi San Juan Seoul Singapore Sydney Toronto Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved. 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Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise. To my wife, Sylvia and to the memory of Dr. Elda E. Anderson and Dr. Thomas Parran This page intentionally left blank CONTENTS Preface/ XI 1. Introduction/1 2. Review of Physical Principles/3 Mechanics/3 Relativistic Effects/6 Electricity/13 Energy Transfer/26 Quantum Theory/44 Summary/51 3. Atomic and Nuclear Structure/59 Atomic Structure/59 The Nucleus/73 Summary/81 4. Radiation Sources/85 Radioactivity/85 Transformation Mechanisms/85 Transformation Kinetics/98 Activity/103 Naturally Occurring Radiation/109 Serial Transformation/117 Summary/134 5. Interaction of Radiation with Matter/143 Beta Particles/143 Alpha Particles/160 Gamma Rays/165 Neutrons/181 Summary/194 vii viii Contents 6. Radiation Dosimetry/203 Units/203 External Exposure/205 Internally Deposited Radionuclides/233 External Exposure: Neutrons/266 Summary/270 7. Biological Basis for Radiation Safety/279 Dose–Response Characteristics/280 The Physiological Basis for Internal Dosimetry/285 Radiation Effects: Deterministic/307 Radiation Effects: Stochastic/315 Radiation-Weighted Dose Units: The Sievert and The Rem/331 Summary/332 8. Radiation Safety Guides/337 Organizations That Set Standards/337 Philosophy of Radiation Safety/342 ICRP Basic Radiation Safety Criteria/347 United States Nuclear Regulatory Program/407 Ecological Radiation Safety/421 Summary/421 9. Health Physics Instrumentation/427 Radiation Detectors/427 Particle-Counting Instruments/428 Dose-Measuring Instruments/447 Neutron Measurements/465 Calibration/477 Counting Statistics/485 Summary/505 10. External Radiation Safety/513 Basic Principles/513 Optimization/571 Summary/574 11. Internal Radiation Safety/583 Internal Radiation/583 Principles of Control/584 Contents ix Surface Contamination Limits/592 Waste Management/595 Assessment of Hazard/618 Optimization/627 Summary/630 12. Criticality/639 Criticality Hazard/639 Nuclear Fission/639 Criticality/645 Nuclear Reactor/651 Criticality Control/658 Summary/661 13. Evaluation of Radiation Safety Measures/667 Medical Surveillance/667 Estimation of Internally Deposited Radioactivity/668 Individual Monitoring/680 Radiation and Contamination Surveys/681 Air Sampling/685 Continuous Environmental Monitoring/706 Combined Exposures/706 Source Control/708 Summary/709 14. Nonionizing Radiation Safety/721 Units/722 UV Light/723 Lasers/728 Radiofrequency Radiation and Microwaves/759 Principles of Radiation Safety/792 Summary/794 Appendix A Values of Some Useful Constants/803 Appendix B Table of the Elements/805 Appendix C The Reference Person Overall Speciﬁcations/809 Appendix D Source in Bladder Contents/815 Appendix E Total Mass Attenuation Coefﬁcients, μ/ρ, cm 2/g/851 Appendix F Mass Energy Absorption Coefﬁcients, μa /ρ, cm 2/g/853 Answers To Problems/855 Index/861 This page intentionally left blank PREFACE The practice of radiation safety is a continually evolving activity. Many of the changes in the practice of ionizing and nonionizing radiation safety, in calculation method- ology, and in the methods for demonstrating compliance with the safety standards that have occurred since the publication of the previous edition of Introduction to Health Physics are incorporated in the fourth edition. Since their inception in 1928, the Recommendations of the International Com- mission on Radiological Protection have formed the scientiﬁc basis for ionizing ra- diation safety standards issued by regulatory authorities throughout the world. Gen- erally, earlier recommendations were successively more restrictive than the previous ones. The 2006 recommendations, however, are essentially the same as the previous recommendations made in 1990. The main difference is that the 2006 recommen- dations are made on the basis of the increased knowledge acquired since 1990. This is not surprising, since no harmful radiation effects have been observed among the population of radiation workers whose doses had been within the previous stan- dards. The new recommendations continued to stress that all unnecessary exposure be avoided and that all exposures should be kept as low as reasonably achievable, economic and social factors being taken into account. A reasonable question, there- fore, that is raised by the ICRP recommendations is “How safe is safe?” This question lies in the ﬁeld that Dr. Alvin Weinberg, the late director of the Oak Ridge National Laboratory, called transscience. Transscientiﬁc questions have a scientiﬁc basis, but they cannot be answered by science alone. Safety is a subjective concept that can be interpreted only within the context of its application. Policy decisions regarding matters of health and safety should be made in the context of public health. In the practice of public health, we ﬁnd that numerous diseases and threats to health are always present in every community. The cost of controlling these threats to health is borne by the community. Since the community has limited resources, it must set priorities regarding which of the many real or perceived health threats to control. One of the techniques for quantifying the likelihood of the expression of a potential risk is called quantitative risk assessment. In the area of radiation safety, this usually deals with two main risks: (1) failure of a large technological system, such as a nu- clear power plant, and (2) the long-term effects of low-level radiation. The results of quantitative risk assessment are often perceived as the determination of a real threat to life or limb, no matter how small the calculated chance of occurrence. However, quantitative risk assessment is a calculation that almost always assumes the most pessimistic, and in many cases entirely unrealistic, values for parameters whose magnitudes include several different uncertainties. In addition to statistical uncer- tainties, for example, we must choose among several different equally reasonable models to which to apply the statistical data. One of the purposes of this edition is xi to provide the technical background needed to understand the calculation and use of quantitative risk assessment for radiation hazards in order to help us to allocate our limited resources. Although it has been a number of years since the ICRP recommended that health physics quantities be expressed in the meter–kilogram–second (MKS) units of the SI system rather than the traditional units based on the centimeter–gram–second (cgs) system, the change to the SI units has not yet been universally implemented. For example, the U.S. Nuclear Regulatory Commission continues to use the traditional system of units in its regulations. For this reason, this edition continues to use both systems, with one or the other equivalent quantity given in parentheses. I wish to thank the many persons, too numerous to mention by name, for their helpful suggestions. I also owe a debt of gratitude to my former student and now colleague, Thomas Johnson, for his authorship of Chapter 14 and for checking the text of the other chapters, and to his wife, Melissa, for giving up her time with her husband so that he could contribute to this book. Herman Cember INTRODUCTION TO Health Physics This page intentionally left blank 1 INTRODUCTION Health physics, radiological health, or radiological engineering are synonymous terms for that area of public health and environmental health engineering that deals with the safe use of ionizing and nonionizing radiation in order to prevent harmful effects of the radiation to individuals, to population groups, and to the biosphere. The health physicist is responsible for safety aspects in the design of processes, equipment, and facilities utilizing radiation sources and for the safe disposal of radioactive waste so that radiation exposure to personnel will be minimized and will at all times be within acceptable limits; he or she must keep personnel and the environment under constant surveillance in order to ascertain that these designs are indeed effective. If control measures are found to be ineffective or if they break down, the health physicist must be able to evaluate the degree of hazard and make recommendations regarding remedial action. Public policy vis-à-vis radiation safety is based on political, economic, moral, and ethical considerations as well as on scientiﬁc and engineering principles. This text- book deals only with the scientiﬁc and engineering bases for the practice of health physics. The scientiﬁc and engineering aspects of health physics are concerned mainly with (1) the physical measurements of different types of radiation and radio- active materials, (2) the establishment of quantitative relationships between radi- ation exposure and biological damage, (3) the movement of radioactivity through the environment, and (4) the design of radiologically safe equipment, processes, and environments. Clearly, health physics is a professional ﬁeld that cuts across the basic physical, life, and earth sciences as well as such applied areas as toxicology, in- dustrial hygiene, medicine, public health, and engineering. The professional health physicist, therefore, in order to perform effectively, must have an appreciation of the complex interrelationships between humans and the physical, chemical, biological, and even social components of the environment. He or she must be competent in the wide spectrum of disciplines that bridge the ﬁelds between industrial operations and technology on one hand and health science, including epidemiology, on the other. In addition to these general prerequisites, the health physicist must be technically competent in the subject matter unique to health physics. The main purpose of this book is to lay the groundwork for attaining techni- cal competency in health physics. Radiation safety standards undergo continuing 1 2 CHAPTER 1 change as new knowledge is gained and as the public’s perception of radiation’s beneﬁts and risks evolve. Radiation safety nomenclature too changes in order to accommodate changing standards. Because of the nature of the subject matter and the topics covered, however, it is hoped that the book will be a useful source of information to workers in environ- mental health as well as to those who will use radiation as a tool. For the latter group, it is also hoped that this book will impart an appreciation for radiation safety as well as an understanding of the philosophy of environmental health. 2 REVIEW OF PHYSICAL PRINCIPLES MECHANICS Units and Dimensions Health physics is a science and hence is a systematic organization of knowledge about the interaction between radiation and organic and inorganic matter. Quite clearly, the organization must be quantitative as well as qualitative since the control of radiation hazards implies knowledge of the dose–response relationship between radiation exposure and the biological effects of radiation. Quantitative relationships are based on measurements, which, in reality, are com- parisons of the attribute under investigation to a standard. A measurement includes two components: a number and a unit. In measuring the height of a person, for example, the result is given as 70 inches (in.) if the British system of units is used or as 177.8 centimeters (cm) if the metric system is used. The unit inches in the ﬁrst case and centimeters in the second tell us what the criterion for comparison is, and the number tells us how many of these units are included in the quantity being measured. Although 70 in. means exactly the same thing as 177.8 cm, it is clear that without an understanding of the units the information contained in the number above would be meaningless. In the United States, the British system of units is used chieﬂy in engineering, while the metric system is widely used in science. Three physical quantities are considered basic in the physical sciences: length, mass, and time. In the British system of units, these quantities are measured in feet, slugs (a slug is that quantity of mass that is accelerated at a rate of one foot per second per second by a force of one pound; a mass of 1 slug weighs 32.2 pounds), and seconds, respectively, while the metric system is divided into two subsystems: the mks—in which the three quantities are speciﬁed in meters, kilograms, and seconds— and the cgs—in which centimeters, grams, and seconds are used to designate length, mass, and time. By international agreement, the metric system is being replaced by a third and new system—the Système International, the International System of Units, or simply 3 4 CHAPTER 2 the SI system. Although many familiar metric units are employed in SI, it should be emphasized that SI is a new system and must not be thought of as a new form of the metric system. All the other units such as force, energy, power, and so on are derived from the three basic units of mass in kilograms (kg), length in meters (m), and time in seconds (s), plus the four additional basic units: electric current in amperes (A), temperature in Kelvin (K) or degrees Celsius (◦ C), where 1 K = 1◦ C, amount of a substance in moles (mol), and luminous intensity in candelas (cd). For example, the unit of force, the newton (N), is deﬁned as follows: One newton is the unbalanced force that will accelerate a mass of one kilogram at a rate of one meter per second per second. Expressed mathematically: Force = mass × acceleration, that is, F = m × a, (2.1) and the dimensions are m/s F = kg ×. s Since dimensions may be treated algebraically in the same way as numbers, the dimension for acceleration is written as m/s2. The dimensions for force in units of newton (N), therefore, are kg · m N=. s2 In the cgs system, the unit of force is called the dyne. The dyne is deﬁned as follows: One dyne is the unbalanced force that will accelerate a mass of one gram at a rate of one cm per second per second. For health physics applications, the magnitude of cgs units were closer to the magnitudes being measured than the mks units and, therefore, the cgs system was universally used. However, despite the long history of cgs-based units, the cgs system is being replaced by SI units in order to be consistent with most of the other sciences that have adopted SI units. All the international bodies that deal with radiation safety base their recommendations on SI units. However, the U.S. Nuclear Regulatory Commission continues to use the traditional cgs units in its regulatory activities. Work and Energy Energy is deﬁned as the ability to do work. Since all work requires the expenditure of energy, the two terms are expressed in the same units and consequently have the same dimensions. Work W is done, or energy expended, when a force f is exerted through some distance r : W = f × r. (2.2) REVIEW OF PHYSICAL PRINCIPLES 5 In the SI system, the joule (J) (named after the British scientist who measured the mechanical equivalent of heat energy) is the unit of work and energy and is deﬁned as follows: One joule of work is done when a force of one newton is exerted through a distance of one meter. Since work is deﬁned as the product of a force and a distance, the dimensions for work and energy are as follows: joule = newton × meter kg · m kg · m2 = × m =. (2.3) s2 s2 The unit of work or energy in the cgs system is called the erg and is deﬁned as follows: One erg of work is done when a force of one dyne is exerted through a distance of one centimeter. The joule is a much greater amount of energy than an erg. 1 joule = 107 ergs. Although the erg is very much smaller than a joule, it nevertheless is very much greater than the energies encountered in the submicroscopic world of the atom. When working on the atomic scale, a more practical unit called the electron volt (eV) is used. The electron volt is a unit of energy and is deﬁned as follows: 1 eV = 1.6 × 10−19 J = 1.6 × 10−13 erg. When work is done on a body, the energy expended in doing the work is added to the energy of the body. For example, if a mass is lifted from one elevation to another, the energy that was expended during the performance of the work is converted to potential energy. On the other hand, when work is done to accelerate a body, the energy that was expended appears as kinetic energy in the moving body. In the case where work was done in lifting a body, the mass possesses more potential energy at the higher elevation than it did before it was lifted. Work was done, in this case, against the force of gravity and the total increase in potential energy of the mass is equal to its weight, which is the force with which the mass is attracted to the earth, multiplied by the height through which the mass was raised. Potential energy is deﬁned as energy that a body possesses by virtue of its position in a force ﬁeld. Kinetic energy is deﬁned as energy possessed by a moving body as result of its motion. For bodies of mass m, moving “slowly” with a velocity v less than about 3 × 107 m/s, the kinetic energy, E k , is given by 1 2 Ek = mv , (2.3a) 2 and the total energy of the body is equal to the sum of its potential energy and its kinetic energy E t = E pe + E k. (2.3b) 6 CHAPTER 2 When the speed of a moving body increases beyond about 3 × 107 m/s, we ob- serve interesting changes in their behavior—changes that were explained by Albert Einstein. RELATIVISTIC EFFECTS According to the system of classical mechanics that was developed by Newton and the other great thinkers of the Renaissance period, mass is an immutable property of matter; it can be changed in size, shape, or state but it can neither be created nor be destroyed. Although this law of conservation of mass seems to be true for the world that we can perceive with our senses, it is in fact only a special case for conditions of large masses and slow speeds. In the submicroscopic world of the atom, where masses are measured on the order of 10−27 kg, where distances are measured on the order of 10−10 m, and where velocities are measured in terms of the velocity of light, classical mechanics is not applicable. Einstein, in his special theory of relativity, postulated that the velocity of light in a vacuum is constant at 3 × 108 m/s relative to every observer in any reference frame. He also postulated that the speed of light is an upper limit of speed that a material body can asymptotically approach, but never can attain. Furthermore, according to Einstein, the mass of a moving body is not constant, as was previously thought, but rather a function of the velocity with which the body is moving. As the velocity increases, the mass increases, and when the velocity of the body approaches the velocity of light, the mass increases very rapidly. The mass m of a moving object whose velocity is v is related to its rest mass m 0 by the equation m0 m= , (2.4) v2 1− 2 c where c is the velocity of light, 3 × 108 m/s. W Example 2.1 Compute the mass of an electron moving at 10% and 90% of the speed of light. The rest mass of an electron is 9.11 × 10−31 kg. Solution At v = 0.1c , 9.11 × 10−31 kg m= = 9.16 × 10−31 kg 2 (0.1 c ) 1− c2 REVIEW OF PHYSICAL PRINCIPLES 7 and at v = 0.99 c , 9.11 × 10−31 kg m= = 64.6 × 10−31 kg 2 (0.99 c ) 1− c2 Example 2.1 shows that whereas an electron suffers a mass increase of only 0.5% when it is moving at 10% of the speed of light, its mass increases about sevenfold when the velocity is increased to 99% of the velocity of light. Kinetic energy of a moving body can be thought of as the income from work put into the body, or energy input, in order to bring the body up to its ﬁnal velocity. Expressed mathematically, we have 1 2 W = Ek = f × r = mv. (2.5) 2 However, the expression for kinetic energy in Eqs. (2.3) and (2.5) is a special case since the mass is assumed to remain constant during the time that the body is undergoing acceleration from its initial to its ﬁnal velocity. If the ﬁnal velocity is sufﬁciently high to produce observable relativistic effects (this is usually taken as v ≈ 0.1c = 3 × 107 m/s, then Eqs. (2.3) and (2.5) are no longer valid. As the body gains velocity under the inﬂuence of an unbalanced force, its mass continuously increases until it attains the value given by Eq. (2.4). This particular value for the mass is thus applicable only to one point during the time that body was undergoing acceleration. The magnitude of the unbalanced force, therefore, must be continuously increased during the accelerating process to compensate for the increasing inertia of the body due to its continuously increasing mass. Equations (2.2) and (2.5) assume the force to be constant and therefore are not applicable to cases where relativistic effects must be considered. One way of overcoming this difﬁculty is to divide the total distance r into many smaller distances, r 1 , r 2 ,... , r n , as shown in Figure 2-1, multiply each of these small distances by the average force exerted while traversing the small distance, and then sum the products. This process may be written as W = f 1 r 1 + f 2 r 2 + · · · · + f n r n (2.6a) f r0 Δr1 Δr2 Δr3 Δrn-1 Δrn rn r n Figure 2-1. Diagram illustrating that the total work done in accelerating a body is W = fn r n. n=1 8 CHAPTER 2 and symbolized by n W= f n r n (2.6b) n =1 As r is successively divided into smaller and smaller lengths, the calculation of the work done, using Eq. (2.6), becomes more accurate. A limiting value for W may be obtained by letting each small distance r n in Eq. (2.6) approach zero, that is, by considering such small increments of distance that the force remains approxi- mately constant during the speciﬁed interval. In the notation of the calculus, such an inﬁnitesimally small quantity is called a differential and is speciﬁed by preﬁxing the symbol for the quantity with the letter “d.” Thus, if r represents distance, dr represents an inﬁnitesimally small distance and the differential of work done, which is the product of the force and the inﬁnitesimally small distance, is given by dW = f dr. (2.7) The total energy expended in going from the point r 0 to point r n , then, is merely the sum of all the products of the force and the inﬁnitesimally small distances through which it acted. This sum is indicated by the mathematical notation r =n W= f dr. (2.8) r =0 The ratio of two differentials, dW /dr, for example, is called a derivative, and the process in which a derivative is obtained is called differentiation. Since acceleration is deﬁned as the rate of change of velocity with respect to time, v2 − v1 v a= = , (2.9) t2 − t 1 t where v1 and v2 are the respective velocities at times t1 and t2. Then Eq. (2.1) may be written as v f =m , (2.10) t and by letting t approach zero we obtain the instantaneous rate of change of velocity or the derivative of velocity with respect to time. Using the differential notation, we have dv f =m. (2.11) dt This is the expression of Newton’s second law of motion for the nonrelativistic case where the mass remains constant. Newton’s second law states that the rate of change of momentum of an accelerating body is proportional to the unbalanced force acting on the body. For the general case, where mass is not constant, Newton’s second law is written as d (mv) f =. (2.12) dt REVIEW OF PHYSICAL PRINCIPLES 9 Substitution of the value of f from Eq. (2.12) into Eq. (2.8) gives r d (mv) W= dr. (2.13) dt 0 Since v = dr /dt, Eq. (2.13) can be written as t mv d (mv) W= v dt = v d (mv), (2.14) dt 0 0 m0 and substituting m = 1/2 , we have v2 1− 2 c ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ v ⎪ ⎨ ⎪ ⎬ m 0v W= vd . (2.15) ⎪ ⎪ v2 1/2 ⎪ ⎪ 0 ⎪ ⎩ 1− ⎪ ⎭ c2 Differentiating the term in the parenthesis gives ⎧ ⎫ ⎪ ⎪ v 3 ⎪ ⎪ v ⎪ ⎨ ⎪ ⎬ v c 2 W = m0 + dv. (2.16) ⎪ ⎪ v2 1/2 v2 3/2 ⎪ ⎪ 0 ⎪⎩ 1− 1− 2 ⎪ ⎭ c2 c Now, multiply the numerator and denominator of the ﬁrst term in Eq. (2.16) by 1 − v 2 /c 2 to obtain ⎧ ⎫ ⎪ ⎪ v3 v3 ⎪ ⎪ v ⎪ ⎨ v− 2 ⎪ ⎬ c c 2 W = m0 3/2 + 3/2 ⎪ dv (2.17) ⎪ ⎪ v2 v2 ⎪ 0 ⎪ ⎩ 1− 2 1− 2 ⎪ ⎭ c c v v v 1 = m0 dv = m 0. 2 3/2 3/2 (2.18) v v2 0 1− 2 0 v 1− 2 dv c c The integrand in Eq. (2.18) is almost in the form b b un+1 un du = , (2.19) n + 1 a a 10 CHAPTER 2 where 1 2v un = du = − 2 3/2 and dv. v c2 1− c2 To convert Eq. (2.18) into the form for integration given by Eq. (2.19), it is neces- sary only to complete du. This is done by multiplying the integrand by –2/c 2 and the entire expression by –c 2 /2 in order to keep the total value of Eq. (2.18) un- changed. The solution of Eq. (2.18), which gives the kinetic energy of a body that was accelerated from zero velocity to a velocity v, is ⎡ ⎤ ⎢ ⎥ 2⎢ 1 ⎥ 1 E k = W = m0c ⎢ 1/2 − 1⎥ = m0c 2 1/2 − 1 , (2.20) ⎣ v2 ⎦ 1 − β2 1− 2 c where β = v/c. Equation (2.20) is the exact expression for kinetic energy and must be used whenever the moving body experiences observable relativistic effects. W Example 2.2 (a) What is the kinetic energy of the electron in Example 2.1 that travels at 99% of the velocity of light? Solution 1 E k = m0c 2 1/2 − 1 1 − β2 −31 m 2 1 −13 = 9.11 × 10 kg 3 × 10 8 1/2 − 1 = 4.99 × 10 J s 1 − (0.99)2 (b) How much additional energy is required to increase the velocity of this electron to 99.9% of the velocity of light, an increase in velocity of only 0.91%? Solution The kinetic energy of an electron whose velocity is 99.9% of the speed of light is 2 −31 8 m 1 −13 E k = 9.11 × 10 kg 3 × 10 − 1 = 17.52 × 10 J s 2 1/2 1 − (0.999) REVIEW OF PHYSICAL PRINCIPLES 11 The additional work necessary to increase the kinetic energy of the electron from 99% to 99.9% of the velocity of light is W = (17.52 − 4.99) × 10−13 J = 12.53 × 10−13 J. (c) What is the mass of the electron whose β is 0.999? Solution m0 9.11 × 10−31 kg −31 m= 1/2 = 1/2 = 204 × 10 kg. 1 − β2 1 − (0.999)2 The relativistic expression for kinetic energy given by Eq. (2.20) is rigorously true for particles moving at all velocities while the nonrelativistic expression for kinetic energy, Eq. (2.3), is applicable only to cases where the velocity of the moving particle is much less than the velocity of light. It can be shown that the relativistic expression reduces to the nonrelativistic expression for low velocities by expanding −1/2 the expression 1 − β 2 in Eq. (2.20) according to the binomial theorem and then dropping higher terms that become insigniﬁcant when v c. According to the binomial theorem, n (n − 1) a n−2 b 2 (a + b)n = a n + na n−1 b + + · · ·· (2.21) 2! −1/2 The expansion of 1 − β 2 according to Eq. (2.21), is accomplished by letting a = 1, b = −β 2 , and n = −1/2. −1/2 1 3 1 − β2 = 1 + β2 + β4 + · · · · (2.22) 2 8 Since β = v/c , then, if v c , terms from β 4 and higher will be insigniﬁcantly small and may therefore be dropped. Then, after substituting the ﬁrst two terms from Eq. (2.22) into Eq. (2.20), we have 1 v2 1 E k = m0c 2 1 + − 1 = m 0v2 , 2 c2 2 which is the nonrelativistic case. Equation (2.3) is applicable when v c. In Example 2.2, it was shown that at a very high velocity (β = 0.99) a kinetic energy increase of 253% resulted in a velocity increase of the moving body by only 0.91%. In nonrelativistic cases, the increase in velocity is directly proportional to the square root of the work done on the moving body or, in other words, to the kinetic energy of the body. In the relativistic case, the velocity increase due to additional energy is smaller than in the nonrelativistic case because the additional energy serves to increase the mass of the moving body rather than its velocity. This equivalence of mass and energy is one of the most important consequences of Einstein’s special 12 CHAPTER 2 theory of relativity. According to Einstein, the relationship between mass and energy is E = mc 2 , (2.23) where E is the total energy of a piece of matter whose mass is m and c is the velocity of light in vacuum. The principle of relativity tells us that all matter contains potential energy by virtue of its mass. It is this energy source that is tapped to obtain nuclear energy. The main virtue of this energy source is the vast amount of energy that can be derived from conversion into its energy equivalent of small amounts of nuclear fuel. W EXAMPLE 2.3 (a) How much energy can be obtained from 1 g of nuclear fuel? Solution m 2 E = mc 2 = 1 × 10−3 kg × 3 × 108 = 9 × 1013 J. s Since there are 2.78 × 10−7 kilowatt-hours (kW h) per joule, 1 g of nuclear fuel yields J kW · h kW · h E = 9 × 1013 × 2.78 × 10−7 = 2.5 × 107 g J g (b) How much coal, whose heat content is 13,000 Btu/lb, must be burned to liberate the same amount of energy as 1 g of nuclear fuel? Solution 1 Btu = 2.93 × 10−4 kW h. Therefore, the amount of coal required is Btu kW · h 2.5 × 107 kW · h = 1.3 × 104 × 2.93 × 10−4 lb Btu lb ×2 × 103 × C tons ton Therefore, C = 3280 tons (2981 metric tons) The loss in mass accompanying ordinary energy transformations is not de- tectable because of the very large amount of energy released per unit mass and the REVIEW OF PHYSICAL PRINCIPLES 13 consequent very small change in mass for ordinary reactions. In the case of coal, for example, the above example shows a loss in mass of 1 g per 3280 tons. The fractional mass loss is m 1g f = = = 3.3 × 10−10. m lb g 3.28 × 10 tons × 2 × 10 3 3 × 4.54 × 10 2 ton lb Such a small fractional loss in mass is not detectable by any of our ordinary weighing techniques. ELECTRICITY Electric Charge: The Coulomb All the elements are electrical in nature and, except for hydrogen, are constructed of multiples of two charged particles and one uncharged particle. Their electrical properties are due to extremely small, charged particles called protons and electrons. The mass of the proton is 1.6726 × 10−27 kg (1.6726 × 10−24 g) and the mass of the electron is 9.1085 × 10−31 kg (9.1085 × 10−28 g). These two particles have charges of exactly the same magnitude but are qualitatively different. A proton is said to have a positive charge and an electron has a negative charge. Under normal con- ditions, matter is electrically neutral because the positive and negative charges are homogeneously (on a macroscopic scale) dispersed in equal numbers in a manner that results in no net charge. However, it is possible, by suitable treatment, to induce either net positive or negative charges on bodies. For example, combing the hair with a hard rubber comb transfers electrons to the comb from the hair, leaving a net negative charge on the comb. The uncharged component in every element is called the neutron; it has a mass of 1.67492 × 10−27 kg (1.67492 × 10−24 g). For health physics purposes, these three particles—electron, proton, and neutron—may be considered the basic building blocks of matter (although we now believe that protons and neutrons themselves are made of still smaller particles called quarks). It should be pointed out here that high-energy accelerators produce—in addition to protons, neutrons, and electrons—a number of different extremely short-lived unstable particles. In the context of health physics, the most important of these particles are charged and uncharged pions (pi-mesons) and muons (mu-mesons) because they give rise to very high-energy electrons and gamma rays when they decay. Muons are also produced by cosmic radiation and contribute to the dose from cosmic radiation. Charged bodies exert forces on each other by virtue of their electric ﬁelds. Bodies with like charges repel each other while those with unlike charges attract each other. In the case of point charges, the magnitude of these electric forces is proportional to the product of the charges and inversely proportional to the square of the distance between the charged bodies. This relationship was described by Coulomb and is known as Coulomb’s law. Expressed algebraically, it is q 1q 2 f =k. (2.24) r2 where k, the constant of proportionality, depends on the nature of the medium that separates the charges. In the SI system, the unit of electric charge, called the coulomb 14 CHAPTER 2 (C), is deﬁned in terms of electric current rather than by Coulomb’s law. For this reason, the constant of proportionality has a value not equal to 1 but rather N · m2 k0 = 9 × 109 (2.25) C2 when the two charges are in a vacuum or in air (air at atmospheric pressure exerts very little inﬂuence on the force developed between charges and thus may be considered equivalent to a vacuum). The subscript 0 signiﬁes the value of k in a vacuum. If the charges are separated by materials, other than air, that are poor conductors of electricity (such materials are called dielectrics), the value of k is different and depends on the material. It is convenient to deﬁne k 0 in terms of another constant, ε0 , called the permittivity: 1 N · m2 k0 = = 9 × 109 , (2.26) 4πε0 C2 1 1 C2 ε0 = = = 8.85 × 10−12 , 4π k0 N·m 2 N · m2 4π × 9 × 10 9 C2 where ε 0 is the permittivity of a vacuum. The permittivity of any other medium is designated by ε. The relative permittivity, K e , of a substance is deﬁned by ε Ke = (2.27) ε0 and is called the dielectric coefﬁcient. For all dielectric materials, the dielectric coefﬁcient has a value greater than 1. The permittivity, or the dielectric coefﬁcient, is a measure of the amount of electric energy that can be stored in a medium when the medium is placed into a given electric ﬁeld. If everything else is held constant, a higher dielectric coefﬁcient leads to a greater amount of stored electric energy. The smallest natural quantity of electric charge is the charge on the electron or proton, ±1.6 × 10−19 C. The reciprocal of the electronic charge, 6.25 × 1018 , is the number of electrons whose aggregate charge is 1 C. In the cgs system, the unit of charge is the statcoulomb (sC) and the electronic charge is 4.8 × 10−10 sC. There are 3 × 109 sC in 1 C. W EXAMPLE 2.4 Compare the electrical and gravitational forces of attraction between an electron and a proton separated by 5 × 10−11 m. Solution The electrical force is given by Eq. (2.24): q 1q 2 9 N·m 2 1.6 × 10−19 C × 1.6 × 10−19 C f = k0 = 9 × 10 × 2 r2 C2 5 × 10−11 m = 9.2 × 10−8 N. REVIEW OF PHYSICAL PRINCIPLES 15 The gravitational force between two bodies follows the same mathematical formu- lation as Coulomb’s law for electrical forces. In the case of gravitational forces, the force is always attractive. The gravitational force is given by G m1m 2 F =. (2.28) r2 G is a universal constant that is equal to 6.67 × 10−11 N · m2 /kg2 and must be used because the unit of force, the newton, was originally deﬁned using “inertial” mass, ac- cording to Newton’s second law of motion, given by Eq. (2.1). The mass in Eq. (2.28) is commonly called “gravitational” mass. Despite the two different designations, it should be emphasized that inertial mass and gravitational mass are equivalent. It should also be pointed out that F in Eq. (2.28) gives the weight of an object of mass m 1 when m 2 represents the mass of the earth and r is the distance from the object to the center of the earth. Weight is merely a measure of the gravitational attractive force between an object and the earth and therefore varies from point to point on the surface of the earth, according to the distance of the point from the earth’s center. On the surface of another planet, the weight of the same object would be different from that on earth because of the different size and mass of that planet and its consequent different attractive force. In outer space, if the object is not under the gravitational inﬂuence of any heavenly body, it must be weightless. Mass, on the other hand, is a measure of the amount of matter and its numerical value is therefore independent of the point in the universe where it is measured. The gravitational force between the electron and the proton is N · m2 6.67 × 10−11 × 9.11 × 10−31 kg × 1.67 × 10−27 kg kg2 F = 2 5 × 10−11 m = 4.1 × 10−47 N. It is immediately apparent that in the interaction between charged particles, grav- itational forces are extremely small in comparison with the electrical forces acting between the particles and may be completely neglected in most instances. Electrical Potential: The Volt If one charge is held rigidly and another charge is placed in the electric ﬁeld of the ﬁrst charge, it will have a certain amount of potential energy relative to any other point within the electric ﬁeld. In the case of electric potential energy, the reference point is taken at an inﬁnite distance from the charge that sets up the electric ﬁeld, that is, at a point far enough from the charge so that its effect is negligible. As a consequence of the great separation, these charges do not interact electrically. Therefore, a value of zero is arbitrarily assigned to the potential energy in the system of charges; the charge at an inﬁnite distance from the one that sets up the electric ﬁeld has no electric potential energy. If the two charges are of the same sign, bringing them closer together requires work (or the expenditure of energy) in 16 CHAPTER 2 b a b Figure 2-2. Diagram illustrating work done in moving a charge between two points of differ- ent potential in an electric ﬁeld. a order to overcome the repulsive force between the two charges. Since work was done in bringing the two charges together, the potential energy in the system of charges is now greater than it was initially. On the other hand, if the two charges are of opposite signs, then a decrease in distance between them occurs spontaneously because of the attractive forces, and work is done by the system. The potential energy of the system consequently decreases, that is, the potential energy of the freely moving charge with respect to the rigidly held charge, decreases. This is exactly analogous to the case of a freely falling mass whose potential energy decreases as it approaches the surface of the earth. In the case of the mass in the earth’s gravitational ﬁeld, however, the reference point for potential energy of the mass is arbitrarily set on the surface of the earth. This means that the mass has no potential energy when it is lying right on the earth’s surface. All numerical values for potential energy of the mass, therefore, are positive numbers. In the case of electric potential energy, however, as a consequence of the arbitrary convention that the point of the zero numerical value is at an inﬁnite distance from the charge that sets up the electric ﬁeld, the numerical values for the potential energy of a charge, owing to attractive electrical forces, must be negative. The quantitative aspects of electric potential energy may be investigated with the aid of Figure 2-2, which shows a charge +Q that sets up an electric ﬁeld extending uniformly in all directions. Another charge, +q , is used to explore the electric ﬁeld set up by Q. When the exploring charge is at point a, at a distance r a cm from Q, it has an amount of potential energy that depends on the magnitudes of Q, q , and r a. If the charge q is now to be moved to point b, which is closer to Q, then, because of the repulsive force between the two charges, work is done in moving the charge from point a to point b. The amount of work that is done in moving charge q from point a to point b may be calculated by multiplying the force exerted on the charge q by the distance through which it was moved, in accordance with Eq. (2.2). From Eq. (2.24), however, it is seen that the force is not constant but varies inversely with the square of the distance between the charges. The magnitude of the force, therefore, increases rapidly as the charge q approaches Q, and increasingly greater amounts of work are done when the exploring charge q is moved a unit distance. The movement of the exploring charge may be accomplished by a series of inﬁnitesimally small movements, during each of which an inﬁnitesimally small amount of work is done. The total energy expenditure, or increase in potential energy of the exploring charge, is then merely equal to the sum of all the inﬁnitesimal increments of work. This inﬁnitesimal energy increment is given by Eq. (2.7): dW = − f dr REVIEW OF PHYSICAL PRINCIPLES 17 (the minus sign is used here because an increase in potential energy results from a decrease in distance between the charges) and, if the value for f from Eq. (2.24) is substituted into Eq. (2.7), we have Qq dW = −k0 dr, (2.29) r2 r b dr W = −k0 Qq. (2.30) r2 ra Integration of Eq. (2.30) gives 1 1 W = k0 Qq −. (2.31) rb ra If the distances a and b are measured in meters and if the charges are given in coulombs, then the energy W is given in joules. W EXAMPLE 2.5 If, in Figure 2-2, Q is +44.4 μC, q is +5 μC, and r a and r b are 2 m and 1 m re- spectively, then calculate the work done in moving the 5 μC charge from point a to point b. Solution The work done is, from Eq. (2.31), N · m2 −6 −6 1 1 W = 9 × 10 9 × 44.4 × 10 C × 5 × 10 C − C2 1m 2m = 1 N · m = 1 J. In this example, 1 J of energy was expended in moving the 5 μC of charge from a to b. The work per unit charge is W 1J J = −6 = 200, 000 q 5 × 10 C C 18 CHAPTER 2 We therefore say that the potential difference between points a and b is 200,000 V, since, by deﬁnition: One volt of potential difference exists between any two points in an electric ﬁeld if one joule of energy is expended in moving a charge of one coulomb between the two points. Expressed more concisely, the deﬁnition of a volt is J 1V=1. C In Example 2.5, point b is the point of higher potential with respect to point a, because work had to be done on the charge to move it to b from a. The electrical potential at any point due to an electric ﬁeld from a point charge Q is deﬁned as the potential energy that a unit positive exploring charge +q would have if it were brought from a point at an inﬁnite distance from Q to the point in question. The electrical potential at point b in Figure 2-2 can be computed from Eq. (2.30) by setting distance r a equal to inﬁnity. The potential at point b, Vb , which is deﬁned as the potential energy per unit positive charge at b, is, therefore: W Q Vb = = k0. (2.32) q rb W EXAMPLE 2.6 (a) What is the potential at a distance of 5 × 10−11 m from a proton? Solution Q N · m2 1.6 × 10−19 C N·m V = k0 = 9 × 109 × −11 = 28.8 r C 2 5 × 10 m C J = 28.8 = 28.8 V C (b) What is the potential energy of another proton at this point? Solution According to Eq. (2.32), the potential energy of the proton is equal to the product of its charge and the potential of its location. Therefore, E p = q V = 1.6 × 10−19 C × 28.8 V = 4.6 × 10−18 J. REVIEW OF PHYSICAL PRINCIPLES 19 Electrical Current: The Ampere A ﬂow of electrically charged particles constitutes an electric current. The unit for the amount of current is the ampere (A), which is a measure of the time rate of ﬂow of charge. The ampere is deﬁned in the SI system by the interaction between a current ﬂowing through a conductor and a magnetic ﬁeld. However, a useful working deﬁnition is that one ampere represents a ﬂow rate of charge of one coulomb per second. Ordinarily, the charge carrier in an electric current is the electron. Since the charge on an electron is 1.6 × 10−19 C, a current of 1 A represents an electron ﬂow rate of 1 C/s 1 electrons/s × = 6.25 × 1018. A C A 1.6 × 10−19 electron A 100-μA electron beam in an X-ray tube represents an electron ﬂow rate of electrons/s 100 × 10−6 A × 6.25 × 1018 = 6.25 × 1014 electrons/s. A Current is determined only by the ﬂow rate of charge. For example, in the case of a beam of alpha particles, whose charge = 2 × 1.6 × 10−19 C = 3.2 × 10−19 C, 1 A corresponds to 3.125 × 1019 alpha particles. The direction of current ﬂow was arbitrarily determined, before the discovery of the electron, to be from the positive electrode to the negative electrode of a closed circuit. In fact, the electrons ﬂow in the opposite direction. However, conventional current ﬂow still goes from positive to negative. When we are interested in the actual direction of ﬂow, we use the term “electron current” to indicate current ﬂow from negative to positive. The Electron Volt: A Unit of Energy If two electrodes are connected to the terminals of a source of voltage, as shown in Figure 2-3, then a charged particle anywhere in the electric ﬁeld between the two plates will have an amount of potential energy given by Eq. (2.32), W = qV, where V is the electrical potential at the point occupied by the charged particle. If, for example, the cathode in Figure 2-3 is 1-V negative with respect to the anode and the charged particle is an electron on the surface of the cathode, then the potential Figure 2-3. Diagram showing the potential energy of an electron in an electric ﬁeld. 20 CHAPTER 2 energy of the electron with respect to the anode is W = q V = −1.6 × 10−19 C × (−1 V) = 1.6 × 10−19 J This amount of energy, 1.6 × 10−19 J, is called an electron volt and is symbolized by eV. Since the magnitude of the electron volt is convenient in dealing with the energetics of atomic and nuclear mechanics, this quantity of energy is taken as a unit and is frequently used in health physics. Multiples of the electron volt are the keV (103 eV), the MeV (106 eV), and the GeV (109 eV). W EXAMPLE 2.7 How many electron volts of energy correspond to the mass of a resting electron? Solution E = mc 2 m 2 = 9.11 × 10−31 kg × 3 × 108 s = 81.99 × 10−15 J J Since there are 1.6 × 10−19 , eV 81.99 × 10−15 J E = = 0.51 × 106 eV. J 1.6 × 10−19 eV It should be emphasized that, although the numerical value for the electron volt was calculated by computing the potential energy of an electron at a potential of 1 V, the electron volt is not a unit of electrons or volts; it is a unit of energy and may be interchanged (after numerical correction) with any other unit of energy. W EXAMPLE 2.8 How many electron volts of heat must be added to change 1 L of water whose temperature is 50◦ C to completely dry steam? REVIEW OF PHYSICAL PRINCIPLES 21 Solution cal The speciﬁc heat of water is 1 , and the heat of vaporization of water is 539 g cal. Therefore, g ! " cal cal heat energy added = 1000 g 1 × (100 − 50) deg + 539 g · deg g = 589,000 cal J J Since there are 4.186 and 1.6 × 10−19 , we have cal eV J 5.89 × 105 cal × 4.186 heat energy added = cal = 1.54 × 1025 eV J 1.6 × 10−19 eV The answer to Example 2.8 is an astronomically large number (but not very much energy on the scale of ordinary physical and chemical reactions) and shows why the electron volt is a useful energy unit only for reactions in the atomic world. W EXAMPLE 2.9 An alpha particle, whose charge is +(2 × 1.6 × 10−19 ) C and whose mass is 6.645 × 10−27 kg, is accelerated across a potential difference of 100,000 V. What is its kinetic energy, in joules and in electron volts, and how fast is it moving? Solution The potential energy of the alpha particle at the moment it starts to accelerate is, from Eq. (2.32), W = q V = 2 × 1.6 × 10−19 C × 105 V = 3.2 × 10−14 J In terms of electron volts, 3.2 × 10−14 J W= = 2 × 105 eV J 1.6 × 10−19 eV 22 CHAPTER 2 Since all the alpha particle’s potential energy is converted into kinetic energy after it falls through the 100,000 V (100 kV) potential difference, the kinetic energy must then be 200,000 eV (200 keV). The velocity of the alpha particle may be computed by equating its potential and kinetic energies, 1 2 qV = mv , (2.33) 2 and solving for v: 1/2 # $1/2 2q V 2 × 105 V × 3.2 × 10−19 C v= = m 6.645 × 10−27 kg m = 3.1 × 106. s Electric Field The term electric ﬁeld was used in the preceding sections of this chapter without an explicit deﬁnition. Implicit in the use of the term, however, was the connotation by the context that an electric ﬁeld is any region where electric forces act. “Electric ﬁeld” is not merely a descriptive term; deﬁning an electric ﬁeld requires a number in order to specify the magnitude of the electric forces that act in the electric ﬁeld and a direction in which these forces act, and, thus, it is a vector quantity. The strength of an electric ﬁeld is called the electric ﬁeld intensity and may be deﬁned in terms of the force (magnitude and direction) that acts on a unit exploring charge that is placed into the electric ﬁeld. Consider an isolated charge +Q that sets up an electric ﬁeld and an exploring charge +q that is used to investigate the electric ﬁeld, as shown in Figure 2-4. The exploring charge will experience a force in the direction shown and of a magnitude given by Eq. (2.24): Qq f = k0. r2 The force per unit charge at the point r meters from charge Q is the electric ﬁeld intensity at that point and is given by the equation f N N · m2 QC ε= = k0 × 2 2. (2.34) q C C2 r m Figure 2-4. The force on an exploring charge +q in the electric ﬁeld of charge +Q. REVIEW OF PHYSICAL PRINCIPLES 23 According to Eq. (2.34), electric ﬁeld intensity is expressed in units of force per unit charge, that is, in newton per coulomb. It should be emphasized that ε is a vector quantity, that is, it has direction as well as magnitude. W EXAMPLE 2.10 (a) What is the electric ﬁeld intensity at point P due to the two charges +6 C and +3 C, shown in Figure 2-5(A)? Solution The electric ﬁeld intensity at point P due to the +6 C charge is Q1 N · m2 6C N ε1 = k0 2 = 9 × 109 2 × 2 = 1.35 × 1010 r1 C (2 m) C and acts in the direction shown in Figure 2-5(A). (The magnitude of the ﬁeld intensity is shown graphically by a vector whose length is proportional to the ﬁeld intensity. In Figure 2-5(A), the scale is 1 cm = 1 × 1010 N/C. ε1 is therefore drawn 1.35-cm long). ε2 , the electric ﬁeld intensity at P due to the +3 C charge, is Q2 N · m2 3C N ε2 = k 0 2 = 9 × 109 2 × 2 = 2.7 × 1010 , r2 C (1 m) C and acts along the line Q 2 P , as shown in the illustration. The resultant electric intensity at point P is the vector sum of ε1 and ε2. If these two vectors are accurately drawn in magnitude and direction, the resultant may be obtained graphically by completing the parallelogram of forces and drawing the diagonal εR. The length of ε ε ε ε ε ε A B Figure 2-5. Resultant electric ﬁeld from (A) two positive charges and (B) two opposite charges. 24 CHAPTER 2 the diagonal is proportional to the magnitude of the resultant electric ﬁeld intensity and its direction shows the direction of the electric ﬁeld at point P. In this case, since 1 × 1010 N/C is represented by 1 cm, the resultant electric ﬁeld intensity is found to be about 4 × 1010 N/C and it acts in a direction 30◦ clockwise from the vertical. The value of ε R may also be determined from the law of cosines a 2 = b 2 + c 2 − 2bc cos A, (2.35) where b and c are two adjacent sides of a triangle, A is the included angle, and a is the side opposite angle A. In this case, b is 2.7 × 1010 , c is 1.35 × 1010 , angle A is 120◦ , and a is the resultant εR , the electric ﬁeld intensity whose magnitude is to be calculated. From Eq. (2.35), we ﬁnd 2 2 εR2 = 2.7 × 1010 + 1.35 × 1010 − 2 2.7 × 1010 1.35 × 1010 cos 120◦ N εR = 3.57 × 1010 C (b) What is the magnitude and direction of εR if the 3 C charge is negative and the 6 C charge is positive? Solution In this case, the magnitudes of ε 1 and ε2 would be exactly the same as in part (a) of this example; the direction of ε1 would also remain unchanged, but the direction of ε 2 would be toward the −3 C charge, as shown in Figure 2-5(B). From the geometric arrangement, it is seen that the resultant intensity acts in a direction 120◦ clockwise from the vertical. The magnitude of εR , from Eq. (2.35), is 2 2 εR2 = 2.7 × 1010 + 1.35 × 1010 − 2 2.7 × 1010 1.35 × 1010 cos 60◦ N εR = 2.34 × 1010. C Point charges result in nonuniform electric ﬁelds. A uniform electric ﬁeld may be produced by applying a potential difference across two large parallel plates made of electrical conductors separated by an insulator, as shown in Figure 2-6. Figure 2-6. Conditions for producing a relatively uniform electric ﬁeld. The ﬁeld will be quite uniform throughout the region between the plates, but will be distorted at the edges of the plates. REVIEW OF PHYSICAL PRINCIPLES 25 The electric intensity throughout the region between the two plates is ε newtons per coulomb. The force acting on any charge within this ﬁeld therefore is f = εq N. (2.36) If the charge q happens to be positive, then to move it across the distance d, from the negative to the positive plates, against the electric force in the uniform ﬁeld requires the expenditure of energy given by the equation W = f d = εq d. (2.37) However, since potential difference (V ) is deﬁned as work per unit charge, Eq. (2.37) may be expressed as W V = = εd, (2.38) q or V V ε=. (2.39) d m Equation (2.39) expresses electric ﬁeld intensity in the units most commonly used for this purpose—volts per meter. A nonuniform electric ﬁeld that is of interest to the health physicist (in instrument design) is that due to a potential difference applied across two coaxial conductors, as shown in Figure 2-7. If the radius of the inner conductor is a meters and that of the outer conductor is b meters, then the electric intensity at any point between the two conductors, r meters from the center, is given by 1 V V ε= × , (2.40) r b m ln a where V is the potential difference between the two conductors. b a V Figure 2-7. Conditions for the nonuni- form electric ﬁeld between two coaxial r conductors given by Eq. (2.40). 26 CHAPTER 2 W EXAMPLE 2.11 A Geiger-Müller counter is constructed of a wire anode whose diameter is 0.1 mm and a cathode, coaxial with the anode, whose diameter is 2 cm. If the voltage across the tube is 1000 V, what is the electric ﬁeld intensity (a) at a distance of 0.03 mm from the surface of the anode and (b) at a point midway between the center of the tube and the cathode? Solution (a) We know, 1 V ε= × . r b ln a Letting r = 1 2 (0.01) + 0.003 = 0.008 cm = 8 × 10−5 m, we have, 1 1000 V V ε= × = 2.36 × 106. 8 × 10−5 m ln 1 m 0.005 (b) At r = 0.005 m, 1 1000 V V ε= × = 3.78 × 104. 0.005 m ln 1 m 0.005 It should be noted that in the case of coaxial geometry, extremely intense electric ﬁelds may be obtained with relatively small potential differences. Such large ﬁelds require mainly a small ratio of outer to inner electrode radii. ENERGY TRANSFER In a quantitative sense, the biological effects of radiation depend on the amount of energy absorbed by living matter from a radiation ﬁeld and by the spatial distribu- tion in tissue of the absorbed energy. In order to comprehend the physics of tissue irradiation, some pertinent mechanisms of energy transfer must be understood. REVIEW OF PHYSICAL PRINCIPLES 27 Elastic Collision An elastic collision is deﬁned as a collision between two bodies in which kinetic energy and momentum are conserved; that is, the sum of the kinetic energy of the two bodies before the collision is equal to their sum after the collision, and the sums of their momenta before and after the collision are the same. In an elastic collision, the total kinetic energy is redistributed between the colliding bodies; one body gains energy at the expense of the other. A simple case is illustrated in the example below. W EXAMPLE 2.12 A block of mass 10 kg, made of a perfectly elastic material, slides on a frictionless surface with a velocity of 2 m/s and strikes a stationary elastic block whose mass is 2 kg (Fig. 2-8). How much energy was transferred from the large block (M) to the small block (m) during the collision? Solution If V1 ,v1 , andV2 and v2 are the respective velocities of the large and small blocks before and after the collision, then, according to the laws of conservation of energy and momentum, we have 1 1 1 1 MV12 + mv12 = MV22 + mv22 (2.41) 2 2 2 2 and MV1 + mv1 = MV2 + mv2 (2.42) Since v1 = 0, Eqs. (2.41) and (2.42) may be solved simultaneously to give m m V2 = 1 13 and v2 = 3 13. s s The kinetic energy transferred during the collision is 1 1 1 16 1 MV12 − MV22 = × 10 4 − = 11 J, 2 2 2 9 9 Figure 2-8. Elastic collision between blocks M and m, in which the sum of both kinetic energy and momenta of the two blocks before and after the collision are the same. 28 CHAPTER 2 and this, of course, is the energy gained by the smaller block: 1 2 1 100 1 mv = × 2 × = 11 J. 2 2 9 9 Note that the magnitude of the force exerted by the larger block on the smaller block during the collision was not considered in the solution of Example 2.12. The reason for not explicitly considering the force in the solution can be seen from Eq. (2.10), which may be written as f × t = m × v. According to Eq. (2.10), the force necessary to change the momentum of a block is dependent on the time during which it acts. The parameter of importance in this case is the product of the force and the time. This parameter is called the impulse; Eq. (2.10) may be written in words as Impulse = change of momentum. The length of time during which the force acts depends on the relative velocity of the system of moving masses and on the nature of the mass. Generally, the more the colliding blocks “give in,” the greater will be the time of application of the force and the smaller, consequently, will be the magnitude of the force. For this reason, for example, a baseball player who catches a ball moves his hand back at the moment of impact, thereby increasing the time during which the stopping force acts and decreasing the shock to his hand. For this same reason, a jumper ﬂexes his knees as his feet strike the ground, thereby increasing the time that his body comes to rest and decreasing the force on his body. For example, a man who jumps down a distance of 1 m is moving with a velocity of 4.43 m/s at the instant that he strikes the ﬂoor. If he weighs 70 kg and if he lands rigidly ﬂat footed and is brought to a complete stop in 0.01 s, then the stopping force, from Eq. (2.10), is 3.1 × 104 N, or 6980 lb. If, however, he lands on his toes and then lowers his heels and ﬂexes his knees as he strikes, thereby increasing his actual stopping time to 0.5 s, the average stopping force is only about 140 lb. In the case of the two blocks in Example 2.12, if the time of contact is 0.01 s, then the average force of the collision during this time interval is m 10 kg × 0.0067 f = s = 6.7 N 0.01 s The instantaneous forces acting on the two blocks vary from zero at the instant of impact to a maximum value at some time during the collision, then to zero again as the second block leaves the ﬁrst one. This may be graphically shown in Figure 2-9, a curve of force versus time during the collision. The average force during the collision is the area under the curve divided by the time that the two blocks are in contact. REVIEW OF PHYSICAL PRINCIPLES 29 Force Figure 2-9. The time variation of the force between colliding Time bodies. In the case of a collision between two masses, such as that described above, one block exerts a force on the other only while the two blocks are in “contact.” During “contact,” the two blocks seem to be physically touching each other. Actually, how- ever, the two blocks are merely very close together, too close, in fact, for us to be able to perceive any space between them. Under this condition, the two blocks repel each other by very short-range forces that are thought to be electrical in nature. (These forces will be discussed again in Chapter 3.) This concept of a “collision” without actual contact between the colliding masses may be easily demonstrated with the aid of magnets. If magnets are afﬁxed to the two blocks in Example 2.12, as shown in Figure 2-10, then the magnetic force, which acts over relatively long distances, will repel the two blocks, and the smaller block will move. If the total mass of each block, including the magnet, remains the same as in Example 2.12, then the calculations and results of Example 2.12 are applicable. The only difference between the physical “collision” and the magnetic “collision” is that the magnitude of the force in the for- mer case is greater than in the latter instance, but the time during which the forces are effective is greater in the case of the magnetic “collision.” In both instances, the product of average force and time is exactly the same. Inelastic Collision If the conditions in Figure 2-8 are modiﬁed by fastening the 2-kg block, block B, to the ﬂoor with a rubber band, then, in order to break the rubber band and cause the block to slide freely, the 10-kg block, block A, must transfer at least sufﬁcient energy to break the rubber band. Any additional energy transferred would then appear as kinetic energy of block B. If the energy necessary to break the rubber band is called the binding energy of block B, then the kinetic energy of block B after it is struck by block A is equal to the difference between the energy lost by A and the binding energy of B. Algebraically, this may be written as E B = E A − φ, (2.43) Figure 2-10. “Collision” between two magnetic ﬁelds. 30 CHAPTER 2 where E A is the energy lost by block A and φ is the binding energy of block B. In a collision of this type, where energy is expended to free one of the colliding bodies, kinetic energy is not conserved and the collision is therefore not elastic, that is, it is inelastic. W EXAMPLE 2.13 A stationary block B, whose mass is 2 kg, is held by an elastic cord whose elastic constant is 10 N/m and whose ultimate strength is 5 N. Another block A, whose mass is 10 kg, is moving with a velocity of 2 m/s on a frictionless surface. If block A strikes block B, with what velocity will block B move after the collision? Solution From Example 2.12, it is seen that the energy lost by block A in this collision is 11 19 J. The energy expended in breaking the rubber band may be calculated from the product of the force needed to break the elastic cord and the distance that the elastic cord stretches before breaking. In the case of a spring, rubber band, or any other substance that is elastically deformed, the deforming force is opposed by a restoring force whose magnitude is proportional to the deformation. That is, f = k × r, (2.44) where f is the force needed to deform the elastic body by an amount r, and k is the “spring constant” or the force per unit deformation. Since Eq. (2.44) shows that the force is not constant but rather is proportional to the deformation of the rubber band, the work done in stretching the rubber band must be computed by the application of calculus. The inﬁnitesimal work, dW , done in stretching the rubber band through a distance dr is dW = f dr, and the total work done in stretching the rubber band from r = 0 to r is given by r W= f dr 0 Substituting Eq. (2.44) for f , we have r W= kr dr (2.45) 0 and solving Eq. (2.45) shows the work done in stretching the rubber band to be kr 2 W=. (2.46) 2 Since in this example k is equal to 10 N/m, the ultimate strength of the elastic cord, REVIEW OF PHYSICAL PRINCIPLES 31 5 N, is reached when the rubber band is extended to 0.5 m. With these numerical values, Eq. (2.46) may be solved: N 10 × (0.5 m)2 N W= m = 1.25 = 1.25 J. 2 m Therefore, of the 11 19 J lost by block A in its collision with block B, 1.25 J are dissipated in breaking the elastic cord (the binding energy) that holds block B. The kinetic energy of block B, using Eq. (2.43), is E B = 11.11 − 1.25 = 9.86 J. 1 9.86 J = mv 2 2 1 9.86 J = (2 kg)v 2 2 m v = 3.14 s If block A had less than 1.25 J of kinetic energy, the elastic cord would not have been broken; the restoring force in the elastic cord would have pulled block B back and caused it to oscillate about its equilibrium position. (For this oscillation to actually occur, block A would have to be withdrawn immediately after the collision, otherwise block B, on its rebound, would transfer its energy back to block A and send it back with the same velocity that it had before the ﬁrst collision. The net effect of the two collisions, then, would have been only the reversal of the direction in which block A traveled.) Waves Energy may be transmitted by disturbing a “medium,” permitting the disturbance to travel through the medium, and then collecting the energy with a suitable receiver. For example, if work is done in raising a stone and the stone is dropped into water, the potential energy of the stone before being dropped is converted into kinetic energy, which is then transferred to the water when the stone strikes. The energy gained by the water disturbs it and causes it to move up and down. This disturbance spreads out from the point of the initial disturbance at a velocity characteristic of the medium (in this case, the water). The energy can be “received” at a remote distance from the point of the initial disturbance by a bob that ﬂoats on the water. The wave, in passing by the bob, will cause the bob to move up and down, thereby imparting energy to it. It should be noted here that the water moves only in a vertical direction while the disturbance moves in the horizontal direction. Displacement of water upward from the undisturbed surface produces a crest, while downward displacement results in a trough. The amplitude of a wave is a mea- sure of the vertical displacement, and the distance between corresponding points on adjacent disturbances is called the wavelength (Fig. 2-11). (The wavelength is usually represented by the Greek letter lambda, λ.) The number of disturbances per second 32 CHAPTER 2 Figure 2-11. Graphical representation of a wave. at any point in the medium is called the frequency. The velocity with which a wave (disturbance) travels is equal to the product of the wavelength and the frequency, v = f × λ. (2.47) W EXAMPLE 2.14 Sound waves, which are disturbances in the air, travel through air at a velocity of 344 m/s. Middle C has a frequency of 264 Hz (cycles per second). Calculate the wavelength of this note. Solution m v 344 λ= = s = 1.3 m. f 1 264 s If more than one disturbance passes through a medium at the same time, then, where the respective waves meet, the total displacement of the medium is equal to the algebraic sum of the two waves. For example, if two rocks are dropped into a pond, then, if the crests of the two waves should coincide as the waves pass each other, the resulting crest is equal to the height of the two separate crests and the trough is as deep as the sum of the two individual troughs, as shown in Figure 2-12. If, on the other hand, the two waves are exactly out of phase, that is, if the crest of one Figure 2-12. The addition of two waves of equal frequency and in phase. REVIEW OF PHYSICAL PRINCIPLES 33 Figure 2-13. The addition of two waves of equal frequency but different amplitude and 180◦ out of phase. coincides with the trough of the other, then the positive and negative displacements cancel each other, as shown in Figure 2-13. If, in Figure 2-13, wave 1 and wave 2 are of exactly the same amplitude as well as the same frequency, there would be no net disturbance. For the more general case, in which the component waves are of different frequencies, different amplitudes, and only partly out of phase, complex waveforms may be formed, as seen below in Figure 2-14. Electromagnetic Waves In 1820, Christian Oersted, a Danish physicist, observed that a compass needle de- ﬂected whenever it was placed in the vicinity of a current-carrying wire. He thus discovered the intimate relationship between electricity and magnetism and found that a magnetic ﬂux coaxial with the wire is always induced in the space around a current-carrying wire. Furthermore, he found that the direction of deﬂection of the compass needle depended on the direction of the electric current, thus showing that the induced magnetic ﬂux has direction as well as magnitude. The direction of the induced magnetic ﬂux can be determined by the “right-hand rule”: If the ﬁngers of the right hand are curled around the wire, as though grasping the wire, with the thumb outstretched and pointing in the direction of conventional current ﬂow, then the curled ﬁngers point in the direction of the induced magnetic ﬂux. If two parallel, current-carrying wires are near each other, they either attract or repel one another, depending on whether the currents ﬂow in the same or in opposite directions. The attractive or repulsive force F per unit length l of wire, as shown in Figure 2-15, is proportional to the product of the currents and inversely proportional to the distance between the wires. F i1 × i2 i1 × i2 ∝ = km ×. (2.48) l r r Figure 2-14. Complex wave formed by the algebraic addition of two different pure waves. 34 CHAPTER 2 Figure 2-15. Force between two parallel current-carrying wires. The force, attractive in this case, is due to the magnetic ﬁelds (shown by the circular lines in the end view) that are generated by the electric current. If the current-carrying wires are in free space (or in air) and if i 1 and i 2 are 1 A each and if the distance r between the wires is 1 m, then the force per unit length of wire is found to be F N = 2 × 10−7. l m The constant of proportionality km , therefore, is equal to 2 × 10−7 N/A2. It is convenient to deﬁne km in terms of another constant, μ0 : μ0 N km = = 2 × 10−7 2 (2.49a) 2π A N μ0 = 4π × 10−7. (2.49b) A2 μ0 is called the permeability of free space. Permeability is a property of the medium in which magnetic ﬂux is established. The permeabilit

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