Radioactive Decay PDF

83 4 Radioactive Decay 4.1 Activity The rate of decay, or transformation, of a radionuclide is described by its activity, that is, by the number of atoms that decay per unit time. The unit of activity is...

83 4 Radioactive Decay 4.1 Activity The rate of decay, or transformation, of a radionuclide is described by its activity, that is, by the number of atoms that decay per unit time. The unit of activity is the becquerel (Bq), defined as one disintegration per second: 1 Bq = 1 s–1. The traditional unit of activity is the curie (Ci), which was originally the activity ascribed to 1 g of 226 Ra. The curie is now defined as 1 Ci = 3.7 × 1010 Bq, exactly. 4.2 Exponential Decay The activity of a pure radionuclide decreases exponentially with time, as we now show. If N represents the number of atoms of a radionuclide in a sample at any given time, then the change dN in the number during a short time dt is propor- tional to N and to dt. Letting λ be the constant of proportionality, we write dN = –λN dt. (4.1) The negative sign is needed because N decreases as the time t increases. The quan- tity λ is called the decay, or transformation, constant; it has the dimensions of in- verse time (e.g., s–1 ). The decay rate, or activity, A, is given by dN A=– = λN. (4.2) dt We separate the variables in Eq. (4.1) by writing dN = –λ dt. (4.3) N Integration of both sides gives ln N = –λt + c, (4.4) Atoms, Radiation, and Radiation Protection. James E. Turner Copyright © 2007 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim ISBN: 978-3-527-40606-7 84 4 Radioactive Decay where c is an arbitrary constant of integration, fixed by the initial conditions. If we specify that N0 atoms of the radionuclide are present at time t = 0, then Eq. (4.4) implies that c = ln N0. In place of (4.4) we write ln N = –λt + ln N0 , (4.5) N ln = –λt (4.6) N0 or N = e–λt. (4.7) N0 Equation (4.7) describes the exponential radioactive decay law. Since the activity of a sample and the number of atoms present are proportional, activity follows the same rate of decrease, A = e–λt , (4.8) A0 where A0 is the activity at time t = 0. The dose rate at a given location in the neigh- borhood of a fixed radionuclide source also falls off at the same exponential rate. The function (4.8) is plotted in Fig. 4.1. During successive times T, called the half-life of the radionuclide, the activity drops by factors of one-half, as shown. To Fig. 4.1 Exponential radioactivity decay law, showing relative activity, A/A0 , as a function of time t; λ is the decay constant and T the half-life.  4.2 Exponential Decay 85 find T in terms of λ, we write from Eq. (4.8) at time t = T, 1 2 = e–λT. (4.9) Taking the natural logarithm of both sides gives ! " –λT = ln 12 = – ln 2, (4.10) and therefore ln 2 0.693 T= =. (4.11) λ λ Written in terms of the half-life, the exponential decay laws (4.7) and (4.8) become N A = = e–0.693t/T. (4.12) N0 A0 The decay law (4.12) can be derived simply on the basis of the half-life. If, for example, the activity decreases to a fraction A/A0 of its original value after passage of time t/T half-lives, then we can write # $t/T A 1 =. (4.13) A0 2 Taking the logarithm of both sides of Eq. (4.13) gives A t 0.693t ln = – ln 2 = – , (4.14) A0 T T from which Eq. (4.12) follows. Example 24 Calculate the activity of a 30-MBq source of 11 Na after 2.5 d. What is the decay con- stant of this radionuclide? Solution The problem can be worked in several ways. We first find λ from Eq. (4.11) and then the activity from Eq. (4.8). The half-life T = 15.0 h of the nuclide is given in Appendix D. From (4.11), 0.693 0.693 λ= = = 0.0462 h–1. (4.15) T 15.0 h With A0 = 30 MBq and t = 2.5 d × 24 h d–1 = 60.0 h, –1 ×60 h) A = 30 e–(0.0462 h = 1.88 MBq. (4.16) Note that the time units employed for λ and t must be the same in order that the exponential be dimensionless. 86 4 Radioactive Decay Example A solution contains 0.10 µCi of 198 Au and 0.04 µCi of 131 I at time t = 0. What is the total beta activity in the solution at t = 21 d? At what time will the total activity decay to one-half its original value? Solution Both isotopes decay to stable daughters, and so the total beta activity is due to these isotopes alone. (A small fraction of 131 I decays into 131m Xe, which does not contribute to the beta activity.) From Appendix D, the half-lives of 198 Au and 131 I are, respectively, 2.70 days and 8.05 days. At the end of 21 days, the activities AAu and AI of the nuclides are, from Eq. (4.12), AAu = 0.10e–0.693×21/2.70 = 4.56 × 10–4 µCi (4.17) and AI = 0.04e–0.693×21/8.05 = 6.56 × 10–3 µCi. (4.18) The total activity at t = 21 days is the sum of these two activities, 7.02 × 10–3 µCi. To find the time t in days at which the activity has decayed to one-half its original value of 0.10 + 0.04 = 0.14 Ci, we write 0.07 = 0.1e–0.693t/2.70 + 0.04e–0.693t/8.05. (4.19) This is a transcendental equation, which cannot be solved in closed form for t. The solution can be found either graphically or by trial and error, focusing in between two values of t that make the right-hand side of (4.19) >0.07 and T. 88 4 Radioactive Decay 4.3 Specific Activity The specific activity of a sample is defined as its activity per unit mass, for example, Bq g–1 or Ci g–1. If the sample is a pure radionuclide, then its specific activity SA is determined by its decay constant λ, or half-life T, and by its atomic weight M as follows. Since the number of atoms per gram of the nuclide is N = 6.02 × 1023 /M, Eq. (4.2) gives for the specific activity 6.02 × 1023 λ 4.17 × 1023 SA = =. (4.24) M MT If T is in seconds, then this formula gives the specific activity in Bq g–1. In practice, using the atomic mass number A in place of M usually gives sufficient accuracy. Example Calculate the specific activity of 226 Ra in Bq g–1. Solution From Appendix D, T = 1600 y and M = A = 226. Converting T to seconds, we have 4.17 × 1023 SA = (4.25) 226 × 1600 × 365 × 24 × 3600 = 3.66 × 1010 s–1 g–1 = 3.7 × 1010 Bq g–1. (4.26) This, by definition, is an activity of 1 Ci. The fact that 226 Ra has unit specific activity in terms of Ci g–1 can be used in place of Eq. (4.24) to find SA for other radionuclides. Compared with 226 Ra, a nuclide of shorter half-life and smaller atomic mass number A will have, in direct proportion, a higher specific activity than 226 Ra. The specific activity of a nuclide of half-life T and atomic mass number A is therefore given by 1600 226 SA = × Ci g–1 , (4.27) T A where T is expressed in years. (The equation gives SA = 1 Ci g–1 for 226 Ra.) Example What is the specific activity of I4 C? Solution With T = 5730 y and A = 14, Eq. (4.27) gives 1600 226 SA = × = 4.51 Ci g–1. (4.28) 5730 14  4.4 Serial Radioactive Decay 89 Alternatively, we can use Eq. (4.24) with T = 5730 × 365 × 24 × 3600 = 1.81 × 1011 s, obtaining 4.17 × 1023 SA = = 1.65 × 1011 Bq g–1 (4.29) 14 × 1.81 × 1011 1.65 × 1011 Bq g–1 = = 4.46 Ci g–1 , (4.30) 3.7 × 1010 Bq Ci–1 in agreement with (4.28). Specific activity need not apply to a pure radionuclide. For example, 14 C produced by the 14 N(n,p)14 C reaction can be extracted chemically as a “carrier-free” radionu- clide, that is, without the presence of nonradioactive carbon isotopes. Its specific activity would be that calculated in the previous example. A different example is af- forded by 60 Co, which is produced by neutron absorption in a sample of 59 Co (100% abundant), the reaction being 59 Co(n,γ )60 Co. The specific activity of the sample de- pends on its radiation history, which determines the fraction of cobalt atoms that are made radioactive. Specific activity is also used to express the concentration of activity in solution; for example, µCi mL–1 or Bq L–1. 4.4 Serial Radioactive Decay In this section we describe the activity of a sample in which one radionuclide pro- duces one or more radioactive offspring in a chain. Several important cases will be discussed. Secular Equilibrium (T1 ≫ T2 ) First, we calculate the total activity present at any time when a long-lived parent (1) decays into a relatively short-lived daughter (2), which, in turn, decays into a sta- ble nuclide. The half-lives of the two radionuclides are such that T1 ≫ T2 ; and we consider intervals of time that are short compared with T1 , so that the activity A1 of the parent can be treated as constant. The total activity at any time is A1 plus the activity A2 of the daughter, on which we now focus. The rate of change, dN2 /dt, in the number of daughter atoms N2 per unit time is equal to the rate at which they are produced, A1 , minus their rate of decay, λ2 N2 : dN2 = A1 – λ2 N2. (4.31) dt To solve for N2 , we first separate variables by writing dN2 = dt, (4.32) A1 – λ2 N2 90 4 Radioactive Decay where A1 can be regarded as constant. Introducing the variable u = A1 – λ2 N2 , we have du = –λ2 dN2 and, in place of Eq. (4.32), du = –λ2 dt. (4.33) u Integration gives ln(A1 – λ2 N2 ) = –λ2 t + c, (4.34) where c is an arbitrary constant. If N20 represents the number of atoms of nuclide (2) present at t = 0, then we have c = ln(A1 – λ2 N20 ). Equation (4.34) becomes A1 – λ2 N2 ln = –λ2 t, (4.35) A1 – λ2 N20 or A1 – λ2 N2 = (A1 – λ2 N20 )e–λ2 t. (4.36) Since λ2 N2 = A2 , the activity of nuclide (2), and λ2 N20 = A20 is its initial activity, Eq. (4.36) implies that A2 = A1 (1 – e–λ2 t ) + A20 e–λ2 t. (4.37) In many practical instances one starts with a pure sample of nuclide (1) at t = 0, so that A20 = 0, which we now assume. The activity A2 then builds up as shown in Fig. 4.4. After about seven daughter half-lives (t ! 7T2 ), e–λ2 t ≪ 1 and Eq. (4.37) reduces to the condition A1 = A2 , at which time the daughter activity is equal to Fig. 4.4 Activity A2 of relatively short-lived radionuclide daughter (T2 ≪ T1 ) as a function of time t with initial condition A20 = 0. Activity of daughter builds up to that of the parent in about seven half-lives (∼7T2 ). Thereafter, daughter decays at the same rate it is produced (A2 = A1 ), and secular equilibrium is said to exist.  4.4 Serial Radioactive Decay 91 that of the parent. This condition is called secular equilibrium. The total activity is 2A1. In terms of the numbers of atoms, N1 and N2 , of the parent and daughter, secular equilibrium can be also expressed by writing λ1 N1 = λ2 N2. (4.38) A chain of n short-lived radionuclides can all be in secular equilibrium with a long- lived parent. Then the activity of each member of the chain is equal to that of the parent and the total activity is n + 1 times the activity of the original parent. General Case When there is no restriction on the relative magnitudes of T1 and T2 , we write in place of Eq. (4.31) dN2 = λ1 N1 – λ2 N2. (4.39) dt With the initial condition N20 = 0, the solution to this equation is λ1 N10 –λ1 t –λ2 t N2 = (e –e ), (4.40) λ 2 – λ1 as can be verified by direct substitution into (4.39). This general formula yields Eq. (4.38) when λ2 ≫ λ1 and A20 = 0, and hence also describes secular equilibrium. Transient Equilibrium (T1 ! T2 ) Another practical situation arises when N20 = 0 and the half-life of the parent is greater than that of the daughter, but not greatly so. According to Eq. (4.40), N2 and hence the activity A2 = λ2 N2 of the daughter initially build up steadily. With the continued passage of time, e–λ2 t eventually becomes negligible with respect to e–λ1 t , since λ2 > λ1. Then Eq. (4.40) implies, after multiplication of both sides by λ2 , that λ2 λ1 N10 e–λ1 t λ2 N2 =. (4.41) λ 2 – λ1 Since A1 = λ1 N1 = λ1 N10 e–λ1 t is the activity of the parent as a function of time, this relation says that λ2 A1 A2 =. (4.42) λ 2 – λ1 Thus, after initially increasing, the daughter activity A2 goes through a maximum and then decreases at the same rate as the parent activity. Under this condition, illustrated in Fig. 4.5, transient equilibrium is said to exist. The total activity also 92 4 Radioactive Decay Fig. 4.5 Activities as functions of time when T1 is somewhat larger than T2 (T1 ! T2 ) and N20 = 0. Transient equilibrium is eventually reached, in which all activities decay with the half-life T1 of the parent. reaches a maximum, as shown in the figure, at a time earlier than that of the max- imum daughter activity. Equation (4.42) can be differentiated to find the time at which the daughter activity is largest. The result is (Problem 25) 1 λ2 t= ln , for maximum A2. (4.43) λ 2 – λ1 λ 1 The total activity is largest at the earlier time (Problem 26) 1 λ22 t= ln , for maximum A1 + A2. (4.44) λ2 – λ1 2λ1 λ2 – λ21 The time at which transient equilibrium is established depends on the individual magnitudes of T1 and T2. Secular equilibrium can be viewed as a special case of transient equilibrium in which λ2 ≫ λ1 and the time of observation is so short that the decay of the activity A1 is negligible. Under these conditions, the curve for A1 in Fig. 4.5 would be flat, A2 would approach A1 , and the figure would resemble Fig. 4.4.  4.4 Serial Radioactive Decay 93 No Equilibrium (T1 < T2 ) When the daughter, initially absent (N20 = 0), has a longer half-life than the parent, its activity builds up to a maximum and then declines. Because of its shorter half- life, the parent eventually decays away and only the daughter is left. No equilibrium occurs. The activities in this case exhibit the patterns shown in Fig. 4.6. Example Starting with a 10.0-GBq (= 1010 Bq) sample of pure 90 Sr at time t = 0, how long will it take for the total activity (90 Sr + 90 Y) to build up to 17.5 GBq? Solution Appendix D shows that 90 – 90 38 Sr β decays with a half-life of 29.12 y into 39 Y, which β – 90 decays into stable 40 Zr with a half-life of 64.0 h. These two isotopes illustrate a long- lived parent (T1 = 29.12 y) decaying into a short-lived daughter (T2 = 64.0 h). Secular equilibrium is reached in about seven daughter half-lives, that is, in 7 × 64 = 448 h. At the end of this time, the 90 Sr activity A1 has not diminished appreciably, the 90 Y activity A2 has increased to the level A2 = A1 = 10.0 GBq, and the total activity is 20.0 GBq. In the present problem we are asked, in effect, to find the time at which the 90 Y activity reaches 7.5 GBq. The answer will be less than 448 h. Equation (4.37) with A20 = 0 applies here.1) The decay constant for 90 Y is λ2 = 0.693/T2 = 0.693/64.0 = 0.0108 h–1. With A1 = 10.0 GBq, A2 = 7.5 GBq, and A20 = 0, Eq. (4.37) gives 7.5 = 10.0(1 – e–0.0108t ), (4.45) where t is in hours. Rearranging, we have e–0.0108t = 14 , (4.46) giving t = 128 h. (In this example note that the 90 Y activity increases in an in- verse fashion to the way a pure sample of 90 Y would decay. It takes two half-lives, 2T2 = 128 h, for the activity to build up to three-fourths its final value at secular equi- librium.) Example How many grams of 90 Y are in secular equilibrium with 1 mg of 90 Sr? Solution The amount of 90 Y will be that having the same activity as 1 mg of 90 Sr. The specific activity, SA, of 90 Sr (T1 = 29.12 y) is [ from Eq. (4.27)] 1600 226 SA1 = × = 138 Ci g–1. (4.47) 29.12 90 1 Equation (4.40), describing the general case working, one will obtain the same numerical without restriction on the relative magnitudes answer from the simplified Eq. (4.37), which of T1 and T2 , can always be applied. To the already contains the appropriate degree of accuracy with which we are approximations. 94 4 Radioactive Decay Fig. 4.6 Activities as functions of time when T2 > T1 and N20 = 0. No equilibrium conditions occur. Eventually, only the daughter activity remains. Therefore, the activity of the 1 mg sample of 90 Sr is A1 = 10–3 g × 138 Ci g–1 = 0.138 Ci, (4.48) which is also equal to the activity A2 of the 90 Y. The latter has a specific activity 1600 y 226 SA2 = × (4.49) 1 d 1 y 90 64.0 h × × 24 h 365 d = 5.50 × 105 Ci g–1. (4.50) Therefore, the mass of 90 Y in secular equilibrium with 1 mg of 90 Sr is 0.138 Ci = 2.51 × 10–7 g = 0.251 µg. (4.51) 5.50 × 105 Ci g–1 Example A sample contains 1 mCi of 191 Os at time t = 0. The isotope decays by β – emission into metastable 191m Ir, which then decays by γ emission into 191 Ir. The decay and half-lives can be represented by writing 191 β– γ 76 Os ––––––––→ 191m → 191 77 Ir –––––––– 77 Ir. (4.52) 15.4 d 4.94 s (a) How many grams of 191 Os are present at t = 0? (b) How many millicuries of 191m Ir are present at t = 25 d?  4.4 Serial Radioactive Decay 95 (c) How many atoms of 191m Ir decay between t = 100 s and t = 102 s? (d) How many atoms of 191m Ir decay between t = 30 d and t = 40 d? Solution As in the last two examples, the parent half-life is large compared with that of the daughter. Secular equilibrium is reached in about 7 × 4.9 = 34 s. Thereafter, the ac- tivities A1 and A2 of the 191 Os and 191m Ir remain equal, as they are in secular equilib- rium. During the periods of time considered in (b) and in (d), however, the osmium will have decayed appreciably; and so one deals with an example of transient equilib- rium. The problem can be solved as follows. (a) The specific activity of 191 Os is, from Eq. (4.27), 1600 × 365 226 SA1 = × = 4.49 × 104 Ci g–1. (4.53) 15.4 191 The mass of the sample, therefore, is 10–3 Ci = 2.23 × 10–8 g. (4.54) 4.49 × 104 Ci g–1 (b) At t = 25 d, A2 = A1 = 1 × e–0.693×25/15.4 = 0.325 mCi. (4.55) (c) Between t = 100 s and 102 s secular equilibrium exists with the osmium source essentially still at its original activity. Thus the 191m Ir decay rate at t = 100 s is A2 = 1 mCi = 3.7 × 107 s–1. During the next 2 s the number of 191m Ir atoms that decay is 2 × 3.7 × 107 = 7.4 × 107. (d) This part is like (c), except that the activities A1 and A2 do not stay constant during the time between 30 and 40 d. Since transient equilibrium exists, the numbers of atoms of 191m Ir and 191 Os that decay are equal. The number of 191m Ir atoms that decay, therefore, is equal to the integral of the 191 Os activity during the specified time (t in days): % & 7 40 –0.693t/15.4 3.7 × 107 –0.0450t &&40 3.7 × 10 e dt = e & (4.56) 30 –0.0450 30 = –8.22 × 108 (0.165 – 0.259) = 7.73 × 107. (4.57) 96 4 Radioactive Decay 4.5 Natural Radioactivity All of the heavy elements (Z > 83) found in nature are radioactive and decay by al- pha or beta emission. The nuclide 209 83 Bi is the only one with atomic number greater than that of lead (82) that is stable. The heaviest elements decay into successive ra- dioactive daughters, forming series of radionuclides that end when a stable species is produced. It is found that all of the naturally occurring heavy radionuclides be- long to one of three series. Since the atomic mass number can change by only four units (viz., when alpha emission occurs), a given nuclide can be easily identified as belonging to one series or another by noting the remainder obtained when its mass number is divided by four. The uranium series, for example, begins with 238 91 U and 206 ends with stable 82 Pb. When divided by four the mass number of every member of the uranium series has remainder two. The thorium series, starting with 232 90 Th and 208 ending with 82 Pb, has remainder zero. The third group, the actinium series, which 207 begins with 235 92 U and ends with 82 Pb, has remainder three. A fourth series, with remainder one, is the neptunium series. However, its longest-lived member, 237 93 Np, has a half-life of 2.2 × 106 years, which is short on a geological time scale. Neptu- nium is not found in nature, but has been produced artificially, starting with 241 94 Pu and ending with 209 82 Pb. All four series contain one gaseous member (an isotope of Rn) and end in a stable isotope of Pb. Primordial 238 U would be found in secular equilibrium with its much shorter- lived daughters, if undisturbed by physical or chemical processes in nature. It is more likely, however, that secular equilibrium will be found only among certain subsets of nuclides in the series. In this regard, a significant change occurs when 226 Ra decays into 222 Rn. The daughter, radon, is a noble gas, not bound chemically in the material where its parents resided. The half-life of 222 Rn is long enough for much of the gas to work its way out into the atmosphere. As seen from Table 1.1, radon (more precisely, its short-lived daughters) contributes an average of about one-half the effective dose to persons from natural background radiation. This im- portant source of human exposure is discussed in the next section. Several lighter elements have naturally occurring, primordial radioactive iso- topes. One of the most important from the standpoint of human exposure is 40 K, which has an isotopic abundance of 0.0118% and a half-life of 1.28 × 109 years. The nuclide decays by β – emission (89%) or EC (11%). The maximum β – energy is 1.312 MeV. This isotope is an important source of human internal and external ra- diation exposure, because potassium is a natural constituent of plants and animals. In addition to the beta particle, 40 K emits a penetrating gamma ray (1.461 MeV) fol- lowing electron capture (11%). Other naturally occurring radionuclides are of cosmogenic origin. Only those produced as a result of cosmic-ray interactions with constituents of the atmosphere result in any mentionable exposure to man: 3 H, 7 Be, 14 C, and 22 Na. The reaction 14 N(n, p)14 C with atmospheric nitrogen produces radioactive 14 C, which has a half- life of 5730 y. The radioisotope, existing as CO2 in the atmosphere, is utilized by  4.6 Radon and Radon Daughters 97 plants and becomes fixed in their structure through photosynthesis. The time at which 14 C was assimilated in a previously living specimen, used to make furniture or paper, for example, can be inferred from the relative amount of the isotope re- maining in it today. Thus the age of such objects can be determined by radiocarbon dating. In modern times, the equilibrium of natural 14 C and 3 H in the atmosphere has been upset by the widespread burning of fossil fuels and by the testing of nu- clear weapons in the atmosphere. Example How many alpha and beta particles are emitted by a nucleus of an atom of the ura- nium series, which starts as 238 206 92 U and ends as stable 82 Pb? Solution Nuclides of the four heavy-element radioactive series decay either by alpha or beta emission. A single disintegration, therefore, either (1) reduces the atomic number by 2 and the mass number by 4 or (2) increases the atomic number by 1 and leaves the mass number unchanged. Since the atomic mass numbers of 238 206 92 U and 82 Pb differ by 32, it follows that 8 alpha particles are emitted in the series. Since this alone would reduce the atomic number by 16, as compared with the actual reduction of 10, a total of 6 beta particles must also be emitted. 4.6 Radon and Radon Daughters As mentioned in the last section, the noble gas 222 Rn produced in the uranium series can become airborne before decaying. Soil and rocks under houses are or- dinarily the principal contributors to indoor radon, which is typically four or five times more concentrated than radon outdoors, where greater air dilution occurs. Additional contributions to indoor radon come from outside air, building materi- als, and the use of water and natural gas. Circumstances vary widely in time and place, and so exceptions to generalizations are frequent. Airborne radon itself poses little health hazard. As an inert gas, inhaled radon is not retained in significant amounts by the body. The potential health hazard arises when radon in the air decays, producing nongaseous radioactive daughters. When inhaled, the airborne daughters can be trapped in the respiratory system, where they are likely to decay before being removed by normal lung-clearing mech- anisms of the body. Some of the daughter atoms in air are adsorbed onto micron- or submicron-sized aerosols or dust particles (the “attached fraction”). Others (the “unattached fraction”) remain in the air as essentially free ions or in small mole- cular agglomerates (e.g., with several water molecules). Still other decay products plate out on various surfaces. In assessing hazard, it is common to characterize the radon daughters in an atmosphere by specifying the unattached fraction of each. When inhaled, this fraction is trapped efficiently, especially in the upper respiratory tract. 98 4 Radioactive Decay As shown in Fig. 4.7, 222 Rn decays into a series of short-lived daughters, two of which, 218 Po and 214 Po, are alpha emitters.2) When an alpha particle is emitted in the lung, it deposits all of its energy locally within a small thickness of adjacent tissue. An alpha particle from 214 Po, for example, deposits its 7.69 MeV of energy within about 70 µm. A 1-MeV beta particle from 214 Bi, on the other hand, deposits its energy over a much larger distance of about 4000 µm. The dose to the cells of the lung from the beta (and gamma) radiation from radon daughters is very small compared with that from the alpha particles. The “radon problem,” technically, is that of alpha-particle irradiation of sensitive lung tissue by the short-lived daughters of radon and the associated risk of lung cancer. The health hazard from radon is thus closely related to the air concentration of the potential alpha-particle energy of the short-lived daughters. Depending on local conditions, the daughters will be in various degrees of secular equilibrium with one another and with the parent radon in an atmosphere. Rather than using individual concentrations of the various progeny, one can characterize an atmosphere radio- logically by means of a collective quantity: the potential alpha-energy concentration (PAEC). The PAEC is defined as the amount of alpha energy per unit volume of Fig. 4.7 Radon and radon daughters. Alpha emission is represented by an arrow slanting downward toward the right; beta emission, by a vertical arrow. Alpha-particle and average beta-particle energies and half-lives are shown in the boxes. 2 In earlier terminology, the successive short-lived daughters, 218 Po through 210 Pb, were called RaA, RaB, RaC, RaC′ , and RaD, respectively.  4.6 Radon and Radon Daughters 99 undisturbed air that would ultimately be released from the particular mixture of short-lived daughters in their decay to 210 Pb. The PAEC can be expressed in J m–3 or MeV m–3. For a given PAEC, the equilibrium-equivalent decay-product concen- tration (EEDC) is defined as the concentration of each decay product that would be present if secular equilibrium existed. The ratio of the EEDC and the concen- tration of radon is called the equilibrium factor. By definition, this factor is equal to unity if the radon and all of its short-lived daughters are in secular equilibrium. Equilibrium factors for most indoor atmospheres are in the range of 0.2 to 0.6, a factor of 0.5 often being assumed as a rule of thumb. A limitation of the quantities described in this paragraph is that they do not distinguish between the attached and unattached fractions. Until now, we have discussed only 222 Rn, which is a member of the uranium se- ries. Radon is also generated in the other two series of naturally occurring radionu- clides. However, these isotopes of radon are of lesser radiological importance. The thorium series generates 220 Rn, which is also called thoron. The parent nuclide, 232 Th, is somewhat more abundant than 238 U, but has a longer half-life. As a re- sult, the average rate of production of 220 Rn in the ground is about the same as that of 222 Rn. However, the shorter half-life of 220 Rn, 56 s, as compared with 3.82 d for 222 Rn, gives it a much greater chance to decay before becoming airborne. The con- tributions of the daughters of 220 Rn to lung dose are usually negligible compared with 222 Rn. The third (actinium) series produces 219 Rn, also called actinon, after several transformations from the relatively rare original nuclide 235 U. Its half-life is only 4 s, and its contribution to airborne radon is insignificant. Example Measurements of room air show the nuclide activity concentrations given in Table 4.1. Calculate the PAEC for this case. Solution The PAEC (and EEDC) pertain to the short-lived decay products and do not involve the radon itself, which is not retained by the lungs. To obtain the PAEC, we need to calculate the number of daughter atoms of each type per unit volume of air; multiply these numbers by the potential alpha-particle energy associated with each type of Table 4.1 Activity Concentration Nuclide (Bq m–3 ) 222 Rn 120 218 Po 93 214 Pb 90 2l4 Bi 76 214 Po 76 100 4 Radioactive Decay Table 4.2 A λ N E NE Nuclide (Bq m–3 ) (s–1 ) (m–3 ) (MeV) (MeV m–3 ) 2l8 Po 93 3.79 × 10–3 2.45 × 104 13.69 3.35 × 105 214 Pb 90 4.31 × 10–4 2.09 × 105 7.69 1.61 × 106 214 Bi 76 5.83 × 10–4 1.30 × 105 7.69 1.00 × 106 214 Po 76 4.23 × 103 1.80 × 10–2 7.69 1.38 × 10–1 atom; and then sum. The number of atoms N of a radionuclide associated with an activity A is given by Eq. (4.2), N = A/λ, where λ is the decay constant. For the first daughter, 218 Po, for example, we find from the half-life T = 183 s given in Fig. 4.7 (or Appendix D) that λ = 0.693/T = 3.79 × 10–3 s–1. From the activity density A = 93 Bq m–3 given in Table 4.1, it follows that the number density of 218 Po atoms is A 93 Bq m–3 N= = = 2.45 × 104 m–3 , (4.58) λ 3.79 × 10–3 s–1 where the units Bq and s–1 cancel. Each atom of 218 Po will emit a 6.00-MeV alpha par- ticle. Each will also lead to the emission later of a 7.69-MeV alpha particle with the de- cay of its daughter 214 Po into 210 Pb. Thus, the presence of one 218 Po atom represents a potential alpha-particle energy E = 6.00 + 7.69 = 13.69 MeV from the short-lived radon daughters. Using N from Eq. (4.58), we find for the potential alpha-particle en- ergy per unit volume contributed by the atoms of 218 Po, NE = 3.35 × 105 MeV m–3. Similar calculations can be made for the contributions of the other three daughters in Table 4.1 to the PAEC. The only modification for the others is that the potential alpha energy associated with each atom is 7.69 MeV (Fig. 4.7). The complete calculation is summarized in Table 4.2. The individual nuclide contributions in the last column can be added to give the final answer, PAEC = 2.95 × 106 MeV m–3. Note that the half-life of 214 Po, which is in secular equilibrium with 214 Bi (equal activity densities), is so short that very few atoms are present. Its contribution to the PAEC is negligible. Example Calculate the EEDC in the last example. What is the equilibrium factor? Solution By definition, the EEDC is the activity concentration of the short-lived radon daugh- ters that would give a specified value of the PAEC under the condition of secular equilibrium. For the last example, the EEDC is the (equal) concentration that would appear in the second column of Table 4.2 for each nuclide that would result in the given value, PAEC = 2.95 × 106 MeV m–3. The solution can be set up in more than one way. We compute the PAEC for secular equilibrium at unit activity density  4.6 Radon and Radon Daughters 101 Table 4.3 NE/A Nuclide (MeV Bq–1 ) 218 Po 3.60 × 103 214 Pb 1.79 × 104 214 Bi 1.32 × 104 214 Po 1.82 × 10–3 (1 Bq m–3 ), from which the answer follows immediately. The contribution per unit activity from 218 Po, for example, is obtained from Table 4.2: NE 3.35 × 105 MeV m–3 = = 3.60 × 103 MeV Bq–1. (4.59) A 93 Bq m–3 Values for the four nuclides are shown in Table 4.3. Adding the numbers in the sec- ond column gives a total of 3.47 × 104 MeV Bq–1. This is the PAEC (MeV m–3 ) per unit activity concentration (Bq m–3 ) of each daughter in secular equilibrium. There- fore, for the last example we have PAEC EEDC = 3.47 × 104 MeV Bq–1 (4.60) 2.95 × 106 MeV m–3 = 4 –1 = 85.0 Bq m–3. 3.47 × 10 MeV Bq The equilibrium factor is the ratio of this activity concentration and that of the radon (Table 4.1): 85.0/120 = 0.708. This example shows the useful relationship between an equilibrium concentra- tion of 1 Bq m–3 of the short-lived radon daughters and the associated potential alpha-energy concentration: 3.47 × 104 MeV m–3 = 3.47 × 104 MeV Bq–1. (4.61) 1 Bq m–3 Note that the potential alpha energy per unit activity of the daughters in secular equilibrium is independent of the actual concentration of the daughters in air. An older unit for the PAEC is the working level (WL), defined as a potential alpha- particle energy concentration of 1.3 × 105 MeV L–1 of air for the short-lived radon daughters. This value corresponds to the presence of 100 pCi L–1 = 3.7 Bq L–1 of the daughters in secular equilibrium, that is, to an EEDC of 3.7 Bq L–1. The exposure of persons to radon daughters is often expressed in working-level months (WLM), with a working month defined as 170 h. The WLM represents the integrated expo- sure of an individual over a specified time period. Concentrations of radon itself are sometimes reported, rather than PAECs or WLs, which pertain to the daugh- ters. As a rule of thumb, 1 WL of radon daughters is often associated with a radon concentration of 200 pCi L–1 , corresponding to an equilibrium factor of 0.5. 102 4 Radioactive Decay Example A person spends an average of 14 hours per day at home, where the average concen- tration of radon is 1.5 pCi L–1 (a representative value for many residences). What is his exposure in WLM over a six-month period? Solution Using the rule of thumb just given, we estimate the daughter concentration to be (1.5/200) (1 WL) = 0.0075 WL. The exposure time in hours for the six months (= 183 d) is 183 × 14 = 2560 h. Since there are 170 h in a working month, the ex- posure time is 2560/170 = 15.1 working months. The person’s exposure to radon daughters over the six-month period is therefore 0.0075 × 15.1 = 0.11 WLM. 4.7 Suggested Reading 1 Attix, F. H., Introduction to Radiolog- [This book discusses a number of en- ical Physics and Radiation Dosimetry, vironmental and human radiological Wiley, New York, NY (1986). [Chap- health issues. It contains extensive ter 6 discusses radioactive and serial nuclear data. A CD-ROM is included.] decay.] 7 National Research Council, Health 2 Bodansky, D., Robkin, M. A., and Risks of Radon and Other Internally Stadler, D. R., eds., Indoor Radon and Deposited Alpha-Emitters: BEIR IV, Na- Its Hazards, University of Washington tional Academy Press, Washington, Press, Seattle, WA (1987). DC (1988). [A comprehensive report 3 Cember, H., Introduction to Health (602 pp.) from the Committee on the Physics, 3rd ed., McGraw-Hill, New Biological Effects of Ionizing Radia- York, NY (1996). [Chapter 4 treats tion.] radioactive transformations, serial 8 Nazaroff, W. W., and Nero, A. V., Jr., decay, and naturally occurring radioac- eds., Radon and its Decay Products tivity.] in Indoor Air, Wiley, New York, NY 4 Evans, R. D., The Atomic Nucleus, (1988). McGraw-Hill, New York, NY (1955). 9 NCRP Report No. 97, Measurement [Chapter 15 describes serial decay, de- of Radon and Radon Daughters in Air, cay schemes, equilibrium, and other National Council on Radiation Pro- topics.] tection and Measurements, Bethesda, 5 Faw, R. E., and Shultis, J. K., Radio- MD (1988). [Contains basic data on logical Assessment, Prentice-Hall, En- the three natural decay series and glewood Cliffs, NJ (1993). [Chapter 4 characteristics of radon daughters dis- gives and excellent coverage of expo- cussed here. Main emphasis of the sure to natural sources of radiation, Report is on measurements.] including radon.] 10 NCRP Report No. 103, Control of 6 Magill, Joseph, and Galy, Jean, Radon in Houses, National Council Radioactivity-Radionuclides-Radiation, on Radiation Protection and Measure- Springer Verlag, New York, NY (2005). ments, Bethesda, MD (1989).