Radioactive Decay Calculations PDF

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...

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.