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.