Radioactive Decay, Specific Activity, and Serial Radioactive Decay PDF

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...

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.