RadioactiveDecay

RadioactiveDecay © 2006, K.E. Holbert Page 1 of 9 RADIOACTIVE DECAY Radioactive materials may either originate from natural sources or be created through technological processes. Naturally radioactive materials include carbon-14, potassium-40, and thorium and uranium isotopes and their progeny. Neutron transmutation of stable isotopes into radioisotopes is a method of artificially creating radioactive material. Of interest to the study of soft errors in circuits are the heavy decay chains of uranium and thorium. To begin our study of radioactivity, we first examine simple radioactive decay and we define terms such as half-life, decay constant, and activity. Basic Balance Equation The basic balance equation is a useful starting point for many analyses: Rate of Change = Production (Inflow) – Losses (Outflow) Simple Decay Chain For a simple radioactive decay chain, the parent radionuclide decays to a stable product. Let N(t) represent the parent radionuclide at time t, where N could be in units of total atoms (n) or atom density (N). We assume in this discussion that no production of the radionuclide occurs after t = 0, so there is an initial number of atoms equal to N(0). These radioactive atoms then decay according the decay constant (λ), which is a probability per unit time that an individual atom decays, and which can be expressed in terms of the half-life (t½) of the substance, ½)2ln( t=λ . The decay rate is a nuclear property independent of (1) temperature, (2) pressure, (3) chemical form of the isotope, and (4) physical state of the substance. Using the basic balance equation above, a first-order differential equation describing N(t) is established )(tN dt Nd λ−= (1) Laplace transforming the differential equation yields: )()0()( sNNsNs λ−=− (2) The above expression is algebraically manipulated to isolate the variable of interest, N(s): )( )0()( )0()()( λ λ += =+ s NsN NsNs (3) Finally, the inverse Laplace transform is taken to determine the time dependent concentration of the parent radionuclide for t ≥ 0: ( ) ½/21)0()0()( ttt NeNtN == −λ (4) where the decay constant (λ) and half-life of the radionuclide (t½) are related by λ )2ln( 2 1 =t (5) The average (or mean) life of a radionuclide is λλτ 1)( )0( 1 0 == ∫ ∞ dttNt N (6) The buildup of a stable decay (daughter) product, which is not initially present, would follow RadioactiveDecay © 2006, K.E. Holbert Page 2 of 9 ( )teN λ−−1)0( (7) Activity The activity is the number of decays or disintegrations per unit time [Becquerels (Bq) or Curies (Ci)] tentntA λλλ −=≡ )0()()( (8) where a Becquerel is the SI unit defined as one transformation per second, and 1 Ci = 3.7×1010 Bq. Figure 1 shows the activity of a (parent) radionuclide where the time scale (ordinate) is expressed in term of the number of half-lives of the radionuclide and the abscissa is measured in comparison to the initial activity (A0). The y-axis could equivalently be stated in terms of N(t) or n(t) as measured in reference to N0 and n0, respectively. A ct iv ity , A (t) Figure 1. Activity of a radionuclide undergoing simple decay. The radionuclide activity is normalized to the initial activity, A(0), and time is in terms of the number of half-lives. Example: Given 1 gram-mole of potassium (K) today, compute: (a) the activity today in Bq, and (b) the number of K-40 atoms one billion years from now. Solution: A gram-mole of any element is 6.022×1023 atoms, and in this case equals the number of potassium (K) atoms. Natural potassium is composed of two stable isotopes, K-39 and K-41, and a radioisotope, K-40, which is only 0.0117 atom percent. Hence, today there are atoms40-K107.046atoms)K10022.6)(000117.0( 1923atomsK atoms40-K 40K ×=×=−n The half-life of K-40 is 1.277×109 years, such that the decay constant is sec 117 9 ½ 10721.1 sec3600 hr1 hr24 day1 day365 yr1 yr10277.1 )2ln()2ln( −×=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×== tλ (a) The activity of the K-40 today is Bq1213)atoms10046.7)(10721.1( 19sec 117 00 =××== −nA λ (b) The number of K-40 atoms in 109 years may be found using Eq. (4): RadioactiveDecay © 2006, K.E. Holbert Page 3 of 9 ( ) ( ) atoms10095.4)atoms10046.7()( 19yr10277.1/yr102119/210 99½ ×=×== ×ttntn Specific Activity The specific activity is the activity per unit mass (m) of the radionuclide [Bq/g or Ci/g] M N M Nm mm n m ASA AvAv λλλ ===≡ (9) where M is the atomic weight. This expression shows that the specific activity is independent of the actual mass and is a fixed value (i.e., time independent) for a particular radionuclide. Example: Compute the specific activity of cobalt-60. Solution: The half-life of Co-60 is 5.27 years. The specific activity is computed using Eq. (9): g Bq1019.4 sec3600 hr1 hr24 day1 day365 yr1 mole)-g/g60( )mole-atoms/g10022.6( )yrs27.5( )2ln( 1323 60-Co ×=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛×=SA Effective Half-life The effective half-life takes into account both the radioactive decay and the biological removal of a radioisotope. It is the combination of the radioactive half-life and the biological half-life (like resistors in a parallel circuit). biorad biorad eff eff bioradeff tt tt t ½,½, ½,½, ½, )2ln( +== += λ λλλ (10) Example: Iodine-131 has a radioactive half-life of 8 days and a biological half-life of 120 days as it tends to collect in the thyroid. What is the effective half-life of I-131? Solution: days5.7 )days1208( )days120)(days8( ½, =+=efft Table I. Natural Heavy Decay Chains Series Decay Chain Parent Parent Half-life (yrs) Stable End Product (4n+0) Thorium Th23290 1.405 × 10 10 Pb20882 (4n+1) Neptunium Np23793 2.14 × 10 6 Bi20983 (4n+2) Uranium U23892 4.468 × 10 9 Pb20682 (4n+3) Actinium U23592 7.038 × 10 8 Pb20782 RadioactiveDecay © 2006, K.E. Holbert Page 4 of 9 Heavy Decay Chains The natural heavy decay chains consist of four series of radionuclides as summarized in Table I. The (4n+b) expression describes the mass number of any member in the series. The numeral "4" occurs because during an alpha transition there is a change in the nucleus of four mass units. The values of "b" (b=0,1,2,3) indicate the number of neutron and/or proton departures from the thorium series (4n) where n is an integer. Hence, in the thorium series the parent and each of the daughter products has a mass number perfectly divisible by 4. Note that the series with a parent half-life of t½>1010 yrs have decayed very little while those with a half-life of t½<108 yrs are gone. The series still present are detailed in Figure 3. These heavy decay chains undergo compound (serial) and complex (branching) decay schemes, which are explored next. Compound Decay (n1→ n2→ n3) The earlier equation for simple decay can be extended to the case in which a radionuclide (n1) decays to a daughter product (n2) that is also radioactive, and which subsequently decays to a stable end product (n3). The differential equation and time domain solution for n1 are the same as the simple decay situation above. tentnsnnsnstn dt nd 1)0()()()0()()( 11111111 1 λλλ −=⇒−=−⇒−= (11) The differential equation for n2, however, includes the production of n2 from the decay of n1 )()( 2211 2 tntn dt nd λλ −= (12) The solution to this differential equation may also be accomplished with Laplace transforms, and substituting for n1(s) using an expression extracted from Eq. (11) above: )()( )0( )( )0( )( )()0( )( )()()0()( 21 11 2 2 2 112 2 221122 λλ λ λ λ λ λλ ++++= + += −=− ss n s n s snn sn snsnnsns (13) Inverse Laplace transforming this expression yields [ ]ttt eenentn 212 12 11 22 )0( )0()( λλλ λλ λ −−− −−+= (14) The differential equation for the end product (granddaughter) n3 consists only of a production term since there are no losses because the end product is stable. )(22 3 tn dt nd λ= (15) The solution may be found by integrating this expression and substituting Eq. (14) ( ) ⎟⎟⎠⎞⎜⎜⎝⎛ −−−++−+= += −−− ∫ ttt t eenenn dnntn 212 21 1 21 2 123 0 2233 1)0(1)0()0( )()0()( λλλ λλ λ λλ λ ττλ (16) Alternatively, the solution may be determined using Laplace transforms )()0()()( 223322 3 snnsnstn dt nd λλ =−⇒= (17) RadioactiveDecay © 2006, K.E. Holbert Page 5 of 9 Compound Radioactive Decay There are three cases of interest for compound decay. 1. The non-equilibrium or general case (λ1>λ2) requires the use of the full equation (i.e., Eq. (14)) for n2(t). For 0)0(2 =n , the full expression is reduced to [ ]tt eentn 21 12 11 2 )0( )( λλλλ λ −− −−= (18) The activities of the parent and daughter are graphed below where it can be seen that eventually the total activity is dominated by the daughter’s activity, that is 1½,221 7 ~for ttAAA >≅+ (19) Non-Equilibrium Compound Decay 0.0 0.2 0.4 0.6 0.8 1.0 0 2 4 6 8 10 Time (no. of half-lives of parent, A1) R el at iv e A ct iv ity A1(t)+A2(t) A2(t) A1(t) 2. Secular equilibrium occurs when the parent is very long-lived compared to the daughter, λ1<<λ2. After about seven half-lives of the daughter, the parent’s and daughter’s activities are equal as shown in the equations and figure below for 0)0(2 =n [ ] 2½,111222 2 11 2 7~for)0()0()()( 1 )0( )( 2 ttAntntA e n tn t >==≡ −≈ − λλ λ λ λ (20) Secular Equilibrium Compound Decay 0.0 0.2 0.4 0.6 0.8 1.0 0 2 4 6 8 10 Time (no. of half-lives of daughter, A2) R el at iv e A ct iv ity A1(t) A2(t) RadioactiveDecay © 2006, K.E. Holbert Page 6 of 9 3. Transient equilibrium occurs when the parent is long-lived (λ1<λ2) since eventually all the activities decay with the half-life of the parent as illustrated in the equations and figure below for 0)0(2 =n 2½, 12 2 1222 1 12 1 12 11 2 7~for)()()( )( )0( )( 1 tttAtntA tne n tn t >−=≡ −=−≈ − λλ λλ λλ λ λλ λ λ (21) Transient Equilibrium Compound Decay 0.0 0.5 1.0 1.5 0 2 4 6 8 10 Time (no. of half-lives of daughter, A2) R el at iv e A ct iv ity A1(t)+A2(t) A2(t) A1(t) Bateman Equation (n1→ n2→ n3→ …→ ni→) Bateman developed a general equation for serial decay chains*, such as the heavy decay chains of Th-232, U-235, and U-238. Assuming that the concentrations of all the daughters are initially zero (i.e., 0)0( =in for i>1), the concentration of the i-th radionuclide can be determined from ∑ ∏= ≠= − − − = i j i jk k jk t ii jentn 1 1 1121 )( )0()( λλ λλλ λ L (22) Example: Natural uranium is composed by atomic percent of 99.2745% U-238, 0.72% U-235, and 0.0055% U-234. Confirm the relative fractions of U-238 and U-234, that is, verify 050,18)0055.0/()2745.99(/ 234U238U ==−− nn Solution: We note that secular equilibrium is eventually established between U-238 and its great- grandchild, U-234, such that their activities are equal: 234-U238-U 234-U238-U )()( nn AA λλ = = From Figure 3, the half-lives of U-238 and U-234 are 4.468×109 yrs and 2.445×105 yrs, respectively. Substituting these values into the above expression yields: 274,18 yrs10445.2 yrs10468.4 /)2ln( /)2ln( 5 9 234-U½, 238-U½, 238-U½, 234-U½, 238-U 234-U 234-U 238-U =× ×==== t t t t n n λ λ Since the fraction of U-234 is only known to two significant figures, the relative fractions are confirmed. * H. Bateman, “The solution of a system of differential equations occurring in the theory of radio-active transformations,” Proc. Cambridge Phil. Soc., 15, p. 423 (1910). RadioactiveDecay © 2006, K.E. Holbert Page 7 of 9 Example: An impurity level of 5 ppm of Th-232 is present in the ceramic packaging material to be used as a direct top covering (lid) for an integrated circuit (IC), as shown below. The ceramic has a density of 4.7 g/cm3 and an effective atomic weight of 43.5 amu. Determine the maximum alpha flux into the IC. Solution: The atomic density of the ceramic is 322 233 ceramic atoms/cm1051.6mole-g/g43.5 )mole-atoms/g10022.6)(g/cm7.4( ×=×== M N N Av ρ The thorium concentration is 317226 ceramic 6 Th atoms/cm1025.3)1051.6)(105()105( ×=××=×= −− NN The decay constant of Th-232 is sec/10564.1 s3600 hr1 hr24 d1 d365 yr1 yr10405.1 )2ln()2ln( 18 10 ½ 232-Th −×=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛ ×== tλ The activity of the thorium is ( )( ) 3 cm atoms17 sec 118 232-Th Bq/cm5088.01025.310564.1 3 =××== −NA λ We must find the depth into the lid from which the alphas will have sufficient energy to escape the lid, that is, we determine an active region of the lid in terms of an alpha range into the lid. We find that Th-232 emits alphas at two different energies: 4.016 MeV (77%) and 3.957 MeV (23%). The range of these ~ 4 MeV alphas in air is cm34.262.2)MeV 4)(24.1(62.224.1 =−=−= αERair The corresponding range in the ceramic lid is computed using the Bragg-Kleeman rule ( ) ( ) μm7.55cm000755.0cm34.2 7.4 5.43103.2103.2 44 ==⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛×=×= −− airRMR ρ Assuming a one-dimensional geometry, only half the alphas at the most move toward the IC such that the maximum alpha emission flux into the IC is seccm alphas00023.0 Bq alpha/sec1)cm000755.0( cm Bq6115.0 2 1 232 1 ⋅=⎟⎟⎠ ⎞ ⎜⎜⎝ ⎛⎟⎠ ⎞⎜⎝ ⎛== dAαφ Of course, a thinner lid than 7.55 µm would emit fewer alphas. The above analysis does not include any of the alpha emissions from progeny of Th-232. Complex Radioactive Decay A radionuclide may also decay by multiple means, for example, by both α and β decay. As examples, Figure 3 shows two such complex or branching decay schemes: (1) in the Th-232 series, Bi-212 decays to either Po-212 or Tl-208, and (2) in the U-235 series, Ac-227 decays to Th-227 or Fr-223 Figure 2. Complex decay scheme in which radionuclide A decays to either B or C. RadioactiveDecay © 2006, K.E. Holbert Page 8 of 9 Suppose that radionuclide A decays proportionally to B and C according to fractions fB and fC, respectively, as depicted in Figure 2. Since λA represents the decay probability for A, then the probability of decay from A to B is ABf λλ =1 , and likewise to C is ACf λλ =2 . The overall decay probability λA is the sum of the individual probabilities, that is, 21 λλλ +=A (i.e., a joint probability from the union of the two decay paths). The balance equation for radionuclide A is therefore )()()( 21 tntndt nd AAA A λλλ +−=−= (23) The activity of A can be shown to be t AA t A t AA AAAA eneAeAtA tAtAtntntA A )()( 2121 2121 )0()0()0()( and )()()()()()( λλλλλ λ λλλ +−+−− === +=+== (24) If B and C are stable, then intuitively from Eq. (7): ( )( )tACCC t ABBB A A enfntn enfntn λ λ − − −+= −+= 1)0()0()( 1)0()0()( (25) Example: The decay of Bi-212 involves a complex decay scheme whose daughters both decay to the same grandchild (stable Pb-208) as illustrated below. Determine whether the time to decay from Bi-212 to Pb- 208 differs based on the decay branch taken. 212Bi 212Po 208Tl fβ=64% , λβ fα=36%, λα 208Pb α α β β Solution: We begin by finding the decay constant of Bi-212 /min01145.0)min 55.60/()2ln(/)2ln( ½212-Bi === tλ Next, determine the decay constants associated with the initial two (α and β) decay branches /min00411.0) /min01145.0)(3593.0( /min00734.0) /min01145.0)(6407.0( 212-Bi 212-Bi === === λλ λλ αα ββ f f From Figure 3, the half-lives of Po-212 and Tl-208 are 305 ns and 3.07 min, respectively. Using Eq. (6), the corresponding average lives ( )2ln(//1 ½t== λτ ) of the radionuclides are Upper branch: ns440)2ln(/)ns305()2ln(/ min2.136 /min)00734.0/(1/1 2/1212-Po 212,-Bi === === tτ λτ ββ Lower branch: min43.4)2ln(/min)07.3()2ln(/ min 3.243 /min)00411.0/(1/1 2/1208-Tl 212,-Bi === === tτ λτ αα This implies that the average time in the upper (Bi-212→Po-212→Pb-208) branch is shorter than in the Bi- 212→Tl-208→Pb-208 path. Noteworthy is that both of these paths represent secular equilibrium behavior. RadioactiveDecay © 2006, K.E. Holbert Page 9 of 9 35 .93 % 1. 38 % Figure 3. Thorium-232, Uranium-235 and U-238 decay chains referenced to atomic (left) and neutron (top) numbers.