advertisement

1 Radiation Sources and Radioactive Decay 1.1 Definitions and Equations 1.1.1 Radioactivity and Decay Equations Activity Activity is defined as A= dN = λN Unit : 1 Bq (becquerel) = 1 s−1 dt (1.1.1) where A is activity, dN/dt is the number of spontaneous nuclear transformations, dN, from a particular energy state in a time interval dt. λ is the decay constant (s−1 ) and N is the number of radioactive nuclei. The specific activity is defined as the activity of a certain radionuclide per mass unit (Bq kg−1 ). C= A m (1.1.2) Radioactive decay A radionuclide decays according to the equation N1 (t) = N0 e−λt = N0 e−t ln 2/T (1.1.3) where N1 (t) is the number of radioactive nuclides after a time t, N0 = N1 (0) is the number of radioactive nuclides at time 0 and T is the half-life (T = ln 2/λ). The equation may also be expressed as A1 (t) = A0 e−λt = A0 e−t ln 2/T (1.1.4) Sometimes the daughter nuclides are also radioactive and a chain of radioactive nuclides is obtained. A general solution for the activity of a radionuclide in the chain is given by the Bateman equations. In this compilation only the first three radionuclides in the chain will be treated. N1 (t) = N0 e−λ1 t (1.1.5) λ1 (e−λ1 t − e−λ2 t ) λ2 − λ1 (1.1.6) e−λ1 t e−λ2 t e−λ3 t + + ] (λ3 − λ1 )(λ2 − λ1 ) (λ3 − λ2 )(λ1 − λ2 ) (λ2 − λ3 )(λ1 − λ3 ) (1.1.7) N2 (t) = N0 N3 (t) = N0 λ1 λ2 [ © 2015 Bo Nilsson This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 License. Unauthenticated Download Date | 2/4/20 3:47 PM 2 | Radiation Sources and Radioactive Decay All these equations assume that N2 (0) and N3 (0) are equal to zero. If not, corrections have to be made, by adding to the equation above, the activity of the separate radionuclides at t = 0 corrected for the decay to the time t. E.g. N20 (t) = N20 (0)e−λ2 t (1.1.8) where N20 (0) is the number of N20 radionuclides at time, t=0. In some situations Eq. (1.1.6) may be simplified, as shown below. If λ1 << λ2 , then λ (1.1.9) N2 (t) = N0 1 (1 − e−λ2 t ) λ2 For large values of t, this equation is reduced to (secular equilibrium) λ1 N1 = λ2 N2 (1.1.10) If there is a chain of radionuclides with a long lived parent radionuclide, then this simplification may be extended to the whole chain. λ1 N1 = λ2 N2 = λ3 N3 = ... = λ k N k = ... (1.1.11) If λ1 < λ2 the equation for large t is reduced to (transient equilibrium) N2 (t) = N0 λ1 e−λ1 t λ1 − λ2 (1.1.12) Production of radionuclides The number of produced radionuclides at time t is given by the equation N(t) = σ Φ̇mNA (e−σ Φ̇t − e−λt ) ma (λ − σ Φ̇) (1.1.13) where σ is the activation cross section (m2 ), Φ̇ is the particle fluence rate (m−2 s−1 ), NA =6.022·1026 (kmol−1 ) is Avogadro’s number, ma is the atomic mass of the target atom, and m is the mass of the target (kg). If the decrease of the number of target nuclei during activation can be neglected the equation is reduced to N(t) = σ Φ̇mNA (1 − e−λt ) ma λ (1.1.14) The corresponding equation for the activity is A(t) = σ Φ̇mNA (1 − e−λt ) ma (1.1.15) Unauthenticated Download Date | 2/4/20 3:47 PM Definitions and Equations | 3 1.1.2 Disintegration Schematics In many situations it is important to know the number of particles emitted per decay and their energies, e.g. when determining the activity of a sample using spectroscopic measurements, or in internal dosimetry. In particular when there are nuclei with excited levels, the de-excitation can lead to both photons and electrons of different energies with different probabilities. A full calculation including all particles is often complicated and in practice one should use tables such as the Table of Radioactive Isotopes (Browne and Firestone, 1986), MIRD: Radionuclide Data and Decay schemes (Eckerman and Endo, 2008), the Table of Radioactive Isotopes (ICRP, 1986) or Nuclear Data Tables (ICRP 2001), but for illustration a simplified example of the decay of 137 Cs is discussed here (Fig 1.1). 137 Cs is a β-emitting radionuclide. As 137 Cs has a half life of 30 y and 137m Ba a half life of 2.55 min, it is often assumed that 137 Cs and 137m Ba are in radioactive equilibrium. Figure 1.1: Decay scheme for 137 Cs. I. Decay of 137 Cs 137 Cs emits two β-particles with different energies and frequencies. 5.4% of the decays go to the stable level of 137 Ba, and 94.6% go to the metastable level 137m Ba. The β-particles are emitted giving a spectrum of energies, as part of the decay energy goes to the neutrino, and both the maximum energy and the mean energy are of interest. The maximum energy is obtained from the energy released in the decay, Qβ , which is obtained from the difference in mass of the parent nucleus and the daughter nucleus. As the β-particle has a mass that is very small compared to the parent nu- Unauthenticated Download Date | 2/4/20 3:47 PM 4 | Radiation Sources and Radioactive Decay cleus, the total decay energy can be assumed to be distributed between the β-particle and the neutrino. The Qβ -value of the 137 Cs-decay is 1.175 MeV. Thus the maximum energy for β1 resulting in the daughter nucleus in a stable state is 1.175 MeV. The maximum energy of β2 , going to the excited state of 137m Ba is obtained by subtracting the energy of the excited level, 0.6616 MeV giving 1.175 − 0.6616 = 0.513 MeV. The shape of the β-particle spectrum is dependent on the type of the decay, the decay energy and the atomic number (See Table 1.1). Data for the mean energy for the different types of decay may be found e.g. in MIRD (Eckerman and Endo, 2008). Table 1.1: Frequency, mean and maximum energies of the β-particles emitted in the 137 Cs-decay. Particle Frequency β1 β2 0.054 0.946 Mean energy Mev 1.175·0.361=0.423 0.513·0.34=0.175 Maximum energy Mev 1.175 0.513 Type of decay 2nd forbidden Unique 1st forbidden II. De-excitation of 137m Ba The excitation energy can be released either by emission of a γ -photon or an internal conversion electron. In the internal conversion electron process, the excitation energy is transferred to an electron in an electron shell, mainly the K-shell. The probability that an electron will be emitted is given by the internal conversion coefficient α defined as α=Ne /Nγ . αK =NeK /Nγ gives only the internal conversion electrons from the K-shell and so on. Ne is the number of internal conversion electrons per de-excitation and Nγ is the number of γ -photons per de-excitation. The electron will obtain the excitation energy minus the binding energy of the electron. The emission of internal conversion electrons will result in vacancies in the electron shells. These vacancies will be filled by electrons from outer shells resulting in either emission of a fluorescence x-ray, also called characteristic x-ray, or an Auger electron. The probability that the vacancy results in an x-ray is called the fluorescence yield. The energy of the fluorescent x-ray is given by the difference in the binding energy of the shell the electron comes from and the binding energy of the shell where the vacancy is. Previously fluorescence x rays had different notations according to a system developed by Siegbahn, a Swedish physicist. Lately it has become more common to just indicate the shells. E.g., if the vacancy is in the K-shell and the electron comes from the L3 -shell, the x-ray was called a Kα1 -x-ray in the Siegbahn notation. However, it is more clear if it is just called xK−L3 . The energies and the fluorescence yields of different fluorescence x rays can e.g. be found in the Table of Radioactive Isotopes (Browne and Firestone, 1986). Unauthenticated Download Date | 2/4/20 3:47 PM Definitions and Equations | 5 As the de-excitation results in either a γ -ray or an internal conversion electron, the following equations are obtained α= Ne Nγ (1.1.16) Ne + Nγ = 1 (1.1.17) This gives 1 (1.1.18) 1+α The number of internal conversion electrons from the different shells are obtained through the relation N e K,L,M... = α K,L,M... · Nγ . From Tables of Isotopes (Lederer and Shirley, 1978) the following data are obtained for 137m Ba. (There are several different sets of data for the internal conversion coefficients in the table and these are just one example). Nγ = αK = 0.0916, αL = 1.162 · 10−2 , αM = 4.2 · 10−3 , αN = 0 α = αK + αL + αM giving α=0.1124 This gives Nγ = 1 = 0.899 1 + 0.1124 and NeK = 0.0916·0.899=0.0823, NeL 10−3 ·0.899=3.8·10−3 . = 0.0162·0.899=0.0146, NeM = 4.2 · The energies of the internal conversion electrons are given by EeK = Eγ − BK =0.6616 − 0.0374 = 0.6242 MeV EeL = Eγ − BK =0.6616 − 0.0052 = 0.6560 MeV (L2 -shell) EeM = Eγ − BK =0.6616 − 0.0011 = 0.6605 MeV (M2 -shell). For eL and eM only one energy is included as the difference in energy between the subshells is small. The data are summarized in Table 1.2. The number of x rays and their energies are obtained by e.g. Table 7 in Table of Radioactive Isotopes (Eckerman and Endo, 1986). Table 1.3 shows data taken from that table giving the number of xK -rays per 100 vacancies and their energies. Only the most frequent x rays are included in the table. The last column shows the number of x rays Unauthenticated Download Date | 2/4/20 3:47 PM 6 | Radiation Sources and Radioactive Decay Table 1.2: Frequency and energy of the photons and internal conversion electrons from the 137 Csdecay. Particle γ eK eL eM Frequency 0.899 0.0823 0.015 0.0038 Energy (MeV) 0.6616 0.624 0.656 0.660 from the de-excitation of 137m Ba, where column two has been multiplied with 0.0823, the number of K-shell vacancies. As shown the probabilities are larger for K-L transitions except for the transition K-L1 which is called forbidden. Table 1.3: Frequency and energy of the fluorescence x rays emitted in the 137 Cs-decay. x-ray K − L3 K − L2 K − L1 K − M3 K − N2 N3 K − M2 Frequency 0.467 0.256 0.0000334 0.0863 0.0273 0.0447 Energy (keV) 32.194 31.817 31.452 36.378 37.255 36.304 x rays per de-excitation 0.0384 0.0211 0 0.0071 0.0022 0.0037 There are also x rays from vacancies in the L-shell. These vacancies are due to both the internal conversion processes in the L-shell and to vacancies produced when vacancies in the K-shell are filled with electrons from the L-shell, resulting in new vacancies in the L-shell. Table of Radioactive Isotopes, Table 7 includes data for both these possibilities. However, due to the sub-shells in the outer electron shells, there will now be a fine structure of x-ray photons with different energies. The energy of the L x rays is normally rather low, in this case around 5 keV, and often it is possible to average all energies with one energy in particular as the frequency also is low. In this de-excitation there will be 0.0067 (0.0823·0.081) L x rays due to the vacancies in the K-shell and 0.0014 (0.0146·0.095) L x rays from the L-vacancies from the internal conversion per de-excitation. The energies are between 4 and 6 keV. Note that the decay started with a 137 Cs nuclide, but the data for the x rays must be calculated for 137m Ba, as it is from the Ba-nuclide the de-excitation is due. An alternative to fluorescence x rays are Auger electrons. These are produced when an electron from an outer shell goes to an inner shell and the excess energy obtained is used to emit another electron. If the process is due to energy transfers between subshells the emitted electrons are often called Coster-Cronig electrons. When there is a vacancy in the K-shell, the most common Auger process is when an electron goes from the L-shell and there is an Auger-electron emitted also from the L-shell (AKLL -electron). Unauthenticated Download Date | 2/4/20 3:47 PM Exercises in Radiation Sources and Radioactive Decay | 7 The energy of the Auger electrons are obtained as the energy difference in the binding energies between the electron shells. It is important to include the binding energy of the emitted Auger electron. E.g. in the process with an AKLL -electron, the electron energy will be EKLL =BK − BL − BL . The Auger process is an alternative to the emission of fluorescence x rays. The probability for the two different processes varies with the atomic number. For low atomic numbers the Auger electrons are dominating and for high atomic numbers the fluorescence x rays. Data for the probabilities are found in the Table of Radioactive Isotopes, Table 8. In Table 1.4 data for the five most common Auger electrons from 137m Ba de-excitation are shown. In the original Table 8 there are probabilities for 38 different combinations of shell transitions through vacancies in the K-shells giving Auger electrons. Table 1.4: Frequency and energy of the Auger electrons emitted in the 137 Cs-decay. Auger electron Frequency Energy (keV) K − L1 L1 K − L1 L2 K − L1 L3 K − L2 L3 K − L3 L3 (x10−2 ) 0.79 0.97 1.11 2.35 1.17 25.46 25.83 26.21 26.57 26.95 Auger electrons per de-excitation (x10−2 ) 0.065 0.080 0.091 0.19 0.096 The Auger electrons and also the fluorescence x rays will produce new vacancies in the outer shells and will give rise to a cascade of mainly Auger- and Coster-Cronigelectrons with low energies. In the calculations we have separated the decay from 137 Cs and the de-excitation from 137m Ba. As mentioned above normally there is radiation secular equilibrium, and when the decay from 137 Cs is described the de-excitation from 137m Ba is included. Then all figures above for 137m Ba have to be multiplied with the probability that 137 Cs goes to the excited level 137m Ba, that is 0.946. Thus e.g. the number of 0.6616 MeV γ rays per 137 Cs decay is 0.946. 0.899=0.85. When using published data of the 137 Cs decay it is important to check how this is handled as this could be different in different tables. 1.2 Exercises in Radiation Sources and Radioactive Decay Exercise 1.1. The α-decay of a nucleus with a mass number of 200 has two components with energies 4.687 MeV and 4.650 MeV. Neither of the decays goes to the basic level of the daughter, but is accompanied with emission of γ -radiation, with the Unauthenticated Download Date | 2/4/20 3:47 PM 8 | Radiation Sources and Radioactive Decay respective energies 266 and 305 keV. No other γ -energies are found. Construct from this information the decay scheme. Exercise 1.2. A radioactive source (210 At) with unknown activity is positioned in a vacuum chamber with the volume 10.0 dm3 . The activity is determined by measuring the amount of He that is obtained when the emitted α-particles attract electrons and produce stable He-atoms. Assume that all α-particles are transformed to He-gas. After 24.0 d the He-mass is determined to 1.84 ng. Calculate the initial activity of 210 At. Exercise 1.3. 239 Pu is α-radioactive and the α-particles are emitted with a kinetic energy of 5.144 MeV. When a sphere of 239 Pu, with a mass of 120.06 g, is placed in a calorimeter with liquid nitrogen, the same mass of nitrogen is vaporized per time unit as when 0.231 W is added by electric means. Calculate the half life of 239 Pu. Exercise 1.4. A radioactive source of pure 210 Bi, is at time t=0, placed in a container of lead, that absorbs all emitted radiation including the produced bremsstrahlung, when the electrons are absorbed. All energy, that is absorbed in the radioactive source and the container is transferred to heat. The heat power is measured with a calorimeter. Calculate the time it takes until the heat power has decreased to 3.60 mW. At time t=0 the 210 Bi activity is 90.0 GBq. Assume that all decays go to the ground state of the daughter nuclide and that only one particle is emitted in every decay. 210 Bi → 210 Po → 206 Pb (stable) Bi : T1/2 =5.01 d, E βmax =1.16 MeV, E βmean =0.344 MeV 210 Po : T1/2 =138.4 d, E α =5.2497 MeV 210 Exercise 1.5. A 137 Cs-source is measured and a count rate of 0.45·106 pulses per minute is obtained for photons with the energy 0.662 MeV. How many counts per minute can be expected for photons with an energy between 30 and 40 keV? Assume the detector measures every 10th photon independent of energy. Calculate also the activity of the source. Exercise 1.6. The γ -photons and the fluorescence x rays obtained when 125 I disintegrates, may be used for x-ray imaging. How many photons are obtained per decay? ECL =0.23, ECM = 0 ECK α=13.8, αK =11.7, αM =0 where ECL /ECK is the ratio between the probabilities that a nucleus captures an L- and a K-electron and α is the internal conversion coefficient. Unauthenticated Download Date | 2/4/20 3:47 PM Exercises in Radiation Sources and Radioactive Decay | 9 Figure 1.2: Decay scheme for 125 I. Exercise 1.7. 111 In is a radionuclide which is used for investigations in nuclear medicine. It is then possible to use both the emitted γ -radiation and the fluorescence x rays. Calculate the ratio between the count rates obtained with one energy window at 245±10 keV and one at 25±5 keV. Assume that the detector efficiency for 245 keV is 12% and for 25 kev 15%. 85% of the electron capture belongs to the K-shell. γ1 : αK = 0.100, αK /αL = 6.22, α M = 0 γ2 : αK = 0.0540, αK /αL = 5.20, α M = 0 αK,L,M = internal conversion coefficients for K, L and M shells respectively αK = NeK /Nγ NeK =number of emitted internal conversion electrons from the K-shell per decay Nγ =number of emitted γ -photons per decay. Figure 1.3: Decay scheme for 111 In. Unauthenticated Download Date | 2/4/20 3:47 PM 10 | Radiation Sources and Radioactive Decay Exercise 1.8. 241 Pu can be produced at nuclear reprocessing plants and contaminate the environment. A sample from the biosphere is measured at a certain date and the ratio of α-activity of 241 Am/239 Pu is determined to 2.00·10−3 . The sample was collected 5.0 years ago. The corresponding ratio was then 0.90·10−3 . Calculate the α-activity ratio 241 Pu/239 Pu at time of collection. Exercise 1.9. 104 Rh can be obtained through irradiation of neutrons. The following nuclear reactions are obtained 103 Rh(n,γ )104m Rh (σ=1.2·10−27 m2 ) and 103 Rh(n,γ )104 Rh (σ=14.0·10−27 m2 ). 103 Rh with thermal 104m Rh decays through internal transition to 104 Rh. The half lives are 104m Rh: T1/2 =4.40 min and 104 Rh: T1/2 =42 s. Calculate the activity of 104 Rh in percentage of the saturation activity 60 s and 600 s after start of irradiation. Exercise 1.10. The radionuclide 169 Yb is obtained through neutron irradiation of 168 Yb. The interaction cross section is large, 1.10·10−24 m2 . The half-life of the produced radionuclide is 32.0 d. Calculate the time when the activity is maximal, if the neutron fluence rate is 1.00·1018 m−2 s−1 . Exercise 1.11. A 99 Mo-99m Tc generator has the activity 12.0 GBq Monday morning at 8:00. Thursday morning at 8:00 the generator is completely emptied of 99m Tc. Later the same day the generator is eluted again and completely emptied of 99m Tc. By mistake the time for the elution was not observed. The activity of the eluted 99m Tc is measured later and is 2.10 GBq at 16:00 the same day. When was the generator eluted? Exercise 1.12. When running a reactor the fission product This disintegrates to 149 Sm according to the reactions below. 149 149 Nd is obtained. Nd(β,T1/2 =1.70 h)→149 Pm(β,T1/2 =47.0 h)→149 Sm 149 Sm is stable but disappears during the drift of the reactor through the reaction 149 Sm+n→150 Sm When the reactor is shut down, this reaction will not continue and the amount of 149 Sm will increase. This may be a problem because the neutron capture cross- Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay | 11 section of 149 Sm is much larger than the fission cross-section for 235 U and it may be difficult to restart the reactor if there is a lot of samarium in the fuel. If there is 300 TBq 149 Nd and 1000 TBq 149 Pm when the reactor is shut down, what mass of 149 Sm will be produced during the first 48 h after the shut down? (The activity numbers are just examples and not meant to be realistic). 1.3 Solutions in Radiation Sources and Radioactive Decay Solution exercise 1.1. In an α-decay the transition energy, Q, is divided between the α-particle and the daughter nuclide. If the daughter nuclide has an excited state with the energy Eγ , not all transition energy will be kinetic energy of the two particles but instead, Q0 = Q−Eγ . The relation between Q0 and the kinetic energy of the α-particle, E α , is given by m α + mD Eα (1.3.1) Q0 = mD Data: m α =4.00 (mass of α-particle) mD =196 (mass of daughter nuclide) E α1 = 4.687 MeV (kinetic energy of α-particle 1) E α2 = 4.650 MeV (kinetic energy of α-particle 2) Eγ 1 = 0.266 MeV (energy of γ ray 1) Eγ 2 = 0.305 MeV (energy of γ ray 2) Inserting the masses and energies in Eq. 1.3.1 gives 0 0 Q1 =4.783 MeV, Q2 =4.745 MeV From this information it is now possible to construct the decay-scheme (Fig. 1.4). Solution exercise 1.2. 210 At is decaying to 210 Po by electron capture. 210 Po is also radioactive and decays by α-decay to 206 Pb, which is stable. The activity of 210 Po is given by the equation A (0)λPo −λAt t e − e−λPo t (1.3.2) APo (t) = At λPo − λAt The number of 210 Po disintegrations during a time T is obtained by integrating equation (1.3.2). −λAt t T ZT A (0)λPo e e−λPo t N = APo (t) dt = At − + (1.3.3) λPo − λAt λAt λPo 0 0 N= AAt (0)λPo λPo − λAt e−λPo T e−λAt T 1 1 − − + λPo λAt λPo λAt (1.3.4) Unauthenticated Download Date | 2/4/20 3:47 PM 12 | Radiation Sources and Radioactive Decay Figure 1.4: Proposed decay scheme. This is equal to the number of α-particles, as there is one α-particle per decay. Every αparticle is capturing electrons and is transferred to a He-nucleus. The relation between the number of He-nuclei NHe and mass is given by the equation NHe = mHe NA ma (1.3.5) where NA =6.022·1026 atoms/kmol (Avogadro’s number) mHe =1.84·10−12 kg (He-mass) ma =4.00 (atomic mass of He) Decay data: λAt =ln2/8.10=0.08557 h−1 =2.0538 d−1 λPo =ln2/138.38=5.009·10−3 d−1 T=24 d Data inserted gives NHe = 1.84 · 10−12 · 6.022 · 1026 = 2.770 · 1014 4.00 and AAt (0) · 5.009 · 10−3 · 24 · 3600 5.009 · 10−3 − 2.0538 " # −5.009·10−3 ·24 e e−2.0538·24 1 1 × − − + 2.0538 5.009 · 10−3 5.009 · 10−3 2.0538 2.770 · 1014 = Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay | 13 AAt (0)=59.2 GBq Answer: The initial activity of 210 At is 59 GBq. Solution exercise 1.3. In a radioactive decay the energy released per time unit, P, is given by the relation P = λNQ (1.3.6) where λ is the decay constant, N is (NA m)/(ma ) (number of nuclides) and Q is the disintegration energy. This gives an expression for the half life T1/2 (=ln 2/λ) T1/2 = ln 2 NA m Q P ma (1.3.7) Data: m =120.06 g NA = 6.022 · 1026 atoms/kmol (Avogadro’s number) m αPu =239.05 (239 Pu) P=0.231 W (power by electric means, that should be the same as by radioactive decay) When a radionuclide disintegrates through α-decay, the disintegration energy is divided between the α-particle and the daughter nuclide (see Fig. 1.5). From the kinetic energy and the momentum relations Q can be obtained: Figure 1.5: Mechanics of the α-decay. Conservation of momentum must hold in the decay process. Kinetic energy: Q= mD v2D m α v2α + 2 2 (1.3.8) Momentum: mD vD = m α v α (1.3.9) Equations (1.3.8) and (1.3.9) give Q= mD v2α m2α m α v2α m α + = Eα + Eα 2 mD 2m2D (1.3.10) Unauthenticated Download Date | 2/4/20 3:47 PM 14 | Radiation Sources and Radioactive Decay Observe that it is the atomic mass of the daughter nuclide that should be inserted in Eq. 1.3.10, and thus m α =4.003 and mD =235.04. The kinetic energy of the α-particle is given in the problem, Eα =5.144 MeV. Data inserted in Eq. 1.3.10 gives Q= 4.003 · 5.144 + 5.144 = 5.2316 MeV 235.04 Data inserted in Eq. (1.3.7) gives T1/2 = ln 2 · 6.022 · 1026 · 5.2316 · 1.602 · 10−13 · 0.12006 0.231 · 239.05 T1/2 =7.606·1011 s=24.1 ·103 years. Answer: The half life of 239 Pu is 24.1· 103 years Solution exercise 1.4. The activity of 210 Bi, ABi (t), varies with time according to the relation ABi (t) = ABi (0)e−λBi t (1.3.11) The activity of 210 Po, APo (t), varies with time according to the relation APo (t) = λPo ABi (0) −λBi t (e − e−λPo t ) λPo − λBi (1.3.12) The power from the radioactive nuclides is obtained from the product of the activity and the energy released per decay. At time t there is activity both from 210 Bi and 210 Po present. Thus the total power at time t is given by P = ABi (0)Ē β e−λBi t + λPo ABi (0)Q −λBi t (e − e−λPo t ) λPo − λBi (1.3.13) Q is the total energy released in the α decay. The kinetic energy both from the α-particle and the daughter nucleus must be included in the calculations. The total energy is obtained from the equation Q = ED + Eα = Eα mα + Eα mD (1.3.14) Data: ABi (0)=90.0 GBq m α =4.00 (atomic mass of He) mD =206 (atomic mass of 206 Pb) E α =5.2497 MeV ⇒ Q =5.3516 MeV (disintegration energy) Ē β =0.344 MeV (mean energy of β-radiation from 210 Bi) P =3.60 mW (electric power) Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay | 15 λBi =ln2/5.01=0.1384 d−1 (decay constant for 210 Bi) λPo =ln2/138.4=5.009·10−3 d−1 (decay constant for 210 Po) Data inserted in Eq. (1.3.13) gives 3.60 · 10−3 = 90.0 · 109 · 0.344 · 1.602 · 10−13 e−0.1384t + −3 5.009 · 10−3 · 90.0 · 109 · 5.3516 · 1.602 · 10−13 −0.1384t (e − e−5.009·10 t ) 5.009 · 10−3 − 0.1384 This is simplified to −3 3.60 · 10−3 = 2.06 · 10−3 e−0.1384t + 2.90 · 10−3 e−5.009·10 t This equation can be solved numerically giving, t=6.84 d. Answer: The time for the heat power to decrease to 3.60 mW is 6.84 d. Solution exercise 1.5. 137 Cs disintegrates to 137m Ba. The de-excitation from the metastable level results in the measured photons. The result is given in Table 1.5. 137m Ba Figure 1.6: Decay scheme for 137 Cs. The count rate r30−40 in channels 30-40 keV is given by r30−40 = rγ · f30−40 /fγ Unauthenticated Download Date | 2/4/20 3:47 PM 16 | Radiation Sources and Radioactive Decay Assume the same efficiency for the two photon energies. Data inserted gives Table 1.5: De-excitation data for 137m Ba and measured count rate. Data for the decay are taken from Table of Radionuclide Transformations (ICRP, 1986). Photons Gamma-rays Fluorescence x rays Frequency(photons per de-excitation) fγ =0.898 f30−40 =0.0605 (Kα +Kβ ) r30−40 = Count rate rγ =0.45·106 min−1 ?? 0.45 · 106 · 0.0605 = 30317 min−1 0.898 To calculate the activity of 137 Cs one has to know how many excited 137m Ba-nuclei there are for each 137 Cs decay. Table of Radionuclide Transformations (ICRP, 1986) gives that the relative number of decays going to the excited level 137m Ba is 0.946. This gives fγ0 =0.946·0.898 γ -rays per 137 Cs decay. The count rate is given by the relation rγ = Afγ0 nγ (1.3.15) where A is the activity of the 137 Cs source and ηγ =0.10 is the efficiency of the detector. This gives the activity A= rγ fγ0 ηγ = 0.45 · 106 60 · 0.10 · 0.946 · 0.898 A = 88.3 · 103 Bq. Answer: The count rate for photons between 30 and 40 keV is 30.3·103 min−1 . The 137 Cs-activity is 88 kBq. Solution exercise 1.6. 125 I disintegrates by electron capture to an excited state of 125 Te. The electron capture will give rise to an electron vacancy in the K- or L-electron shells followed by emission of fluorescence x rays and/or Auger electrons. The de-excitation from the excited state of 125 Te results in emission of γ -radiation and/or internal conversion electrons. The vacancies obtained after the emission of internal electrons will also produce emission of fluorescence x rays and/or Auger electrons. The decay data are given in Fig. 1.7 and ECL /ECK is the ratio between the probabilities that a nucleus captures an L- or a K-electron and α is the total internal conversion coefficient. Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay | 17 Figure 1.7: Decay scheme for 125 I. Electron capture: Combining the relations (as ECM =0, the electron capture is either from the K- or the L-shell) ( ECL ECK = 0.23 ECK + ECL = 1 results in ECK =0.813 and ECL =0.187. This means that on average there are 0.813 electron captures from the K-shell and 0.187 from the L-shell. De-excitation: Combining the relations (there is either a gamma ray or an internal conversion electron) gives ( α = NNγe N e + Nγ = 1 and Nγ = 1 α+1 Data inserted gives Nγ =0.0676. The number of internal conversion electrons from the K-shell NeK is then given Unauthenticated Download Date | 2/4/20 3:47 PM 18 | Radiation Sources and Radioactive Decay by NeK = αK Nγ = 11.7 · 0.0676 = 0.791 Total: The total number of K-electrons, NeK , and thus vacancies in the K-shell, from both the electron capture and the de-excitation is then given by NeK,tot =0.813+0.791=1.604 The number of fluorescence K x rays is given by the relation NXK = NeK,tot ωK ωK =0.877 (fluorescence yield) Observe that ωK shall be taken for tellurium as this is the atom after the decay. This gives NXK =0.877·1.604=1.407 To obtain the total number of photons per decay, the number of gamma rays, Nγ , and the number of fluorescence K x rays, NXK , are added. N=1.407+0.0676=1.4746 Answer: The total number of photons per 100 decays is 147. Solution exercise 1.7. The photons in the energy interval 25±5 keV are fluorescence K-x rays from cadmium. The photons in the energy interval 245±10 keV come from the de-excitation from the 0.2454 MeV level. The de-excitation can be performed either by emitting a gamma photon or an internal conversion electron. The electron will leave a vacancy in a electron shell, which will be filled by electrons from outer shells and give rise to emission of fluorescence x rays and/or Auger electrons. The relation between gamma photons and internal conversion electrons is given by the internal conversion coefficient α = Ne /Nγ . Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay | 19 Figure 1.8: Decay scheme for 111 In. Energy interval 245±10 keV The number of Nγ photons per decay is obtained from the equations ( α = αK + αL = Ne + Nγ = 1 Ne Nγ which gives 1 Nγ = 1+α Energy interval 25±5 keV The number of electron vacancies in the K-shell is obtained from internal conversion both for γ1 and γ2 . The number of internal conversion electrons from the K-shell is obtained from the relation NeKγ = αK Nγ . There are also electron vacancies from the electron capture, NeKEC . This number is given in the exercise text. The total number of electron vacancies per decay, NeK , is now obtained by adding the different components. NeK = NeKγ1 + NeKγ2 + NeKEC The number of K-fluorescence x rays is obtained by multiplying with the fluorescence yield ωK . N X K = ω K N eK Data: Unauthenticated Download Date | 2/4/20 3:47 PM 20 | Radiation Sources and Radioactive Decay γ1 : αK1 = 0.100, αK1 /αL1 = 6.22, αM1 = 0 γ2 : αK2 = 0.0540, αK2 /αK2 = 5.20, αM2 = 0 NeKEC =0.85 ωK (Cd)=0.843 Data inserted in the relations above gives α1 =0.100+0.100/6.22=0.1161 and Nγ 1 =0.896 This will give the number of electron vacancies in the K-shell. NeK1 =0.896·0.100=0.0896 Correspondingly for γ2 is obtained α2 =0.054+0.054/5.20=0.06438 and Nγ 2 =0.9395 NeK2 =0.9395·0.054=0.0507 Total number of K-vacancies is thus NeK =0.0896+0.0507+0.85=0.9903 Total number of K x rays NXK =0.9903·0.843=0.8348 The count ratio between the number of photons of energy 245 keV and energy 25 keV is given by the relation N γ 2 η245 (1.3.16) N XK η25 where η245 =0.12 and η25 =0.15 are the detector efficiencies for the different energies. This gives the ratio in count rate 0.9395 · 0.12 = 0.900 0.8348 · 0.15 Answer: The count rate ratio is 0.90. Solution exercise 1.8. 241 Pu decays to 241 Am by β-decay. 241 Am is also radioactive and decays by α-decay. Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay 241 | 21 Pu(β,T1/2 =14.4 y)→241 Am(α,T1/2 =433 y)→237 Np→... 239 Pu is radioactive with a half-life T1/2 =2.41·104 y. This means that the decay of 239 Pu can be neglected in the calculations as the time period considered is five years. The activity of 241 Am produced from 241 Pu after t years is given by the relation AAm,1 (t) = λAm APu (0) −λPu t (e − e−λAm t ) λAm − λPu (1.3.17) where APu (0)= activity of 241 Pu at t=0 λPu =ln2/14.4=4.8135·10−2 y−1 (decay constant for 241 Pu) λAm =ln2/433=1.601·10−3 y−1 (decay constant for 241 Am) According to information in the problem, the activity of by the relation 241 Am at t=0 is given AAm (0) = 0.90 · 10−3 A3 A3 =activity of 239 Pu (assumed to be constant). After t years the activity of cay. 241 Am, AAm , has decreased due to radioactive de- AAm,2 (t) = 0.90 · 10−3 A3 · e−λAm t Thus the total activity of 241 Am after t years is AAm (t) = λAm APu (0) −λPu t (e − e−λAm t ) + 0.90 · 10−3 A3 · e−λAm t λAm − λPu (1.3.18) If t= 5.0 y then AAm (5) = 2.0 · 10−3 A3 according to measurements. Inserting λPu , λAm and t=5.0 y in Eq. (1.3.18) gives 2.0 · 10−3 A3 = −2 −3 1.601 · 10−3 APu (0) (e−5·4.8135·10 − e−5·1.601·10 ) −3 −2 1.601 · 10 − 4.8135 · 10 + 0.90 · 10−3 A3 · e−5·1.601·10 −3 Solving Eq. (1.3.18) gives APu (0) = 0.156A3 Answer: The α-activity ratio A(241 Pu)/A(239 Pu) is 0.16 at time of collection of the sample. Unauthenticated Download Date | 2/4/20 3:47 PM 22 | Radiation Sources and Radioactive Decay Solution exercise 1.9. Production of 104m Rh-nuclei (N1 ) is given by Eq. (1.3.19), if it is assumed that the decrease in the number of the 103 Rh-nuclei may be neglected. dN1 = U1 − λ1 N1 dt (1.3.19) where dN1 /dt is the change in the number of N1 -nuclei per time unit, U1 is the production rate in the reaction giving 104m Rh and λ1 is the decay constant of nuclide N1 . The solution of Eq. (1.3.19) is N1 = U1 (1 − e−λ1 t ) λ1 (1.3.20) The production of 104 Rh-nuclei (N2 ) through activation and decay from the excited state of 104m Rh is given by the equation dN2 = U2 + λ1 N1 − λ2 N2 dt (1.3.21) Eq. (1.3.20) combined with Eq. (1.3.21) gives dN2 = U2 − λ2 N2 + U1 (1 − e−λ1 t ) dt (1.3.22) Multiply with integrating factor e λ2 t dN2 λ2 t e = U2 e λ2 t − λ2 N2 e λ2 t + U1 (1 − e−λ1 t )e λ2 t dt Eq. (1.3.23) can be rewritten as d(N2 e λ2 t ) = U2 e λ2 t + U1 (1 − e−λ1 t )e λ2 t dt Eq. (1.3.24) is integrated Z Z h i d(N2 e λ2 t ) dt = U2 + U1 (1 − e−λ1 t ) e λ2 t dt dt (1.3.23) (1.3.24) (1.3.25) The indefinite solution of the integral gives N 2 e λ2 t = (U2 + U1 )e λ2 t U1 e(λ2 −λ1 )t − +C λ2 λ2 − λ1 (1.3.26) C can be determined by assuming that at t=0, N2 (0)=0. This inserted in Eq. (1.3.26) gives (U + U1 ) U1 0= 2 − +C (1.3.27) λ2 λ2 − λ1 Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay C=− (U2 + U1 ) U1 + λ2 λ2 − λ1 | 23 (1.3.28) Combining this with Eq. (1.3.26) results in N 2 e λ2 t = (U2 + U1 )e λ2 t U1 e(λ2 −λ1 )t (U2 + U1 ) U1 − − + λ2 λ2 − λ1 λ2 λ2 − λ1 (1.3.29) Multiplying with λ2 and dividing with e λ2 t gives λ2 N2 = U2 + U1 − λ2 U1 e−λ1 t λ U − (U2 + U1 )e−λ2 t + 2 1 e−λ2 t λ2 − λ1 λ2 − λ1 (1.3.30) The production rate of radionuclides is given by U = Φ̇σN0 (1.3.31) where Φ̇ is the neutron fluence rate, σ is the reaction cross section and N0 is the number of target nuclei at t=0. Combining Eq. (1.3.30) with (1.3.31) gives λ σ A2 = λ2 N2 = Φ̇N0 (σ2 + σ1 )(1 − e−λ2 t ) + 2 1 (e−λ2 t − e−λ1 t ) λ2 − λ1 (1.3.32) When t → ∞ then A2 (∞) = Φ̇N0 (σ1 + σ2 ) (1.3.33) Data: σ1 = 1.20 · 10−27 m2 (reaction cross section for producing 104m Rh) σ2 = 14.0 · 10−27 m2 (reaction cross section for producing 104 Rh) λ1 =ln2/(4.40·60) s−1 (decay constant for 104m Rh) λ2 =ln2/42 s−1 (decay constant for 104 Rh) t1 =60 s (irradiation time) t2 =600 s (irradiation time) Data inserted in Eq. (1.3.32) gives for t=60 s A2 =Φ̇N0 [(14.0 · 10−27 + 1.2 · 10−27 )(1 − e− ln 2·60/42 ) + ln 2/42 · 1.2 · 10−27 (e− ln 2·60/42 − e− ln 2·60/(4.4·60) )] ln 2/42 − ln 2/(4.40 · 60) = 8.864 · 10−27 Φ̇N0 Data inserted in Eq. (1.3.33) gives A2 (∞) = 15.2 · 10−27 Φ̇N0 Unauthenticated Download Date | 2/4/20 3:47 PM 24 | Radiation Sources and Radioactive Decay This gives A2 (60)=0.58A2 (∞) Data for t=600 s gives A2 (600)=0.98A2 (∞) Answer: The 104 Rh activity is 58% of the saturation activity after 60 s and 98% of the saturation activity after 600 s. Solution exercise 1.10. The change in the number of active 169 Yb-nuclei is given by the equations dN = U(t) − λN dt U(t) = (1.3.34) mNA Φ̇σ −Φ̇σt e ma (1.3.35) where U(t) is the production rate of radioactive nuclides, m is the mass of target (kg), NA is Avogadro’s number, Φ̇ is the neutron fluence rate, σ is the activation cross section, ma is the atomic mass and t is the irradiation time. The equations are solved by multiplying with the integrating factor eλt e λt ( d Ne λt dt dN + λN) = U(t)e λt dt = U(t)e λt = mNA Φ̇σ −Φ̇σt λt e e ma (1.3.36) (1.3.37) Integration gives Ne λt = mNA Φ̇σ t(λ−Φ̇σ) e +C ma (λ − Φ̇σ) (1.3.38) If t=0 then N=0 mNA Φ̇σ ma (λ − Φ̇σ) (1.3.39) mNA Φ̇σ −Φ̇σt e − e−λt ma (λ − Φ̇σ) (1.3.40) C=− N= The activity is given by A = λN. The maximal activity is obtained when ( dA/ dt) = 0 −Φ̇σe−Φ̇σt + λe−λt = 0 (1.3.41) Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay λ = e(λ−Φ̇σ)t Φ̇σ λ ln Φ̇σ ⇒t= λ − Φ̇σ | 25 (1.3.42) (1.3.43) Data: λ=ln2/32 d−1 =2.507·10−7 s−1 (decay constant for 169 Yb) Φ̇ = 1.00 · 1018 m−2 s−1 (neutron fluence rate) σ=1.10·10−24 m2 (reaction cross section) Data inserted in Eq. (1.3.43) gives t=1.741·106 s=20.15 d Answer: The maximal activity of 169 Yb is obtained after 20.2 days. Solution exercise 1.11. The activity of 99 Mo on Monday 8:00 is AMo (0). The activity of 99 Mo on Thursday 8:00, AMo (tMo ), is AMo (tMo ) = AMo (0)e−λMo tMo . The generator is then eluted again at a time t after the first elution. The activity of the 99m Tc is measured at time t1 . The activity of 99m Tc at time t1 is obtained from the relation fλ A (0)e−λMo tMo −λMo t − e−λTc t e−(t1 −t)λTc (1.3.44) ATc = Tc Mo e λTc − λMo where tMo =72 h (time from Monday 8:00 to Thursday 8:00) t1 =8.0 h (time from first elution at 8:00 to when measuring the activity at 16:00) AMo (0) =12.0 GBq (activity of 99 Mo Monday morning 8:00) ATc =2.10 GBq (activity of 99m Tc at Thursday 16:00) f =0.876 (fraction of 99 Mo disintegrations that will give 99m Tc activity) λMo =ln2/66.0 h−1 (decay constant for 99 Mo) λTc =ln2/6.02 h−1 (decay constant for 99m Tc) Data inserted in Eq. (1.3.44) gives 2.10 = 0.876 · (ln 2/6.02) · 12 · e−(ln 2/6.02)·72 ln 2/6.02 − ln 2/66.0 Unauthenticated Download Date | 2/4/20 3:47 PM 26 | Radiation Sources and Radioactive Decay × e−(ln 2/66.0)·t − e−(ln 2/6.02)·t e−(8.0−t)·ln 2/6.02 t =6.49 h Answer: The elution was made at 14:30 h (t is calculated from 8:00). Solution exercise 1.12. When the reactor is shut down there are 149 Pm and 149 Nd radionuclides in the reactor. Both neodymium and promethium will end up in samarium through the reaction 149 Nd→149 Pm→149 Sm (stable) Samarium is thus obtained in two ways, either form promethium directly or from neodymium through promethium. a) The number of 149 Sm nuclides that are obtained from the directly is given by ZT N1,Sm = APm e−λPm t dt = 149 Pm radionuclides APm 1 − e−λPm T λPm (1.3.45) 0 b) The number of 149 Sm nuclides that are obtained from 149 Nd radionuclides is obtained from the equations APm = and ZT N2,Sm = 0 N2,Sm = λPm ANd −λPm t e − e−λNd t λNd − λPm (1.3.46) λPm t T λPm ANd e e λNd t APm dt = − + λNd − λPm λPm λNd 0 λPm ANd λNd − λPm e−λNd T e−λPm T 1 1 − − + λNd λPm λNd λPm (1.3.47) (1.3.48) The mass is obtained from the relation m= ma N NA (1.3.49) where APm =1000 TBq (activity of 149 Pm when the reactor is shut down) ANd =300 TBq (activity of 149 Nd when the reactor is shut down) T =48 h (time after reactor shut down) λNd =ln2/1.7 h−1 (decay constant for 149 Nd) Unauthenticated Download Date | 2/4/20 3:47 PM Solutions in Radiation Sources and Radioactive Decay | 27 λPm =ln2/47 h−1 (decay constant for 149 Pm) ma =149 (atomic mass) NA =6.022·1026 kmol−1 (Avogadro’s number) Data inserted in Eq. (1.3.45) and Eq. (1.3.48) gives N1,Sm = 1000 · 1012 · 3600 1 − e−(ln 2/47)·48 ln 2/47 and N2,Sm = 300 · 1012 (ln 2/47) · 3600 ln 2/1.7 − ln 2/47 −(ln 2/1.7)48 e e−(ln 2/47)48 1 1 × − − + ln 2/1.7 ln 2/47 ln 2/1.7 ln 2/47 a) N1,Sm = 1.2384·1020 nuclei b) N2,Sm = 1.295·1018 nuclei In total there are 1.251·1020 nuclei. This gives the mass m= 3.095·10−5 kg. Answer: The mass of 149 Sm after 48 h is 31 mg. Unauthenticated Download Date | 2/4/20 3:47 PM