Chaines de décroissance radioactive et Equilibre Radioactivité -2 Chaines de décroissance radioactive et Equilibre Now let’s discuss radioactive decay, including important terms and concepts Jour 1 – Leçon 5

Objectif Discuter les chaines de décroissance radioactive (le père et le produit unique de désintégration) et les situations d’équilibre

Contenu Equilibre Séculaire Equilibre transitoire Cas de non-équilibre Séries de décroissance radioactive Croissance d'un produit de désintégration à partir d'un radionucléide père This lesson includes activity and how it relates to number of atoms of a radioactive isotope, the equation for radioactive decay (including terms), half-life and how it relates to the decay constant, definition of mean life and how it relates to half-life, and important units used to measure radioactive decay.

T1/2 père >>>T1/2 fils Types d’équilibres Radioactifs Séculaire La demi-vie du père est beaucoup plus grande (> 100 fois) que celle de produit de désintégration (fils) T1/2 père >>>T1/2 fils The first major type of radioactive equilibrium situation is “secular”. This occurs when a very long half-life parent decays to a relatively short half-life decay product. There is no precise definition of how much longer the parent half-life must be relative to the half-life of the daughter product. However, secular equilibrium can be attained if the parent half-life is at least 100 times greater than that of the decay product.

Types d’équilibres Radioactifs Transitoire La demi-vie du père est seulement 10 fois plus grande que celle du produit de désintégration If the half-life of the parent is only greater than (but not much greater than) the decay product half-life, secular equilibrium cannot be attained. However, an equilibrium situation called “transient” can be attained. For example, if the half-life of the parent is than that of the decay product, transient equilibrium occurs. The differences in secular and transient equilibrium will become more apparent in the new few slides in this session. T1/2 père ~ 10 T1/2 fils

90Sr  90Y  90Zr Exemple de séries de désintégration Radioactive où 90Sr est le père (demi-vie = 28 ans) et 90Y est le produit de décroissance (demi-vie = 64 heures) As an example to introduce series decay, we will look at Sr-90 decaying to Y-90, which in turn decays to stable (i.e. non-radioactive) Zr-90. Since the half-life of Sr-90 is more than 100 times that of Y-90, this will turn out to be an example of secular equilibrium. Sr-90 decays by beta emission to Y-90, which also decays by beta emission to Zr-90.

Equation Différentielle pour les Séries de décroissance radioactive Le père et le produit unique de décroissance = Sr NSr - Y NY dNY dt This shows the differential equation for production of Y-90 by Sr-90 decay. The instantaneous rate of change of Y-90 is made up of two terms: the production rate, which is equal to the Sr-90 decay rate; and the rate of loss, which is the decay rate of Y-90. These two terms are in fact the instantaneous activities of the two radionuclides. N refers to the number of atoms of each radionuclide at time t.  refers to the decay constant of each radionuclide. Le taux instantané de variation de Y-90 est composé de deux termes: le taux de production, qui est égale à la vitesse de décroissance du Sr-90; et le taux de perte, ce qui est le taux de décroissance de Y-90.

Equation Différentielle pour les Séries de décroissance radioactive Le père et le produit unique de désintégration NY(t) = (e-  t - e-  t) Sr Y SrNSr Y - Sr o The differential equation on the previous slide can be integrated to yield the above equation. Note that this equation is expressed in terms of number of atoms of Y-90 at any time t, assuming that there is no Y-90 initially present. If there were some Y-90 initially present, then a second term would be added to this equation to account for it. This second term would take the form of the equation NY(t) = Nyo exp (-t), where NY(t) is the number of Y-90 atoms at time t and Nyo is the number of Y-90 atoms initially present at time t = 0. Rappelons que Sr NoSr = AoSr qui est égal à l’activité initiale du 90Sr au temps t = 0

Equation Générale pour Les séries de décroissance radioactive Activité du 90Sr au temps t = 0 YNY(t) = (e- t - e- t) Sr Y Y - Sr Y SrNSr o Activité du 90Y au temps t ou AY(t) To express the equation on the previous slide in terms of activity instead of atoms, simply multiply both sides of the equation by the decay constant of Y-90. This does not change the value of the equation. The result is an expression for the value of the Y-90 activity at any time t, based on decay of Sr-90. This equation is known as the general equation of radioactive series decay. It works in all equilibrium situations, including secular and transient. Although the equation has been derived using Sr-90 and Y-90 as examples, this equation works for any parent/daughter pair.

AY(t) = ASr (1 - e- t) Equilibre Séculaire Les Conditions d'accumulation d'un produit de décroissance dans les conditions d'équilibre séculaire AY(t) = (1 - e- t) Y ASr For the case of secular equilibrium (very long half-life parent compared to decay product half-life), some simplifications can be made to the general equation to make it more useable. In this case, the parent decay constant is very small compared to that of the decay product and can be ignored. Also, with the passing of time, the first exponential term is very close to 1. When we make these simplifying assumptions and express the equation in terms of activity, we get the equation shown in the slide above. AY(t) is the activity of Y-90 at any time t and Asr is the initial activity of Sr-90. Note that since the half-life of Sr-90 is so long compared with the half-life of Y-90, for all practical purposes the initial activity of Sr-90 remains a constant over the time period of interest. This is a characteristic of secular equilibrium. It should be noted that the general equation will always provide a rigorous solution to any series decay situation and could of course be used instead of the above equation.

SrNSr = YNY ASr = AY Equilibre Séculaire With long decay time, the exponential term in the equation on the previous slide approaches 0, so that the whole term in the parentheses approaches 1. Thus, at secular equilibrium the activities of the parent and decay product are equal and constant with time. The total activity of the sample is the sum of the activities of the parent and decay product. A l’équilibre séculaire, l’activité du père et celle du fils sont égales et constantes dans le temps

Equilibre Séculaire ARn (t) = Ao (1 - e- t ) Ra Décroissance du 226Ra au 222Rn ARn (t) = Ao (1 - e- t ) Rn This behavior is shown graphically in this slide, using Ra-226 decaying to Rn-222. Ra-226 has about a 1,600 year half-life, whereas Rn-222 has an approximately 3 day half-life. So this represents a secular equilibrium situation. Recall that the initial amount of Rn-222 is assumed to be zero. Beginning with zero activity, the activity of the decay product becomes equal to the activity of the parent within 7 or so half-lives of the decay product. For example, in a uranium mill tailings pile, which initially has Ra-226 but essentially no Rn-222, the activity of Rn-222 in the pile will reach that of the Ra-226 within about 30 days! This is a principal cause of exposure (and concern) from U mill tailings piles, since inhalation of Rn-222 has been linked with lung cancer. Ra Commençant par une activité zéro, l’activité du fils devient égale à l’activité du père dans les 7 ou plus demi-vies du fils

Exemple - Problème 1 226Ra (demi-vie = 1600 ans) se désintègre pour donner le 222Rn (demi-vie = 3.8 jours). Si on a initialement 100 µCi du 226Ra dans un échantillon et pas de 222Rn, calculer l’activité produite du 222Rn: a. Après 7 demi-vies du 222Rn À l’équilibre Ce problème va illustrer l’équilibre séculaire This problem will illustrate secular equilibrium.

Solution du Problème dNRn = Ra NRa - Rn NRn dt RaNRa (1 - e- t) Le nombre d’atome de 222Rn au temps t est donné par: = Ra NRa - Rn NRn dNRn dt Solution: NRn(t) = (1 - e- t) Rn RaNRa Rn

Solution du Problème ARn(t) = ARa (1 - e- t) Multiplions les deux membres de l’équation par Rn: ARn(t) = ARa (1 - e- t) Rn Si t = 7 TRn Rnt = (0.693/TRn) x 7 TRn = 0.693 * 7 = 4.85 e-4.85 = 0.00784 ARn (7 demi-vies) = 100 µCi * (1 - 0.00784 ) Note that as stated before, the activity of the decay product Rn-222 has essentially become equal to that of its parent Ra-226 after only 7 half-lives of Rn-222. = 100 * (0.992) = 99.2 µCi du 222Rn

Solution du Problème Maintenant à l’équilibre Séculaire: RnNRn = RaNRa ou ARn = ARa = 100 µCi Noter que l’activité totale dans ce cas est: RnNRn + RaNRa ou ARn + ARa = People often forget that the total activity in the sample is the sum of the activities of the parent and decay product. Please do not forget this fact. It could cause underestimation of radiation doses and health effects and also affect radioactive waste shipments, since shipping manifests require accounting for all radioactivity present. 100 µCi + 100 µCi = 200 µCi

 fils Père NPère filsNfils = fils - Père Equilibre Transitoire For the case of transient equilibrium, the general equation for radioactive series decay reduces to the above equation. This equation is valid only during the time when the parent and daughter radionuclides are decaying at the same rate. This will be more readily seen when the plot of the decay curves is shown in a few slides. The actual derivation of the above equation is beyond the scope of this session and will not be shown here. All terms are as defined previously. Pour le cas de l'équilibre transitoire, l'équation générale pour une série de décroissances radioactives se réduit à l'équation ci-dessus

APère fils Afils = fils - Père Equilibre Transitoire The previous equation can be simplified even further by expressing it in terms of activities of parent and decay product. Exprimant la en terme d’activités du père et celle du fils

l’activité maximale du fils Equilibre Transitoire Le temps max. pour avoir l’activité maximale du fils Tm fils = fils - Père ln fils Père Starting with zero initial decay product activity, the decay product activity will reach a maximum level, after which it will decrease. This equation describes this situation mathematically and allows calculation of the time at which the maximum decay product activity occurs. tmD stands for the time at which the decay product “D” reaches its maximum activity.

Equilibre Transitoire Exemple de l’équilibre transitoire 132Te se désintègre à 132I Te-132 - 78.2 h (demi-vie) I -132 - 2.2 h (demi-vie) Noter que: I-132 atteint une activité maximale, après quoi il semble décroître avec la demi-vie du père Te-132. L’activité du fils ne jamais être plus élevée que l’activité initiale de son père. This graph shows an example of transient equilibrium, starting with zero initial amount of the decay product I-132. Note that I-132 reaches a maximum activity, after which it appears to decay with the half-life of the parent Te-132. The two activities track each other and are not equal but rather are in a constant ratio to each other. The equation for transient equilibrium shown on slides 18-19 is valid when the two decay curves are parallel. It does not apply when the daughter activity is increasing nor when the daughter activity is maximum. Note that surprisingly enough, the activity of the decay product I-132 is actually greater than that of its parent Te-132! However, as shown by the curve, the activity of the decay product can never be higher than the initial activity of its parent.

Exemple: Problème Le principe de l'équilibre transitoire est illustré par le générateur de molybdène-technétium, radio-isotopes utilisés dans des applications en médecine nucléaire. On considère initialement que le Générateur contient 100 mCi de 99Mo (demi-vie 66 h) et pas de 99mTc (demi-vie 6 h) calculer: a. Le temps nécessaire pour que le 99mTc atteigne le maximum de son activité b. l’activité de 99Mo à ce temps, et c. l’activité de 99mTc à ce temps

Exemple: Problème Noter que seulement 86% transformations de 99Mo produisent le 99mTc. Le reste qui est 14% se transforme en 99Tc qui n’est pas utilisable en médecine nucléaire. This slide shows that not all Mo-99 decays produce Tc-99m. The direct decays to Tc-99 are wasted, as far as this problem is concerned. Tc-99 is not used in nuclear medicine. It does show up elsewhere in the nuclear industry (e.g. it is a contaminant in the cascades of U gaseous diffusion plants, due to previous reprocessing of spent nuclear fuel many years ago).

Solution du Problème Tc ln Mo tmTc = Tc - Mo ln tmTc = a) = 21.9 h Tc = 0.693/(6 h) = 0.12 h-1 Mo = 0.693/(66 h) = 0.011 h-1 tmTc = 0.12 – 0.011 ln 0.12 0.011 = 21.9 h

Solution du Problème (b) l’activité de 99Mo est donnée par: A(t) = Ao e-t = 100 mCi e(-0.011/hr * 21.9 hr) = 100 * (0.79) = 79 mCi

Solution du Problème c) L’activité du 99mTc à t = 21.9 h est donnée par: ATc(t) = (e- t - e- t ) Mo Tc Tc - Mo TcAMo (voir diapos 10) ATc(t) = (e-(0.011)(21.9) - e-(0.12)(21.9)) (0.12 – 0.011) (0.12)(100 mCi)(0.86) Note that only 86% of the Mo-99 decays actually produce Tc-99m and this must be explicitly accounted for in the above equation. If the transient equilibrium equation from slide 19 is used, the result is about 74 mCi of Tc-99m. This shows that the transient equilibrium equation does not apply at this point since the two decay curves for the parent and the daughter are not yet in transient equilibrium. = (94.7) (0.785 - 0.071) = 67.6 mCi du 99mTc

Solution du Problème L’activité maximale du 99mTc est atteinte à 21,9 h qui est proche de 1 jour. This graph illustrates the results for problem 2. Note that the activity of Tc-99m does not quite reach that of Mo-99, since only 86% of Mo-99 decays produce Tc-99m. After 2 days in this example, the molybdenum cow is “milked”, the Tc-99m activity goes to zero, and then begins to grow in again due to Mo-99 decay. As calculated previously, the first maximum of Tc-99m activity occurs at about 22 hours.

Types d’équilibres radioactifs Pas d’équilibre La demi-vie du père est inférieure à celle du fils This situation is not nearly as interesting from a health physics point of view as compared with the first two equilibrium situations. A plot of parent vs daughter activity would show that in this case, the two activities do not track each other but diverge from each other with time.

Pas d’Equilibre Dans ce cas, la demi-vie du père est inférieure à celle du fils et l’équilibre ne sera établi. The case of no equilibrium is of less interest from a health physics standpoint relative to secular and transient but we should still be aware of it, since such radionuclides can contribute to exposure. The above graph shows a parent A1 decaying to a decay product A2. In this case, the half-life of the parent is less than that of the decay product and no equilibrium can be established. In fact, as the curves show, the activities of the parent and daughter actually diverge with time. The activity of the parent monotonically decreases with time, whereas the activity of the daughter (assuming no initial daughter activity was present) increases to a maximum and then decays.

Résumé L’activité a été définie et les unités ont été discutés la constante de décroissance a été définie La demi-vie a été définie en fonction de la constante de décroissance l’équation de la décroissance radioactive a été dérivée dérivée de la vie moyenne en fonction de la demi-vie L’équilibre a été défini La cas où il n’y a pas d’équilibre a été aussi défini

Où trouver plus d’informations Cember, H., Johnson, T. E, Introduction to Health Physics, 4th Edition, McGraw-Hill, New York (2009) International Atomic Energy Agency, Postgraduate Educational Course in Radiation Protection and the Safety of Radiation Sources (PGEC), Training Course Series 18, IAEA, Vienna (2002)