Question 10.3.Q3: Figure 10.4 shows three activity A(t) curves plotted against......
Figure 10.4 shows three activity A(t) curves plotted against time t for the radioactive series decay: Molybdenum-99 (Mo-99) → Technetium99m (Tc-99m) → Technetium-99 (Tc-99) starting with a pure 10 mCi (0.37 GBq) Mo-99 source. \text { Half-lives }\left(t_{1 / 2}\right)_{\mathrm{P}}=\left(t_{1 / 2}\right)_{\mathrm{Mo}-99}=66 \mathrm{~h} and \left(t_{1 / 2}\right)_{\mathrm{D}}=\left(t_{1 / 2}\right)_{\mathrm{Tc}-99 \mathrm{~m}}=6.02 \mathrm{~h} \text { correspond to decay constants } \lambda_{\mathrm{P}}=\lambda_{\mathrm{Mo}-99}= 1.05 \times 10^{-2} \mathrm{~h}^{-1} \text { and } \lambda_{\mathrm{D}}=\lambda_{\mathrm{Tc}-99 \mathrm{~m}}=0.115 \mathrm{~h}^{-1}, respectively. The following three points should be noted:
(1) Curve 1 is exponential and represents the parent P radionuclide Mo99 decay into daughter D radionuclide Tc-99m starting with activity \mathcal{A}_P(0) = 10 mCi at time t = 0.
(2) Curve 2 shows the activity \mathcal{A}_D(t) of daughter radionuclide for initial condition \mathcal{A}_D(0) = 0.
(3) Curve 3 represents the total activity \mathcal{A}_{tot}(t) of the Mo-Tc sample as the sum of the parent and daughter activity as a function of time starting with \mathcal{A}_{tot}(0) = 10 mCi at time t = 0.
(a) State or derive equations for the three activity curves shown in Fig. 10.4.
(b) For curve 2 of Fig. 10.4 verify that \lim _{t \rightarrow 0} \mathcal{A}_{\mathrm{D}}(t)=0 \text { and } \lim _{t \rightarrow \infty} \mathcal{A}_{\mathrm{D}}(t) = 0 and determine the characteristic time \left(t_{max}\right)_D at which activity \mathcal{A}_D(t) reaches its maximum value.
(c) For curve 3 of Fig. 10.4 verify that \lim _{t \rightarrow 0} \mathcal{A}_{\mathrm{tot}}(t)=10 \mathrm{mCi} and \lim _{t \rightarrow \infty} \mathcal{A}_{\text {tot }}(t)=0.
(d) Derive a general expression for calculation of the characteristic time T_{max} at which curve 3 of Fig. 10.4 representing the total activity \mathcal{A}_{tot}(t) of the Mo-Tc sample attains its maximum value.
(e) Calculate T_{max} for curve 3 of Fig. 10.4.
(f) Calculate the maximum total activity \mathcal{A}_{\mathrm{tot}}(t) \text { at } t=T_{\max } for curve 3 of Fig. 10.4.
(a) Since the daughter product Tc-99m of the Mo-99 → Tc-99m → Tc-99 radioactive decay series is a metal, it remains in the molybdenum sample unless it is separated for medical purpose from the sample by means of a solvent (elution process). When left in the Mo-99 sample, the total activity \mathcal{A}_{tot}(t) of the sample is the sum of the parent (Mo-99) activity \mathcal{A}_P(t) and the daughter (Tc-99m) activity \mathcal{A}_D(t). The daughter Tc-99m decays into granddaughter G (Tc-99) which is a radionuclide with a very long half-life of 211000 years, so that we assume that G in our decay series is stable.
Expressions for the three decay curves of Fig. 10.4 are as follows:
(1) Curve 1 represents a simple exponential decay of the parent nucleus expressed as (T10.10)
\mathcal{A}_{\mathrm{P}}(t)=\mathcal{A}_{\mathrm{P}}(0) e^{-\lambda_{\mathrm{P}} t}, (10.76)
with the initial condition \mathcal{A}_P(0) = 10 mCi.
(2) Curve 2 represents the activity \mathcal{A}_D(t) of the daughter radionuclide and is expressed as follows (T10.35)
\mathcal{A}_{\mathrm{D}}(t)=\mathcal{A}_{\mathrm{P}}(0) \frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}}\left[e^{-\lambda_{\mathrm{p}} t}-e^{-\lambda_{\mathrm{D}} t}\right] (10.77)
with the initial condition \mathcal{A}_D(0) = 0.
(3) Curve 3 represents the total activity \mathcal{A}_{tot}(t) of the Mo-Tc sample and is given as a sum of \mathcal{A}_P(t) of (10.76) and \mathcal{A}_D(t) of (10.77) as follows
(b) Daughter activity \mathcal{A}_D(t) is given by (10.77) and the limits for (1) t → 0 and (2) t → ∞ are
(1)
\lim _{t \rightarrow 0} \mathcal{A}_{\mathrm{D}}(t)=\mathcal{A}_{\mathrm{P}}(0) \frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}} \lim _{t \rightarrow 0}\left[e^{-\lambda_{\mathrm{p}} t}-e^{-\lambda_{\mathrm{D}} t}\right]=\mathcal{A}_{\mathrm{P}}(0) \frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}}[1-1]=0 (10.79)
and
(2)
\lim _{t \rightarrow \infty} \mathcal{A}_{\mathrm{D}}(t)=\mathcal{A}_{\mathrm{P}}(0) \frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}} \lim _{t \rightarrow \infty}\left[e^{-\lambda_{\mathrm{p}} t}-e^{-\lambda_{\mathrm{D}} t}\right]=\mathcal{A}_{\mathrm{P}}(0) \frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}}[0-0]=0 (10.80)
(3) Characteristic time \left(t_{max}\right)_D at which \mathcal{A}_D(t) reaches its maximum is determined by setting \mathrm{d} \mathcal{A}_{\mathrm{D}}(t) / \mathrm{d} t=0 \text { at } t=\left(t_{\max }\right)_{\mathrm{D}} \text { and solving for }\left(t_{\max }\right)_{\mathrm{D}} to get the following result for Tc-99m (T10.37) using the following decay constants for \text { Tc-99m: } \lambda_{\mathrm{P}}=\lambda_{\mathrm{Mo}-99}=1.05 \times 10^{-2} \mathrm{~h}^{-1} \text { and } \lambda_{\mathrm{D}}=\lambda_{\mathrm{Tc}-99 \mathrm{~m}}=0.115 \mathrm{~h}^{-1}.
(c) Total activity \mathcal{A}_{tot}(t) is given by (10.78) and the limits for t → 0 and t → ∞ are as follows
and
\lim _{t \rightarrow \infty} \mathcal{A}_{\mathrm{tot}}(t)=\frac{\mathcal{A}_{\mathrm{P}}(0)}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}} \lim _{t \rightarrow \infty}\left[\left(2 \lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}\right) e^{-\lambda_{\mathrm{p}} t}-\lambda_{\mathrm{D}} e^{-\lambda_{\mathrm{D}} t}\right]=\frac{\mathcal{A}_{\mathrm{P}}(0)}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}}[0-0]=0 (10.83)
(d) As curve 3 of Fig. 10.4 shows, \mathcal{A}_{\text {tot }}(t) \text { is equal to } \mathcal{A}_{\mathrm{p}}(0)=10 \mathrm{mCi} \text { at } t=0 where \mathcal{A}_D(0) = 0, and then rises with increasing time t until it reaches a peak at a characteristic time t=T_{\max } \text {. After the peak, total activity } \mathcal{A}_{\mathrm{tot}}(t) decreases toward 0 with increasing t following the decrease of both \mathcal{A}_{\mathrm{P}}(t) \text { and } \mathcal{A}_{\mathrm{D}}(t) \text { toward } 0 \text { at } t=\infty \text {. } Characteristic time T_{\max } \text { should not be confused with }\left(t_{\max }\right)_{\mathrm{D}}, the characteristic time at which the daughter activity \mathcal{A}_D(t) attains its maximum value, since the two characteristic times are not equal.
(e) T_{max} is determined by setting \mathrm{d} \mathcal{A}_{\text {tot }} / \mathrm{d} t=0 \text { at } t=T_{\max } \text { and solving for } T_{\max } and solving for Tmax as follows
\left.\frac{\mathrm{d} \mathcal{A}_{\mathrm{tot}}}{\mathrm{d} t}\right|_{t=T_{\max }}=\frac{\mathcal{A}_{\mathrm{P}}(0)}{\lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}}\left[-\left(2 \lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}\right) \lambda_{\mathrm{P}} e^{-\lambda_{\mathrm{p}} T_{\max }}+\lambda_{\mathrm{D}}^2 e^{-\lambda_{\mathrm{D}} T_{\max }}\right]=0 . (10.84)
Equation (10.84) can be simplified to read
\left(2 \lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}\right) \lambda_{\mathrm{P}} e^{-\lambda_{\mathrm{p}} T_{\max }}=\lambda_{\mathrm{D}}^2 e^{-\lambda_{\mathrm{D}} T_{\max }}, (10.85)
resulting in the following general expression for T_{max}.
T_{\max }=\frac{\ln \frac{\left(2 \lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}\right) \lambda_{\mathrm{P}}}{\lambda_{\mathrm{D}}^2}}{\lambda_{\mathrm{P}}-\lambda_{\mathrm{D}}} (10.86)
and the following solution for our problem with decay constants \lambda_P=\lambda_{\mathrm{Mo-99}}=1.05 \times 10^{-2} \mathrm{~h}^{-1} \text { and } \lambda_{\mathrm{D}}=\lambda_{\mathrm{Tc}-99 \mathrm{~m}}=0.115 \mathrm{~h}^{-1}.
T_{\max }=\frac{\ln \frac{\left(2 \lambda_{\mathrm{D}}-\lambda_{\mathrm{P}}\right) \lambda_{\mathrm{P}}}{\lambda_{\mathrm{D}}^2}}{\lambda_{\mathrm{P}}-\lambda_{\mathrm{D}}}=\frac{\ln \frac{\left(2 \times 0.115-1.05 \times 10^{-2}\right) \times 1.05 \times 10^{-2}}{0.115 \times 0.115}}{\left(1.05 \times 10^{-2}-0.115\right) \mathrm{h}^{-1}}=16.72 \mathrm{~h} . (10.87)
(f) The maximum in total activity \mathcal{A}_{\text {tot }} \text { occurs at } T_{\max }=16.72 \mathrm{~h} \text {, } as determined in (e). At t = T_{max} the total activity has the following magnitude
Equation (10.88) shows that the total activity \mathcal{A}_{tot}(t) of a pure molybdenum-99 source increases with time from t = 0 until it reaches peak activity \mathcal{A}_{\text {tot }}\left(T_{\max }\right) at a characteristic time T_{max} and then it decreases with time until it reaches zero activity at time t → ∞. In contrast, the daughter activity \mathcal{A}_D reaches its maximum value at its own characteristic time \left(t_{max}\right)_D that is larger than the characteristic time T_{max}.
For initial activity of a pure molybdenum source of 10 mCi, the total sample activity reaches its maximum of 16.02 mCi at a characteristic time T_{max} of 16.72 hours. The daughter activity, on the other hand, reaches its maximum at a characteristic time \left(t_{max}\right)_D of 23 hours.