Question 10.7.Q1: An unknown mixture of two or more independently decaying rad......
An unknown mixture of two or more independently decaying radionuclides, each with its own half-life and decay constant, will produce a composite decay curve that does not result in a straight line when plotted on a semi-logarithmic plot, unless, of course, all radionuclides have identical or very similar halflives. In principle, the decay curves of the individual radionuclides can be resolved graphically, if their half-lives differ sufficiently and if at most three radioactive components are present.
A solution contains an unknown amount of gold-198 (Au-198) and iodine-131 (I-131) beta emitters. If the total activity \mathcal{A}(t) of the solution at time t = 0 is 0.140 µCi (5.18 kBq) and drops to half of its initial value in 3 days,
(a) Calculate the initial activities \mathcal{A}_{Au}(0)\ and\ \mathcal{A}_I(0) of Au-198 and I-131, respectively, in the solution. Half-lives of Au-198 and I-131 are 2.70 d and 8.05 d, respectively.
(b) Calculate the total activity \mathcal{A}(t) of the solution at time t = 15 d.
(c) Calculate the time t as well as the total activity \mathcal{A}(t) at which the activities of Au-198 and I-131 in the solution are equal.
(d) Plot on a semi-logarithmic graph the activities \mathcal{A}_{Au}(t)\ and\ \mathcal{A}_I(t) of Au198 and I-131, respectively, and the total activity A(t) in the solution as a function of time. Show on the plot the activities at t = 3 d, t = 15 d, and at time t determined in (c).
(a) Activities \mathcal{A}_{Au}(t)\ and\ \mathcal{A}_I(t) of Au-198 and I-131 beta emitters, respectively, in a solution can be described with the standard exponential decay law governing radioactivity. We thus express activity \mathcal{A}_{Au}(t) as a function of time t in the solution as
\mathcal{A}_{\mathrm{Au}}(t)=\mathcal{A}_{\mathrm{Au}}(0) e^{-\lambda_{\mathrm{Au}} t}=\mathcal{A}_{\mathrm{Au}}(0) e^{-\frac{\ln 2}{\left(I_{1 / 2}\right)_{\mathrm{Au}}} t} (10.183)
and activity \mathcal{A}_I(t) as a function of time t in the solution as
\mathcal{A}_{\mathrm{I}}(t)=\mathcal{A}_{\mathrm{I}}(0) e^{-\lambda_{\mathrm{I}} t}=\mathcal{A}_{\mathrm{I}}(0) e^{-\frac{\ln 2}{\left(t_{1 / 2}\right)_{\mathrm{I}}} t}, (10.184)
where the decay constant λ is related to half-life t_{1/2} through the relationship λ = (ln 2)/t_{1/2} and
\mathcal{A}_{Au}(0)\ and\ \mathcal{A}_I(0) are the initial activities at t = 0 of Au-198 and I-131, respectively.
λ_{Au}\ and\ λ_I are the decay constants of Au-198 and I-131, respectively.
\left(t_{1/2}\right)_{Au}\ and\ \left(t_{1/2}\right)_I are the half-lives of Au-198 and I-131, respectively.
Total activity \mathcal{A}(t) of the radioactive solution as a function of time t is in general the sum of \mathcal{A}_{Au}(t) given in (10.183) and \mathcal{A}_I(t) given in (10.184), i.e.,
\mathcal{A}(t)=\mathcal{A}_{\mathrm{Au}}(t)+\mathcal{A}_{\mathrm{I}}(t)=\mathcal{A}_{\mathrm{Au}}(0) e^{-\frac{\ln 2}{\left(t_{1 / 2}\right)_{Au}} t}+\mathcal{A}_{\mathrm{I}}(0) e^{-\frac{\ln 2}{\left(t_{1 / 2}\right)_1} t} (10.185)
Data for our problem stipulate that the initial activity \mathcal{A}(0) at t = 0 of our radioactive solution is 0.140 µCi, while its activity \mathcal{A}(t) at t = 3 days is 0.070 µCi. Using (10.184) we now express \mathcal{A}(0) at t = 0 and \mathcal{A}(t) at t = 3 d, respectively, as follows
\mathcal{A}(0)=\mathcal{A}(t=0)=\mathcal{A}_{\mathrm{Au}}(0)+\mathcal{A}_{\mathrm{I}}(0)=0.140 \mu \mathrm{Ci} (10.186)
and
In (10.186) and (10.187) we have two equations for two unknowns: \mathcal{A}_{Au}(0)\ and\ \mathcal{A}_I(0). Solving (10.186) for \mathcal{A}_{Au}(0) and inserting \mathcal{A}_{Au}(0) = 0.140 µCi − \mathcal{A}_I(0) into (10.187) we obtain the following results:
\mathcal{A}_{\mathrm{Au}}(0)=0.123 \mu \mathrm{Ci} \quad \text { and } \quad \mathcal{A}_{\mathrm{I}}(0)=0.017 \mu \mathrm{Ci} \text {. } (10.188)
(b) To calculate the total activity of the solution at time t = 15 d we use (10.185) in conjunction with the initial activities \mathcal{A}_{Au}(0)\ and\ \mathcal{A}_I(0) at time t = 0 given in (10.188) and get
(c) To determine the time t=t_{\mathrm{eq}} \text { at which the activities } \mathcal{A}_{\mathrm{Au}}(t) \text { and } \mathcal{A}_{\mathrm{I}}(t) are equal we write (10.183) and (10.184) in the following form
\mathcal{A}_{\mathrm{Au}}\left(t_{\mathrm{eq}}\right)=\mathcal{A}_{\mathrm{Au}}(0) e^{-\frac{\ln 2}{\left(t_{1 / 2}\right)_{\mathrm{Au}}} t_{\mathrm{eq}}}=\mathcal{A}_{\mathrm{I}}\left(t_{\mathrm{eq}}\right)=\mathcal{A}_{\mathrm{I}}(0) e^{-\frac{\ln 2}{\left(t_{1 / 2}\right)_I} t_{\mathrm{eq}}} . (10.190)
Solving (10.190) for t_{eq} we first get the general result
t_{\mathrm{eq}}=\frac{\left(t_{1 / 2}\right)_{\mathrm{Au}}\left(t_{1 / 2}\right)_{\mathrm{I}} \ln \frac{\mathcal{A}_{\mathrm{Au}}(0)}{\mathcal{A}_{\mathrm{I}}(0)}}{\left[\left(t_{1 / 2}\right)_{\mathrm{B}}-\left(t_{1 / 2}\right)_{\mathrm{A}}\right] \ln 2} (10.191)
and after inserting the initial activities \mathcal{A}_{Au}(0)\ and\ \mathcal{A}_I(0) from (10.188) we determine t_{eq} as
t_{\mathrm{eq}}=\frac{(2.70 \mathrm{~d}) \times(8.05 \mathrm{~d}) \times \ln \frac{0.123}{0.017}}{(8.05 \mathrm{~d}-2.70 \mathrm{~d}) \times \ln 2}=11.6 \mathrm{~d} (10.192)
Activity \mathcal{A}(t) of the radioactive solution at t = t_{eq} = 11.6 d is calculated from (10.185)
(d) The radioactive solution investigated in this problem consists of two beta emitters: gold-198 and iodine-131 of unknown initial activities. Based on known solution’s total initial activity \mathcal{A}(0) and total activity at t = 3 h, activities \mathcal{A}_{Au}(t),\ \mathcal{A}_I(t),\ and\ \mathcal{A}(t) were determined and are plotted against time t in Fig. 10.12.
