Question 10.4.Q1: The parent P and daughter D activities AP(t) and AD(t), resp......

The general variables x, y_P, and y_D as well as the decay factor m are defined as follows

x=\frac{1}{\left(t_{1 / 2}\right) \mathrm{P}},          (10.106)

y_{\mathrm{P}}=\frac{\mathcal{A}_{\mathrm{P}}(t)}{\mathcal{A}_{\mathrm{P}}(0)}          (10.107)

y_{\mathrm{D}}=\frac{\mathcal{A}_{\mathrm{D}}(t)}{\mathcal{A}_{\mathrm{P}}(0)},           (10.108)

m=\frac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{D}}}            (10.109)

(a) The standard form of parent activity \mathcal{A}_P(t), expressed as follows (T10.10) irrespective of the status of the daughter (stable or radioactive)

\mathcal{A}_{\mathrm{P}}(t)=\mathcal{A}_{\mathrm{P}}(0) e^{-\lambda_{\mathrm{P}} t}=\mathcal{A}_{\mathrm{P}}(0) e^{-\frac{(\ln 2) t}{\left(I_{1 / 2}\right) \mathrm{P}}}          (10.110)

is, after incorporating (10.106) and (10.107), written as (T10.47)

y_{\mathrm{P}}=\frac{\mathcal{A}_{\mathrm{P}}(t)}{\mathcal{A}_{\mathrm{P}}(0)}=e^{-\frac{(\ln 2) t}{\left(t_{1 / 2}\right)_{\mathrm{P}}}}=e^{-x \ln 2^{-x}}=2^{-x}=\frac{1}{2^x}        (10.111)

(b) The standard form of daughter activity \mathcal{A}_D(t), expressed as follows (T10.35) irrespective of the stability status of the granddaughter (stable or radioactive)

\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],          (10.112)

is, after incorporating (10.106), (10.108), and (10.110), written as (T10.45)

(c) Equation (10.113) for y_D as a function of x has physical meaning for all positive values of decay factor m except for m = 1 for which y_D is not defined. However, since for m = 1, (10.113) gives y_D = 0/0, we can apply the l’Hôpital rule and determine the function that governs y_D at m = 1 as follows

\left.y_{\mathrm{D}}\right|_{m=1}=\lim _{m \rightarrow 1} \frac{\frac{\mathrm{d}}{\mathrm{d} m}\left[\frac{1}{2^x}-\frac{1}{2^{x / m}}\right]}{\frac{\mathrm{d}}{\mathrm{d} m}(1-m)}=\lim _{m \rightarrow 1} \frac{-2^{-\frac{x}{m}}(\ln 2) \frac{x}{m^2}}{-1}=(\ln 2) \frac{x}{2^x}           (10.114)

(d) The characteristic time \left(t_{max}\right)_D which represents the time of maximum daughter activity and is written as (T10.37)

\left(t_{\max }\right)_{\mathrm{D}}=\frac{\ln \frac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{D}}}}{\lambda_{\mathrm{P}}-\lambda_{\mathrm{D}}}           (10.115)

can be expressed in general form incorporating (10.106) and (10.108) into (10.115) as follows

\left(x_{\mathrm{D}}\right)_{\max }=\frac{\left(t_{\max }\right)_{\mathrm{D}}}{\left(t_{1 / 2}\right)_{\mathrm{P}}}=\frac{\lambda_{\mathrm{P}} \ln \frac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{D}}}}{(\ln 2)\left(\lambda_{\mathrm{P}}-\lambda_{\mathrm{D}}\right)}=\frac{\ln \frac{\lambda_{\mathrm{P}}}{\lambda_{\mathrm{D}}}}{(\ln 2)\left(1-\frac{\lambda_{\mathrm{D}}}{\lambda_{\mathrm{P}}}\right)}=\frac{m \ln m}{(\ln 2)(m-1)} .           (10.116)

Equation (10.116) has physical meaning for all positive m except for m = 1 for which it is not defined, since it gives \left(x_D\right)_{max} = 0/0. Again we apply l’Hôpital rule to get a finite value for \left.\left(x_{\mathrm{D}}\right)_{\max }\right|_{m=1} as follows

\left.\left(x_{\mathrm{D}}\right)_{\max }\right|_{m=1}=\lim _{m \rightarrow 1} \frac{\frac{\mathrm{d}(m \ln m)}{\mathrm{d} m}}{(\ln 2) \frac{\mathrm{d}(m-1)}{\mathrm{d} m}}=\lim _{m \rightarrow 1} \frac{(\ln m+1)}{\ln 2}=\frac{1}{\ln 2}=1.4427 .          (10.117)

(e) The maximum daughter activity \left(y_D\right)_{max} can be determined by inserting \left(x_D\right)_{max} into y_D given by (10.113). However, since at the point of maximum daughter activity both the parent and the daughter have identical activities equal to \left(y_D\right)_{max}, it is much easier to obtain \left(y_D\right)_{max} by inserting \left(x_D\right)_{max} into y_P given by (10.110)

\left(y_{\mathrm{D}}\right)_{\max }=y_{\mathrm{P}}\left[\left(x_{\mathrm{D}}\right)_{\max }\right]=\frac{1}{2^{\left(x_{\mathrm{D}}\right)_{\max }}}=2^{\frac{m \ln m}{(\ln 2)(1-m)}}=e^{\frac{m}{\mathrm{I}-m} \ln m} .             (10.118)

Equation (10.118) is valid for all positive m except for m = 1 in which case \left(y_D\right)_{max} can be determined by applying the l’Hôpital rule to (10.118) as follows

\left.\left(y_{\mathrm{D}}\right)_{\max }\right|_{m=1}=\lim _{m \rightarrow 1} \exp \frac{\frac{\mathrm{d}(m \ln m)}{\mathrm{d} m}}{\frac{\mathrm{d}(1-m)}{\mathrm{d} m}}=\lim _{m \rightarrow 1} \exp \frac{\ln m+1}{-1}=e^{-1}=\frac{1}{e}=0.368 .             (10.119)

(f) It is easy to show that the relationship for positive m but m ≠ 1 between \left(y_D\right)_{max} given by (10.118) and \left(x_D\right)_{max} given by (10.116) is a simple exponential expression

\left(y_{\mathrm{D}}\right)_{\text {max }}=e^{\frac{m}{1-m t} \ln m}=e^{-\frac{m}{m-\mathrm{I}} \ln m}=e^{-(\ln 2)\left(x_{\mathrm{D}}\right)_{\text {max }}}=2^{-\left(x_{\mathrm{D}}\right)_{\text {max }}},            (10.120)

while for m = 1, \left(y_D\right)_{max} = 1/e = 0.368, as shown in (10.119), and \left(x_D\right)_{max} = 1/ ln 2 = 1.4427, as shown in (10.117).

(g) Data points for y_P(x) are calculated from (10.111) and for y_D(x) from (10.113) and results for 0 ≤ x ≤ 10 in steps of 1 are displayed in Table 10.3 and plotted in Fig. 10.6. The maximum value of y_D for m = 1 is 1/e = 0.368 and occurs at x = 1/ ln 2 = 1.4427 which is also the intersection point of y_P\ and\ y_D curves.