Question

Consider the decay scheme-
$A\to B\to C$ with ${\lambda }_A<{\lambda }_B.$ After transient equilibrium is established between $A$ and $B$, show that the interval of time $\Delta t$, such that activity of $A$ at $t-\Delta t$ is equal to activity of $B$ at $t$ is given by
$\Delta t={\tau }_{A}ln\left(\frac{{\lambda }_B}{{\lambda }_{B-A}}\right)$ and that this approaches $T_B$, as $\left(\frac{T_B}{T_A}\right)\to 0$

Answer 1

Let $N_1$ be the amount of parent atom of $A$ at any instant $t-\Delta t$
Therefore, $N_1=N_0^1{e}^{-{\lambda }_A(t-\Delta t)}$ . . . ($1$)
Where $N_1^0$ is the number of parent atom of $A$ at $t=0$. The number of daughter atoms at any time $t$ is given by
$N_2=\frac{{\lambda }_A-N_1^0}{{\lambda }_B-{\lambda }_A}[e^{-{\lambda }_{A}t}-e^{{-\lambda }_{B}t}]$ . . . ($2$)
For transient atom ${\lambda }_A<{\lambda }_B$, hence $e^{-{\lambda }_B}t$ tern can be neglected in $\left(2\right)$
$\therefore N_2=\frac{{\lambda }_A-N_1^0}{{\lambda }_B-{\lambda }_A}{e}^{-{\lambda }_{A}t}$ . . . ($3$)
Given the activity of $A$ at ($t-\Delta$)= activity of $B$ at $t$.
$\Rightarrow {\lambda }_A{N}_0^1{e}^{-{\lambda }_A(t-\Delta t)}={\lambda }_B\frac{{\lambda }_A-N_1^0}{{\lambda }_B-{\lambda }_A}{e}^{-{\lambda }_{A}t}$
$\Rightarrow e^{{\lambda }_A\Delta t}=\frac{{\lambda }_B}{{\lambda }_B-{\lambda }_A}$
$\Rightarrow \Delta =\frac{1}{{\lambda }_A}in\left[\frac{{\lambda }_B}{{\lambda }_B-{\lambda }_A}\right]$
The average life of ${\tau }_A=\frac{1}{{\lambda }_A}$ and that of $B$ is ${\tau }_B=\frac{1}{{\lambda }_B}$
$\therefore \Delta t={\tau }_{A}ln\left[\frac{1}{1-{\lambda }_B/{\lambda }_A}\right]$
$\Rightarrow \Delta t\cong {\tau }_A.\frac{{\tau }_B}{{\tau }_A}$ as $\frac{{\tau }_B}{{\tau }_A}\to 0$ (given)
$\Rightarrow \Delta t={\tau }_B$