Question 107: Atoms & Nuclei

Question

In a radioactive decay process, the activity is defined as $A=-\frac{d N}{d t}$, where $N(t)$ is the number of radioactive nuclei at time $t$. Two radioactive sources, $S_1$ and $S_2$ have same activity at time $t=0$. At a later time, the activities of $S_1$ and $S_2$ are $A_1$ and $A_2$, respectively. When $S_1$ and $S_2$ have just completed their $3^{	ext {rd }}$ and $7^{	ext {th }}$ half-lives, respectively, the ratio $A_1 / A_2$ is _________.

Accepted Answer

$$
A_1=A_0 e^{-\lambda_1 t_1}
$$
also $A_2=A_0 e^{-\lambda_2 t_2}$
At $t_1=\frac{3 \ln 2}{\lambda_1}, $
$A_1=A_0 e^{-\lambda_1 \frac{3 \ln 2}{\lambda_1}}$
$$
=A_0 e^{-3 \ln 2}
$$ ...........(i)
Similarly, at
$$
\begin{aligned}
&t_2=  \frac{7 \ln 2}{\lambda_2}, \\& A_2=A_0 e^{-\lambda_2 \frac{7 \ln 2}{\lambda_2}} \\
& =A_0 e^{-7 \ln 2} ...........(ii)
\end{aligned}
$$
From (i) and (ii)
$$
\frac{A_1}{A_2}=\frac{A_0 e^{-3 \ln 2}}{A_0 e^{-7 \ln 2}}=\frac{2^{-3}}{2^{-7}}=\frac{1}{2^{-4}}=2^4=16
$$
$	herefore \frac{A_1}{A_2}=16$

Atoms & Nuclei

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