JEE Advance - Physics (2023 - Paper 2 Online - No. 12)
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 $3^{\text {rd }}$ and $7^{\text {th }}$ half-lives, respectively, the ratio $A_1 / A_2$ is _________.
Answer
16
Explanation
$$
A_1=A_0 e^{-\lambda_1 t_1}
$$
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)
$$ \frac{A_1}{A_2}=\frac{A_0 e^{-3 \ln 2}}{A_0 e^{-7 \ln 2}}=\frac{2^{-3}}{2^{-7}}=\frac{1}{2^{-4}}=2^4=16 $$
$\therefore \frac{A_1}{A_2}=16$
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)
$$ \frac{A_1}{A_2}=\frac{A_0 e^{-3 \ln 2}}{A_0 e^{-7 \ln 2}}=\frac{2^{-3}}{2^{-7}}=\frac{1}{2^{-4}}=2^4=16 $$
$\therefore \frac{A_1}{A_2}=16$
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