Search any question & find its solution
Question:
Answered & Verified by Expert
In a radioactive material the activity at time $t_1$, is $A_1$ and at a later time $t_2$, it is $A_2$. If the decay constant of the material is $\lambda$, then
Options:
Solution:
2331 Upvotes
Verified Answer
The correct answer is:
$A_1=A_2 e^{-\lambda\left(t_1-t_2\right)}$
$A_1=A_2 e^{-\lambda\left(t_1-t_2\right)}$
From radioactive decay law,
$$
-\frac{d N}{d t} \propto N \text { or }-\frac{d N}{d t}=\lambda N
$$
Thus,
$$
A=-\frac{d N}{d t} \text { or } A=\lambda N
$$
$$
\Rightarrow \quad A=\lambda N_0 e^{-\lambda t}
$$
Where, $A_0=\lambda N_0$ is the activity of the radioactive material at time, $t=0$
At time, $t_1 \quad A_1=A_0 e^{-\lambda t_1}$
At time, $t_2 \quad A_2=A_0 e^{-\lambda t_2}$
On dividing Eq. (ii) by Eq. (iii), we have
$$
\begin{aligned}
\quad \frac{A_1}{A_2} & =\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{-\lambda\left(t_1-t_2\right)} \\
\Rightarrow \quad \quad A_1 & =A_2 e^{-\lambda\left(t_1-t_2\right)}
\end{aligned}
$$
$$
-\frac{d N}{d t} \propto N \text { or }-\frac{d N}{d t}=\lambda N
$$
Thus,
$$
A=-\frac{d N}{d t} \text { or } A=\lambda N
$$
$$
\Rightarrow \quad A=\lambda N_0 e^{-\lambda t}
$$
Where, $A_0=\lambda N_0$ is the activity of the radioactive material at time, $t=0$
At time, $t_1 \quad A_1=A_0 e^{-\lambda t_1}$
At time, $t_2 \quad A_2=A_0 e^{-\lambda t_2}$
On dividing Eq. (ii) by Eq. (iii), we have
$$
\begin{aligned}
\quad \frac{A_1}{A_2} & =\frac{e^{-\lambda t_1}}{e^{-\lambda t_2}}=e^{-\lambda\left(t_1-t_2\right)} \\
\Rightarrow \quad \quad A_1 & =A_2 e^{-\lambda\left(t_1-t_2\right)}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.