Search any question & find its solution
Question:
Answered & Verified by Expert
In a radioactive material the activity at time $t_1$ is $R_1$ and at a later time $t_2$, it is $R_2$. If the decay constant of the material is $\lambda$, then:
Options:
Solution:
2220 Upvotes
Verified Answer
The correct answer is:
\(R_1=R_2 e^{-\lambda\left(t_1-t_2\right)}\)
$R_1=R_0 e^{-\lambda_4}$ and $R_2=R_0 e^{-\lambda \lambda_2}$
$$
\begin{aligned}
\Rightarrow \quad \frac{R_1}{R,} & =\frac{e^{-\lambda \lambda_1}}{e^{-\lambda \lambda_2}} \\
& =e^{-\lambda\left(t_1-t_2\right)} \\
\Rightarrow \quad R_1 & =R_2 e^{-\lambda\left(t_1-t_2\right)}
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow \quad \frac{R_1}{R,} & =\frac{e^{-\lambda \lambda_1}}{e^{-\lambda \lambda_2}} \\
& =e^{-\lambda\left(t_1-t_2\right)} \\
\Rightarrow \quad R_1 & =R_2 e^{-\lambda\left(t_1-t_2\right)}
\end{aligned}
$$
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.