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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
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The correct answer is:
$R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}$
The decay rate $R$ of a radioactive material is the number of decays per second. From radioactive decay law.
$$
-\frac{d N}{d t} \propto N \quad \text { or }-\frac{d N}{d t}=\lambda N
$$
Thus, $\quad R=-\frac{d N}{d t}$
Or $R=\lambda N \quad$ or $\quad R=\lambda N_{0} e^{-\lambda t} \ldots$..(i)
where $R_{0}=\lambda N_{0}$ is the activity of the radioactive material at time $t=0$. At time $t_{1}, \quad R_{1}=R_{0} e^{-\lambda t_{1}}$
At time $t_{2}, \quad R_{2}=R_{0} e^{-\lambda t_{2}}$
Dividing Eq. (ii) by (iii), we have
or
$$
\begin{array}{l}
\frac{R_{1}}{R_{2}}=\frac{e^{-\lambda t_{1}}}{e^{-\lambda t_{2}}}=e^{-\lambda\left(t_{1}-t_{2}\right)} \\
R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}
\end{array}
$$
$$
-\frac{d N}{d t} \propto N \quad \text { or }-\frac{d N}{d t}=\lambda N
$$
Thus, $\quad R=-\frac{d N}{d t}$
Or $R=\lambda N \quad$ or $\quad R=\lambda N_{0} e^{-\lambda t} \ldots$..(i)
where $R_{0}=\lambda N_{0}$ is the activity of the radioactive material at time $t=0$. At time $t_{1}, \quad R_{1}=R_{0} e^{-\lambda t_{1}}$
At time $t_{2}, \quad R_{2}=R_{0} e^{-\lambda t_{2}}$
Dividing Eq. (ii) by (iii), we have
or
$$
\begin{array}{l}
\frac{R_{1}}{R_{2}}=\frac{e^{-\lambda t_{1}}}{e^{-\lambda t_{2}}}=e^{-\lambda\left(t_{1}-t_{2}\right)} \\
R_{1}=R_{2} e^{-\lambda\left(t_{1}-t_{2}\right)}
\end{array}
$$
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