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A radioactive nuclei with decay constant $0.5 / \mathrm{s}$ is being produced at a constant rate of 100 nuclei/s. If at $t=0$ there were no nuclei, the time when there are 50 nuclei is:
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Verified Answer
The correct answer is:
$2 \ln \left(\frac{4}{3}\right) \mathrm{s}$
$2 \ln \left(\frac{4}{3}\right) \mathrm{s}$
Let $N$ be the number of nuclei at any time $t$ then,
$$
\begin{aligned}
&\frac{d N}{d t}=100-\lambda N \\
&\text { or } \int_0^N \frac{d N}{(100-\lambda N)}=\int_0^t d t \\
&-\frac{1}{\lambda}[\log (100-\lambda N)]_0^N=t \\
&\log (100-\lambda N)-\log 100=-\lambda t \\
&\log \frac{100-\lambda N}{100}=-\lambda t \\
&\frac{100-\lambda N}{100}=e^{-\lambda t} \\
&1-\frac{\lambda N}{100}=e^{-\lambda t}
\end{aligned}
$$
$$
\begin{aligned}
&N=\frac{100}{\lambda}\left(1-e^{-\lambda} t\right) \\
&\text { As, } N=50 \text { and } \lambda=0.5 / \mathrm{sec} \\
&\therefore 50=\frac{100}{0.5}\left(1-e^{-0.5 t}\right)
\end{aligned}
$$
Solving we get,
$$
t=2 \ln \left(\frac{4}{3}\right) \mathrm{sec}
$$
$$
\begin{aligned}
&\frac{d N}{d t}=100-\lambda N \\
&\text { or } \int_0^N \frac{d N}{(100-\lambda N)}=\int_0^t d t \\
&-\frac{1}{\lambda}[\log (100-\lambda N)]_0^N=t \\
&\log (100-\lambda N)-\log 100=-\lambda t \\
&\log \frac{100-\lambda N}{100}=-\lambda t \\
&\frac{100-\lambda N}{100}=e^{-\lambda t} \\
&1-\frac{\lambda N}{100}=e^{-\lambda t}
\end{aligned}
$$
$$
\begin{aligned}
&N=\frac{100}{\lambda}\left(1-e^{-\lambda} t\right) \\
&\text { As, } N=50 \text { and } \lambda=0.5 / \mathrm{sec} \\
&\therefore 50=\frac{100}{0.5}\left(1-e^{-0.5 t}\right)
\end{aligned}
$$
Solving we get,
$$
t=2 \ln \left(\frac{4}{3}\right) \mathrm{sec}
$$
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