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A radioactive nuclei with decay constant $0.5/s$ 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
$1\,s$
$2\ln \left( {\frac{4}{3}} \right)s$
$ln\, 2\, s$
$\ln \left( {\frac{4}{3}} \right)s$
Solution
Let $N$ be the number of nucleiat any time $t$ then,
$\frac{d N}{d t}=100-\lambda N$
or $\int_0^N {\frac{{dN}}{{(100 – \lambda N)}}} = \int_0^t d t$
$-\frac{1}{\lambda}[\log (100-\lambda N)]_{0}^{N}=t$
$\log (100-\lambda N)-\log 100=-\lambda t$
$\log \frac{100-\lambda N}{100}=-\lambda t$
$\frac{100-\lambda N}{100}=e^{-\lambda t}$
$1 – \frac{{\lambda N}}{{100}} = {e^{ – \lambda t}}$
$N=\frac{100}{\lambda}\left(1-e^{-\lambda} t\right)$
As, $N=50$ and $\lambda=0.5\,/sec$
$\therefore 50=\frac{100}{0.5}\left(1-e^{-0.5}\right)$
Solving we get,
$t=2 \ln \left(\frac{4}{3}\right) \sec$
