Question

Half life of a radioactive substance is $2.34\, \min$. It is produced at a constant rate of $10^{8}$ nuclei per second. How soon (in minute) after the beginning of production will its activity be equal to disintegration per second?

Answer 1

$\frac{d N}{d t}=q-\lambda N\,\, \left(q=10^{8}\right.$ nuclei $\left./ s \right)$
$\Rightarrow \frac{d N}{q-\lambda N}=d t$
$\Rightarrow \int\limits_{0}^{N} \frac{d N}{q-\lambda N}=\int\limits_{0}^{t} d t$
$\Rightarrow -\frac{1}{\lambda} \ln \left(\frac{q-\lambda N}{q}\right)=t$
$\Rightarrow t=\frac{t_{1 / 2}}{\ln 2} \cdot \ln \left(\frac{q}{q-A}\right)$
$\left(\because \lambda=\frac{\ln 2}{t_{1 / 2}}\right.$ and $\left.A=\lambda N\right)$
$=\frac{t_{1 / 2}}{\ln 2} \cdot \ln \frac{10^{8}}{\left(10^{8}-7.5 \times 10^{7}\right)}$
$=\frac{t_{1 / 2}}{\ln 2} \cdot \ln \left(\frac{10}{10-7.5}\right)$
$=2 t_{1 / 2}=4.68\, \min$