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

Answer: 4.68

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