Question

Half life of a radioactive substance is $1min$ . It is produced at a constant rate $R_0=10^8$ nuclei per sec. How soon after (in $min$ , to the nearest integer), the beginning of production will its activity be equal to $7.5\times 10^7$ disintegration per sec.

Answer 1

Answer: 2

The half life of the radioactive substance is given as $1\text{min}$ . Therefore, decay constant $\lambda =\frac{ln2}{60}{\text{s}}^{-1}=0.012{\text{disintegration s}}^{-1}$
For activity to be equal to $7.5\times 10^7{\text{disintegration s}}^{-1}$
$-\frac{dN}{dt}=\lambda N_0\Rightarrow 7.5\times 10^7=0.012\times N_0\Rightarrow N_0=6.25\times 10^9$
Now the radioactive element is being created at the rate of $R_0=10^8{\text{nuclei s}}^{-1}$ and simultaneously they are disintegrating too, if at any time $t$ , the no of nuclei are $N$ then,
$\frac{dN}{dt}=R_0-\lambda N$
$\Rightarrow \frac{dN}{R_0-\lambda N}=dt$
integrating it from $0\text{to}t\text{and}0\text{to}{N}_0$
${\int }_0^{N_0}\frac{dN}{R_0-\lambda N}={\int }_0^{t}dt\Rightarrow -\frac{1}{\lambda }ln\frac{R_0-\lambda N_0}{R_0}=t$
$\Rightarrow t=\frac{1}{\lambda }ln\frac{R_0}{R_0-\lambda N_0}=\frac{1}{0.012}ln\frac{{\left(10\right)}^8}{\left({\left(10\right)}^8-0.012\times 6.25\times {\left(10\right)}^9\right)}=\frac{1}{0.012}ln4=115\text{s}$
Therefore, the time for activity to reach $7.5\times 10^7{\text{disintegration s}}^{-1}$ is $115\text{s}\approx 2\text{min}$ .