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

The half life of the radioactive substance is given as $1\text{min}$ . Therefore, decay constant $\lambda =\frac{ln 2}{60}\text{s}^{- 1}=0.012\text{disintegration s}^{- 1}$
For activity to be equal to $7.5\times 10^{7}\text{disintegration s}^{- 1}$
$-\frac{d N}{d t}=\lambda N_{0}\Rightarrow 7.5\times 10^{7}=0.012\times N_{0}\Rightarrow N_{0}=6.25\times 10^{9}$
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,
$\frac{d N}{d t}=R_{0}-\lambda N$
$\Rightarrow \frac{d N}{R_{0} - \lambda N}=dt$
integrating it from $0\text{to}t\text{and}0\text{to}N_{0}$
$\displaystyle \int _{0}^{N_{0}}\frac{d N}{R_{0} - \lambda N}=\displaystyle \int _{0}^{t}dt\Rightarrow -\frac{1}{\lambda }ln\frac{R_{0} - \lambda N_{0}}{R_{0}}=t$
$\Rightarrow t=\frac{1}{\lambda }ln\frac{R_{0}}{R_{0} - \lambda N_{0}}=\frac{1}{0 . 012}ln\frac{\left(10\right)^{8}}{\left(\left(10\right)^{8} - 0 . 012 \times 6 . 25 \times \left(10\right)^{9}\right)}=\frac{1}{0 . 012}ln4=115\text{s}$
Therefore, the time for activity to reach $7.5\times 10^{7}\text{disintegration s}^{- 1}$ is $115\text{s}\approx2\text{min}$ .