Question

A radioactive substance is being produced at a constant rate of $200$ nuclei. The decay constant of the substance is $1{\text{s}}^{-1}.$ Assuming that initially there are no nuclei present, the time (in second) after which the number of nuclei will become $100$ is

• A. $1\text{s}$
• B. $\ln \left(2\right)\text{s}$
• C. $\frac{1}{\ln \left(2\right)}\text{s}$
• D. $2\text{s}$

Answer 1

The correct option is B $\ln \left(2\right)\text{s}$

Let $N$ be the number of nuclei at any time $t.$ Then,

$\frac{dN}{dt}=P-\lambda N$

$\Rightarrow {\int }_0^N\frac{dN}{P-\lambda N}={\int }_0^{t}dt$

$\Rightarrow {\left[\ln \right(P-\lambda N\left)(-\frac{1}{\lambda })\right]}_0^N=t$

$\Rightarrow \left[\ln \frac{P-\lambda N}{P}\right]=-\lambda t$

$\Rightarrow \lambda N=P-P{e}^{-\lambda t}$

$\Rightarrow N=\frac{P}{\lambda }(1-e^{-\lambda t})$

Here, $P=200;N=100;\lambda =1{\text{s}}^{-1}$

$\Rightarrow 100=200(1-e^{-t})$

$\Rightarrow e^{-t}=\left(\frac{1}{2}\right)$

$\Rightarrow t=\ln \left(2\right)\text{s}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.