Question

A radioactive nucleus with decay constant $0.5{{s}^{-1}}$ is being produced at a constant rate of $100nuclei/s$. If at $t=0$ there were no nuclei, the time when there are $50$ nuclei is
$\begin{align}
  & A\text{)1s} \\
 & \text{B)2ln}\left( \dfrac{\text{4}}{\text{3}} \right)\text{ s} \\
 & \text{C)ln2 s} \\
 & \text{D)ln}\left( \dfrac{4}{3} \right)\text{ s} \\
\end{align}$

Answer 1

Hint: From the radio-active decay law, rate of disintegration of a radioactive-nuclei at any instant is directly proportional to the number of radioactive nuclei present in the sample at that particular instant. An expression for radioactive decay is deduced from the given information by taking into consideration the production rate of nuclei in that element. To determine the time when there are $50$ nuclei, we make use of integration.
Formula used:
$\left( -\dfrac{dN}{dt} \right)=\lambda N$

Complete answer:
We know that the rate of disintegration of a radioactive-nuclei at any instant is directly proportional to the number of radioactive nuclei present in the sample at that particular instant. Mathematically, rate of disintegration is given by
$\left( -\dfrac{dN}{dt} \right)=\lambda N$
where
$\lambda $ is the decay constant
$N$ is the number of nuclei present in the sample at a particular instant $t$
$dN$ is the number of disintegrated nuclei during a time interval $dt$
Let this be equation 1.
Coming to our question, we are provided that
$\lambda =0.5{{s}^{-1}}$
and that
$\dfrac{dN}{dt}+\lambda N=100$
Let this be equation 2.
Rearranging equation 2, we have
$\dfrac{dN}{dt}+\lambda N=100\Rightarrow \dfrac{dN}{dt}=100-\lambda N\Rightarrow \dfrac{dN}{100-\lambda N}=dt$
Now, integrating the above expression to determine $t$, we have
\[\int\limits_{0}^{50}{\dfrac{dN}{\left( 100-\lambda N \right)}}=\int\limits_{0}^{t}{dt}\Rightarrow -\dfrac{1}{\lambda }\ln \left( 100-\lambda N \right)_{0}^{50}=t\Rightarrow -\dfrac{1}{\lambda }\ln \left( 100-\lambda \times 50 \right)+\dfrac{1}{\lambda }\ln \left( 100-0 \right)=t\]
Solving further by substituting the value of decay constant, we have
\[t=-2\times \ln \left( 100-25 \right)+2\times \ln \left( 100 \right)\Rightarrow t=2\times \left( \ln 100-\ln 75 \right)\Rightarrow t=2\times \ln \dfrac{100}{75}=2\ln \dfrac{4}{3}s\]
Let this be equation 3.

Therefore, from equation 3, we can conclude that the correct answer is option $B$.

Note:
Radioactivity refers to the phenomenon of spontaneous emission of radiation by the nucleus of a radio-active element. The three types of radiations, usually emitted by radioactive elements are alpha rays, beta rays and gamma rays. Students should take care in deducing equation 2. What is given in the question is the rate of production of nuclei, which is a consequence of both disintegration as well as generation. Hence deduced, is this equation.