At time t = 0, a material is composed of two radioactive atoms A and B, where ${\mathrm{N}}_{\mathrm{A}}\left(0\right)=2{\mathrm{N}}_{\mathrm{B}}\left(0\right)$. The decay constant of both kind of radioactive atoms is λ. However, A disintegrates to B and B disintegrates to C. Which of the following figures represents the evolution of ${\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right)/{\mathrm{N}}_{\mathrm{B}}\left(0\right)$ with respect to time t ?

$\left[\begin{array}{l}{\mathrm{N}}_{\mathrm{A}}\left(0\right)=\text{ Number of }\mathrm{A}\text{ atoms at }\mathrm{t}=0\\ {\mathrm{N}}_{\mathrm{B}}\left(0\right)=\text{ Number of }\mathrm{B}\text{ atoms at }\mathrm{t}=0\end{array}\right]$

Given, at time $\mathrm{t}=0, {\mathrm{N}}_{\mathrm{A}}\left(0\right)=2{\mathrm{N}}_{\mathrm{B}}\left(0\right)$

Decay constant is same for both radioactive atoms as λ.

For A → B,

         $\frac{{\mathrm{dN}}_{\mathrm{B}}\left(\mathrm{t}\right)}{\mathrm{dt}}={\mathrm{\lambda N}}_{\mathrm{A}}\left(\mathrm{t}\right)-{\mathrm{\lambda N}}_{\mathrm{B}}\left(\mathrm{t}\right)$

Substituting ${\mathrm{N}}_{\mathrm{A}}\left(0\right){\mathrm{e}}^{-\mathrm{\lambda t}}\text{ for }{\mathrm{N}}_{\mathrm{A}}\left(\mathrm{t}\right)$ in above expression, we get

         $\begin{array}{l} \frac{{\mathrm{dN}}_{\mathrm{B}}\left(\mathrm{t}\right)}{\mathrm{dt}}={\mathrm{\lambda N}}_{\mathrm{A}}\left(0\right){\mathrm{e}}^{-\mathrm{\lambda t}}-{\mathrm{\lambda N}}_{\mathrm{B}}\left(\mathrm{t}\right)\\ =2{\mathrm{\lambda N}}_{\mathrm{B}}\left(0\right){\mathrm{e}}^{-\mathrm{\lambda t}}-{\mathrm{\lambda N}}_{\mathrm{B}}\left(\mathrm{t}\right)\\ \Rightarrow \frac{{\mathrm{dN}}_{\mathrm{B}}\left(\mathrm{t}\right)}{\mathrm{dt}}+{\mathrm{\lambda N}}_{\mathrm{B}}\left(\mathrm{t}\right)=2{\mathrm{\lambda N}}_{\mathrm{B}}\left(0\right){\mathrm{e}}^{-\mathrm{\lambda t}}\end{array}$

Multiplying both sides by e^λt, we get

         $\begin{array}{l} {\mathrm{e}}^{\mathrm{\lambda t}}\left[\frac{{\mathrm{dN}}_{\mathrm{B}}\left(\mathrm{t}\right)}{\mathrm{dt}}+{\mathrm{\lambda N}}_{\mathrm{B}}\left(\mathrm{t}\right)\right]=2{\mathrm{\lambda N}}_{\mathrm{B}}\left(0\right){\mathrm{e}}^{-\mathrm{\lambda t}}\times {\mathrm{e}}^{\mathrm{\lambda t}}\\ \Rightarrow \frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{\lambda t}}\right]=2{\mathrm{\lambda N}}_{\mathrm{B}}\left(0\right)\end{array}$

Integrating both sides,

         $\left[{\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{\lambda t}}\right]=2{\mathrm{\lambda N}}_{\mathrm{B}}\left(0\right)\mathrm{t}+\mathrm{C} ...\left(\mathrm{i}\right)$

Putting t = 0 in above expression,

          $\begin{array}{l} \left[{\mathrm{N}}_{\mathrm{B}}\left(0\right){\mathrm{e}}^{\mathrm{\lambda }\times 0}\right]=2{\mathrm{\lambda N}}_{\mathrm{B}}\left(0\right)\times 0+\mathrm{C}\\ \Rightarrow \mathrm{C}={\mathrm{N}}_{\mathrm{B}}\left(0\right)\end{array}$

Putting value of C in Eq. (i)

$\begin{array}{l}\Rightarrow \left[{\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right){\mathrm{e}}^{\mathrm{\lambda t}}\right]=2{\mathrm{\lambda N}}_{\mathrm{B}}\left(0\right)\mathrm{t}+{\mathrm{N}}_{\mathrm{B}}\left(0\right) ....\left(\mathrm{ii}\right)\\ \Rightarrow {\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right)={\mathrm{N}}_{\mathrm{B}}\left(0\right)[1+2\mathrm{\lambda t}]{\mathrm{e}}^{-\mathrm{\lambda t}}\\ \Rightarrow \frac{{\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right)}{{\mathrm{N}}_{\mathrm{B}}\left(0\right)}=[1+2\mathrm{\lambda t}]{\mathrm{e}}^{-\mathrm{\lambda t}}\\ \Rightarrow {\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right)={\mathrm{N}}_{\mathrm{B}}\left(0\right)[1+2\mathrm{\lambda t}){\mathrm{e}}^{-\mathrm{\lambda t}}\\ \Rightarrow {\mathrm{N}}_{\mathrm{B}}\left(\mathrm{t}\right)=\mathrm{C}[1+2\mathrm{\lambda t}]{\mathrm{e}}^{-\mathrm{\lambda t}} ....\left(\mathrm{iii}\right)\end{array}$

The maximum value of N_B(t) is obtained at

         $\frac{{\mathrm{dN}}_{\mathrm{B}}\left(\mathrm{t}\right)}{\mathrm{dt}}=0$

From Eq. (ii),

             $\frac{{\mathrm{dN}}_{\mathrm{B}}\left(\mathrm{t}\right)}{\mathrm{dt}}=-\mathrm{\lambda C}[1+2\mathrm{\lambda t}]{\mathrm{e}}^{-\mathrm{\lambda t}}+2{\mathrm{C\lambda e}}^{-\mathrm{\lambda t}}=0$

Solving the above expression,

              $\mathrm{t}=\frac{1}{2\mathrm{\lambda }}\mathrm{s}$

Thus, maximum value of function will be at $\mathrm{t}=\frac{1}{2\mathrm{\lambda }}\mathrm{s}$.

Hence, graph (c) is correct option.