JEE Main 26 Aug 2021 Shift 2 Solved Paper

Given, at time $t=0, N_{A}(0)=2 N_{B}(0)$ Decay constant is same for both radioactive atoms as λ. For A → B , $\frac{d N_{B}(t)}{d t}=\lambda N_{A}(t)-\lambda N_{B}(t)$ Substituting $N_{A}(0) e^{-\lambda t}$ for $N_{A}(t)$ in above expression, we get $\frac{d N_{B}(t)}{d t}=\lambda N_{A}(0) e^{-\lambda t}-\lambda N_{B}(t)$ $=2 \lambda N_{B}(0) e^{-\lambda t}-\lambda N_{B}(t)$ $\Rightarrow \frac{d N_{B}(t)}{d t}+\lambda N_{B}(t)=2 \lambda N_{B}(0) e^{-\lambda t}$ Multiplying both sides by $e^{\lambda t}$, we get $e^{\lambda t}\left[\frac{d N_{B}(t)}{d t}+\lambda N_{B}(t)\right]=2 \lambda N_{B}(0) e^{-\lambda t} \times e^{\lambda t}$ $\Rightarrow \frac{d}{d t}\left[N_{B}(t) e^{\lambda t}\right]=2 \lambda N_{B}(0)$ Integrating both sides, $\left[N_{B}(t) e^{\lambda t}\right]=2 \lambda N_{B}(0) t+C$ ...(i) Putting t = 0 in above expression, $\left[N_{B}(0) e^{\lambda \times 0}\right]=2 \lambda N_{B}(0) \times 0+C$ $\Rightarrow C=N_{B}(0)$ Putting value of C in Eq. (i) $\left[N_{B}(t) e^{\lambda t}\right]=2 \lambda N_{B}(0) t+N_{B}(0)$ ...(ii) $\Rightarrow N_{B}(t)=N_{B}(0)\left[1+2 \lambda t\right] e^{-\lambda t}$ $\Rightarrow \frac{N_{B}(t)}{N_{B}(0)}=\left[1+2 \lambda t\right] e^{-\lambda t}$ As, $N_{B}(t)=N_{B}(0)[1+2 \lambda t) e^{-\lambda t}$ $\Rightarrow N_{B}(t)=C\left[1+2 \lambda t\right] e^{-\lambda t}$ ...(iii) The maximum value of $N_{B}(t)$ is obtained at $\frac{d N_{B}(t)}{d t}=0$ From Eq. (ii), $t=\frac{1}{2 \lambda} \text{s}$ Solving the above expression, $t=\frac{1}{2 \lambda} \text{s}$ Thus, maximum value of function will be at $t = \frac{1}{2\lambda}s$. Hence, graph (c) is correct option.