At time $t = 0$, a material is composed of two radioactive atoms $A$ and $B$, where $N_A(0) = 2N_B(0)$. The decay constant of both kind of radioactive atoms is $\lambda$. However, $A$ disintegrates to $B$ and $B$ disintegrates to $C$. Which of the following figures represents the evolution of $\frac{N_B(t)}{N_B(0)}$ with respect to time $t$?
Answer: (C) see figure
Let $N_B(0) = N_0$, so $N_A = 2N_0e^{-\lambda t}$ and
$$\frac{dN_B}{dt} = \lambda N_A - \lambda N_B$$
The solution with $N_B(0) = N_0$ is $N_B = N_0e^{-\lambda t}(1 + 2\lambda t)$, so
$$\frac{N_B(t)}{N_B(0)} = e^{-\lambda t}(1 + 2\lambda t)$$
It starts at $1$. Setting the derivative to zero: $\lambda e^{-\lambda t}(2 - 1 - 2\lambda t) = 0 \Rightarrow t = \dfrac{1}{2\lambda}$, where the ratio is $2e^{-1/2} \approx 1.21$. After that it decays to zero.
So the ratio rises from $1$ to a maximum at $t = \dfrac{1}{2\lambda}$ and then falls.
Solution by Sreeraj P, M.Sc Physics