Question

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 $N_B(t)/N_B(0)$ with respect to time $t$ ?

• A.
• B.
• C.
• D.

Answer 1

Answer: (C)

$A\to B,B\to C$
$\frac{d{N}_B}{dt}=\lambda N_A-\lambda N_B$
$\frac{d{N}_B}{dt}=2\lambda N_{B_0}{e}^{-\lambda t}-\lambda N_B$
$e^{-\lambda t}\left(\frac{d{N}_B}{dt}+\lambda N_B\right)$
$=2\lambda N_{B_0}{e}^{-\lambda t}\times e^{\lambda t}$
$\frac{d}{dt}\left(N_B{e}^{\lambda t}\right)=2\lambda N_{B_0}$,
on integrating
$N_B{e}^{\lambda t}=2\lambda t{N}_{B_0}+N_{B_0}$
$N_B=N_{B_0}[1+2\lambda t]e^{-\lambda t}$
$\frac{d{N}_B}{dt}=0$
at $-\lambda [1+2\lambda t)e^{-\lambda t}+2\lambda e^{-\lambda t}=0$
$N_{B_{\max }}$ at $t=\frac{1}{2\lambda }$