Question

At time $t =0$, a material is composed of two radioactive atoms $A$ and $B$, where $N _{ A }(0)=2 N _{ B }(0)$. 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)

$A \rightarrow B , B \rightarrow C$
$\frac{ d N _{ B }}{ dt }=\lambda N _{ A }-\lambda N _{ B }$
$\frac{ d N _{ B }}{ dt }=2 \lambda \,N _{ B _{0}} e ^{-\lambda t }-\lambda N _{ B }$
$e ^{-\lambda t }\left(\frac{ d N _{ B }}{ dt }+\lambda N _{ B }\right)$
$=2 \lambda N _{ B _{0}} \,e ^{-\lambda t } \times e ^{\lambda t }$
$\frac{ d }{ dt }\left( N _{ B } \,e ^{\lambda t }\right)=2 \lambda N _{ B _{0}}$,
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 $t =\frac{1}{2 \lambda}$