Question

Nucleus $A$ decays into $B$ with a decay constant ${\lambda }_1$ and $B$ further decays into $C$ with a decay constant ${\lambda }_2$ . Initially, at $t=0$ , the number of nuclei of $A$ and $B$ were $3N_0$ and $N_0$ respectively. If at t = t₀ number of nuclei of $B$ becomes constant and equal to $2N_0$ , then

• A. $t_0=\frac{1}{{\lambda }_1}ln\left[\frac{3{\lambda }_1}{2{\lambda }_2}\right]$
• B. $t_0=\frac{1}{{\lambda }_1}ln\left[\frac{{\lambda }_1}{{\lambda }_2}\right]$
• C. $t_0=\frac{1}{{\lambda }_1}ln\left[\frac{2{\lambda }_1}{3{\lambda }_2}\right]$
• D. $t_0=\frac{1}{{\lambda }_2}ln\left[\frac{3{\lambda }_1}{2{\lambda }_2}\right]$

Answer 1

Answer: (A)

$N_A=3N_0{e}^{-{\lambda }_{1}t}$
$\frac{d{N}_A}{dt}=-3{\lambda }_1{N}_0{e}^{-{\lambda }_{1}t}$
$\frac{d{N}_B}{dt}=3{\lambda }_1{N}_0{e}^{-{\lambda }_{1}t}-{\lambda }_2{N}_B$
At $t=t_0$ , $\frac{d{N}_B}{dt}=0$
$\Rightarrow 3{\lambda }_1{N}_0{e}^{-{\lambda }_{1}t}-{\lambda }_2{N}_B=0$
$\Rightarrow 3{\lambda }_1{N}_0{e}^{-{\lambda }_{1}t}={\lambda }_2\left(2N_0\right)$
$e^{{\lambda }_1{t}_0}=\frac{3{\lambda }_1}{2{\lambda }_2}$
$\Rightarrow t_0=\frac{1}{{\lambda }_1}ln\left[\frac{3{\lambda }_1}{2{\lambda }_2}\right]$