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)

$N_{A}=3N_{0}e^{- \lambda _{1} t}$
$\frac{d N_{A}}{d t}=-3\lambda _{1}N_{0}e^{- \lambda _{1} t}$
$\frac{d N_{B}}{d t}=3\lambda _{1}N_{0}e^{- \lambda _{1} t}-\lambda _{2}N_{B}$
At $t=t_{0}$ , $\frac{d N_{B}}{d t}=0$
$\Rightarrow 3\lambda _{1}N_{0}e^{- \lambda _{1} t}-\lambda _{2}N_{B}=0$
$\Rightarrow 3 \lambda_1 N_0 e^{-\lambda_1 t}=\lambda_2\left(2 N_0\right)$
$e^{\lambda _{1} t_{0}}=\frac{3 \lambda _{1}}{2 \lambda _{2}}$
$\Rightarrow t_{0}=\frac{1}{\lambda _{1}}ln\left[\frac{3 \lambda _{1}}{2 \lambda _{2}}\right]$