Question

A radioactive nucleus $A$ decays into another radioactive nucleus $B$ (with decay constant $4\lambda$ for the process).$B$ further disintegrates into stable nucleus C(decay constant for the process being $\lambda$) and into stable nucleus $D$ (by another process for which decay constant is $2\lambda$). $N_A,N_B,N_C$ and $N_D$ are no. of nuclei of $A,B,C,D$ respectively at any general instant $t$

• A. Considering that at $t=0,N_A=N_0,N_B=N_C=N_D=0,$ variation of $N_B$ as function of time will be as shown by the graph here
  
• B. Consider that at $t=0,N_A=N_0,N_B=N_C=N_D=0,$ theoretical value of maximum no. of nuclei of B during the decay process is $N_0{\left(\frac{4}{3}\right)}^3$
• C. Considering that at $t=0,N_A=N_0,N_B=2N_0,N_C=N_D=0.$ At $t=\infty$, theoretical value of number of nuclei $C$ equals $N_0$
• D. Rate of accumulation of nuclei B at any general instant is $4\lambda N_A-3\lambda N_B$
  

Answer 1

Answer: (A), (B), (D)

Rate of accumulation of nuclei B at any general instant is $4\lambda N_A-3\lambda N_B$


$\frac{d{N}_b}{dt}=4\lambda N_A-\lambda N_B-2\lambda N_B$
$=4\lambda N_A-3\lambda N_B$
Also, $N_A=N_o{e}^{-4\lambda t}$

$\frac{d{N}_b}{dt}=4\lambda N_A-3\lambda N_B$
$=4\lambda \left[N_0{e}^{-4\lambda t}\right]-3\lambda N_B$
$\Rightarrow N_B=4N_0[e^{3\lambda t}-e^{4\lambda t}]$
For $N_B$ to be maximum,
$\frac{d{N}_B}{dt}=-3\lambda e^{3\lambda t}+4\lambda e^{-4\lambda t}=0$
$\Rightarrow \lambda t=ln\left(\frac{4}{3}\right)$
$\Rightarrow t=\frac{ln\left(\frac{4}{3}\right)}{\lambda }$
$\Rightarrow N_B=4N_0[e^{-ln}{\left(\frac{4}{3}\right)}^3-e{\left(\frac{4}{3}\right)}^4]l$
$=N_0{\left(\frac{3}{4}\right)}^3$
Also, $\frac{d{N}_0}{dt}=\lambda N_B,\frac{d{N}_D}{dt}=2\lambda N_B$
$\Rightarrow \frac{N_c}{N_D}=\frac{1}{2}$ at $t=\infty$
Also $N_C+N_D=3N_O$
$\Rightarrow N_C=N_O$