Question

A radioactive sample contains two different types of radioactive nuclei A-with half-life $5.0$ days and B-with half-life $30.0$ days initially the decay rate of the A-type nuclei is $64$ time that of B type of nuclei. Their decay rates will be equal when time is $9n$ days Find the value of $n$

Answer 1

Answer: 4

$(\lambda {)}_A=\frac{\ln (2)}{5}{d}^{-1}{y}^{-1}$ and $(\lambda {)}_B=\frac{\ln (2)}{30}$ day ${}^{-1}$
Given $\left(\frac{d{N}_{OA}}{dt}\right)=64\left(\frac{d{N}_{OB}}{dt}\right)$
$\therefore {\lambda }_A{N}_{OA}=64{\lambda }_B{N}_{OB}$
At time $t,\frac{d{N}_A}{dt}=\frac{d{N}_B}{dt}$
$\therefore {\lambda }_A{N}_A={\lambda }_B{N}_B$
Dividing equation $(1)$ and $(2)\frac{N_{OA}}{N_A}=64\left(\frac{N_{OB}}{N_B}\right)$
But $N_A=N_{OA}{e}^{-{\lambda }_{A}t}$ and $N_B=N_{OB}{e}^{-{\lambda }_{B}t}$
$\therefore$ equation $(3)$ becomes $e^{{\lambda }_{A}t}=64e^{{\lambda }_{B}t}$
$\therefore e\left({\lambda }_A-{\lambda }_B\right)t=64$
$\therefore \left({\lambda }_A-{\lambda }_B\right)t=\ln (64)$
$\therefore t=\frac{\ln (64)}{\frac{\ln (2)}{5}-\frac{\ln (2)}{30}}=36$ days $=9\times 4$ days