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

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