Question

Half-life of a radioactive substance A is two times the half-life of another radioactive substance B. Initially, the number of A and B are $N_A$ and $N_B$, respectively. After three half-lives of A, number of nuclei of both are equal. Then, the ratio $N_A/N_B$ is

• A. $1/4$
• B. $1/8$
• C. $1/3$
• D. $1/6$

Answer 1

The correct option is B $1/8$

$t_{1/2}=\frac{\ln 2}{\lambda }$
So, if ${\lambda }_A=\lambda$ then ${\lambda }_B=2\lambda$
$t_{1/2A}=\frac{\ln 2}{\lambda }$
$N_A\left(t\right)=N_A{e}^{-\lambda t}$
$N_B\left(t\right)=N_B{e}^{-2\lambda t}$

Given after $3t_{1/2A}$, both are equal
$N_A\left(t\right)=N_B\left(t\right)$
$N_A{e}^{-3\lambda \frac{\ln 2}{\lambda }}=N_B{e}^{-6\lambda \frac{\ln 2}{\lambda }}$
$\frac{N_A}{N_B}=\frac{1}{8}$