Question

A radioactive sample has equal numbers of ${}^{15}O$ and ${}^{19}O$ nuclei. The half life of ${}^{15}O$ is $120\text{sec}$. and the half life of ${}^{19}O$ is $30\text{sec}$. How long it will take (in sec.) before there are twice as many ${}^{15}O$ nuclei as ${}^{19}O$. Then the value of $\frac{t}{10}$ in seconds is (where $t$ is the time taken).

Answer 1

$N_1=N_0{e}^{-{\lambda }_{1}t}$
$N_2=N_0{e}^{-{\lambda }_{2}t}$
where $N_1$ corresponds to ${}^{15}O$ nuclei and $N_2$ corresponds to ${}^{19}O$ nuclei.
According to the problem
$N_1=2N_2$
$2e^{-{\lambda }_{2}t}=e^{-{\lambda }_{1}t}$
$\text{ln}2-\frac{\text{ln}2}{30}t=-\frac{\text{ln}2}{120}t$
$t=40\text{sec}.$
So,
$\frac{t}{10}=\frac{40}{10}=4.$