In a radioactive material, fraction of active material remaining after time t is 9 / 16. The fraction that was remaining after t / 2 is :
Question:
In a radioactive material, fraction of active material remaining after time $t$ is $9 / 16$. The fraction that was remaining after $t / 2$ is :
Correct Option: , 3
Solution:
(3) As we know, for first order decay, $N(t)=N_{0} e^{-\lambda t}$
According to question,
$\frac{N(t)}{N_{0}}=\frac{9}{16}=e^{-\lambda t}$
After time, $t / 2 ;$
$N(t / 2)=N_{0} e^{-\lambda(t / 2)}$
$\frac{N(t / 2)}{N_{0}}=\sqrt{e^{-\lambda t}}=\sqrt{\frac{9}{16}}$
$\therefore N(t / 2)=\frac{3}{4} N_{0}$