In a radioactive material, fraction of active material remaining after time t is 9 / 16. The fraction that was remaining after t / 2 is :

(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, $t / 2 ;$

$N(t / 2)=N_{0} e^{-\lambda(t / 2)}$

$\frac{N(t / 2)}{N_{0}}=\sqrt{e^{-\lambda t}}=\sqrt{\frac{9}{16}}$

$\therefore N(t / 2)=\frac{3}{4} N_{0}$