Question

A radioactive isotope have a half life of $T$. after how much time is its activity reduced to $6.25$ of its original activity

Answer 1

given,

$T_n=t$

$N=\frac{N_0}{2^n}$

$N=6.25$% of $N_0$

$\frac{6.25N_0}{100}=\frac{N_0}{2^n}$

$\frac{1}{16}=\frac{1}{2^n}$

$\frac{1}{2^4}=\frac{1}{2^n}$

$\therefore n=4$

$\frac{t}{T_n}=4$

$\Rightarrow t=4T_n$