Question

A radioactive isotope has a half life of $T$. After how much time is its activity reduced to $6.25\%$ of its original activity?

Answer 1

Half life is $T$

if initial activity is $N_0$ and the final activity after time $t$ is $N$ then we will have following relation $N=N_0(\frac{1}{2}{)}^{t/T}$...... (1)

But as given in the question that $N=6.25$% OF $N_0$ i.e $N=N_0\times \frac{6.25}{100}$ or $N=\frac{N_0}{16}$

using above relation in equation-1 we get $\frac{1}{16}=(\frac{1}{2}{)}^{t/T}$

or $(\frac{1}{2}{)}^4=(\frac{1}{2}{)}^{t/T}$

So $t/T=4$ or $t=4T$ or required time is $4$ times of the half life.