Question

The half-life of radon is $3.8$ days. In how many days will its activity drop to $6.25\%$ of its initial value?

Answer 1

Suppose after time $t$ the initial amount $\left(N_0\right)$ becomes only $N$ where $N=\frac{6.25}{100}{N}_0=N_0/16$

Using above relation in the following formula,

$N=N_0(\frac{1}{2}{)}^n$

we get $\frac{N_0}{16}=N_0(\frac{1}{2}{)}^n$ or $n=4$

Where $n=\frac{t}{t_{1/2}}$

so $t=n{t}_{1/2}=4\times 3.8days=15.2days$