Question

A radioactive isotope has a half-life of 5 years. After how much time is its activity reduced to $3.125\%$ of its original activity ?

Answer 1

We know $\frac{R}{R_0}={\left(\frac{1}{2}\right)}^n$

Given $\frac{R}{R_0}=3.125\%=\frac{3.125}{100}$

$\therefore \frac{3.125}{100}={\left(\frac{1}{2}\right)}^{n}or\frac{1}{32}={\left(\frac{1}{2}\right)}^n$

or ${\left(\frac{1}{2}\right)}^5={\left(\frac{1}{2}\right)}^n\Rightarrow n=5$

Given $T=5$ years

As $n=\frac{t}{T}$

$\therefore \frac{t}{T}=5$

or $t=5\times 5=25years$