A radioactive isotope has half-life of 'K' years. How long will it take, so that activity reduces to 6.25% of its original value?

4 K

The correct answer is Option (2) → 4 K

The activity of a radioactive substance,

$A=A_0\left(\frac{1}{2}\right)^n$

where,

A → Remaining Activity

$A_0$ → Initial activity

n → Number of Half life elipsed

if, $\frac{A}{A_0}=0.0625$ [∵ Activity reduces to $\frac{6.25}{100}$]

$∴ \left(\frac{1}{2}\right)^n=0.0625$

$n=\log_2(0.0625)$

$n=4$

Hence, time taken for half life = $n.k=4k$