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?
Answer & explanation
Correct answer: option 2
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 = $n.k=4k$