The half life of a radioactive substance is 10 days. How many days will it take to disintegrate 3/4 of its initial value?

20 days

The correct answer is Option (3) → 20 days

formula for radioactive decay is,

$N(t)=N_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

$⇒\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}=\left(\frac{1}{4}\right)=\left(\frac{1}{2}\right)^2$

$⇒\frac{t}{T_{1/2}}=2$

$⇒t=2×T_{1/2}=2×10=20days$