A radioactive element decays to 1/64th of its initial activity in 30…

The correct answer is Option (3) → 5 days

Using exponential decay law -

$A(t)=A_0\left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}}$

and,

$\frac{A(t)}{A_0}=\frac{1}{64}$ for $t=30$ days [given]

$∴\frac{1}{64}=\left(\frac{1}{2}\right)^{\frac{30}{T_{1/2}}}$

$⇒\left(\frac{1}{2}\right)^6=\left(\frac{1}{2}\right)^{\frac{30}{T_{1/2}}}$

$⇒6=\frac{30}{T_{1/2}}$

$⇒T_{1/2}=5\,days$