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The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\frac{1}{16}\right)^{th}$ of its initial value?
Explanation
$\frac{1}{16}=\left(\frac{1}{2}\right)^4$, so $t=4T_{1/2}=80$ minutes.
Detailed Solution
$\frac{N}{N_0}=\frac{1}{16}=\left(\frac{1}{2}\right)^{t/T_{1/2}}$
$t=4T_{1/2}=20\times4=80$ minutes