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NEET PHYSICSEasy

The half-life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $(1/16)^{th}$ of its initial value?

A

80 minutes

B

20 minutes

C

40 minutes

D

60 minutes

Step-by-Step Solution

Identify the Relationship: Radioactive decay follows first-order kinetics. The remaining activity $A$ after $n$ half-lives is given by $A = A_0 (\frac{1}{2})^n$ , where $A_0$ is the initial activity .
Determine Number of Half-lives ( $n$ ):

We are given that the final activity is $\frac{1}{16}$ of the initial activity.
$\frac{A}{A_0} = \frac{1}{16} = \left(\frac{1}{2}\right)^4$ .
Comparing exponents, the number of half-lives $n = 4$ .

Calculate Total Time ( $t$ ):

The total time elapsed is the number of half-lives multiplied by the half-life period ( $T_{1/2}$ ).
$t = n \times T_{1/2} = 4 \times 20 \text{ minutes}$ .
$t = 80 \text{ minutes}$ .

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