The half-life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $

Question

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?

Accepted Answer

1. **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 .
2. **Determine Number of Half-lives ($n$):**
   - We are given that the final activity is $\frac{1}{16}$ of the initial activity.
   - $\frac{A}{A_0} = \frac{1}{16} = \left(\frac{1}{2}ight)^4$.
   - Comparing exponents, the number of half-lives $n = 4$.
3. **Calculate Total Time ($t$):**
   - The total time elapsed is the number of half-lives multiplied by the half-life period ($T_{1/2}$).
   - $t = n 	imes T_{1/2} = 4 	imes 20 	ext{ minutes}$.
   - $t = 80 	ext{ minutes}$.

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