The half-life of `._(38)^(90)Sr` is 20 years. If its sample having initial activity of 800 dis/min is taken, what would be its activity after 80 years

Options

A. 500 dis/min
B. 800 dis/min
C. 1000 dis/min
D. 1600 dis/min

Explanation

To solve the problem, we need to determine the activity of the radioactive isotope Strontium-90 (Sr-90) after 80 years, given its half-life and initial activity.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Half-life of Sr-90 = 20 years
- Initial activity (A₀) = 800 disintegrations per minute (DPM)
- Total time (t) = 80 years

2. **Calculate the Number of Half-Lives:**
- To find out how many half-lives fit into the total time of 80 years, we divide the total time by the half-life:
\[ \text{Number of half-lives} = \frac{t}{\text{Half-life}} = \frac{80 \text{ years}}{20 \text{ years}} = 4 \]

3. **Determine the Remaining Activity After Each Half-Life:**
- The activity of a radioactive substance decreases by half after each half-life. We can calculate the remaining activity after each half-life:
- After 1st half-life (20 years):
\[ A_1 = \frac{A_0}{2} = \frac{800 \text{ DPM}}{2} = 400 \text{ DPM} \]
- After 2nd half-life (40 years):
\[ A_2 = \frac{A_1}{2} = \frac{400 \text{ DPM}}{2} = 200 \text{ DPM} \]
- After 3rd half-life (60 years):
\[ A_3 = \frac{A_2}{2} = \frac{200 \text{ DPM}}{2} = 100 \text{ DPM} \]
- After 4th half-life (80 years):
\[ A_4 = \frac{A_3}{2} = \frac{100 \text{ DPM}}{2} = 50 \text{ DPM} \]

4. **Final Activity After 80 Years:**
- Therefore, the activity of the Sr-90 sample after 80 years is:
\[ A = 50 \text{ DPM} \]

### Conclusion:
The activity of the Sr-90 sample after 80 years is **50 disintegrations per minute (DPM)**.

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