A sample of `.^(210)Po` which is `alpha` -emitter with `T_(1/2)=138` days is observed by a student to have 200 disintegration (2000 Bq) . The activity in `muCi` for this source is

Options

A. `0.050 muCi`
B. `0.051 muCi`
C. `0.055 muCi`
D. `0.054 muCi`

Answer

To solve the problem of converting the activity of a sample of Polonium-210 from Becquerels to microcuries, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The half-life of Polonium-210 (T₁/₂) is 138 days.
- The activity of the sample is given as 2000 disintegrations per second (2000 Bq).

2. **Convert Becquerels to Curie:**
- We know that 1 Curie (Ci) is equivalent to \(3.7 \times 10^{10}\) disintegrations per second (Bq).
- To convert from Becquerels to Curie, we use the formula:
\[ \text{Activity in Curie} = \frac{\text{Activity in Bq}}{3.7 \times 10^{10}} \]

3. **Calculate the Activity in Curie:**
- Substitute the given activity into the formula:
\[ \text{Activity in Curie} = \frac{2000 \text{ Bq}}{3.7 \times 10^{10}} \]
- Performing the calculation:
\[ \text{Activity in Curie} = \frac{2000}{3.7 \times 10^{10}} \approx 5.405 \times 10^{-8} \text{ Ci} \]

4. **Convert Curie to Microcurie:**
- Since 1 microcurie (µCi) is \(10^{-6}\) Curie, we can convert the activity from Curie to microcurie:
\[ \text{Activity in µCi} = \text{Activity in Ci} \times 10^{6} \]
- Substitute the value we calculated:
\[ \text{Activity in µCi} = 5.405 \times 10^{-8} \text{ Ci} \times 10^{6} \approx 0.05405 \text{ µCi} \]

5. **Final Result:**
- Rounding to three significant figures, the activity of the sample in microcuries is approximately:
\[ \text{Activity} \approx 0.054 \text{ µCi} \]

### Conclusion:
The activity of the Polonium-210 sample is approximately **0.054 µCi**.

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