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, we need to convert the activity of the sample from becquerels (Bq) to microcuries (µCi).

**Step 1: Understand the given information.**
- The activity of the sample is given as 2000 Bq (which is equivalent to 200 disintegrations per second).
- We need to convert this activity into microcuries (µCi).

**Step 2: Recall the conversion factor between Bq and curies.**
- 1 Curie (Ci) = 3.7 × 10^10 Bq
- To convert Bq to Ci, we can use the formula:
\[ \text{Activity in Ci} = \frac{\text{Activity in Bq}}{3.7 \times 10^{10}} \]

**Step 3: Calculate the activity in curies.**
- Using the formula:
\[ \text{Activity in Ci} = \frac{2000 \text{ Bq}}{3.7 \times 10^{10}} \]
- Performing the calculation:
\[ \text{Activity in Ci} = \frac{2000}{3.7 \times 10^{10}} \approx 5.4054 \times 10^{-8} \text{ Ci} \]

**Step 4: Convert the activity from curies to microcuries.**
- Since 1 Ci = 10^6 µCi, we can convert the activity in Ci to µCi:
\[ \text{Activity in µCi} = 5.4054 \times 10^{-8} \text{ Ci} \times 10^6 \text{ µCi/Ci} \]
- Performing the multiplication:
\[ \text{Activity in µCi} \approx 0.054054 \text{ µCi} \]

**Step 5: Round the final answer.**
- The final answer can be rounded to three decimal places:
\[ \text{Activity in µCi} \approx 0.054 \text{ µCi} \]

Thus, the activity of the sample in microcuries is approximately **0.054 µCi**.