A radioactive sample is `alpha-`emitter with half life `138.6` days is observed by a student to have `2000` disintegration/sec. The number of radioactive nuclei for given activity are.

Options

A. `3.45xx10^(10)`
B. `1xx10^(10)`
C. `3.45XX10^(15)`
D. `2.75xx10^(11)`

Answer

To solve the problem of finding the number of radioactive nuclei in a sample given its activity and half-life, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Activity (A) = 2000 disintegrations/second
- Half-life (T_half) = 138.6 days

2. **Convert Half-life from Days to Seconds**:
- Since we need to work in SI units, we convert the half-life from days to seconds.
- 1 day = 24 hours
- 1 hour = 3600 seconds
- Therefore,
\[ T_{half} = 138.6 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \]
- Calculating this gives:
\[ T_{half} = 138.6 \times 24 \times 3600 = 11986224 \text{ seconds} \]

3. **Calculate the Decay Constant (λ)**:
- The decay constant (λ) is related to the half-life by the formula:
\[ T_{half} = \frac{\ln(2)}{\lambda} \]
- Rearranging gives:
\[ \lambda = \frac{\ln(2)}{T_{half}} \]
- Substituting the value of T_half:
\[ \lambda = \frac{0.693}{11986224} \]
- Calculating this gives:
\[ \lambda \approx 5.78 \times 10^{-8} \text{ s}^{-1} \]

4. **Calculate the Number of Radioactive Nuclei (n)**:
- The relationship between activity (A), decay constant (λ), and the number of radioactive nuclei (n) is given by:
\[ A = n \lambda \]
- Rearranging gives:
\[ n = \frac{A}{\lambda} \]
- Substituting the values of A and λ:
\[ n = \frac{2000}{5.78 \times 10^{-8}} \]
- Calculating this gives:
\[ n \approx 3.46 \times 10^{10} \]

### Final Answer:
The number of radioactive nuclei for the given activity is approximately \( n \approx 3.45 \times 10^{10} \).

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