A radioactive sample contains `1.00xx10^15` atoms and has an activity of `6.00xx10^11` Bq. What is its half-life?

Explanation

To find the half-life of a radioactive sample, we can use the relationship between activity (A), the number of radioactive atoms (N), and the decay constant (λ). The formula for activity is given by:

\[ A = \lambda N \]

Where:
- \( A \) is the activity in Becquerels (Bq),
- \( \lambda \) is the decay constant (in s\(^{-1}\)),
- \( N \) is the number of radioactive atoms.

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of atoms, \( N = 1.00 \times 10^{15} \) atoms
- Activity, \( A = 6.00 \times 10^{11} \) Bq

2. **Rearranging the formula to find the decay constant (λ):**
\[ \lambda = \frac{A}{N} \]

3. **Substituting the known values into the equation:**
\[ \lambda = \frac{6.00 \times 10^{11} \text{ Bq}}{1.00 \times 10^{15} \text{ atoms}} = 6.00 \times 10^{-4} \text{ s}^{-1} \]

4. **Using the relationship between half-life (T₁/₂) and decay constant (λ):**
The formula for half-life is:
\[ T_{1/2} = \frac{\ln(2)}{\lambda} \]
Where \( \ln(2) \approx 0.693 \).

5. **Substituting the decay constant into the half-life formula:**
\[ T_{1/2} = \frac{0.693}{6.00 \times 10^{-4}} \text{ s} \]

6. **Calculating the half-life:**
\[ T_{1/2} \approx 1155 \text{ s} \]

7. **Converting seconds into minutes:**
\[ T_{1/2} \approx \frac{1155 \text{ s}}{60} \approx 19.25 \text{ minutes} \]

### Final Answer:
The half-life of the radioactive sample is approximately **1155 seconds** or **19.25 minutes**.

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