The half life of polonium is 140 days. In what time will 15 g of polonium be disintegrated out of its initial mass of 16 g ?

Options

A. 230 days
B. 560 days
C. 730 days
D. 160 days

Explanation

To solve the problem, we need to determine how long it will take for 15 g of polonium to disintegrate from an initial mass of 16 g, given that the half-life of polonium is 140 days.

### Step-by-Step Solution:

1. **Identify Initial and Final Mass:**
- Initial mass (M₀) = 16 g
- Final mass (M) = 15 g
- Mass disintegrated (ΔM) = M₀ - M = 16 g - 15 g = 1 g

2. **Determine the Fraction of Mass Remaining:**
- The fraction of the initial mass remaining after disintegration is:
\[ \text{Fraction remaining} = \frac{M}{M₀} = \frac{15 \text{ g}}{16 \text{ g}} = \frac{15}{16} \]

3. **Relate Mass Remaining to Number of Half-Lives:**
- The relationship between the remaining mass and half-lives can be expressed as:
\[ \text{Remaining mass} = M₀ \left(\frac{1}{2}\right)^n \]
- Where \( n \) is the number of half-lives. We need to find \( n \) such that:
\[ \frac{15}{16} = \left(\frac{1}{2}\right)^n \]

4. **Calculate the Number of Half-Lives:**
- We can rewrite the equation:
\[ \left(\frac{1}{2}\right)^n = \frac{15}{16} \]
- Taking logarithm on both sides:
\[ n \log\left(\frac{1}{2}\right) = \log\left(\frac{15}{16}\right) \]
- Solving for \( n \):
\[ n = \frac{\log\left(\frac{15}{16}\right)}{\log\left(\frac{1}{2}\right)} \]

5. **Calculate \( n \):**
- Using logarithm values:
\[ \log\left(\frac{15}{16}\right) \approx -0.09691 \quad \text{and} \quad \log\left(\frac{1}{2}\right) \approx -0.30103 \]
- Therefore:
\[ n \approx \frac{-0.09691}{-0.30103} \approx 0.322 \]

6. **Calculate Total Time:**
- Each half-life is 140 days, so the total time \( T \) is:
\[ T = n \times \text{half-life} = 0.322 \times 140 \text{ days} \approx 45.08 \text{ days} \]

### Final Answer:
The time required for 15 g of polonium to disintegrate from an initial mass of 16 g is approximately **45.08 days**.