To analyze the decay of a radioactive sample with a decay constant of \( \lambda = 4.9 \times 10^{-18} \, \text{s}^{-1} \), we will calculate the average life, half-life, and the changes in activity over specified time intervals.
### Step-by-Step Solution:
1. **Calculate Average Life (Mean Life)**:
The average life \( t_{\text{average}} \) is given by the formula:
\[
t_{\text{average}} = \frac{1}{\lambda}
\]
Substituting the value of \( \lambda \):
\[
t_{\text{average}} = \frac{1}{4.9 \times 10^{-18}} \approx 2.04 \times 10^{17} \, \text{s}
\]
2. **Convert Average Life to Years**:
To convert seconds into years, we use the conversion factor:
\[
1 \text{ year} = 365 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} = 3.156 \times 10^7 \text{ seconds/year}
\]
Therefore, converting average life:
\[
t_{\text{average}} \text{ (in years)} = \frac{2.04 \times 10^{17} \, \text{s}}{3.156 \times 10^7 \, \text{s/year}} \approx 6.49 \times 10^9 \, \text{years}
\]
3. **Calculate Half-Life**:
The half-life \( t_{1/2} \) is given by:
\[
t_{1/2} = \frac{0.693}{\lambda}
\]
Substituting the value of \( \lambda \):
\[
t_{1/2} = \frac{0.693}{4.9 \times 10^{-18}} \approx 1.41 \times 10^{17} \, \text{s}
\]
4. **Convert Half-Life to Years**:
Converting half-life to years:
\[
t_{1/2} \text{ (in years)} = \frac{1.41 \times 10^{17} \, \text{s}}{3.156 \times 10^7 \, \text{s/year}} \approx 4.45 \times 10^9 \, \text{years}
\]
5. **Activity Decrease Over Time**:
The activity of the sample decreases by half every half-life. In \( 9 \times 10^{-9} \, \text{years} \):
- Calculate how many half-lives fit into \( 9 \times 10^{-9} \, \text{years} \):
\[
\text{Half-lives in } 9 \times 10^{-9} \text{ years} = \frac{9 \times 10^{-9}}{4.45 \times 10^9} \approx 0.00202 \text{ half-lives}
\]
Since this is less than one half-life, the activity does not decrease by a factor of 4.
6. **Calculate Decay Over 13.5 Billion Years**:
In \( 13.5 \times 10^9 \, \text{years} \):
- Number of half-lives:
\[
\text{Half-lives in } 13.5 \times 10^9 \text{ years} = \frac{13.5 \times 10^9}{4.45 \times 10^9} \approx 3.03 \text{ half-lives}
\]
After 3 half-lives:
- Remaining nuclei:
\[
N = N_0 \left(\frac{1}{2}\right)^3 = N_0 \times \frac{1}{8}
\]
Thus, the percentage of nuclei decayed:
\[
\text{Decayed} = 100\% - 12.5\% = 87.5\%
\]
### Summary of Statements:
1. Average life is approximately \( 6.49 \times 10^9 \) years.
2. Half-life is approximately \( 4.45 \times 10^9 \) years.
3. The activity of the sample decreases by a factor of 4 in \( 9 \times 10^{-9} \) years (this is incorrect).
4. 87.5% of the initial number of nuclei is decayed in \( 13.5 \times 10^9 \) years (this is correct).
