The nucleus (15 8) O has a half-life of 122.2 s; (19 8) O has a half-life of 26.9 s. Assume that at some time a sample contains equal amounts of both isotopes. What is the ratio of (15 8) O to (19 8) O after 3.0 minutes?

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**Half-Life Calculation of Oxygen Isotopes**

This problem involves calculating the ratio of two isotopes of oxygen with distinct half-lives.

**Given:**
- The isotope \(\mathbf{{{}^{15}_{8}O}}\) has a half-life of 122.2 seconds.
- The isotope \(\mathbf{{{}^{19}_{8}O}}\) has a half-life of 26.9 seconds.
- Initially, a sample contains equal amounts of both isotopes.

The task is to determine the ratio of \(\mathbf{{{}^{15}_{8}O}}\) to \(\mathbf{{{}^{19}_{8}O}}\) after 3.0 minutes.

**Steps to Solve:**

1. **Convert Time to Seconds:**
   - 3 minutes = 3 * 60 = 180 seconds

2. **Calculate Number of Half-Lives:**
   - For \(\mathbf{{{}^{15}_{8}O}}\):
     \( \frac{180}{122.2} \approx 1.472 \text{ half-lives} \)
   
   - For \(\mathbf{{{}^{19}_{8}O}}\):
     \( \frac{180}{26.9} \approx 6.690 \text{ half-lives} \)

3. **Exponential Decay Calculation:**
   - Remaining fraction of \(\mathbf{{{}^{15}_{8}O}}\):
     \( (0.5)^{1.472} \approx 0.351 \)
   
   - Remaining fraction of \(\mathbf{{{}^{19}_{8}O}}\):
     \( (0.5)^{6.690} \approx 0.01 \)

4. **Calculate the Ratio:**
   - Ratio of \(\mathbf{{{}^{15}_{8}O}}\) to \(\mathbf{{{}^{19}_{8}O}}\) after 3.0 minutes:
     \( \frac{0.351}{0.01} \approx 35.1 \)

Therefore, the ratio of \(\mathbf{{{}^{15}_{8}O}}\) to \(\mathbf{{{}^{19}_{8}O}}\) after 3.0 minutes is approximately \(\mathbf{35.1}\).

**Conclusion:**
After 3.0 minutes, for every 1 unit of \(\mathbf{{{}^{19}_{8}O}}\
Transcribed Image Text:**Half-Life Calculation of Oxygen Isotopes** This problem involves calculating the ratio of two isotopes of oxygen with distinct half-lives. **Given:** - The isotope \(\mathbf{{{}^{15}_{8}O}}\) has a half-life of 122.2 seconds. - The isotope \(\mathbf{{{}^{19}_{8}O}}\) has a half-life of 26.9 seconds. - Initially, a sample contains equal amounts of both isotopes. The task is to determine the ratio of \(\mathbf{{{}^{15}_{8}O}}\) to \(\mathbf{{{}^{19}_{8}O}}\) after 3.0 minutes. **Steps to Solve:** 1. **Convert Time to Seconds:** - 3 minutes = 3 * 60 = 180 seconds 2. **Calculate Number of Half-Lives:** - For \(\mathbf{{{}^{15}_{8}O}}\): \( \frac{180}{122.2} \approx 1.472 \text{ half-lives} \) - For \(\mathbf{{{}^{19}_{8}O}}\): \( \frac{180}{26.9} \approx 6.690 \text{ half-lives} \) 3. **Exponential Decay Calculation:** - Remaining fraction of \(\mathbf{{{}^{15}_{8}O}}\): \( (0.5)^{1.472} \approx 0.351 \) - Remaining fraction of \(\mathbf{{{}^{19}_{8}O}}\): \( (0.5)^{6.690} \approx 0.01 \) 4. **Calculate the Ratio:** - Ratio of \(\mathbf{{{}^{15}_{8}O}}\) to \(\mathbf{{{}^{19}_{8}O}}\) after 3.0 minutes: \( \frac{0.351}{0.01} \approx 35.1 \) Therefore, the ratio of \(\mathbf{{{}^{15}_{8}O}}\) to \(\mathbf{{{}^{19}_{8}O}}\) after 3.0 minutes is approximately \(\mathbf{35.1}\). **Conclusion:** After 3.0 minutes, for every 1 unit of \(\mathbf{{{}^{19}_{8}O}}\
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