Question
The half-life of Carbon-14 is approximately 5, 730 _________.
iii. Using the image above, how long would it take to have 12.5% of the
original amount. _____________________________
Collaborative Catch-up: Half-Life
Bromine: 84 (84B) has a half-life of 32 minutes.
Complete the following diagram showing the decay of 140 units of 84B.
Figure 2-4. Calculating Half-Life4
3 Created by S. Hyatt - 2024.
4 Created by C. Moore-Crawford - 20242
iii. Using the image above, how long would it take to have 12.5% of the
original amount. _____________________________
Collaborative Catch-up: Half-Life
Bromine: 84 (84B) has a half-life of 32 minutes.
Complete the following diagram showing the decay of 140 units of 84B.
Figure 2-4. Calculating Half-Life4
3 Created by S. Hyatt - 2024.
4 Created by C. Moore-Crawford - 20242
Answers
GPT-4o mini
The half-life of Carbon-14 is approximately 5,730 **years**.
To determine how long it would take for 12.5% of the original amount of Carbon-14 to remain, we need to understand the concept of half-lives.
12.5% is equivalent to \( \frac{1}{8} \) of the original amount, which means that it takes three half-lives to reach this percentage because:
- After 1 half-life, 50% remains.
- After 2 half-lives, 25% remains.
- After 3 half-lives, 12.5% remains.
Since the half-life of Carbon-14 is 5,730 years, the time it would take to reach 12.5% is:
\[
3 \times 5,730 \text{ years} = 17,190 \text{ years}
\]
For the decay of Bromine-84 (84B), which has a half-life of 32 minutes and starts with 140 units:
1. After the 1st half-life (32 minutes):
\( \frac{140}{2} = 70 \) units remaining.
2. After the 2nd half-life (64 minutes total):
\( \frac{70}{2} = 35 \) units remaining.
3. After the 3rd half-life (96 minutes total):
\( \frac{35}{2} = 17.5 \) units remaining.
4. After the 4th half-life (128 minutes total):
\( \frac{17.5}{2} = 8.75 \) units remaining.
5. After the 5th half-life (160 minutes total):
\( \frac{8.75}{2} = 4.375 \) units remaining.
6. After the 6th half-life (192 minutes total):
\( \frac{4.375}{2} = 2.1875 \) units remaining.
7. After the 7th half-life (224 minutes total):
\( \frac{2.1875}{2} = 1.09375 \) units remaining.
8. After the 8th half-life (256 minutes total):
\( \frac{1.09375}{2} = 0.546875 \) units remaining.
This gives you a comprehensive view of the decay of Bromine-84 over several half-lives. If you need a specific illustration or diagram, please provide that, and I can guide you on how to complete it based on the given values.
To determine how long it would take for 12.5% of the original amount of Carbon-14 to remain, we need to understand the concept of half-lives.
12.5% is equivalent to \( \frac{1}{8} \) of the original amount, which means that it takes three half-lives to reach this percentage because:
- After 1 half-life, 50% remains.
- After 2 half-lives, 25% remains.
- After 3 half-lives, 12.5% remains.
Since the half-life of Carbon-14 is 5,730 years, the time it would take to reach 12.5% is:
\[
3 \times 5,730 \text{ years} = 17,190 \text{ years}
\]
For the decay of Bromine-84 (84B), which has a half-life of 32 minutes and starts with 140 units:
1. After the 1st half-life (32 minutes):
\( \frac{140}{2} = 70 \) units remaining.
2. After the 2nd half-life (64 minutes total):
\( \frac{70}{2} = 35 \) units remaining.
3. After the 3rd half-life (96 minutes total):
\( \frac{35}{2} = 17.5 \) units remaining.
4. After the 4th half-life (128 minutes total):
\( \frac{17.5}{2} = 8.75 \) units remaining.
5. After the 5th half-life (160 minutes total):
\( \frac{8.75}{2} = 4.375 \) units remaining.
6. After the 6th half-life (192 minutes total):
\( \frac{4.375}{2} = 2.1875 \) units remaining.
7. After the 7th half-life (224 minutes total):
\( \frac{2.1875}{2} = 1.09375 \) units remaining.
8. After the 8th half-life (256 minutes total):
\( \frac{1.09375}{2} = 0.546875 \) units remaining.
This gives you a comprehensive view of the decay of Bromine-84 over several half-lives. If you need a specific illustration or diagram, please provide that, and I can guide you on how to complete it based on the given values.
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