Question
A.find the half-life of a 25g sample that decayed to 3g after 17527.5 years.
B. Radioactive dating is a common method used to date anything that was once alive ( including plants) and up to a certain number of years. All living things take in carbon from the environment in the form of carbon- 12 and carbon-14. When an organism die, carbon intakes stops and the carbon- 14 begin to decay at a known rate. Scientists can determine how much C-14 remains in an organism by measuring radiation emitted by the C-14 isotopes. Can we use C-14 to calculate the above sample? State your reason.
B. Radioactive dating is a common method used to date anything that was once alive ( including plants) and up to a certain number of years. All living things take in carbon from the environment in the form of carbon- 12 and carbon-14. When an organism die, carbon intakes stops and the carbon- 14 begin to decay at a known rate. Scientists can determine how much C-14 remains in an organism by measuring radiation emitted by the C-14 isotopes. Can we use C-14 to calculate the above sample? State your reason.
Answers
GPT-4o mini
### A. Finding the Half-Life
To find the half-life of the radioactive substance, we can use the formula for exponential decay:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
Where:
- \( N(t) \) is the remaining amount of the substance after time \( t \).
- \( N_0 \) is the initial amount of the substance.
- \( T_{1/2} \) is the half-life.
- \( t \) is the elapsed time.
Given:
- \( N_0 = 25 \) g
- \( N(t) = 3 \) g
- \( t = 17527.5 \) years
We need to rearrange the formula to solve for \( T_{1/2} \):
1. Substitute the known values into the decay formula:
\[ 3 = 25 \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}} \]
2. Divide both sides by 25:
\[ \frac{3}{25} = \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}} \]
3. Take the natural logarithm of both sides:
\[ \ln\left(\frac{3}{25}\right) = \frac{17527.5}{T_{1/2}} \ln\left(\frac{1}{2}\right) \]
4. Rearranging gives:
\[ T_{1/2} = \frac{17527.5 \cdot \ln\left(\frac{1}{2}\right)}{\ln\left(\frac{3}{25}\right)} \]
5. Now, calculate the values:
\[
\ln\left(\frac{1}{2}\right) \approx -0.693147
\]
\[
\ln\left(\frac{3}{25}\right) \approx \ln(3) - \ln(25) \approx 1.098612 - 3.218876 \approx -2.120264
\]
6. Substitute these values into the equation:
\[
T_{1/2} \approx \frac{17527.5 \cdot (-0.693147)}{-2.120264}
\]
\[
T_{1/2} \approx \frac{12162.6952}{2.120264} \approx 5734.7 \text{ years}
\]
**Therefore, the half-life of the substance is approximately 5734.7 years.**
### B. Can We Use C-14 to Calculate the Sample?
**No, C-14 is not appropriate for calculating the sample's age in this specific case for the following reasons:**
1. **Time Limitations:** Carbon-14 dating is effective for dating once-living organisms that died within the last approximately 50,000 years. The half-life of C-14 is about 5,730 years, and after about five half-lives, the amount of C-14 becomes very minimal and difficult to accurately measure.
2. **Sample Age:** The sample in question has decayed over approximately 17,527.5 years, which may be pushing the limits of detectable C-14 if it were originally C-14. However, the total radioactive decay during this time may lead to an insufficient amount of C-14 remaining to make an accurate determination.
3. **Assumptions of C-14 Dating:** Carbon dating assumes a constant atmospheric C-14 level and an organism having a uniform ratio of C-12 to C-14. If the material being measured has undergone significant environmental changes (e.g., exposure to radiation, burial, etc.), results can be skewed.
4. **Nature of the Sample:** If the sample does not derive from a biological origin (e.g., minerals, metals), then C-14 dating is not suitable.
In conclusion, while C-14 dating is a powerful tool for dating organic remains, it is not applicable here given the time frame and the nature of the sample.
To find the half-life of the radioactive substance, we can use the formula for exponential decay:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
Where:
- \( N(t) \) is the remaining amount of the substance after time \( t \).
- \( N_0 \) is the initial amount of the substance.
- \( T_{1/2} \) is the half-life.
- \( t \) is the elapsed time.
Given:
- \( N_0 = 25 \) g
- \( N(t) = 3 \) g
- \( t = 17527.5 \) years
We need to rearrange the formula to solve for \( T_{1/2} \):
1. Substitute the known values into the decay formula:
\[ 3 = 25 \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}} \]
2. Divide both sides by 25:
\[ \frac{3}{25} = \left( \frac{1}{2} \right)^{\frac{17527.5}{T_{1/2}}} \]
3. Take the natural logarithm of both sides:
\[ \ln\left(\frac{3}{25}\right) = \frac{17527.5}{T_{1/2}} \ln\left(\frac{1}{2}\right) \]
4. Rearranging gives:
\[ T_{1/2} = \frac{17527.5 \cdot \ln\left(\frac{1}{2}\right)}{\ln\left(\frac{3}{25}\right)} \]
5. Now, calculate the values:
\[
\ln\left(\frac{1}{2}\right) \approx -0.693147
\]
\[
\ln\left(\frac{3}{25}\right) \approx \ln(3) - \ln(25) \approx 1.098612 - 3.218876 \approx -2.120264
\]
6. Substitute these values into the equation:
\[
T_{1/2} \approx \frac{17527.5 \cdot (-0.693147)}{-2.120264}
\]
\[
T_{1/2} \approx \frac{12162.6952}{2.120264} \approx 5734.7 \text{ years}
\]
**Therefore, the half-life of the substance is approximately 5734.7 years.**
### B. Can We Use C-14 to Calculate the Sample?
**No, C-14 is not appropriate for calculating the sample's age in this specific case for the following reasons:**
1. **Time Limitations:** Carbon-14 dating is effective for dating once-living organisms that died within the last approximately 50,000 years. The half-life of C-14 is about 5,730 years, and after about five half-lives, the amount of C-14 becomes very minimal and difficult to accurately measure.
2. **Sample Age:** The sample in question has decayed over approximately 17,527.5 years, which may be pushing the limits of detectable C-14 if it were originally C-14. However, the total radioactive decay during this time may lead to an insufficient amount of C-14 remaining to make an accurate determination.
3. **Assumptions of C-14 Dating:** Carbon dating assumes a constant atmospheric C-14 level and an organism having a uniform ratio of C-12 to C-14. If the material being measured has undergone significant environmental changes (e.g., exposure to radiation, burial, etc.), results can be skewed.
4. **Nature of the Sample:** If the sample does not derive from a biological origin (e.g., minerals, metals), then C-14 dating is not suitable.
In conclusion, while C-14 dating is a powerful tool for dating organic remains, it is not applicable here given the time frame and the nature of the sample.