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Answer :
- Calculate the elapsed time in minutes: $4 \text{ hours} \times 60 + 9 = 249 \text{ minutes}$.
- Determine the number of half-lives: $n = \frac{\log(\frac{104.8}{13.1})}{\log(2)} = 3$.
- Calculate the half-life of the unknown isotope: $\frac{249}{3} = 83 \text{ minutes}$.
- Compare the calculated half-life to the given options and identify the radioisotope: Barium-139.
$\boxed{\text{Barium-139}}$
### Explanation
1. Problem Analysis
We are given the initial mass (104.8 kg) and final mass (13.1 kg) of a radioisotope sample, along with the time elapsed between the two measurements (from 12:02:00 P.M. to 4:11:00 P.M. on the same day). Our goal is to identify the radioisotope from a list of possibilities (Potassium-42, Nitrogen-13, Barium-139, Radon-220) based on their half-lives.
2. Calculating Elapsed Time
First, we need to determine the time elapsed between the two measurements in minutes. The time difference is 4 hours and 9 minutes. Converting this to minutes, we have:
$4 \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} + 9 \text{ minutes} = 240 + 9 = 249 \text{ minutes}$
3. Calculating Number of Half-Lives
Next, we need to determine how many half-lives have passed during this time. We use the formula $m_f = m_i \times (\frac{1}{2})^n$, where $m_f$ is the final mass, $m_i$ is the initial mass, and $n$ is the number of half-lives. We can rearrange this formula to solve for $n$: $n = \frac{\log(\frac{m_i}{m_f})}{\log(2)}$. Plugging in the given values, we have:
$n = \frac{\log(\frac{104.8}{13.1})}{\log(2)} = \frac{\log(8)}{\log(2)} = 3$
4. Calculating Half-Life
Now we can calculate the half-life of the unknown isotope. The half-life is the elapsed time divided by the number of half-lives: $\text{half-life} = \frac{\text{elapsed time}}{n} = \frac{249 \text{ minutes}}{3} = 83 \text{ minutes}$
5. Identifying the Radioisotope
Finally, we compare the calculated half-life (83 minutes) with the half-lives of the possible isotopes:
- Potassium-42: 12.4 hours = 744 minutes
- Nitrogen-13: 9.97 minutes
- Barium-139: 83.1 minutes
- Radon-220: 55.6 seconds
The calculated half-life (83 minutes) is closest to the half-life of Barium-139 (83.1 minutes). Therefore, the radioisotope in the sample is Barium-139.
6. Final Answer
The radioisotope in the sample is Barium-139.
### Examples
Radioactive isotopes are used in medicine for both diagnostic and therapeutic purposes. For example, Barium-139, if it existed in a form suitable for medical use, could potentially be used as a tracer in imaging studies due to its relatively short half-life. This would allow doctors to monitor certain bodily functions or locate specific tissues without exposing the patient to prolonged radiation.
- Determine the number of half-lives: $n = \frac{\log(\frac{104.8}{13.1})}{\log(2)} = 3$.
- Calculate the half-life of the unknown isotope: $\frac{249}{3} = 83 \text{ minutes}$.
- Compare the calculated half-life to the given options and identify the radioisotope: Barium-139.
$\boxed{\text{Barium-139}}$
### Explanation
1. Problem Analysis
We are given the initial mass (104.8 kg) and final mass (13.1 kg) of a radioisotope sample, along with the time elapsed between the two measurements (from 12:02:00 P.M. to 4:11:00 P.M. on the same day). Our goal is to identify the radioisotope from a list of possibilities (Potassium-42, Nitrogen-13, Barium-139, Radon-220) based on their half-lives.
2. Calculating Elapsed Time
First, we need to determine the time elapsed between the two measurements in minutes. The time difference is 4 hours and 9 minutes. Converting this to minutes, we have:
$4 \text{ hours} \times 60 \frac{\text{minutes}}{\text{hour}} + 9 \text{ minutes} = 240 + 9 = 249 \text{ minutes}$
3. Calculating Number of Half-Lives
Next, we need to determine how many half-lives have passed during this time. We use the formula $m_f = m_i \times (\frac{1}{2})^n$, where $m_f$ is the final mass, $m_i$ is the initial mass, and $n$ is the number of half-lives. We can rearrange this formula to solve for $n$: $n = \frac{\log(\frac{m_i}{m_f})}{\log(2)}$. Plugging in the given values, we have:
$n = \frac{\log(\frac{104.8}{13.1})}{\log(2)} = \frac{\log(8)}{\log(2)} = 3$
4. Calculating Half-Life
Now we can calculate the half-life of the unknown isotope. The half-life is the elapsed time divided by the number of half-lives: $\text{half-life} = \frac{\text{elapsed time}}{n} = \frac{249 \text{ minutes}}{3} = 83 \text{ minutes}$
5. Identifying the Radioisotope
Finally, we compare the calculated half-life (83 minutes) with the half-lives of the possible isotopes:
- Potassium-42: 12.4 hours = 744 minutes
- Nitrogen-13: 9.97 minutes
- Barium-139: 83.1 minutes
- Radon-220: 55.6 seconds
The calculated half-life (83 minutes) is closest to the half-life of Barium-139 (83.1 minutes). Therefore, the radioisotope in the sample is Barium-139.
6. Final Answer
The radioisotope in the sample is Barium-139.
### Examples
Radioactive isotopes are used in medicine for both diagnostic and therapeutic purposes. For example, Barium-139, if it existed in a form suitable for medical use, could potentially be used as a tracer in imaging studies due to its relatively short half-life. This would allow doctors to monitor certain bodily functions or locate specific tissues without exposing the patient to prolonged radiation.
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