Answer :
- Calculate the elapsed time: 4 hours and 9 minutes = 14940 seconds.
- Calculate the decay constant ($\lambda$) for each radioisotope using $\lambda = \frac{ln(2)}{t_{1/2}}$.
- Calculate the mass remaining for each radioisotope after 14940 seconds using $m(t) = m_0 e^{-\lambda t}$.
- Compare the calculated masses with the measured mass (13.1 kg) to identify the radioisotope: Barium-139 matches closely, so the answer is $\boxed{Barium-139}$.
### Explanation
1. Problem Analysis
First, let's analyze the problem. We have an initial mass of a radioisotope (104.8 kg) and a final mass (13.1 kg) after a certain time has passed. We need to determine which of the given radioisotopes (Potassium-42, Nitrogen-13, Barium-139, or Radon-220) matches this decay pattern. The key is to use the half-life of each isotope to calculate the expected mass after the elapsed time and compare it to the measured final mass.
2. Calculate Elapsed Time
The initial time is 12:02:00 P.M. and the final time is 4:11:00 P.M. The elapsed time is 4 hours and 9 minutes, which is equal to $4 \times 60 + 9 = 249$ minutes. In seconds, this is $249 \times 60 = 14940$ seconds.
3. Decay Formula
Now, let's consider each radioisotope and calculate the mass remaining after 14940 seconds, using the decay formula: $m(t) = m_0 e^{-\lambda t}$, where $m_0 = 104.8$ kg and $\lambda = \frac{ln(2)}{t_{1/2}}$.
4. Potassium-42 Calculation
For Potassium-42, the half-life is 12.36 hours, which is $12.36 \times 3600 = 44496$ seconds. Therefore, $\lambda = \frac{ln(2)}{44496} \approx 0.00001557$. The mass remaining is $m(14940) = 104.8 \times e^{-0.00001557 \times 14940} \approx 104.8 \times e^{-0.2326} \approx 104.8 \times 0.7925 \approx 83.04$ kg.
5. Nitrogen-13 Calculation
For Nitrogen-13, the half-life is 9.97 minutes, which is $9.97 \times 60 = 598.2$ seconds. Therefore, $\lambda = \frac{ln(2)}{598.2} \approx 0.001159$. The mass remaining is $m(14940) = 104.8 \times e^{-0.001159 \times 14940} \approx 104.8 \times e^{-17.31} \approx 104.8 \times 0.000000318 \approx 0.0000333$ kg. This is very small.
6. Barium-139 Calculation
For Barium-139, the half-life is 83 minutes, which is $83 \times 60 = 4980$ seconds. Therefore, $\lambda = \frac{ln(2)}{4980} \approx 0.0001392$. The mass remaining is $m(14940) = 104.8 \times e^{-0.0001392 \times 14940} \approx 104.8 \times e^{-2.08} \approx 104.8 \times 0.1254 \approx 13.1$ kg.
7. Radon-220 Calculation
For Radon-220, the half-life is 54.5 seconds. Therefore, $\lambda = \frac{ln(2)}{54.5} \approx 0.01272$. The mass remaining is $m(14940) = 104.8 \times e^{-0.01272 \times 14940} \approx 104.8 \times e^{-190.0} \approx 0$ kg. This is essentially zero.
8. Conclusion
Comparing the calculated masses to the measured mass of 13.1 kg, we see that the calculated mass for Barium-139 (13.1 kg) matches the measured mass almost perfectly. Therefore, 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 were suitable for medical use, could be used as a tracer to follow the flow of a fluid through the body. By understanding the half-life and decay rate, doctors can accurately administer the correct dosage and monitor its progress, ensuring effective treatment while minimizing potential harm. This same principle applies in industrial applications, such as gauging the thickness of materials or tracing leaks in pipelines.
- Calculate the decay constant ($\lambda$) for each radioisotope using $\lambda = \frac{ln(2)}{t_{1/2}}$.
- Calculate the mass remaining for each radioisotope after 14940 seconds using $m(t) = m_0 e^{-\lambda t}$.
- Compare the calculated masses with the measured mass (13.1 kg) to identify the radioisotope: Barium-139 matches closely, so the answer is $\boxed{Barium-139}$.
### Explanation
1. Problem Analysis
First, let's analyze the problem. We have an initial mass of a radioisotope (104.8 kg) and a final mass (13.1 kg) after a certain time has passed. We need to determine which of the given radioisotopes (Potassium-42, Nitrogen-13, Barium-139, or Radon-220) matches this decay pattern. The key is to use the half-life of each isotope to calculate the expected mass after the elapsed time and compare it to the measured final mass.
2. Calculate Elapsed Time
The initial time is 12:02:00 P.M. and the final time is 4:11:00 P.M. The elapsed time is 4 hours and 9 minutes, which is equal to $4 \times 60 + 9 = 249$ minutes. In seconds, this is $249 \times 60 = 14940$ seconds.
3. Decay Formula
Now, let's consider each radioisotope and calculate the mass remaining after 14940 seconds, using the decay formula: $m(t) = m_0 e^{-\lambda t}$, where $m_0 = 104.8$ kg and $\lambda = \frac{ln(2)}{t_{1/2}}$.
4. Potassium-42 Calculation
For Potassium-42, the half-life is 12.36 hours, which is $12.36 \times 3600 = 44496$ seconds. Therefore, $\lambda = \frac{ln(2)}{44496} \approx 0.00001557$. The mass remaining is $m(14940) = 104.8 \times e^{-0.00001557 \times 14940} \approx 104.8 \times e^{-0.2326} \approx 104.8 \times 0.7925 \approx 83.04$ kg.
5. Nitrogen-13 Calculation
For Nitrogen-13, the half-life is 9.97 minutes, which is $9.97 \times 60 = 598.2$ seconds. Therefore, $\lambda = \frac{ln(2)}{598.2} \approx 0.001159$. The mass remaining is $m(14940) = 104.8 \times e^{-0.001159 \times 14940} \approx 104.8 \times e^{-17.31} \approx 104.8 \times 0.000000318 \approx 0.0000333$ kg. This is very small.
6. Barium-139 Calculation
For Barium-139, the half-life is 83 minutes, which is $83 \times 60 = 4980$ seconds. Therefore, $\lambda = \frac{ln(2)}{4980} \approx 0.0001392$. The mass remaining is $m(14940) = 104.8 \times e^{-0.0001392 \times 14940} \approx 104.8 \times e^{-2.08} \approx 104.8 \times 0.1254 \approx 13.1$ kg.
7. Radon-220 Calculation
For Radon-220, the half-life is 54.5 seconds. Therefore, $\lambda = \frac{ln(2)}{54.5} \approx 0.01272$. The mass remaining is $m(14940) = 104.8 \times e^{-0.01272 \times 14940} \approx 104.8 \times e^{-190.0} \approx 0$ kg. This is essentially zero.
8. Conclusion
Comparing the calculated masses to the measured mass of 13.1 kg, we see that the calculated mass for Barium-139 (13.1 kg) matches the measured mass almost perfectly. Therefore, 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 were suitable for medical use, could be used as a tracer to follow the flow of a fluid through the body. By understanding the half-life and decay rate, doctors can accurately administer the correct dosage and monitor its progress, ensuring effective treatment while minimizing potential harm. This same principle applies in industrial applications, such as gauging the thickness of materials or tracing leaks in pipelines.
