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Sagot :
Let's analyze the given data and answer the questions step by step.
### Given Data
| Time (h) | Mass Remaining of the Isotope (g) |
|----------|-----------------------------------|
| 0.0 | 40.00 |
| 3.0 | 20.00 |
| 6.0 | 10.00 |
| 9.0 | 5.00 |
| 12.0 | 2.50 |
| 15.0 | 1.25 |
| 18.0 | 0.63 |
### Decay Curve
To graph the data, plot the given points on a graph with the time (in hours) on the x-axis and the mass remaining (in grams) on the y-axis. You will observe that the mass decreases exponentially over time, forming a decay curve.
### Exponential Decay Relationship
The mass of a radioactive substance undergoing exponential decay can be modeled by the formula:
[tex]\[ M(t) = M_0 \cdot e^{-\lambda t} \][/tex]
where:
- [tex]\( M(t) \)[/tex] is the mass remaining at time [tex]\( t \)[/tex],
- [tex]\( M_0 \)[/tex] is the initial mass,
- [tex]\( \lambda \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is the time.
From the data, we see that the mass halves every 3 hours. This information is crucial for determining the decay constant [tex]\( \lambda \)[/tex].
### a. Mass Remaining After 10.0 Hours
To find the mass remaining after 10.0 hours, we use the exponential decay formula. Given:
- Initial mass [tex]\( M_0 = 40.00 \)[/tex] g,
- Decay constant [tex]\( \lambda \)[/tex],
- Time [tex]\( t = 10.0 \)[/tex] hours.
First, determine the decay constant [tex]\( \lambda \)[/tex]:
The half-life [tex]\( T_{\text{half}} \)[/tex] is 3 hours, and the relationship with [tex]\( \lambda \)[/tex] is given by:
[tex]\[ T_{\text{half}} = \frac{\ln(2)}{\lambda} \][/tex]
Solving for [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{\ln(2)}{3} \approx 0.231 \, \text{h}^{-1} \][/tex]
Now we can calculate the mass remaining after 10.0 hours:
[tex]\[ M(10) = 40.00 \cdot e^{-0.231 \cdot 10} \approx 3.968 \, \text{g} \][/tex]
So, approximately 3.97 grams remains after 10.0 hours.
### b. Mass Remaining After 28.0 Hours
Similarly, for 28.0 hours, we use the same exponential decay formula with [tex]\( t = 28.0 \)[/tex] hours:
[tex]\[ M(28) = 40.00 \cdot e^{-0.231 \cdot 28} \approx 0.062 \, \text{g} \][/tex]
Thus, approximately 0.062 grams remains after 28.0 hours.
### c. Half-Life of the Isotope
The half-life [tex]\( T_{\text{half}} \)[/tex] of the isotope is determined using the decay constant [tex]\( \lambda \)[/tex]:
[tex]\[ T_{\text{half}} = \frac{\ln(2)}{\lambda} \][/tex]
Since we previously calculated [tex]\( \lambda \)[/tex] from the data:
[tex]\[ T_{\text{half}} = \frac{\ln(2)}{0.231} \approx 3.0 \, \text{hours} \][/tex]
Therefore, the half-life of this isotope is 3.0 hours.
### Given Data
| Time (h) | Mass Remaining of the Isotope (g) |
|----------|-----------------------------------|
| 0.0 | 40.00 |
| 3.0 | 20.00 |
| 6.0 | 10.00 |
| 9.0 | 5.00 |
| 12.0 | 2.50 |
| 15.0 | 1.25 |
| 18.0 | 0.63 |
### Decay Curve
To graph the data, plot the given points on a graph with the time (in hours) on the x-axis and the mass remaining (in grams) on the y-axis. You will observe that the mass decreases exponentially over time, forming a decay curve.
### Exponential Decay Relationship
The mass of a radioactive substance undergoing exponential decay can be modeled by the formula:
[tex]\[ M(t) = M_0 \cdot e^{-\lambda t} \][/tex]
where:
- [tex]\( M(t) \)[/tex] is the mass remaining at time [tex]\( t \)[/tex],
- [tex]\( M_0 \)[/tex] is the initial mass,
- [tex]\( \lambda \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is the time.
From the data, we see that the mass halves every 3 hours. This information is crucial for determining the decay constant [tex]\( \lambda \)[/tex].
### a. Mass Remaining After 10.0 Hours
To find the mass remaining after 10.0 hours, we use the exponential decay formula. Given:
- Initial mass [tex]\( M_0 = 40.00 \)[/tex] g,
- Decay constant [tex]\( \lambda \)[/tex],
- Time [tex]\( t = 10.0 \)[/tex] hours.
First, determine the decay constant [tex]\( \lambda \)[/tex]:
The half-life [tex]\( T_{\text{half}} \)[/tex] is 3 hours, and the relationship with [tex]\( \lambda \)[/tex] is given by:
[tex]\[ T_{\text{half}} = \frac{\ln(2)}{\lambda} \][/tex]
Solving for [tex]\( \lambda \)[/tex]:
[tex]\[ \lambda = \frac{\ln(2)}{3} \approx 0.231 \, \text{h}^{-1} \][/tex]
Now we can calculate the mass remaining after 10.0 hours:
[tex]\[ M(10) = 40.00 \cdot e^{-0.231 \cdot 10} \approx 3.968 \, \text{g} \][/tex]
So, approximately 3.97 grams remains after 10.0 hours.
### b. Mass Remaining After 28.0 Hours
Similarly, for 28.0 hours, we use the same exponential decay formula with [tex]\( t = 28.0 \)[/tex] hours:
[tex]\[ M(28) = 40.00 \cdot e^{-0.231 \cdot 28} \approx 0.062 \, \text{g} \][/tex]
Thus, approximately 0.062 grams remains after 28.0 hours.
### c. Half-Life of the Isotope
The half-life [tex]\( T_{\text{half}} \)[/tex] of the isotope is determined using the decay constant [tex]\( \lambda \)[/tex]:
[tex]\[ T_{\text{half}} = \frac{\ln(2)}{\lambda} \][/tex]
Since we previously calculated [tex]\( \lambda \)[/tex] from the data:
[tex]\[ T_{\text{half}} = \frac{\ln(2)}{0.231} \approx 3.0 \, \text{hours} \][/tex]
Therefore, the half-life of this isotope is 3.0 hours.
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