Question
If a patient is administered 4.7 nanograms (ng) of technetium-99m, how much will remain 1.251 days (d) later, assuming that no technetium has been eliminated by any other process? The half-life of technetium-99m is 6.007 hours. Be sure your answer has the correct number of significant figures.
Answers
GPT-4o mini
To calculate the remaining amount of technetium-99m after a given time period, we can use the formula from radioactive decay:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( t \) is the elapsed time,
- \( t_{1/2} \) is the half-life of the substance.
Given:
- \( N_0 = 4.7 \, \text{ng} \)
- \( t_{1/2} = 6.007 \, \text{hours} \)
- \( t = 1.251 \, \text{days} \)
First, we need to convert the time \( t \) from days to hours:
\[
t = 1.251 \, \text{days} \times 24 \, \text{hours/day} = 30.024 \, \text{hours}
\]
Now, we can calculate the number of half-lives that have passed during this time:
\[
\text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{30.024 \, \text{hours}}{6.007 \, \text{hours}} \approx 5.000
\]
Now we can substitute these values into the decay formula:
\[
N(t) = 4.7 \, \text{ng} \left( \frac{1}{2} \right)^{5.000}
\]
Calculating \( \left( \frac{1}{2} \right)^{5.000} \):
\[
\left( \frac{1}{2} \right)^{5} = \frac{1}{32} \approx 0.03125
\]
Now substitute this back into the equation:
\[
N(t) = 4.7 \, \text{ng} \times 0.03125 = 0.146875 \, \text{ng}
\]
Finally, we need to round this answer to the correct number of significant figures. The initial quantity \( 4.7 \, \text{ng} \) has two significant figures. Thus, we round \( 0.146875 \, \text{ng} \) to two significant figures:
\[
\text{Final answer: } 0.15 \, \text{ng}
\]
Therefore, the amount of technetium-99m that will remain after 1.251 days is **0.15 ng**.
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{t_{1/2}}} \]
Where:
- \( N(t) \) is the remaining quantity after time \( t \),
- \( N_0 \) is the initial quantity,
- \( t \) is the elapsed time,
- \( t_{1/2} \) is the half-life of the substance.
Given:
- \( N_0 = 4.7 \, \text{ng} \)
- \( t_{1/2} = 6.007 \, \text{hours} \)
- \( t = 1.251 \, \text{days} \)
First, we need to convert the time \( t \) from days to hours:
\[
t = 1.251 \, \text{days} \times 24 \, \text{hours/day} = 30.024 \, \text{hours}
\]
Now, we can calculate the number of half-lives that have passed during this time:
\[
\text{Number of half-lives} = \frac{t}{t_{1/2}} = \frac{30.024 \, \text{hours}}{6.007 \, \text{hours}} \approx 5.000
\]
Now we can substitute these values into the decay formula:
\[
N(t) = 4.7 \, \text{ng} \left( \frac{1}{2} \right)^{5.000}
\]
Calculating \( \left( \frac{1}{2} \right)^{5.000} \):
\[
\left( \frac{1}{2} \right)^{5} = \frac{1}{32} \approx 0.03125
\]
Now substitute this back into the equation:
\[
N(t) = 4.7 \, \text{ng} \times 0.03125 = 0.146875 \, \text{ng}
\]
Finally, we need to round this answer to the correct number of significant figures. The initial quantity \( 4.7 \, \text{ng} \) has two significant figures. Thus, we round \( 0.146875 \, \text{ng} \) to two significant figures:
\[
\text{Final answer: } 0.15 \, \text{ng}
\]
Therefore, the amount of technetium-99m that will remain after 1.251 days is **0.15 ng**.