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.
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