To find out how much Actinium-226 remains after 87 hours, we can use the formula for radioactive decay based on the half-life:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
where:
- \( N(t) \) is the remaining amount of substance after time \( t \),
- \( N_0 \) is the initial amount of substance,
- \( T_{1/2} \) is the half-life of the substance,
- \( t \) is the elapsed time.
Given:
- \( N_0 = 80 \) mg (initial amount),
- \( T_{1/2} = 29 \) hours (half-life),
- \( t = 87 \) hours (elapsed time).
First, we need to compute how many half-lives fit into 87 hours:
\[ \text{Number of half-lives} = \frac{t}{T_{1/2}} = \frac{87}{29} \approx 3 \]
This result means that 87 hours is approximately 3 half-lives.
Now, we can calculate the remaining amount of Actinium-226 using the decay formula:
\[ N(87) = 80 \left( \frac{1}{2} \right)^{3} \]
Calculating \( \left( \frac{1}{2} \right)^{3} \):
\[ \left( \frac{1}{2} \right)^{3} = \frac{1}{8} \]
Now substituting this into the equation:
\[ N(87) = 80 \times \frac{1}{8} = 80 \div 8 = 10 \text{ mg} \]
Therefore, the amount of Actinium-226 remaining after 87 hours is \(\boxed{10}\) mg.
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