To determine how much americium-241 remains after 1,296 years, we can use the half-life formula. The half-life of americium-241 is 432 years, which means that every 432 years, half of the amount will remain.
First, we calculate how many half-lives fit into 1,296 years:
\[ \text{Number of half-lives} = \frac{1296 \text{ years}}{432 \text{ years/half-life}} = 3 \]
Next, we start with 1 gram of americium-241 and apply the half-life reduction for each half-life:
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After the first half-life (432 years): \[ 1 \text{ g} \times \frac{1}{2} = 0.5 \text{ g} \]
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After the second half-life (864 years): \[ 0.5 \text{ g} \times \frac{1}{2} = 0.25 \text{ g} \]
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After the third half-life (1296 years): \[ 0.25 \text{ g} \times \frac{1}{2} = 0.125 \text{ g} \]
Therefore, the amount of americium-241 remaining after 1,296 years is 0.125 g.
The correct answer is c) 0.125 g.
1 answer