Americium-241 is a radioactive substance used in smoke detectors. The half life of americium is 432 years. If a smoke detector initially contains 1 gram of Americium 241, how much will remain in 1,296 years?
a
0.125 g
b
0.25 g
c
0.5 g
d
1.0 g
To determine how much Americium-241 will remain in 1,296 years, we need to calculate the number of half-lives that have passed.
Since the half-life of Americium-241 is 432 years, we can divide 1,296 years by 432 years to find the number of half-lives:
1,296 years / 432 years = 3 half-lives
Each half-life reduces the amount of Americium-241 by half. So, after 3 half-lives, the remaining amount will be:
1 gram / 2^3 = 1 gram / 8 = 0.125 grams
Therefore, the answer is:
a) 0.125 g