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
0.5 g
b
1.0 g
c
0.125 g
d
0.25 g
To determine how much americium-241 will remain in 1,296 years, we need to find the number of half-lives that have occurred in that time period.
Given that the half-life of americium-241 is 432 years, we can calculate the number of half-lives by dividing the total time of 1,296 years by the half-life:
Number of half-lives = 1,296 years / 432 years = 3
Each half-life reduces the amount of americium-241 by half. Therefore, after 3 half-lives, the amount remaining will be:
1 gram / (2^3) = 1 gram / 8 = 0.125 g
Therefore, the correct answer is c) 0.125 g.