Question

By far the largest and most widespread use of americium-241 is as a component in household and industrial smoke detectors, where a small amount is used in an ionization chamber inside the detector. The half life of americium-241 is 432.2 years. If a smoke detector is buried in a land fill, how many years are required for the original amount of americium-241 to decay to less than 1% of the original amount? Round your numerical answer to the nearest tenth of a year.
1,435.7 years


4,322 years


2,871.5 years


43,220 years

Answer 1

Answer:

2871.5 years

Step-by-step explanation:

The number of half-lives can be found by solving ...

... (1/2)^x = (1/100)

Taking logs, you have ...

... x·(-log(2)) = -log(100)

... x = log(100)/log(2) ≈ 6.64386

Then the number of years for the desired decay is ...

... (432.2)×6.64386 ≈ 2871.5 . . . . years