CALCULUS
posted by Anonymous .
A smoke detector contains 0.2 mg of Americium 241 (Am241), a radioactive element that decays in t years according to the relation m = 0.2(0.5)^(t/432.2). Where m is the mass, in milligrams, remaining after t years.
A) The smoke detector will no longer work when the amount of Am241 drops below half it's initial value. Is it likely to fail while you own it? Justify your answer.
B) If you buy a smoke detector today, how much of the Am241 will remain after 50 years?
C) How long will it take for the amount of Am241 to drop to 0.05 mg?

CALCULUS 
Steve
looking at the formula, it's clear that the halflife is 432.2 years. Not likely you'll be around that long
m(50) = .2(.5)^(50/432.2) = 0.18mg
.05 = .2(.5)^(t/432.2)
.25 = .5^(t/432.2)
Now, .25 = .5^2 so we need
t/432.2 = 2
t = 864.4 years
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