Posted by **Anonymous** on Saturday, April 21, 2012 at 1:34am.

A smoke detector contains 0.2 mg of Americium 241 (Am-241), 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 Am-241 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 Am-241 will remain after 50 years?

C) How long will it take for the amount of Am-241 to drop to 0.05 mg?

- CALCULUS -
**Steve**, Saturday, April 21, 2012 at 5:26am
looking at the formula, it's clear that the half-life 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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