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?

A) To determine if the smoke detector is likely to fail while you own it, we need to find the time at which the amount of Americium-241 drops below half its initial value (0.1 mg).

To do this, we can set up the equation m = 0.2(0.5)^(t/432.2), where m is the mass remaining and t is the time in years.

We need to find the value of t when m = 0.1.

0.1 = 0.2(0.5)^(t/432.2)

Dividing both sides by 0.2 gives:

0.5^(t/432.2) = 0.1/0.2

0.5^(t/432.2) = 0.5

Taking the logarithm of both sides with base 0.5:

t/432.2 = log0.5(0.5) [Note: log0.5(0.5) = 1]

t/432.2 = 1

t = 432.2

This means the smoke detector will fail after approximately 432.2 years. Whether it is likely to fail while you own it depends on how long you plan on owning it. If you plan on owning it for more than 432.2 years, then it is likely to fail while you own it. Otherwise, it's unlikely to fail while you own it.

B) To find the amount of Am-241 that will remain after 50 years, we can simply substitute t = 50 into the given equation.

m = 0.2(0.5)^(50/432.2)

Calculating this gives:

m ≈ 0.2(0.5)^0.1157 ≈ 0.2(0.9474) ≈ 0.1895 mg

Therefore, approximately 0.1895 mg of Am-241 will remain after 50 years.

C) To find the time it takes for the amount of Am-241 to drop to 0.05 mg, we can set up the equation m = 0.2(0.5)^(t/432.2) and solve for t.

0.05 = 0.2(0.5)^(t/432.2)

Dividing both sides by 0.2 gives:

0.5^(t/432.2) = 0.05/0.2

0.5^(t/432.2) = 0.25

Taking the logarithm of both sides with base 0.5:

t/432.2 = log0.5(0.25) [Note: log0.5(0.25) = -2]

t/432.2 = -2

t = -2 * 432.2

t ≈ -864.4

Since time cannot be negative in this context, we can conclude that it will not take any time for the amount of Am-241 to drop to 0.05 mg because it will never reach that value.