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 will fail while you own it, we need to find the time it takes for the amount of Am-241 to drop below half its initial value of 0.2 mg.
The relation given in the problem is m = 0.2(0.5)^(t/432.2), where m is the mass remaining after t years. We want to find the value of t when m is less than 0.1 mg (half of 0.2 mg).
By substituting m = 0.1 into the equation, we get:
0.1 = 0.2(0.5)^(t/432.2)
To simplify the equation, divide both sides by 0.2:
0.1/0.2 = (0.5)^(t/432.2)
Now, we need to take the logarithm of both sides to solve for t:
log(0.1/0.2) = log[(0.5)^(t/432.2)]
Using logarithm properties, we can bring the exponent down:
log(0.1/0.2) = (t/432.2) * log(0.5)
Now, divide both sides by log(0.5) to isolate t:
t/432.2 = log(0.1/0.2) / log(0.5)
Finally, multiply both sides by 432.2 to get t:
t = 432.2 * log(0.1/0.2) / log(0.5)
Evaluate this expression using a calculator to find the value of t. If the resulting value of t is greater than the time you plan on owning the smoke detector, it is likely to fail while you own it. Otherwise, it won't fail.
B) To determine how much of the Am-241 will remain after 50 years, we can simply substitute t = 50 into the given equation for m:
m = 0.2(0.5)^(50/432.2)
Evaluate this expression using a calculator to find the remaining mass of Am-241 after 50 years.
C) To find how long it takes for the amount of Am-241 to drop to 0.05 mg, we need to solve the equation m = 0.05:
0.05 = 0.2(0.5)^(t/432.2)
Follow the same steps as in part A to solve for t, substituting 0.05 for m. Again, evaluate the expression using a calculator to find the time it takes for the amount of Am-241 to drop to 0.05 mg.
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