if your smoke detector contains 0.10 mg of 241Am, how long will it take to decay 1.25 * 10^-2mg
I assume you mean how long will it take to decay from 0.10 mg to 1.25E-2 mg?
k = 0.693/t1/2
ln(No/N) = kt
No = 0.10 mg
N = 1.25E-2 mg
k from above.
Substitute and solve for t. You will need to look up the half life of Am241.
To calculate the time it takes for a certain amount of a radioactive substance to decay, we can use the radioactive decay formula:
N = N0 * e^(-λt)
Where:
- N is the amount of the substance remaining after time t
- N0 is the initial amount of the substance
- λ is the decay constant, which depends on the specific radioactive isotope
- e is the base of the natural logarithm (approximately 2.71828)
In this case, we are working with americium-241 (^241Am), so we need to find the decay constant (λ) specific to this isotope. The decay constant (λ) for ^241Am is approximately 0.0123 per year.
First, let's find the initial amount of ^241Am remaining in the smoke detector:
N0 = 0.10 mg
Next, we can calculate the time it takes for the amount to decay to 1.25 * 10^-2 mg:
N = 1.25 * 10^-2 mg
Now, we can rearrange the equation and solve for time (t):
N / N0 = e^(-λt)
Taking the natural logarithm (ln) of both sides:
ln(N / N0) = -λt
Solving for t:
t = -ln(N / N0) / λ
Substituting the given values:
t = -ln((1.25 * 10^-2 mg) / (0.10 mg)) / 0.0123 per year
Using a calculator or computer program:
t ≈ 52.72 years
Therefore, it will take approximately 52.72 years for the 1.25 * 10^-2 mg of ^241Am to decay in the smoke detector.