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The half-life of Americium 241, used to make smoke detectors, is 432.7 years. Use the the formula A(t) = Apekt to model the situation. (a) Find the exact value of k, the rate of decay per year. Round to 6 decimal places or enter an exact answer. k = (b) If a sample contains 0.61 mg, how many mg will remain after 700 years? Round to 2 decimal places. amount: mg (c) How long will it take for 70 % of the material to decay? Round to the nearest tenth of a year

User Phn
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Final answer:

The half-life of Americium-241 is 432.7 years. The rate of decay per year, k, is approximately 0.0016. After 700 years, approximately 0.192 mg of Americium-241 will remain. It takes approximately 682.4 years for 70% of the material to decay.

Step-by-step explanation:

To model the decay of Americium-241, we can use the formula A(t) = A₀e^(-kt), where A(t) represents the amount of Americium-241 remaining after time t, A₀ is the initial amount of Americium-241, and k is the rate of decay per year. To find k, we can use the half-life equation: half-life = ln(2) / k. Plugging in the given half-life of 432.7 years, we can solve for k:

half-life = ln(2) / k

432.7 = ln(2) / k

k = ln(2) / 432.7

Using a calculator, we find that k ≈ 0.0016.

(b) To find the amount of Americium-241 remaining after 700 years, we can use the formula A(t) = A₀e^(-kt) and plug in the values:

A(700) = 0.61e^(-0.0016*700)

Solving this equation, we find that approximately 0.192 mg of Americium-241 will remain after 700 years.

(c) To find the time it takes for 70% of the material to decay, we can set up the equation:

0.3A₀ = A₀e^(-kt)

Simplifying, we have:

e^(-kt) = 0.3

Taking the natural log of both sides:

-kt = ln(0.3)

Solving for t, we find that approximately 682.4 years are needed for 70% of the material to decay.

User Snukker
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