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If it takes 100.0 minutes for a 20.00-g sample of potassium-44 to decay to 2.50 g. What is the half-life of potassium-44?

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Answer:

The half-life of a radioactive substance is the amount of time it takes for half of the substance to decay. We can use the given information to determine the half-life of potassium-44.

Let's start by finding the amount of potassium-44 remaining after one half-life. If a sample decays to half its original amount, then the remaining amount will be:

(1/2) x 20.00 g = 10.00 g

The time it takes for the sample to decay to this amount is the half-life, which we can call t₁/₂.

Now, let's use the given information to find t₁/₂. We can use the equation for radioactive decay:

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, and e is the mathematical constant e.

If we divide both sides of this equation by N₀, we get:

N/N₀ = e^(-λt)

We can use this equation to find the decay constant λ, which is related to the half-life by the following equation:

t₁/₂ = ln(2) / λ

If we substitute the given values into the first equation, we get:

N/N₀ = 2.50 g / 20.00 g = 0.125

t = 100.0 minutes

N₀ = 20.00 g

Substituting these values into the second equation and solving for λ, we get:

0.125 = e^(-λ x 100.0)

ln(0.125) = -λ x 100.0

λ = ln(2) / t₁/₂

We can solve for t₁/₂ by substituting the value we just found for λ:

ln(0.125) = -ln(2) x 100.0 / t₁/₂

t₁/₂ = -100.0 ln(0.125) / ln(2)

t₁/₂ ≈ 20.7 minutes

Therefore, the half-life of potassium-44 is approximately 20.7 minutes.

Step-by-step explanation:

User Marc Van Leeuwen
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