Question

The isotope calcium-41 decays into potassium-41, with a half-life of 1.03 × 10^5 years. There is a sample of calcium-41 containing 5 × 10^9 atoms. How many atoms of calcium-41 and potassium-41 will there be after 4.12 × 10^5 years?

Answer 1

D.

3.125 × 10^8 atoms of calcium-41 and 4.6875 × 10^9 atoms of potassium-41

Explanation:

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Answer 2

The decay equation is of the form

`N(t) = N_(0)e^(-kt)`
where
N₀ = initial amount
k = decay constant
t = time

The half-life is 1.03 x 10⁵ years. Therefore

`e^{-1.03 * 10^(5)k} = (1)/(2) \\-1.03 * 10^(5)k=ln(0.5) \\ k=- (ln(0.5))/(-1.03 * 10^(5)) =6.7296 * 10^(-6)`

Because N₀ = 5 x 10⁹ atoms, the number of atoms remaining when t = 4.12 x 10⁵ years is

`N = (5 * 10^(9)) e^{-(6.7296 * 10^(-6))(4.12 * 10^(5))} = 3.125 * 10^(8)`

Answer: 3.125 x 10⁸ atoms