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

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
$e^{-1.03 \times 10^{5}k} = \frac{1}{2} \\-1.03 \times 10^{5}k=ln(0.5) \\ k=- \frac{ln(0.5)}{-1.03 \times 10^{5}} =6.7296 \times 10^{-6}$

Because N₀ =  5 x 10⁹ atoms, the number of atoms remaining when t  = 4.12 x 10⁵ years is
$N = (5 \times 10^{9}) e^{-(6.7296 \times 10^{-6})(4.12 \times 10^{5})} = 3.125 \times 10^{8}$

Answer: 3.125 x 10⁸ atoms

Answer 2

Answer:

D.

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

Step-by-step explanation:

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