The half-life of ⁴⁰ K is 1.30 10⁹ y . A 1.00 g sample of pure KCl yields 160 counts s ⁻¹ . Determine the relative abundance of ⁴⁰ K (the fraction of ⁴⁰ K present) in natural potassium.

Step-by-step solution

The decay constant of ⁴⁰ K is given by: = 2 T_ 1/2 = 0.693 1.30 10⁹ 365 24 3600 s = 1.69 10⁻¹⁷ s ⁻¹ The activity A is 160 counts s ⁻¹ . The number of ⁴⁰ K atoms in the sample is: N₄₀ = A = 160 1.69 10⁻¹⁷ = 9.47 10¹⁸ atoms The molar mass of KCl is approximately 39.1 + 35.5 = 74.6 g mol ⁻¹ . The total number of potassium atoms in 1.00 g of KCl is: N_ K = 1.00 74.6 6.022 10²³ = 8.07 10²¹ atoms The relative abundance of ⁴⁰ K in natural potassium is the ratio of the number of ⁴⁰ K atoms to the total number of K atoms: R