Answer to Question #96595 in Chemistry for Ryan

Question

Geologists use the decay of potassium-40 in volcanic rocks to determine their age. Potassium-40 has a half-life of 1.26 x 10⁹ years, so it can be used to date very old rocks. If a sample of rock 3.78 x 10⁹ years old contains 2.73 x 10⁻⁷ g of potassium-40 today, how much potassium-40 was originally present in the rock? *

Answer 1

The exponential decay of potassium-40 can be described by the following equation:

N_t = N₀ × e^-λt,

where N_t - the remaining quantity of atoms, N₀ - the initial quantity, λ - the decay constant (λ = 0.693 / t_1/2), t - the mean lifetime, t_1/2 - the half-life.

From here:

N₀ = N_t / e^-λt.

N_t = N_A × (m_K / Mr_K),

where N_A - Avogadro's number, m_K - potassium-40 mass, Mr_K - atomic weight.

As a result:

Nt = 6.022 × 10²³ mol^-1 × (2.73 × 10⁻⁷ g / 40 g/mol) = 0.41 × 10¹⁶

λ = 0.693 / 1.26 × 10⁹ = 0.55 × 10^-9.

Finally, the initial number of potassium-40 atoms:

N₀ = 0.41 × 10¹⁶ / e^{-(0.55 × 10-9 × 3.78 x 10⁹)} = 3.28 × 10¹⁶

From here:

m = Mr × N₀/N_A = 40 × (3.28 × 10¹⁶ / 6.022 × 10²³) = 21.79 × 10^-7 g.

Answer: 21.79 × 10^-7 g of potassium-40 was originaly present in the rock.