Question

The half-life for the radioactive decay of U−238 is 4.5 billion years and is independent of initial concentration.

A) How long will it take for 10% of the U−238 atoms in a sample of U−238 to decay?

Express your answer using two significant figures.

B) If a sample of U−238 initially contained 1.4×10¹⁸ atoms when the universe was formed 13.8 billion years ago, how many U−238 atoms will it contain today?

Express your answer using two significant figures.

Answer 1

(A)

For first order decay process:

ln(C_t/C_o) = -kt ...................(1)

where

Co= initial concentration

C_t = concentration of U-238 remaining at time t

k = decay constant

we know for first order reaction

k = 0.693 / t

k = 0.6931 / 4.5 x 10⁹ = 1.5403 x 10^-10 year^-1

10% decayed means 90% remains

and C_t/Co x 100% = 90%

so

C_t/Co = 0.9

ln(0.9) = -1.5403 x 10^-10 x t

- 0.10536 = -1.5403 x 10^-10 x t

t = - 0.10536 / -1.5403 x 10^-10

t = 0.68 x 10^-9 years

t = 0.68 billion years

(B) given C_o = 1.4 × 10¹⁸ atoms

t = 13.8 billion years = 13.8 x 10⁹ years

Amount of U-238 remaining C_t = C_o e^(-kt)

C_t = 1.4 × 10¹⁸ x e^{(-1.5403 x ₁₀-10 x 13.8 x ₁₀ 9)}

C_t = 1.4 × 10¹⁸ x e^-2.125614

C_t = 1.4 × 10¹⁸ x 0.11935965893247

C_t = 1.67 x 10¹⁷ ≈ 1.7 x 10¹⁷ atoms

so

C_t = 1.7 x 10¹⁷ atoms