Determine the activity of a sample containing initially 1.00 g of ¹⁹⁸ Au after 7 days have elapsed. The half-life of ¹⁹⁸ Au is 2.7 days and its atomic weight is 198 g mol ⁻¹ .

Step-by-step solution

The initial number of atoms N₀ in the sample is given by: N₀ = m M N_A Substituting m = 1.00 g = 10⁻⁶ g , M = 198 g mol ⁻¹ , and N_A = 6.023 10²³ mol ⁻¹ : N₀ = 10⁻⁶ 198 6.023 10²³ 3.04 10¹⁵ atoms The decay constant is: = 2 T_ 1/2 = 0.693 2.7 24 3600 2.97 10⁻⁶ s ⁻¹ The initial activity A₀ is: A₀ = N₀ = (2.97 10⁻⁶) (3.04 10¹⁵) 9.03 10^9 decays/s Converting this to Curies ( 1 Ci = 3.7 10¹⁰ decays/s ): A₀ = 9.03 10^9 3.7 10¹⁰ 0.244 Ci The activity A after t = 7 days is: A = A₀ ( 1 2 )^ t T_ 1/2 A = 0.244 ( 1 2 )^ 7 2.7