Given that the half-life of ¹⁹⁸ Au is 2.7 days , determine the activity of a sample which contains 1.00 g of ¹⁹⁸ Au . Consider the atomic weight of ¹⁹⁸ Au to be 198 g mol ⁻¹ .

Step-by-step solution

The number of atoms in the sample is given by: N = m M N_A = 1.00 10⁻⁶ 198 6.022 10²³ 3.041 10¹⁵ atoms The decay constant is: = 2 T_ 1/2 = 0.693 2.7 24 3600 2.97 10⁻⁶ s ⁻¹ The activity of the sample is: A = N = (2.97 10⁻⁶) (3.041 10¹⁵) 9.03 10^9 Bq Converting the activity to Curies ( 1 Ci = 3.7 10¹⁰ Bq ): A = 9.03 10^9 3.7 10¹⁰ 0.244 Ci