Final answer:
The activity of 2 mg of ¹¹Au is calculated using its half-life of 3 days and atomic weight of 198 g/mol. By relating the number of radioactive nuclei in the sample to the decay constant derived from the half-life, the activity is found to be of the order of 16.18 × 10¹² disintegrations per second.
Step-by-step explanation:
The activity of a radioactive isotope, such as ¹¹Au, can be calculated using its half-life, which in the case of ¹¹Au is 3 days. To find the activity (in disintegrations per second) of 2 mg of ¹¹Au, one can use the decay law, which relates activity (A) to the number of nuclei (N), the decay constant (λ), and the half-life (T½) of the isotope. The formula for activity is A = λN, where λ can be found using the equation λ = ln(2)/T½, and N is the number of atoms in the 2 mg sample, calculated from the atomic weight and Avogadro's number.
To calculate the number of moles in 2 mg of ¹¹Au, you divide the mass by the atomic weight, which is 2 mg / 198 g/mol. Then, you convert the mass in grams to moles by multiplying by 1 mol/198 g. Next, use Avogadro's number to find the number of atoms present. Once you have N, you can calculate the activity using the decay constant λ, derived from its known half-life. The resulting activity level for 2 mg of ¹¹Au would be found to follow the order of magnitude around 16.18 × 10¹² disintegrations per second.