Bromine-88 is radioactive and has a half life of 16.3 seconds. Calculate the activity of a 3.6mg sample of bromine-88. Give your answer in becquerels and in curies. Be sure your answer has the correct number of significant digits.BqCi

To calculate the activity of a radioactive sample, we need to know the number of atoms in the sample and the half-life of the isotope.

Step 1: Calculate the number of atoms in the sample The atomic mass of Bromine-88 is 88 g/mol. Therefore, a 3.6 mg sample is 3.6 mg * (1 g/1000 mg) = 0.0036 g.

The number of moles in the sample is 0.0036 g * (1 mol/88 g) = 0.0000409 mol.

The number of atoms in the sample is 0.0000409 mol * (6.022 x 10^23 atoms/mol) = 2.46 x 10^19 atoms.

Step 2: Calculate the decay constant The decay constant (λ) is related to the half-life (t1/2) by the equation λ = ln(2) / t1/2.

For Bromine-88, λ = ln(2) / 16.3 s = 0.0425 s^-1.

Step 3: Calculate the activity The activity (A) is the product of the decay constant and the number of atoms, A = λN.

For the Bromine-88 sample, A = 0.0425 s^-1 * 2.46 x 10^19 atoms = 1.04 x 10^18 decays/s = 1.04 x 10^18 Bq.

Step 4: Convert to curies 1 curie (Ci) is defined as 3.7 x 10^10 Bq. Therefore, the activity in curies is 1.04 x 10^18 Bq * (1 Ci/3.7 x 10^10 Bq) = 2.81 x 10^7 Ci.

So, the activity of a 3.6 mg sample of Bromine-88 is 1.04 x 10^18 Bq or 2.81 x 10^7 Ci.