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.

To solve this problem, we need to use the formula for radioactive decay, which is:

A = λN

where: A is the activity (in decays per second, or Becquerels), λ is the decay constant (in per second), N is the number of atoms.

First, we need to find the decay constant λ. The relationship between the decay constant and the half-life (T) is:

λ = ln(2) / T

Substituting the given half-life of 16.3 seconds:

λ = ln(2) / 16.3 ≈ 0.0425 s^-1

Next, we need to find the number of atoms N. We can do this by converting the mass of the sample to moles, and then using Avogadro's number to find the number of atoms. The molar mass of bromine-88 is 88 g/mol.

3.6 mg is 0.0036 g, so the number of moles is:

0.0036 g / 88 g/mol ≈ 0.000041 moles

Multiplying by Avogadro's number (6.022 x 10^23 atoms/mol) gives:

N ≈ 0.000041 moles * 6.022 x 10^23 atoms/mol ≈ 2.47 x 10^19 atoms

Finally, we can find the activity A:

A = λN = 0.0425 s^-1 * 2.47 x 10^19 atoms ≈ 1.05 x 10^18 Bq

To convert to curies, use the conversion factor 1 Ci = 3.7 x 10^10 Bq:

A ≈ 1.05 x 10^18 Bq * (1 Ci / 3.7 x 10^10 Bq) ≈ 28,400 Ci

So the activity of the sample is approximately 1.05 x 10^18 Bq or 28,400 Ci.