Indium-116 is radioactive and has a half life of 14.10 seconds. Calculate the activity of a 6.6mg sample of indium-116. Give your answer in becquerels and in curies. Be sure your answer has the correct number of significant digits.BqCi

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 decay constant is related to the half-life (T) of the isotope by the formula:

λ = ln(2) / T

where ln(2) is the natural logarithm of 2 (approximately 0.6931), and T is the half-life in seconds. Substituting the given half-life:

λ = 0.6931 / 14.10 s = 0.0492 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 (6.022 x 10^23 atoms/mole) to convert to atoms. The atomic mass of indium-116 is 116 g/mole, so:

N = (6.6 mg) * (1 g / 1000 mg) * (1 mole / 116 g) * (6.022 x 10^23 atoms/mole) N = 3.42 x 10^19 atoms

Now we can find the activity (A):

A = λN = (0.0492 s^-1)(3.42 x 10^19 atoms) = 1.68 x 10^18 Bq

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

A = (1.68 x 10^18 Bq) * (1 Ci / 3.7 x 10^10 Bq) = 4.54 x 10^7 Ci

So, the activity of a 6.6 mg sample of indium-116 is 1.68 x 10^18 Bq or 4.54 x 10^7 Ci.