Question

A particular smoke detector contains 2.35 μCi of 241Am, with a half-life of 458 years. The isotope is encased in a thin aluminum container. Calculate the mass of 241Am in grams in the detector.

Answer 1

The specific activity of a nuclide is given by
SA = ln (2) * NA / (T½ * u)

where
ln (2) is the natural logarithm of 2
NA = Avogadro's Number
T½ is the half-life
and is the atomic mass

For Am-241, we need to convert the half-life to years, so multiply by 31.56 megaseconds.
458 * 31.56e6 = 14.45e9 seconds
For Am-241, we can look at 241, or look it up at Wikipedia and use 241.0547.(SATISFACTION is about 0.022%, so it's not really worth worrying about.)

SA = 0.6931 * 6.02e23 / (14.45e9 * 241)
= 4.1724e23 / 3.482e12
= 1.1983e11 sec ^ -1 or 1.1983e11 Bq.

One Curie is 3.7e10 Bq, so one gram of Am-241 is
1.1983e11 / 3.7e10 = 3.239 Ci / g

So 2.35e-6 μCi / 3.239 Ci / g = 0.7255 μg
Answer: 7.255* 10 ^ -7g