Question

A particular smoke detector contains 1.05 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. Express your answer numerically in grams.

Answer 1

Solution-

A half-life of 458 years= 458/ln(2)= lifetime of 661 years

661 * 31.56 * 10^6 = 20861 * 10^6 seconds, or 2.086 * 10^10 seconds.

As we know that the decay constant is the reciprocal of the average lifetime,

= 1/2.086 * 10^(-10)

= 0.4794 * 10^(-10)

Each atom of the Am-241 sample has a probability = 4.794 * 10^(-11) of decaying in the next second.

So in one mole,

=6.02 * 10^23 * 4.794 * 10^(-11)

= 28.86 * 10^14 atoms of Am-241 to decay every second.

Specific activity=28.86 * 10^12 / 241

= 0.1198 * 10^12

= 1.198 * 10^11 dps/g

One Ci = 3.7 * 10^10 dps
One gram of Am-241 contains 1.198 * 10^11 / 3.7 * 10^10 = 3.238 Ci = 3.238 * 10^6 uCi.
1.05 uCi * 1 g / (3.238 * 10^6 uCi) =3.24 * 10^(-7) g

= 0.324 ug