In: Chemistry
A particular smoke detector contains 1.95 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.
This involves calculating the specific activity of Am-241.
A half-life of 458 years works out to an average lifetime of 661
years. (Divide by the natural logarithm of 2 to get this.) The
average lifetime is the length of time it would take for the entire
sample to decay to zero, if it continued to decay at the current
rate. (Of course, it doesn't. Rate is proportional to
quantity.)
Since uCi is in units of disintegrations per second (dps), convert
years to seconds. One year is 31.56 million seconds, so 661 * 31.56
* 10^6 = 20861 * 10^6 seconds, or 2.086 * 10^10 seconds.
The decay constant is the reciprocal of the average lifetime, and
in this case, 1/2.086 * 10^(-10) = 0.4794 * 10^(-10)
Each atom of the Am-241 sample has a probability of 4.794 *
10^(-11) of decaying in the next second. In one mole, we expect
6.02 * 10^23 * 4.794 * 10^(-11) = 28.86 * 10^14 atoms of Am-241 to
decay every second.
Specific activity is often specified per gram. Since the atomic
weight of Am-241 is right around 241 grams/mole, we divide our
decay rate by 241 to get
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.
1.95 uCi * 1 g / (3.238 * 10^6 uCi) = 5.94 * 10^(-7) g = 0.0.594 ug
Ans
now your formula becomes:
SA = (NA * MW) / (T * 3.7 * 10^10)
SA = Specific Activity
NA = Avogadro's Number
AW = Atomic Weight
T = average lifetime
A good reasonableness check is to recall that the Curie is defined
as the activity present in one gram of radium-226.
The half-life of radium-226 is 1620 years.
The atomic weight is 226 g/mole.
1620/458 = 3.54 increase in dps/mole
226/241 = 0.938 decrease in moles/gram
3.54*0.938 = 3.32.
This is not very far from our calculated specific activity, so we
can assume there are no major errors in the calculation