Question

Americium is used in household smoke detectors as a source of ? particles. The device actually responds to the presence of ions from a fire, rather than responding to smoke. A schematic is shown in the image. (Figure 1) The entire setup is enclosed in the familiar plastic housing. The 241Am, which is an ?-emitter, is encased in an aluminum container with an opening that aims the ? particles through an air gap between electrically charged plates. The ? particles ionize some of the air molecules, which are attracted to the electrical plates. This produces a current that is detected by the ammeter in the circuit. When smoke is present, a different number of ions form, altering the current and setting off the alarm.

Part A A particular smoke detector contains 2.45 ?Ci 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. View Available Hint(s)   g   Submit Part B Fears of radiation exposure from normal use of such detectors are largely unfounded. Identify reasons why 241Am smoke detectors are perfectly safe. Choose all that apply. View Available Hint(s) Choose all that apply. Ions get trapped by electrodes. The penetrating power of ? radiation is limited. The detector is housed in an aluminum case. The number of ? particles leaving the case is low. The detector has a plastic cover. The amount of americium is very little.

Answer 1

Answer - Part A

we have given information for the answer.

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
u 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 approximate u = 241, or look it up on Wikipedia and use 241.0547. (Note the approximation is off by 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 2.45e-6 µCi / 3.239 Ci/g = 0.7564µg

Part B - Choose all that apply.

The detector is housed in an aluminum case.
Ions get trapped by electrodes.
The detector has a plastic cover.
The penetrating power of ? radiation is limited.
The number of ? particles leaving the case is low.
The amount of americium is very little.

Answer: All are correct. Check all boxes.