Americium is used in household smoke detectors as a source of particles. The dev

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.

A particular smoke detector contains 1.05 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.

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.

Choose all that apply.

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

Explanation / Answer

we know that

activity is given by

A = lamda x N

now

lamda = decay constant = ln2 / t1/2

given

half life ( t1/2) = 458 years

t1/2 = 1.444 x 10^10 sec

so

lamda = ln2 / 1.444 x 10^10

lamda = 4.8 x 10-11

now

given

activity = 1.05 x 10-6 Ci

we know that

1 Ci = 3.7 x 10^10 decays per second

so

activity (A) = 1.05 x 10-6 x 3.7 x 10^10

A = 38850

now

A = lamda x N

38850 = 4.8 x 10-11 x N

N = 8.09375 x 10^14

now

we know that

moles = number of molecules (N ) / 6.023 x 10^23

so

moles = 8.09375 x 10^14 / 6.023 x 10^23

moles = 1.3438 x 10-9

now

mass = moles x molar mass

so

mass of Am = 1.3438 x 10-9 x 241

mass of Am = 3.2386 x 10-7 g

so

mass of 241 Am in the detector is 3.2386 x 10-7 grams

B)

All the given statements are correct

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