Smoke detectors and radioactive isotopes Part A) A particular smoke detector con

Question

Smoke detectors and radioactive isotopes

Part A) A particular smoke detector contains 1.25 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.

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

Explanation / Answer

Part-A

Its rate of decay R(t) is given by:
R(t)=1.25e^(-t/tau) where t and tau are in years, R is in uCi (microCuries)
At t = half life,
R=0.5=1.25e^(-458yr/tau)
0.5/1.25=e^(-458yr/tau)
ln(0.5/1.25)=(-458yr/tau)
tau=-458yr/ln(0.5/1.25)
tau=499.84yr


Integrate curie rate to determine the number of atom dissociations during one half life:

Integrate R(t) from t=0yr to t=458yr
Integral of R(t) =
=1.25(-tau)e^(-458/tau) - 1.25(-tau)e^(-0/tau)
=1.25(-tau) x [e^(-458/499.84)-e^0]
=1.25(--499.84)(0.3999-1)
=374.94 uCi/yr
The above number is in units of microcurie-years. Convert to total number of dissociations (decays). 1 Curie is 3.7x10^10 dissociations per second.

(374.94uCi-yr) x (1Ci/10^6uCi) x (3.7x10^10dis/sec) x (365.24day/yr) x (86400sec/day)
=4.37x10^14 dissociations

The above number represents half of the atoms dissociating, so the total initial mass is twice the above number of atoms, or:
2 x 4.37x10^14 = 8.74x10^14 atoms

Divide by Avogadro's number to find the initial number of moles:
8.74x10^14atoms / 6.0221415×10^23moles/atoms
=7.26x10^(-10)moles
Multiply by relative atomic mass of 241Am to determine the number of grams:
mass = 7.26x10^(-10)moles x 241.056823grams/mole
=1.75x10^(-7)g
=175ng

Part-B

All of them are correct

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.