KinexiaLEARN PHYSICS

Radioactive Activity — Smoke Detector (Worked Example)

Hard
Problem

A smoke detector contains 0.30 μg of americium-241 (half-life 432 years, molar mass 241 g/mol). What is its activity?

Given

  • m = 0.30 μg = 3.0 × 10⁻¹⁰ kg

  • t½ = 432 y

  • M = 0.241 kg/mol

Find

The activity A.

Principle

First find N from the mass (N = m/M × N_A), then A = λN. Long half-life → tiny λ.

Formula

Solution

  1. Half-life in seconds: t1/2=432×3.156×107=1.363×1010 st_{1/2} = 432 \times 3.156 \times 10^7 = 1.363 \times 10^{10}\ \mathrm{s}

  2. λ=0.693/1.363×1010=5.09×10−11 s−1\lambda = 0.693/1.363 \times 10^{10} = 5.09 \times 10^{-11}\ \mathrm{s^{-1}}

  3. Atoms: N=(3.0×10−10/0.241)(6.022×1023)=7.50×1014N = (3.0 \times 10^{-10}/0.241)(6.022 \times 10^{23}) = 7.50 \times 10^{14}

  4. A=(5.09×10−11)(7.50×1014)=3.81×104 BqA = (5.09 \times 10^{-11})(7.50 \times 10^{14}) = 3.81 \times 10^4\ \mathrm{Bq}

3.81 × 10⁴ Bq (≈ 38 kBq)

Unit check

s⁻¹ × atom count = Bq. Activity is decays per second by definition — consistent.

What this means physically

≈38 kBq — matches real smoke detectors (~37 kBq). The alpha particles ionize air; smoke disrupts the current and triggers the alarm. The activity is small and safely shielded.

Common mistake

Converting μg wrong: 0.30 μg = 3.0 × 10⁻¹⁰ kg (not 10⁻⁶ kg). Track micro (10⁻⁶ g) → kg carefully.