Radioactive Activity — Radon in a Basement (Worked Example)

Hard

Problem

A basement air sample contains 1.0 × 10¹⁰ atoms of radon-222 (half-life 3.8 days). What is the activity of the sample in becquerels?

Given

N = 1.0 × 10¹⁰ atoms
t½ = 3.8 days = 328,320 s

Find

The activity A in Bq.

Principle

Activity A = λN with λ = ln2/t½ — decays per second. The half-life must be in seconds.

Formula

A = \lambda N,\quad \lambda = \frac{\ln 2}{t_{1/2}}

A = λN, λ = ln2/t½

Solution

Convert half-life: $t_{1/2} = 3.8 \times 86400 = 328{,}320\ \mathrm{s}$
Decay constant: $\lambda = 0.693/328320 = 2.11 \times 10^{-6}\ \mathrm{s^{-1}}$
Activity: $A = (2.11 \times 10^{-6})(1.0 \times 10^{10})$
Multiply: $A = 2.11 \times 10^4\ \mathrm{Bq}$

2.11 × 10⁴ Bq

Unit check

s⁻¹ × (dimensionless count) = s⁻¹ = Bq — correct.

What this means physically

21,100 decays per second — a significant radon level. Radon seeps from soil and is the second leading cause of lung cancer; testing basements is genuinely important health physics.

Common mistake

Using t½ in days directly. λ needs seconds — 3.8 days is 328,320 s; the day-unit error would inflate activity 86,400-fold.